{"id":598145,"date":"2022-05-02T11:01:34","date_gmt":"2022-05-02T11:01:34","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598145"},"modified":"2022-05-09T06:57:48","modified_gmt":"2022-05-09T06:57:48","slug":"selina-class-7-icse-solutions-mathematics-chapter-2-rational-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-2-rational-numbers\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 2-\u00a0Rational Numbers"},"content":{"rendered":"\n

Class 7: Maths Chapter 2 solutions. Complete Class 7 Maths Chapter 2 Notes.<\/p>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 2- Rational Numbers<\/h2>\n\n\n\n

Selina 7th Maths Chapter 2, Class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Exercise 2A page: 19<\/h4>\n\n\n\n

1. Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the largest two digit number is 99<\/p>\n\n\n\n

So the smallest four digit number is 1000<\/p>\n\n\n\n

Numerator = 99<\/p>\n\n\n\n

Denominator = 1000<\/p>\n\n\n\n

Rational number = 99\/1000<\/p>\n\n\n\n

2. Write the numerator of each of the following rational numbers:<\/strong><\/p>\n\n\n\n

(i) \u2013 125\/127<\/strong><\/p>\n\n\n\n

(ii) 37\/ -137<\/strong><\/p>\n\n\n\n

(iii) \u2013 85\/ 93<\/strong><\/p>\n\n\n\n

(iv) 2<\/strong><\/p>\n\n\n\n

(v) 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u2013 125\/127<\/p>\n\n\n\n

Here the numerator = \u2013 125<\/p>\n\n\n\n

(ii) 37\/ -137<\/p>\n\n\n\n

Here the numerator = 37<\/p>\n\n\n\n

(iii) \u2013 85\/ 93<\/p>\n\n\n\n

Here the numerator = \u2013 85<\/p>\n\n\n\n

(iv) 2 = 2\/1<\/p>\n\n\n\n

Here the numerator = 2<\/p>\n\n\n\n

(v) 0 = 0\/1<\/p>\n\n\n\n

Here the numerator = 0<\/p>\n\n\n\n

3. Write the denominator of each of the following rational numbers:<\/strong><\/p>\n\n\n\n

(i) 7\/ -15<\/strong><\/p>\n\n\n\n

(ii) \u2013 18\/29<\/strong><\/p>\n\n\n\n

(iii) \u2013 3\/4<\/strong><\/p>\n\n\n\n

(iv) \u2013 7<\/strong><\/p>\n\n\n\n

(v) 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/ -15<\/p>\n\n\n\n

Here the denominator = \u2013 15<\/p>\n\n\n\n

(ii) \u2013 18\/29<\/p>\n\n\n\n

Here the denominator = 29<\/p>\n\n\n\n

(iii) \u2013 3\/4<\/p>\n\n\n\n

Here the denominator = 4<\/p>\n\n\n\n

(iv) \u2013 7 = -7\/1<\/p>\n\n\n\n

Here the denominator = 1<\/p>\n\n\n\n

(v) 0 = 0\/1<\/p>\n\n\n\n

Here the denominator = 1<\/p>\n\n\n\n

4. Write down a rational number with numerator (-5) \u00d7 (-4) and with denominator (28 \u2013 27) \u00d7 (8 \u2013 5).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Numerator = (-5) \u00d7 (-4) = 20<\/p>\n\n\n\n

Denominator = (28 \u2013 27) \u00d7 (8 \u2013 5) = 1 \u00d7 3 = 3<\/p>\n\n\n\n

So the rational number = 20\/3<\/p>\n\n\n\n

5. (i) \u2013 15\/1 in integer form is \u2026\u2026.<\/strong><\/p>\n\n\n\n

(ii) 23\/-1 in integer form is \u2026\u2026..<\/strong><\/p>\n\n\n\n

(iii) If 18 = 18\/a then a = \u2026\u2026.<\/strong><\/p>\n\n\n\n

(iv) If \u2013 57 = 57\/a then a = \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u2013 15\/1 in integer form is \u2013 15.<\/p>\n\n\n\n

(ii) 23\/-1 in integer form is \u2013 23.<\/p>\n\n\n\n

(iii) If 18 = 18\/a then a = 18\/18 = 1.<\/p>\n\n\n\n

(iv) If \u2013 57 = 57\/a then a = 57\/-57 = \u2013 1.<\/p>\n\n\n\n

6. Separate positive and negative rational numbers from the following:<\/strong><\/p>\n\n\n\n

-3\/ 5, 3\/-5, -3\/-5, 3\/5, 0, -13\/-3, 15\/-8, -15\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here the positive rational numbers are<\/p>\n\n\n\n

-3\/-5 = 3\/5 as both are negative<\/p>\n\n\n\n

-13\/-3 = 13\/3 as both are negative and 3\/5<\/p>\n\n\n\n

Similarly the negative rational numbers are<\/p>\n\n\n\n

-3\/5, 3\/ -5, 15\/-8 and -15\/8<\/p>\n\n\n\n

0 is neither positive nor negative integer.<\/p>\n\n\n\n

7. Find three rational numbers equivalent to<\/strong><\/p>\n\n\n\n

(i) 3\/5<\/strong><\/p>\n\n\n\n

(ii) 4\/-7<\/strong><\/p>\n\n\n\n

(iii) -5\/9<\/strong><\/p>\n\n\n\n

(iv) 8\/-15<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 2)\/ (5 \u00d7 2) = 6\/10<\/p>\n\n\n\n

3\/5 = (3 \u00d7 3)\/ (5 \u00d7 3) = 9\/15<\/p>\n\n\n\n

3\/5 = (3 \u00d7 4)\/ (5 \u00d7 4) = 12\/20<\/p>\n\n\n\n

Therefore, 6\/10, 9\/15 and 12\/20 are the rational numbers which are equivalent to the given rational number 3\/5.<\/p>\n\n\n\n

(ii) 4\/-7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/-7 = (4 \u00d7 2)\/ (-7 \u00d7 2) = 8\/ -14<\/p>\n\n\n\n

4\/-7 = (4 \u00d7 3)\/ (-7 \u00d7 3) = 12\/-21<\/p>\n\n\n\n

4\/-7 = (4 \u00d7 4)\/ (-7 \u00d7 4) = 16\/ -28<\/p>\n\n\n\n

Therefore, 8\/-14, 12\/-21 and 16\/-28 are the rational numbers which are equivalent to the given rational number 4\/-7.<\/p>\n\n\n\n

(iii) -5\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-5\/9 = (-5 \u00d7 2)\/ (9 \u00d7 2) = -10\/18<\/p>\n\n\n\n

-5\/9 = (-5 \u00d7 3)\/ (9 \u00d7 3) = \u2013 15\/27<\/p>\n\n\n\n

-5\/9 = (-5 \u00d7 4)\/ (9 \u00d7 4) = -20\/36<\/p>\n\n\n\n

Therefore, -10\/18, -15\/27 and -20\/36 are the rational numbers which are equivalent to the given rational number -5\/9.<\/p>\n\n\n\n

(iv) 8\/-15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

8\/-15 = (8 \u00d7 2)\/ (-15 \u00d7 2) = 16\/-30<\/p>\n\n\n\n

8\/-15 = (8 \u00d7 3)\/ (-15 \u00d7 3) = 24\/ -45<\/p>\n\n\n\n

8\/-15 = (8 \u00d7 4)\/ (-15 \u00d7 4) = 32\/-60<\/p>\n\n\n\n

Therefore, 16\/-30, 24\/-45 and 32\/-60 are the rational numbers which are equivalent to the given rational number 8\/-15.<\/p>\n\n\n\n

8. Which of the following are not rational numbers:<\/strong><\/p>\n\n\n\n

(i) \u2013 3<\/strong><\/p>\n\n\n\n

(ii) 0<\/strong><\/p>\n\n\n\n

(iii) 0\/4<\/strong><\/p>\n\n\n\n

(iv) 8\/0<\/strong><\/p>\n\n\n\n

(v) 0\/0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u2013 3 = -3\/1 is a rational number.<\/p>\n\n\n\n

(ii) 0 = 0\/1 is a rational number.<\/p>\n\n\n\n

(iii) 0\/4 is a rational number.<\/p>\n\n\n\n

(iv) 8\/0 is not a rational number.<\/p>\n\n\n\n

(v) 0\/0 is not a rational number as both numerator and denominator are zero.<\/p>\n\n\n\n

9. Express each of the following integers as a rational number with denominator 7:<\/strong><\/p>\n\n\n\n

(i) 5<\/strong><\/p>\n\n\n\n

(ii) \u2013 8<\/strong><\/p>\n\n\n\n

(iii) 0<\/strong><\/p>\n\n\n\n

(iv) \u2013 16<\/strong><\/p>\n\n\n\n

(v) 7<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5<\/p>\n\n\n\n

By multiplying and dividing by 7<\/p>\n\n\n\n

= (5 \u00d7 7)\/ 7<\/p>\n\n\n\n

= 35\/7<\/p>\n\n\n\n

(ii) \u2013 8<\/p>\n\n\n\n

By multiplying and dividing by 7<\/p>\n\n\n\n

= (-8 \u00d7 7)\/ 7<\/p>\n\n\n\n

= \u2013 56\/7<\/p>\n\n\n\n

(iii) 0<\/p>\n\n\n\n

By multiplying and dividing by 7<\/p>\n\n\n\n

= (0 \u00d7 7)\/ 7<\/p>\n\n\n\n

= 0\/7<\/p>\n\n\n\n

(iv) \u2013 16<\/p>\n\n\n\n

By multiplying and dividing by 7<\/p>\n\n\n\n

= (-16 \u00d7 7)\/ 7<\/p>\n\n\n\n

= \u2013 112\/7<\/p>\n\n\n\n

(v) 7<\/p>\n\n\n\n

By multiplying and dividing by 7<\/p>\n\n\n\n

= (7 \u00d7 7)\/ 7<\/p>\n\n\n\n

= 49\/7<\/p>\n\n\n\n

10. Express 3\/5 as a rational number with denominator:<\/strong><\/p>\n\n\n\n

(i) 20<\/strong><\/p>\n\n\n\n

(ii) \u2013 20<\/strong><\/p>\n\n\n\n

(iii) 45<\/strong><\/p>\n\n\n\n

(iv) 25<\/strong><\/p>\n\n\n\n

(v) \u2013 35<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 4)\/ (5 \u00d7 4) = 12\/20<\/p>\n\n\n\n

(ii) \u2013 20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 -4)\/ (5 \u00d7 -4) = -12\/-20<\/p>\n\n\n\n

(iii) 45<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 9)\/ (5 \u00d7 9) = 27\/45<\/p>\n\n\n\n

(iv) 25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 5)\/ (5 \u00d7 5) = 15\/25<\/p>\n\n\n\n

(v) \u2013 35<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3\/5 = (3 \u00d7 -7)\/ (5 \u00d7 -7) = -21\/-35<\/p>\n\n\n\n

11. Express 4\/7 as a rational number with numerator:<\/strong><\/p>\n\n\n\n

(i) 12<\/strong><\/p>\n\n\n\n

(ii) \u2013 12<\/strong><\/p>\n\n\n\n

(iii) \u2013 16<\/strong><\/p>\n\n\n\n

(iv) \u2013 20<\/strong><\/p>\n\n\n\n

(v) 20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/7 = (4 \u00d7 3)\/ (7 \u00d7 3) = 12\/21<\/p>\n\n\n\n

(ii) \u2013 12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/7 = (4 \u00d7 -3)\/ (7 \u00d7 -3) = -12\/-21<\/p>\n\n\n\n

(iii) \u2013 16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/7 = (4 \u00d7 -4)\/ (7 \u00d7 -4) = \u2013 16\/-28<\/p>\n\n\n\n

(iv) \u2013 20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/7 = (4 \u00d7 -5)\/ (7 \u00d7 -5) = -20\/-35<\/p>\n\n\n\n

(v) 20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4\/7 = (4 \u00d7 5)\/ (7 \u00d7 5) = 20\/35<\/p>\n\n\n\n

12. Find x, such that:<\/strong><\/p>\n\n\n\n

(i) \u2013 2\/3 = 6\/ x<\/strong><\/p>\n\n\n\n

(ii) 7\/-4 = x\/8<\/strong><\/p>\n\n\n\n

(iii) 3\/7 = x\/-35<\/strong><\/p>\n\n\n\n

(iv) -48\/x = 6<\/strong><\/p>\n\n\n\n

(v) 36\/x = 3<\/strong><\/p>\n\n\n\n

(vi) \u2013 27\/x = 9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u2013 2\/3 = 6\/ x<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

-2x = 6 \u00d7 3<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = (6 \u00d7 3)\/ -2<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 18\/-2 = -9<\/p>\n\n\n\n

Hence, \u2013 2\/3 = 6\/-9.<\/p>\n\n\n\n

(ii) 7\/-4 = x\/8<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

7 \u00d7 8 = \u2013 4 \u00d7 x<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

56 = -4x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 56\/-4 = -14<\/p>\n\n\n\n

Hence, 7\/-4 = -14\/8.<\/p>\n\n\n\n

(iii) 3\/7 = x\/-35<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

7x = \u2013 35 \u00d7 3<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = (-35 \u00d7 3)\/ 7<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = -15<\/p>\n\n\n\n

Hence, 3\/7 = -15\/-35.<\/p>\n\n\n\n

(iv) -48\/x = 6<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

6x = -48<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = -48\/6 = \u2013 8<\/p>\n\n\n\n

Hence, -48\/-8 = 6.<\/p>\n\n\n\n

(v) 36\/x = 3<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

3x = 36<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = 12<\/p>\n\n\n\n

Hence, 36\/12 = 3.<\/p>\n\n\n\n

(vi) \u2013 27\/x = 9<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

9x = -27<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = -27\/9 = \u2013 3<\/p>\n\n\n\n

Hence, -27\/-3 = 9.<\/p>\n\n\n\n

13. Express each of the following rational numbers to the lowest terms:<\/strong><\/p>\n\n\n\n

(i) 12\/15<\/strong><\/p>\n\n\n\n

(ii) \u2013 120\/144<\/strong><\/p>\n\n\n\n

(iii) \u2013 48\/ \u2013 72<\/strong><\/p>\n\n\n\n

(iv) 14\/-56<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 12\/15<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here dividing by 3 which is the HCF of 12 and 15<\/p>\n\n\n\n

(12 \u00f7 3)\/ (15 \u00f7 3) = 4\/5<\/p>\n\n\n\n

(ii) \u2013 120\/144<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here dividing by 25 which is the HCF of -120 and 144<\/p>\n\n\n\n

(-120 \u00f7 24)\/ (144 \u00f7 24) = \u2013 5\/6<\/p>\n\n\n\n

(iii) \u2013 48\/ \u2013 72<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here dividing by 24 which is the HCF of -48 and \u2013 72<\/p>\n\n\n\n

(-48 \u00f7 24)\/ (-72 \u00f7 24) = -2\/-3 = 2\/3<\/p>\n\n\n\n

(iv) 14\/-56<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here dividing by 14 which is the HCF of 14 and \u2013 56<\/p>\n\n\n\n

(14 \u00f7 14)\/ (-56 \u00f7 14) = 1\/-4 or \u2013 1\/4<\/p>\n\n\n\n

14. Express each of the following rational numbers in the standard form.<\/strong><\/p>\n\n\n\n

(i) -7\/-8<\/strong><\/p>\n\n\n\n

(ii) 5\/ \u2013 12<\/strong><\/p>\n\n\n\n

(iii) \u2013 7\/ \u2013 20<\/strong><\/p>\n\n\n\n

(iv) 4\/ -9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here a rational number is in standard form if its denominator is positive in lowest term.<\/p>\n\n\n\n

(i) -7\/-8 = 7\/8<\/p>\n\n\n\n

(ii) 5\/ \u2013 12 = -5\/12<\/p>\n\n\n\n

(iii) \u2013 7\/ \u2013 20 = 7\/20<\/p>\n\n\n\n

(iv) 4\/ -9 = -4\/9<\/p>\n\n\n\n

Exercise 2B page: 25<\/h4>\n\n\n\n

1. Mark the following pairs of rational numbers on the separate number lines:<\/strong><\/p>\n\n\n\n

(i) 3\/4 and -1\/4<\/strong><\/p>\n\n\n\n

(ii) 2\/5 and -3\/5<\/strong><\/p>\n\n\n\n

(iii) 5\/6 and -2\/3<\/strong><\/p>\n\n\n\n

(iv) 2\/5 and -4\/5<\/strong><\/p>\n\n\n\n

(v) 1\/4 and -5\/4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/4 and -1\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) 2\/5 and -3\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) 5\/6 and -2\/3<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iv) 2\/5 and -4\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(v) 1\/4 and -5\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

2. Compare:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 5\/7<\/strong><\/p>\n\n\n\n

(ii) -7\/2 and 5\/2<\/strong><\/p>\n\n\n\n

(iii) -3 and 2 \u00be<\/strong><\/p>\n\n\n\n

(iv) \u2013 1 \u00bd and 0<\/strong><\/p>\n\n\n\n

(v) 0 and 3\/4<\/strong><\/p>\n\n\n\n

(vi) 3 and -1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 5\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

5\/7 is on the right side of the number line.<\/p>\n\n\n\n

Hence, 3\/5 < 5\/7.<\/p>\n\n\n\n

(ii) -7\/2 and 5\/2<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -7\/2 < 5\/2.<\/p>\n\n\n\n

(iii) -3 and 2 \u00be<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -3 < 11\/4 or \u2013 3 < 2 \u00be.<\/p>\n\n\n\n

(iv) \u2013 1 \u00bd and 0<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -3\/2 < 0 or \u2013 1 \u00bd < 0.<\/p>\n\n\n\n

(v) 0 and \u00be<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, 0 < 3\/4.<\/p>\n\n\n\n

(vi) 3 and -1<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, 3 > \u2013 1.<\/p>\n\n\n\n

3. Compare:<\/strong><\/p>\n\n\n\n

(i) -1\/4 and 0<\/strong><\/p>\n\n\n\n

(ii) 1\/4 and 0<\/strong><\/p>\n\n\n\n

(iii) -3\/8 and 2\/5<\/strong><\/p>\n\n\n\n

(iv) -5\/8 and 7\/-12<\/strong><\/p>\n\n\n\n

(v) 5\/-9 and -5\/-9<\/strong><\/p>\n\n\n\n

(vi) -7\/8 and 5\/-6<\/strong><\/p>\n\n\n\n

(vii) 2\/7 and -3\/-8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -1\/4 and 0<\/p>\n\n\n\n

-1\/4 is a negative rational number which is always less than 0.<\/p>\n\n\n\n

Hence, -1\/4 < 0.<\/p>\n\n\n\n

(ii) 1\/4 and 0<\/p>\n\n\n\n

1\/4 is a positive rational number which is always greater than 0.<\/p>\n\n\n\n

Hence, 1\/4 > 0.<\/p>\n\n\n\n

(iii) -3\/8 and 2\/5<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 3 \u00d7 5 and 2 \u00d7 8<\/p>\n\n\n\n

\u2013 15 < 16<\/p>\n\n\n\n

Hence, -3\/8 < 2\/5.<\/p>\n\n\n\n

(iv) -5\/8 and 7\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-5\/ 8 and -7\/12<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 5 \u00d7 12 and \u2013 7 \u00d7 8<\/p>\n\n\n\n

-60 < \u2013 56<\/p>\n\n\n\n

Hence, -5\/8 < 7\/-12.<\/p>\n\n\n\n

(v) 5\/-9 and -5\/-9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

5 \u00d7 -9 and -5 \u00d7 -9<\/p>\n\n\n\n

\u2013 45 < 45<\/p>\n\n\n\n

Hence, 5\/-9 < -5\/-9.<\/p>\n\n\n\n

(vi) -7\/8 and 5\/-6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-7\/ 8 and -5\/6<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 7 \u00d7 6 and -5 \u00d7 8<\/p>\n\n\n\n

-42 < -40<\/p>\n\n\n\n

Hence, -7\/8 < 5\/-6.<\/p>\n\n\n\n

(vii) 2\/7 and -3\/-8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

2\/7 and 3\/8<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

2 \u00d7 8 and 7 \u00d7 3<\/p>\n\n\n\n

16 < 21<\/p>\n\n\n\n

Hence, 2\/7 < \u2013 3\/ -8.<\/p>\n\n\n\n

4. Arrange the given rational numbers in ascending order:<\/strong><\/p>\n\n\n\n

(i) 7\/10, -11\/-30 and 5\/-15<\/strong><\/p>\n\n\n\n

(ii) 4\/-9, -5\/12 and 2\/-3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/10, -11\/-30 and 5\/-15<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

= 7\/10, -11\/-30 and -5\/-5<\/p>\n\n\n\n

LCM of 10, 30 and 15 = 30<\/p>\n\n\n\n

= (7 \u00d7 3)\/ (10 \u00d7 3), 11\/30 and (-5 \u00d7 2)\/ (15 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 21\/30, 11\/30 and -10\/30<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -10 < 11 < 21<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-10\/ 30 < 11\/30 < 21\/30<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

5\/ -15 < -11\/ -30 < 7\/10<\/p>\n\n\n\n

(ii) 4\/-9, -5\/12 and 2\/-3<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

= -4\/9, -5\/12 and -2\/3<\/p>\n\n\n\n

LCM of 9, 12 and 3 is 36<\/p>\n\n\n\n

= (-4 \u00d7 4)\/ (9 \u00d7 4), (-5 \u00d7 3)\/ (12 \u00d7 3) and (-2 \u00d7 12)\/ (3 \u00d7 12)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -16\/36, -15\/36 and -24\/36<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -24 < -16 < -15<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-24\/36 < -16\/36 < -15\/36<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\/ -3 < 4\/ -9 < -5\/12<\/p>\n\n\n\n

5. Arrange the given rational numbers in descending order:<\/strong><\/p>\n\n\n\n

(i) 5\/8, 13\/-16 and -7\/12<\/strong><\/p>\n\n\n\n

(ii) 3\/-10, -13\/30 and 8\/-20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/8, 13\/-16 and -7\/12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/8, -13\/16 and -7\/12<\/p>\n\n\n\n

LCM of 8, 16 and 12 is 48<\/p>\n\n\n\n

= (5 \u00d7 6)\/ (8 \u00d7 6), (-13 \u00d7 3)\/ (16 \u00d7 3) and (-7 \u00d7 4)\/ (12 \u00d7 4)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 30\/48, -39\/48 and -28\/48<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, 30 > \u2013 28 > \u2013 39<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

30\/48 > -28\/48 > -39\/48<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

5\/8 > -7\/12 > 13\/-16<\/p>\n\n\n\n

(ii) 3\/-10, -13\/30 and 8\/-20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/10, -13\/30 and -8\/20<\/p>\n\n\n\n

LCM of 10, 20 and 30 is 60<\/p>\n\n\n\n

= (- 3 \u00d7 6)\/ (10 \u00d7 6), (-13 \u00d7 2)\/ (30 \u00d7 2) and (-8 \u00d7 3)\/ (20 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -18\/60, -26\/60 and -24\/60<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -18 > -24 > -26<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-18\/60 > -24\/60 > -26\/60<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3\/ -10 > 8\/-20 > -13\/30<\/p>\n\n\n\n

6. Fill in the blanks:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 3\/10 are on the \u2026\u2026. side of zero.<\/strong><\/p>\n\n\n\n

(ii) -5\/8 and 3\/10 are on the \u2026\u2026.. sides of zero.<\/strong><\/p>\n\n\n\n

(iii) -5\/8 and -3\/10 are on the \u2026\u2026. side of zero.<\/strong><\/p>\n\n\n\n

(iv) 5\/8 and -3\/10 are on the \u2026\u2026.. sides of zero.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 3\/10 are on the same side of zero.<\/p>\n\n\n\n

(ii) -5\/8 and 3\/10 are on the opposite sides of zero.<\/p>\n\n\n\n

(iii) -5\/8 and -3\/10 are on the same side of zero.<\/p>\n\n\n\n

(iv) 5\/8 and -3\/10 are on the opposite sides of zero.<\/p>\n\n\n\n

Exercise 2C page: 30<\/h4>\n\n\n\n

1. Add:<\/strong><\/p>\n\n\n\n

(i) 7\/5 and 2\/5<\/strong><\/p>\n\n\n\n

(ii) -4\/9 and 2\/9<\/strong><\/p>\n\n\n\n

(iii) 5\/-12 and 1\/12<\/strong><\/p>\n\n\n\n

(iv) 4\/-15 and -7\/-15<\/strong><\/p>\n\n\n\n

(v) -7\/25 and 9\/-25<\/strong><\/p>\n\n\n\n

(vi) -7\/26 and 7\/-26<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/5 and 2\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/5 + 2\/5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (7 + 2)\/ 5<\/p>\n\n\n\n

= 9\/5<\/p>\n\n\n\n

= 1 4\/5<\/p>\n\n\n\n

(ii) -4\/9 and 2\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -4\/9 + 2\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 + 2)\/ 9<\/p>\n\n\n\n

= -2\/9<\/p>\n\n\n\n

(iii) 5\/-12 and 1\/12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/12 + 1\/12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 + 1)\/ 12<\/p>\n\n\n\n

= -4\/ 12<\/p>\n\n\n\n

= -1\/3<\/p>\n\n\n\n

(iv) 4\/-15 and -7\/-15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 4\/15 + 7\/15<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 + 7)\/ 15<\/p>\n\n\n\n

= 3\/15<\/p>\n\n\n\n

= 1\/5<\/p>\n\n\n\n

(v) -7\/25 and 9\/-25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/25 + -9\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= [(-7) + (-9)]\/ 25<\/p>\n\n\n\n

= -16\/25<\/p>\n\n\n\n

(vi) -7\/26 and 7\/-26<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 7\/26 + -7\/26<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= [(-7) + (-7)]\/ 26<\/p>\n\n\n\n

= \u2013 14\/26<\/p>\n\n\n\n

= -7\/13<\/p>\n\n\n\n

2. Add:<\/strong><\/p>\n\n\n\n

(i) -2\/5 and 3\/7<\/strong><\/p>\n\n\n\n

(ii) -5\/6 and 4\/9<\/strong><\/p>\n\n\n\n

(iii) \u2013 3 and 2\/3<\/strong><\/p>\n\n\n\n

(iv) -5\/9 and 7\/18<\/strong><\/p>\n\n\n\n

(v) -7\/24 and -5\/48<\/strong><\/p>\n\n\n\n

(vi) 1\/-18 and 5\/-27<\/strong><\/p>\n\n\n\n

(vii) -9\/25 and 1\/-75<\/strong><\/p>\n\n\n\n

(viii) 13\/-16 and -11\/24<\/strong><\/p>\n\n\n\n

(ix) -9\/-16 and -11\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/5 and 3\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 \u00d7 7)\/ (5 \u00d7 7) + (3 \u00d7 5)\/ (7 \u00d7 5)<\/p>\n\n\n\n

LCM of 5 and 7 is 35<\/p>\n\n\n\n

= -14\/35 + 15\/35<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 15)\/35<\/p>\n\n\n\n

= 1\/35<\/p>\n\n\n\n

(ii) -5\/6 and 4\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/6 + 4\/9<\/p>\n\n\n\n

LCM of 6 and 9 is 36<\/p>\n\n\n\n

= (-5 \u00d7 6)\/ (6 \u00d7 6) + (4 \u00d7 4)\/ (9 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -30\/36 + 16\/36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-30 + 16)\/ 36<\/p>\n\n\n\n

= -14\/36<\/p>\n\n\n\n

= \u2013 7\/18<\/p>\n\n\n\n

(iii) \u2013 3 and 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/1 + 2\/3<\/p>\n\n\n\n

LCM of 1 and 3 is 3<\/p>\n\n\n\n

= (- 3 \u00d7 3)\/ (1 \u00d7 3) + (2 \u00d7 1)\/ (3 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -9\/3 + 2\/3<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-9 + 2)\/ 3<\/p>\n\n\n\n

= -7\/3<\/p>\n\n\n\n

(iv) -5\/9 and 7\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/9 + 7\/18<\/p>\n\n\n\n

LCM of 9 and 18 is 18<\/p>\n\n\n\n

= (-5 \u00d7 2)\/ (9 \u00d7 2) + (7 \u00d7 1)\/ (18 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -10\/18 + 7\/18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-10 + 7)\/ 18<\/p>\n\n\n\n

= \u2013 3\/18<\/p>\n\n\n\n

= -1\/6<\/p>\n\n\n\n

(v) -7\/24 and -5\/48<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/24 + -5\/48<\/p>\n\n\n\n

LCM of 24 and 48 is 48<\/p>\n\n\n\n

= (-7 \u00d7 2)\/ (24 \u00d7 2) + (-5 \u00d7 1)\/ (48 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -14\/48 + -5\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-14 \u2013 5)\/ 48<\/p>\n\n\n\n

= \u2013 19\/48<\/p>\n\n\n\n

(vi) 1\/-18 and 5\/-27<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 1\/18 + -5\/27<\/p>\n\n\n\n

LCM of 18 and 27 is 54<\/p>\n\n\n\n

= (-1 \u00d7 3)\/ (18 \u00d7 3) + (-5 \u00d7 2)\/ (27 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -3\/54 + -10\/54<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (- 3 \u2013 10)\/ 54<\/p>\n\n\n\n

= -13\/54<\/p>\n\n\n\n

(vii) -9\/25 and 1\/-75<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -9\/25 + -1\/75<\/p>\n\n\n\n

LCM of 24 and 75 is 75<\/p>\n\n\n\n

= (-9 \u00d7 3)\/ (25 \u00d7 3) + (-1 \u00d7 1)\/ (75 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -27\/75 + -1\/75<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-27 \u2013 1)\/ 75<\/p>\n\n\n\n

= -28\/75<\/p>\n\n\n\n

(viii) 13\/-16 and -11\/24<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -13\/16 + -11\/24<\/p>\n\n\n\n

LCM of 16 and 24 is 48<\/p>\n\n\n\n

= (-13 \u00d7 3)\/ (16 \u00d7 3) + (-11 \u00d7 2)\/ (24 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -39\/48 + -22\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-39 \u2013 22)\/ 48<\/p>\n\n\n\n

= -61\/48<\/p>\n\n\n\n

(ix) -9\/-16 and -11\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/16 + -11\/8<\/p>\n\n\n\n

LCM of 16 and 8 is 16<\/p>\n\n\n\n

= (9 \u00d7 1)\/ (16 \u00d7 1) + (-11 \u00d7 2)\/ (8 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 9\/16 + -22\/16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 \u2013 22)\/ 16<\/p>\n\n\n\n

= -13\/16<\/p>\n\n\n\n

3. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -2\/5 + 3\/5 + -1\/5<\/strong><\/p>\n\n\n\n

(ii) -8\/9 + 4\/9 + -2\/9<\/strong><\/p>\n\n\n\n

(iii) 5\/-24 + -1\/8 + 3\/16<\/strong><\/p>\n\n\n\n

(iv) -7\/6 + 4\/-15 + -4\/-30<\/strong><\/p>\n\n\n\n

(v) -2 + 2\/5 + -2\/15<\/strong><\/p>\n\n\n\n

(vi) -11\/12 + 5\/16 + -3\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/5 + 3\/5 + -1\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 + 3 \u2013 1)\/ 5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 0\/5<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

(ii) -8\/9 + 4\/9 + -2\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-8 + 4 \u2013 2)\/ 9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-10 + 4)\/ 9<\/p>\n\n\n\n

= -6\/9<\/p>\n\n\n\n

= -2\/3<\/p>\n\n\n\n

(iii) 5\/-24 + -1\/8 + 3\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/24 + -1\/8 + 3\/16<\/p>\n\n\n\n

LCM of 8, 16 and 24 is 48<\/p>\n\n\n\n

= (-5 \u00d7 2)\/ (24 \u00d7 2) + (-1 \u00d7 6)\/ (8 \u00d7 6) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -10\/ 48 + -6\/48 + 9\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-10 \u2013 6 + 9)\/ 48<\/p>\n\n\n\n

= (-16 + 9)\/ 48<\/p>\n\n\n\n

= -7\/48<\/p>\n\n\n\n

(iv) -7\/6 + 4\/-15 + -4\/-30<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/6 + -4\/15 + 4\/30<\/p>\n\n\n\n

LCM of 6, 15 and 30 is 30<\/p>\n\n\n\n

= (-7 \u00d7 5)\/ (6 \u00d7 5) + (-4 \u00d7 2)\/ (15 \u00d7 2) + (4 \u00d7 1)\/ (30 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -35\/30 + -8\/30 + 4\/30<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-35 \u2013 8 + 4)\/ 30<\/p>\n\n\n\n

= (-43 + 4)\/ 30<\/p>\n\n\n\n

= \u2013 39\/30<\/p>\n\n\n\n

= -13\/10<\/p>\n\n\n\n

(v) -2 + 2\/5 + -2\/15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/1 + 2\/5 + -2\/15<\/p>\n\n\n\n

LCM of 1, 5 and 15 is 15<\/p>\n\n\n\n

= (-2 \u00d7 15)\/ (1 \u00d7 15) + (2 \u00d7 3)\/ (5 \u00d7 3) + (-2 \u00d7 1)\/ (15 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -30\/15 + 6\/15 + -2\/15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-30 + 6 \u2013 2)\/ 15<\/p>\n\n\n\n

= (-32 + 6)\/ 15<\/p>\n\n\n\n

= -26\/15<\/p>\n\n\n\n

(vi) -11\/12 + 5\/16 + -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -11\/12 + 5\/16 + -3\/8<\/p>\n\n\n\n

LCM of 12, 16 and 8 is 48<\/p>\n\n\n\n

= (-11 \u00d7 4)\/ (12 \u00d7 4) + (5 \u00d7 3)\/ (16 \u00d7 3) + (-3 \u00d7 6)\/ (8 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -44\/48 + 15\/48 + -18\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-44 + 15 \u2013 18)\/ 48<\/p>\n\n\n\n

= (-62 + 15)\/ 48<\/p>\n\n\n\n

= -47\/ 48<\/p>\n\n\n\n

4. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -11\/18 + -3\/9 + 2\/-3<\/strong><\/p>\n\n\n\n

(ii) -9\/4 + 13\/3 +25\/6<\/strong><\/p>\n\n\n\n

(iii) \u2013 5 + 5\/-8 + -5\/-12<\/strong><\/p>\n\n\n\n

(iv) -2\/3 + 5\/2 + 2<\/strong><\/p>\n\n\n\n

(v) 5 + -3\/4 + -5\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -11\/18 + -3\/9 + 2\/-3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 11\/18 + -3\/9 + -2\/3<\/p>\n\n\n\n

LCM of 3, 9 and 18 is 18<\/p>\n\n\n\n

= (-11 \u00d7 1)\/ (18 \u00d7 1) + (-3 \u00d7 2)\/ (9 \u00d7 2) + (-2 \u00d7 6)\/ (3 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -11\/18 + -6\/18 + -12\/18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-11 \u2013 6 \u2013 12)\/18<\/p>\n\n\n\n

= -29\/18<\/p>\n\n\n\n

(ii) -9\/4 + 13\/3 +25\/6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -9\/4 + 13\/3 + 25\/6<\/p>\n\n\n\n

LCM of 4, 3 and 6 is 24<\/p>\n\n\n\n

= (-9 \u00d7 6)\/ (4 \u00d7 6) + (13 \u00d7 8)\/ (3 \u00d7 8) + (25 \u00d7 4)\/ (6 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -54\/24 + 104\/24 + 100\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-54 + 104 + 100)\/ 24<\/p>\n\n\n\n

= 150\/24<\/p>\n\n\n\n

= 25\/4<\/p>\n\n\n\n

= 6 1\/4<\/p>\n\n\n\n

(iii) \u2013 5 + 5\/-8 + -5\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/1 + -5\/8 + 5\/12<\/p>\n\n\n\n

LCM of 1, 8 and 12 is 24<\/p>\n\n\n\n

= (-5 \u00d7 24)\/ (1 \u00d7 24) + (-5 \u00d7 3)\/ (8 \u00d7 3) + (5 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -120\/24 + -15\/24 + 10\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-120 \u2013 15 + 10)\/24<\/p>\n\n\n\n

= -125\/24<\/p>\n\n\n\n

(iv) -2\/3 + 5\/2 + 2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/3 + 5\/2 + 2\/1<\/p>\n\n\n\n

LCM of 3, 2 and 1 is 6<\/p>\n\n\n\n

= (-2 \u00d7 2)\/ (3 \u00d7 2) + (5 \u00d7 3)\/ (2 \u00d7 3) + (2 \u00d7 6)\/ (1 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -4\/6 + 15\/6 + 12\/6<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-4 + 15 + 12)\/ 6<\/p>\n\n\n\n

= 23\/6<\/p>\n\n\n\n

= 3 5\/6<\/p>\n\n\n\n

(v) 5 + -3\/4 + -5\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/1 + -3\/4 + -5\/8<\/p>\n\n\n\n

LCM of 1, 4 and 8 is 8<\/p>\n\n\n\n

= (5 \u00d7 8)\/ (1 \u00d7 8) + (-3 \u00d7 2)\/ (4 \u00d7 2) + (-5 \u00d7 1)\/ (8 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 40\/8 + -6\/8 + -5\/8<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (40 \u2013 6 \u2013 5)\/ 8<\/p>\n\n\n\n

= (40 \u2013 11)\/8<\/p>\n\n\n\n

= 29\/8<\/p>\n\n\n\n

= 3 5\/8<\/p>\n\n\n\n

5. Subtract:<\/strong><\/p>\n\n\n\n

(i) 2\/9 from 5\/9<\/strong><\/p>\n\n\n\n

(ii) -6\/11 from -3\/-11<\/strong><\/p>\n\n\n\n

(iii) -2\/15 from -8\/15<\/strong><\/p>\n\n\n\n

(iv) 11\/18 from -5\/18<\/strong><\/p>\n\n\n\n

(v) -4\/11 from -2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 2\/9 from 5\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/9 \u2013 2\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (5 \u2013 2)\/ 9<\/p>\n\n\n\n

= 3\/9<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

(ii) -6\/11 from -3\/-11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/ 11 \u2013 (-6\/11)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 3\/11 + 6\/11<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (3 + 6)\/ 11<\/p>\n\n\n\n

= 9\/11<\/p>\n\n\n\n

(iii) -2\/15 from -8\/15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -8\/15 \u2013 (-2\/15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -8\/15 + 2\/15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-8 + 2)\/ 15<\/p>\n\n\n\n

= -6\/ 15<\/p>\n\n\n\n

= -2\/5<\/p>\n\n\n\n

(iv) 11\/18 from -5\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/18 \u2013 11\/18<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u2013 11)\/ 18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -16\/18<\/p>\n\n\n\n

= -8\/9<\/p>\n\n\n\n

(v) -4\/11 from -2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/1 \u2013 (-4\/11)<\/p>\n\n\n\n

LCM of 1 and 11 is 11<\/p>\n\n\n\n

= (- 2 \u00d7 11)\/ (1 \u00d7 11) + (4 \u00d7 1)\/ (11 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -22\/11 + 4\/11<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-22 + 4)\/ 11<\/p>\n\n\n\n

= -18\/11<\/p>\n\n\n\n

6. Subtract:<\/strong><\/p>\n\n\n\n

(i) -3\/10 from 1\/5<\/strong><\/p>\n\n\n\n

(ii) -6\/25 from -8\/5<\/strong><\/p>\n\n\n\n

(iii) -7\/4 from -2<\/strong><\/p>\n\n\n\n

(iv) -16\/21 from 1<\/strong><\/p>\n\n\n\n

(v) -8\/15 from 0<\/strong><\/p>\n\n\n\n

(vi) 0 from -3\/8<\/strong><\/p>\n\n\n\n

(vii) -2 from -3\/10<\/strong><\/p>\n\n\n\n

(viii) 5\/8 from -5\/16<\/strong><\/p>\n\n\n\n

(ix) 4 from -3\/13<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -3\/10 from 1\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1\/5 \u2013 (-3\/10)<\/p>\n\n\n\n

LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (1 \u00d7 2)\/ (5 \u00d7 2) + 3\/10<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/10 + 3\/10<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (2 + 3)\/ 10<\/p>\n\n\n\n

= 5\/10<\/p>\n\n\n\n

= 1\/2<\/p>\n\n\n\n

(ii) -6\/25 from -8\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -8\/5 \u2013 (-6\/25)<\/p>\n\n\n\n

LCM of 5 and 25 is 25<\/p>\n\n\n\n

= (-8 \u00d7 5)\/ (5 \u00d7 5) + 6\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -40\/25 + 6\/25<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-40 + 6)\/ 25<\/p>\n\n\n\n

= -34\/25<\/p>\n\n\n\n

(iii) -7\/4 from -2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2\/1) \u2013 (-7\/4)<\/p>\n\n\n\n

LCM of 1 and 4 is 4<\/p>\n\n\n\n

= (-2 \u00d7 4)\/ (1 \u00d7 4) + 7\/4<\/p>\n\n\n\n

= -8\/4 + 7\/4<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 7)\/ 4<\/p>\n\n\n\n

= -1\/4<\/p>\n\n\n\n

(iv) -16\/21 from 1<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1\/1 \u2013 (-16\/21)<\/p>\n\n\n\n

= 1\/1 + 16\/21<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (21 + 16)\/ 21<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (21 + 16)\/21<\/p>\n\n\n\n

= 37\/21<\/p>\n\n\n\n

= 1 16\/21<\/p>\n\n\n\n

(v) -8\/15 from 0<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 0 \u2013 (-8\/15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 0 + 8\/15<\/p>\n\n\n\n

= 8\/15<\/p>\n\n\n\n

(vi) 0 from -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/8 \u2013 0<\/p>\n\n\n\n

= -3\/8<\/p>\n\n\n\n

(vii) -2 from -3\/10<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/10 \u2013 (-2\/1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -3\/10 + 2\/1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 + 2 \u00d7 10)\/10<\/p>\n\n\n\n

= 17\/10<\/p>\n\n\n\n

= 1 7\/10<\/p>\n\n\n\n

(viii) 5\/8 from -5\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/16 \u2013 5\/8<\/p>\n\n\n\n

LCM of 8 and 16 is 16<\/p>\n\n\n\n

= -5\/16 \u2013 (5 \u00d7 2)\/ (8 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -5\/16 \u2013 10\/16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u2013 10)\/16<\/p>\n\n\n\n

= \u2013 15\/16<\/p>\n\n\n\n

(ix) 4 from -3\/13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/13 \u2013 4\/1<\/p>\n\n\n\n

LCM of 13 and 1 is 13<\/p>\n\n\n\n

= (- 3 \u2013 4 \u00d7 13)\/ 13<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u2013 52)\/13<\/p>\n\n\n\n

= -55\/13<\/p>\n\n\n\n

7. The sum of two rational numbers is 11\/24. If one of them is 3\/8, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = 11\/24<\/p>\n\n\n\n

One of the rational number = 3\/8<\/p>\n\n\n\n

Other rational number = 11\/24 \u2013 3\/8<\/p>\n\n\n\n

LCM of 24 and 8 is 24<\/p>\n\n\n\n

= 11\/24 \u2013 (3 \u00d7 3)\/ (8 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 11\/24 \u2013 9\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (11 \u2013 9)\/ 24<\/p>\n\n\n\n

= 2\/24<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

8. The sum of two rational numbers is -7\/12. If one of them is 13\/24, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -7\/12<\/p>\n\n\n\n

One of the rational number = 13\/24<\/p>\n\n\n\n

Other rational number = -7\/12 \u2013 13\/24<\/p>\n\n\n\n

LCM of 12 and 24 is 24<\/p>\n\n\n\n

= (-7 \u00d7 2)\/ (12 \u00d7 2) \u2013 13\/24<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -14\/24 \u2013 13\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-14 \u2013 13)\/ 24<\/p>\n\n\n\n

= -27\/24<\/p>\n\n\n\n

= -9\/8<\/p>\n\n\n\n

9. The sum of two rational numbers is -4. If one of them is -13\/12, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -4<\/p>\n\n\n\n

One of the rational number = -13\/12<\/p>\n\n\n\n

Other rational number = \u2013 4 \u2013 (-13\/12)<\/p>\n\n\n\n

LCM of 1 and 12 is 12<\/p>\n\n\n\n

= \u2013 4 + 13\/12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 \u00d7 12 + 13)\/ 12<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-48 + 13)\/ 12<\/p>\n\n\n\n

= -35\/12<\/p>\n\n\n\n

10. What should be added to -3\/16 to get 11\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -3\/16<\/p>\n\n\n\n

Sum of two numbers = 11\/24<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-3\/16 + x = 11\/24<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 11\/24 + 3\/16<\/p>\n\n\n\n

LCM of 16 and 24 is 48<\/p>\n\n\n\n

x = (11 \u00d7 2)\/ (24 \u00d7 2) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 22\/48 + 9\/48<\/p>\n\n\n\n

x = (22 + 9)\/ 48 = 31\/48<\/p>\n\n\n\n

11. What should be added to -3\/5 to get 2?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -3\/5<\/p>\n\n\n\n

Here the sum of two numbers is 2<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-3\/5 + x = 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 2 + 3\/5<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

x = (2 \u00d7 5 + 3)\/ 5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (10 + 3)\/ 5<\/p>\n\n\n\n

= 13\/5<\/p>\n\n\n\n

= 2 3\/5<\/p>\n\n\n\n

12. What should be subtracted from -4\/5 to get 1?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -4\/5<\/p>\n\n\n\n

Here the difference between two numbers is 1<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-4\/5 \u2013 x = 1<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

-4\/5 \u2013 1 = x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = (-4 \u2013 1 \u00d7 5)\/ 5<\/p>\n\n\n\n

x = (-4 \u2013 5)\/ 5 = -9\/5<\/p>\n\n\n\n

13. The sum of two numbers is -6\/5. If one of them is -2, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two numbers = -6\/5<\/p>\n\n\n\n

One of the numbers = -2<\/p>\n\n\n\n

Other number = -6\/5 \u2013 (-2\/1)<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

= \u2013 6\/5 \u2013 (2 \u00d7 5)\/ (1 \u00d7 5)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 + 10)\/ 5<\/p>\n\n\n\n

= 4\/5<\/p>\n\n\n\n

14. What should be added to -7\/12 to get 3\/8?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other rational number = -7\/12<\/p>\n\n\n\n

Sum of two numbers = 3\/8<\/p>\n\n\n\n

Using the question<\/p>\n\n\n\n

-7\/12 + x = 3\/8<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 3\/8 \u2013 (-7\/12)<\/p>\n\n\n\n

LCM of 8 and 12 is 24<\/p>\n\n\n\n

x = (3 \u00d7 3)\/ (8 \u00d7 3) + (7 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 9\/24 + 14\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 + 14)\/ 24 = 23\/34<\/p>\n\n\n\n

15. What should be subtracted from 5\/9 to get 9\/5?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the first number<\/p>\n\n\n\n

Other number is 5\/9<\/p>\n\n\n\n

Here the difference between two numbers is 9\/5<\/p>\n\n\n\n

Using the question<\/p>\n\n\n\n

5\/9 \u2013 x = 9\/5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 5\/9 \u2013 9\/5<\/p>\n\n\n\n

LCM of 9 and 5 is 45<\/p>\n\n\n\n

x = (5 \u00d7 5)\/ (9 \u00d7 5) \u2013 (9 \u00d7 9)\/ (5 \u00d7 9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 25\/45 \u2013 81\/45<\/p>\n\n\n\n

x = (25 \u2013 81)\/ 45 = \u2013 56\/45<\/p>\n\n\n\n

Exercise 2D page: 34<\/h4>\n\n\n\n

1. Evaluate:<\/strong><\/p>\n\n\n\n

(i) 5\/4 \u00d7 3\/7<\/strong><\/p>\n\n\n\n

(ii) 2\/3 \u00d7 -6\/7<\/strong><\/p>\n\n\n\n

(iii) (-12\/5) \u00d7 (10\/-3)<\/strong><\/p>\n\n\n\n

(iv) -45\/39 \u00d7 -13\/ 15<\/strong><\/p>\n\n\n\n

(v) 3 1\/8 \u00d7 (-2 2\/5)<\/strong><\/p>\n\n\n\n

(vi) 2 14\/25 \u00d7 (-5\/16)<\/strong><\/p>\n\n\n\n

(vii) (-8\/9) \u00d7 (-3\/ 16)<\/strong><\/p>\n\n\n\n

(viii) (5\/-27) \u00d7 (-9\/ 20)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/4 \u00d7 3\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5 \u00d7 3)\/ (4 \u00d7 7)<\/p>\n\n\n\n

= 15\/28<\/p>\n\n\n\n

(ii) 2\/3 \u00d7 -6\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 -6)\/ (3 \u00d7 7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (2 \u00d7 -2)\/ 7<\/p>\n\n\n\n

= -4\/7<\/p>\n\n\n\n

(iii) (-12\/5) \u00d7 (10\/-3)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-12 \u00d7 10)\/ (5 \u00d7 -3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 4 \u00d7 2<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(iv) -45\/39 \u00d7 -13\/ 15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-45 \u00d7 -13)\/ (39 \u00d7 15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u00d7 -1)\/ (3 \u00d7 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3\/3<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

(v) 3 1\/8 \u00d7 (-2 2\/5)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (3 \u00d7 8 + 1)\/ 8 \u00d7 (-2 \u00d7 5 + 2)\/ 5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 25\/8 \u00d7 (-12\/5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (25 \u00d7 -12)\/ (8 \u00d7 5)<\/p>\n\n\n\n

On further simplification<\/p>\n\n\n\n

= (5 \u00d7 -3)\/ (2 \u00d7 1)<\/p>\n\n\n\n

= -15\/2<\/p>\n\n\n\n

(vi) 2 14\/25 \u00d7 (-5\/16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 25 + 14)\/ 25 \u00d7 (-5\/16)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 64\/25 \u00d7 (-5\/16)<\/p>\n\n\n\n

= (64 \u00d7 -5)\/ (25 \u00d7 16)<\/p>\n\n\n\n

On further simplification<\/p>\n\n\n\n

= (4 \u00d7 -1)\/ (5 \u00d7 1)<\/p>\n\n\n\n

= -4\/5<\/p>\n\n\n\n

(vii) (-8\/9) \u00d7 (-3\/ 16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-8 \u00d7 -3)\/ (9 \u00d7 16)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -1)\/ (3 \u00d7 2)<\/p>\n\n\n\n

= 1\/6<\/p>\n\n\n\n

(viii) (5\/-27) \u00d7 (-9\/ 20)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5 \u00d7 -9)\/ (-27 \u00d7 20)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1 \u00d7 1)\/ (3 \u00d7 4)<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

2. Multiply:<\/strong><\/p>\n\n\n\n

(i) 3\/25 and 4\/5<\/strong><\/p>\n\n\n\n

(ii) 1 1\/8 and 10 2\/3<\/strong><\/p>\n\n\n\n

(iii) 6 2\/3 and -3\/8<\/strong><\/p>\n\n\n\n

(iv) -13\/15 and -25\/26<\/strong><\/p>\n\n\n\n

(v) 1 1\/6 and 18<\/strong><\/p>\n\n\n\n

(vi) 2 1\/14 and -7<\/strong><\/p>\n\n\n\n

(vii) 5 1\/8 and -16<\/strong><\/p>\n\n\n\n

(viii) 35 and -18\/25<\/strong><\/p>\n\n\n\n

(ix) 6 2\/3 and -3\/8<\/strong><\/p>\n\n\n\n

(x) 3 3\/5 and -10<\/strong><\/p>\n\n\n\n

(xi) 27\/28 and -14<\/strong><\/p>\n\n\n\n

(xii) -24 and 5\/16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/25 and 4\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/25 \u00d7 4\/5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (3 \u00d7 4)\/ (25 \u00d7 5)<\/p>\n\n\n\n

= 12\/125<\/p>\n\n\n\n

(ii) 1 1\/8 and 10 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/8 \u00d7 32\/2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (9 \u00d7 32)\/ (8 \u00d7 3)<\/p>\n\n\n\n

= 3 \u00d7 4<\/p>\n\n\n\n

= 12<\/p>\n\n\n\n

(iii) 6 2\/3 and -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 20\/3 \u00d7 -3\/8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (20 \u00d7 -3)\/ (3 \u00d7 8)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 -1)\/ (1 \u00d7 2)<\/p>\n\n\n\n

= -5\/2<\/p>\n\n\n\n

(iv) -13\/15 and -25\/26<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-13 \u00d7 -25)\/ (15 \u00d7 26)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -5)\/ (3 \u00d7 2)<\/p>\n\n\n\n

= 5\/6<\/p>\n\n\n\n

(v) 1 1\/6 and 18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/6 \u00d7 18<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 7 \u00d7 3<\/p>\n\n\n\n

= 21<\/p>\n\n\n\n

(vi) 2 1\/14 and -7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 14 + 1)\/ 14 \u00d7 (-7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 29\/4 \u00d7 (-7)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (29 \u00d7 -1)\/ 2<\/p>\n\n\n\n

= -29\/2<\/p>\n\n\n\n

(vii) 5 1\/8 and -16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 41\/8 \u00d7 -16<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 41 \u00d7 -2<\/p>\n\n\n\n

= -82<\/p>\n\n\n\n

(viii) 35 and -18\/25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 35 \u00d7 -18\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (35 \u00d7 -18)\/ 25<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 -18)\/ 5<\/p>\n\n\n\n

= -126\/5<\/p>\n\n\n\n

(ix) 6 2\/3 and -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 20\/3 \u00d7 -3\/8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (20 \u00d7 -3)\/ (3 \u00d7 8)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 -1)\/ (1 \u00d7 2)<\/p>\n\n\n\n

= -5\/2<\/p>\n\n\n\n

(x) 3 3\/5 and -10<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (3 \u00d7 5 + 3)\/ 5 \u00d7 -10<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 18\/5 \u00d7 -10<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 18 \u00d7 -2<\/p>\n\n\n\n

= -36<\/p>\n\n\n\n

(xi) 27\/28 and -14<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 27\/28 \u00d7 -14<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (27 \u00d7 -1)\/ 2<\/p>\n\n\n\n

= -27\/2<\/p>\n\n\n\n

(xii) -24 and 5\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-24 \u00d7 5)\/ 16<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u00d7 5)\/ 2<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -15\/2<\/p>\n\n\n\n

3. Evaluate:<\/strong><\/p>\n\n\n\n

(i) (6 \u00d7 5\/18) \u2013 (- 4 2\/9)<\/strong><\/p>\n\n\n\n

(ii) (7\/8 \u00d7 8\/7) + (-5\/9) \u00d7 (6\/-25)<\/strong><\/p>\n\n\n\n

(iii) (11\/-9 \u00d7 21\/44) + (-5\/9) \u00d7 (63\/ -100)<\/strong><\/p>\n\n\n\n

(iv) (-5\/9 \u00d7 6\/-25) + (24\/21 \u00d7 7\/8)
(v) (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/strong><\/p>\n\n\n\n

(vi) (-4\/5 \u00d7 3\/2) + (9\/-5 \u00d7 10\/3) \u2013 (-3\/2 \u00d7 -1\/4)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (6 \u00d7 5\/18) \u2013 (- 4 2\/9)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-1 \u00d7 5\/3) \u2013 [- (4 \u00d7 9 + 2)\/ 9]<\/p>\n\n\n\n

LCM of 3 and 9 is 9<\/p>\n\n\n\n

= -5\/3 \u2013 (-38\/9)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -5\/3 + 38\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u00d7 3)\/ (3 \u00d7 3) + (38 \u00d7 1)\/ (9 \u00d7 1)<\/p>\n\n\n\n

= (-15 + 38)\/ 9<\/p>\n\n\n\n

= 23\/9<\/p>\n\n\n\n

= 2 5\/9<\/p>\n\n\n\n

(ii) (7\/8 \u00d7 8\/7) + (-5\/9) \u00d7 (6\/-25)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (7\/8 \u00d7 8\/7) + (-5\/9 \u00d7 6\/-25)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 1\/1 + (1 \u00d7 2)\/ (3 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 1\/1 + 2\/15<\/p>\n\n\n\n

= (15 + 2)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n

(iii) (11\/-9 \u00d7 21\/44) + (-5\/9) \u00d7 (63\/ -100)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (11\/-9 \u00d7 21\/44) + (5\/9 \u00d7 63\/100)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 7)\/ (3 \u00d7 4) + (1 \u00d7 7)\/ (1 \u00d7 20)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -7\/12 + 7\/20<\/p>\n\n\n\n

LCM of 12 and 20 is 60<\/p>\n\n\n\n

= (-7 \u00d7 5)\/ (12 \u00d7 5) + (7 \u00d7 3)\/ (20 \u00d7 3)<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

= -35\/60 + 211\/60<\/p>\n\n\n\n

= (-35 + 21)\/ 60<\/p>\n\n\n\n

= -14\/60<\/p>\n\n\n\n

= -7\/30<\/p>\n\n\n\n

(iv) (-5\/9 \u00d7 6\/-25) + (24\/21 \u00d7 7\/8)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5\/9 \u00d7 6\/25) + (24\/21 \u00d7 7\/8)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/ (3 \u00d7 5) + 1<\/p>\n\n\n\n

= 2\/15 + 1<\/p>\n\n\n\n

LCM of 15 and 1 is 15<\/p>\n\n\n\n

= (2 + 15)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n


(v) (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u00d7 -1)\/ (3 \u00d7 1) \u2013 (1 \u00d7 -1)\/ (5 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 5\/3 \u2013 (-1\/10)<\/p>\n\n\n\n

LCM of 3 and 10 is 30<\/p>\n\n\n\n

= (5 \u00d7 10)\/ (3 \u00d7 10) + 1\/ (10 \u00d7 3)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= (50 + 3)\/ 30<\/p>\n\n\n\n

= 53\/30<\/p>\n\n\n\n

= 1 23\/30<\/p>\n\n\n\n

(vi) (-4\/5 \u00d7 3\/2) + (9\/-5 \u00d7 10\/3) \u2013 (-3\/2 \u00d7 -1\/4)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 \u00d7 3)\/ (5 \u00d7 1) + (3 \u00d7 2)\/ (-1 \u00d7 1) \u2013 (-3 \u00d7 -1)\/ (2 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -6\/5 + -6\/1 \u2013 3\/8<\/p>\n\n\n\n

LCM of 5, 1 and 8 is 40<\/p>\n\n\n\n

= = (-6 \u00d7 8)\/ (5 \u00d7 8) \u2013 (6 \u00d7 40)\/ (1 \u00d7 40) \u2013 (3 \u00d7 5)\/ (8 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-48 \u2013 240 \u2013 15)\/ 40<\/p>\n\n\n\n

= \u2013 303\/40<\/p>\n\n\n\n

4. Find the cost of 3 \u00bd m cloth, if one metre cloth costs \u20b9 325 \u00bd.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that cost of one metre cloth = \u20b9 325 \u00bd<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 325 + 1)\/ 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (650 + 1)\/ 2<\/p>\n\n\n\n

= \u20b9 651\/2<\/p>\n\n\n\n

Cost of 3 \u00bd m cloth<\/p>\n\n\n\n

(2 \u00d7 3 + 1)\/ 2 = 7\/2 m<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 651\/2 \u00d7 7\/2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (651 \u00d7 7)\/ (2 \u00d7 2)<\/p>\n\n\n\n

= 4557\/4<\/p>\n\n\n\n

= \u20b9 1139 \u00bc<\/p>\n\n\n\n

5. A bus is moving with a speed of 65 \u00bd km per hour. How much distance will it cover in 1 1\/3 hours.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Speed of bus = 65 \u00bd km per hour<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 65 + 1)\/ 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (130 + 1)\/ 2<\/p>\n\n\n\n

= 131\/ 2 km<\/p>\n\n\n\n

Distance covered in 1 1\/3 hour = 4\/3 hour can be written as<\/p>\n\n\n\n

= 131\/2 \u00d7 4\/3<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 131\/1 \u00d7 2\/3<\/p>\n\n\n\n

We know that distance covered = speed \u00d7 time<\/p>\n\n\n\n

= 131\/2 \u00d7 4\/3<\/p>\n\n\n\n

= (131 \u00d7 2)\/ (1 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 262\/3<\/p>\n\n\n\n

= 87 1\/3 km<\/p>\n\n\n\n

6. Divide:<\/strong><\/p>\n\n\n\n

(i) 15\/28 by \u00be<\/strong><\/p>\n\n\n\n

(ii) -20\/9 by -5\/9<\/strong><\/p>\n\n\n\n

(iii) 16\/-5 by -8\/7<\/strong><\/p>\n\n\n\n

(iv) -7 by -14\/5<\/strong><\/p>\n\n\n\n

(v) -14 by 7\/-2<\/strong><\/p>\n\n\n\n

(vi) -22\/9 by 11\/18<\/strong><\/p>\n\n\n\n

(vii) 35 by -7\/9<\/strong><\/p>\n\n\n\n

(viii) 21\/44 by -11\/9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 15\/28 by \u00be<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 15\/28 \u00f7 3\/4<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 15\/28 \u00d7 4\/3<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 5\/7 \u00d7 1\/1<\/p>\n\n\n\n

= 5\/7<\/p>\n\n\n\n

(ii) -20\/9 by -5\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -20\/9 \u00f7 -5\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -20\/9 \u00d7 9\/-5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -4\/-1<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

(iii) 16\/-5 by -8\/7<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 16\/-5 \u00f7 -8\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 16\/-5 \u00d7 7\/-8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/-5 \u00d7 7\/-1<\/p>\n\n\n\n

= (2 \u00d7 7)\/ (-5 \u00d7 -1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 14\/5<\/p>\n\n\n\n

= 2 4\/5<\/p>\n\n\n\n

(iv) -7 by -14\/5<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -7 \u00f7 \u2013 14\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7 \u00d7 5\/-14<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 1 \u00d7 5\/2<\/p>\n\n\n\n

= (1 \u00d7 5)\/ 2<\/p>\n\n\n\n

= 5\/2<\/p>\n\n\n\n

= 2 \u00bd<\/p>\n\n\n\n

(v) -14 by 7\/-2<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -14 \u00f7 7\/-2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -14 \u00d7 -2\/7<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-2 \u00d7 -2)\/ (1 \u00d7 1)<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

(vi) -22\/9 by 11\/18<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= \u2013 22\/9 \u00f7 11\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -22\/9 \u00d7 18\/11<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -2\/1 \u00d7 2\/1<\/p>\n\n\n\n

= (-2 \u00d7 2)\/ (1 \u00d7 1)<\/p>\n\n\n\n

= \u2013 4\/1<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

(vii) 35 by -7\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 35 \u00f7 -7\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 35 \u00d7 9\/-7<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 5 \u00d7 9\/-1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 9)\/ -1<\/p>\n\n\n\n

= 45\/-1<\/p>\n\n\n\n

= -45<\/p>\n\n\n\n

(viii) 21\/44 by -11\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 21\/44 \u00f7 -11\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 21\/44 \u00d7 -9\/11<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (21 \u00d7 -9)\/ (44 \u00d7 11)<\/p>\n\n\n\n

= -189\/484<\/p>\n\n\n\n

7. Evaluate:<\/strong><\/p>\n\n\n\n

(i) 3 5\/12 + 1 2\/3<\/strong><\/p>\n\n\n\n

(ii) 3 5\/12 \u2013 1 2\/3<\/strong><\/p>\n\n\n\n

(iii) (3 5\/12 + 1 2\/3) \u00f7 (3 5\/12 \u2013 1 2\/3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3 5\/12 + 1 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (12 \u00d7 3 + 5)\/ 12 + (3 \u00d7 1 + 2)\/ 3<\/p>\n\n\n\n

= 41\/12 + 5\/3<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 \u00d7 1)\/ (12 \u00d7 1) + (5 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 41\/12 + 20\/12<\/p>\n\n\n\n

= (41 + 20)\/ 12<\/p>\n\n\n\n

= 61\/12<\/p>\n\n\n\n

= 5 1\/12<\/p>\n\n\n\n

(ii) 3 5\/12 \u2013 1 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (12 \u00d7 3 + 5)\/ 12 \u2013 (3 \u00d7 1 + 2)\/ 3<\/p>\n\n\n\n

= 41\/12 \u2013 5\/3<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 \u00d7 1)\/ (12 \u00d7 1) \u2013 (5 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (41 \u2013 20)\/ 12<\/p>\n\n\n\n

= 21\/12<\/p>\n\n\n\n

= 2\/4<\/p>\n\n\n\n

= 1 \u00be<\/p>\n\n\n\n

(iii) (3 5\/12 + 1 2\/3) \u00f7 (3 5\/12 \u2013 1 2\/3)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= [(12 \u00d7 3 + 5)\/ 12 + (3 \u00d7 1 + 2)\/ 3] \u00f7 [(12 \u00d7 3 + 5)\/ 12 \u2013 (3 \u00d7 1 + 2)\/ 3]<\/p>\n\n\n\n

= (41\/12 + 5\/3) \u00f7 (41\/12 \u2013 5\/3)<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 + 20)\/ 12 \u00f7 (41 \u2013 20)\/ 12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 61\/12 \u00f7 21\/12<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 61\/12 \u00d7 12\/21<\/p>\n\n\n\n

= 61\/21<\/p>\n\n\n\n

= 2 19\/21<\/p>\n\n\n\n

8. The product of two numbers is 14. If one of the numbers is -8\/7, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Product of two numbers = 14<\/p>\n\n\n\n

One of the number = -8\/7<\/p>\n\n\n\n

Other number = 14 \u00f7 -8\/7<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 14 \u00d7 -7\/8<\/p>\n\n\n\n

= -98\/8<\/p>\n\n\n\n

= \u2013 49\/4<\/p>\n\n\n\n

9. The cost of 11 pens is \u20b9 24 \u00be. Find the cost of one pen.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Cost of 11 pens = \u20b9 24 \u00be<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (24 \u00d7 4 + 3)\/ 4<\/p>\n\n\n\n

= \u20b9 99\/4<\/p>\n\n\n\n

So the cost of one pen = 99\/4 \u00f7 11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 99\/4 \u00d7 1\/11<\/p>\n\n\n\n

= \u20b9 9\/4<\/p>\n\n\n\n

= \u20b9 2 \u00bc<\/p>\n\n\n\n

10. If 6 identical articles can be bought for \u20b9 2 6\/17. Find the cost of each article.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Cost of 6 articles = \u20b9 2 6\/17<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 17 + 6)\/ 17<\/p>\n\n\n\n

= \u20b9 40\/17<\/p>\n\n\n\n

So the cost of each article = 40\/17 \u00f7 6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 40\/17 \u00d7 1\/6<\/p>\n\n\n\n

= \u20b9 20\/51<\/p>\n\n\n\n

11. By what number should -3\/8 be multiplied so that the product is -9\/16?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Number = -3\/8 \u00f7 (-9\/16)<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= -3\/8 \u00d7 16\/-9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/3<\/p>\n\n\n\n

= 1 \u00bd<\/p>\n\n\n\n

12. By what number should -5\/7 be divided so that the result is -15\/28?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the number as x<\/p>\n\n\n\n

-5\/7 \u00f7 x = -15\/28<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-5\/7 \u00d7 1\/x = -15\/28<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

-5\/7x = -15\/28<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 5\/7 \u00d7 28\/15 = 4\/3<\/p>\n\n\n\n

x = 1 1\/3<\/p>\n\n\n\n

13. Evaluate: (32\/15 + 8\/5) \u00f7 (32\/15 \u2013 8\/5).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

(32\/15 + 8\/5) \u00f7 (32\/15 \u2013 8\/5)<\/p>\n\n\n\n

LCM of 15 and 5 is 15<\/p>\n\n\n\n

= [(32 \u00d7 1)\/ (15 \u00d7 1) + (8 \u00d7 3)\/ (5 \u00d7 3)] \u00f7 [(32 \u00d7 1)\/ (15 \u00d7 1) \u2013 (8 \u00d7 1)\/ (5 \u00d7 1)]<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (32 + 24)\/ 15 \u00f7 (32 \u2013 24)\/ 15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 56\/15 \u00f7 8\/15<\/p>\n\n\n\n

= 56\/15 \u00d7 15\/8<\/p>\n\n\n\n

= 7<\/p>\n\n\n\n

14. Seven equal pieces are made out of a rope of 21 5\/7 m. Find the length of each piece.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Length of 7 pieces of rope = 21 5\/7 m<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (21 \u00d7 7 + 5)\/ 7<\/p>\n\n\n\n

= 152\/7<\/p>\n\n\n\n

So the length of each piece = 152\/7 \u00f7 7<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 152\/7 \u00d7 1\/7<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 152\/49<\/p>\n\n\n\n

= 3 5\/49 m<\/p>\n\n\n\n

Exercise 2E page: 36<\/h4>\n\n\n\n

1. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -2\/3 + 3\/4<\/strong><\/p>\n\n\n\n

(ii) 7\/-27 + 11\/18<\/strong><\/p>\n\n\n\n

(iii) -3\/8 + -5\/12<\/strong><\/p>\n\n\n\n

(iv) 9\/-16 + -5\/-12<\/strong><\/p>\n\n\n\n

(v) -5\/9 + -7\/12 + 11\/18<\/strong><\/p>\n\n\n\n

(vi) 7\/-26 + 16\/39<\/strong><\/p>\n\n\n\n

(vii) -2\/3 \u2013 (-5\/7)<\/strong><\/p>\n\n\n\n

(viii) -5\/7 \u2013 (-3\/8)<\/strong><\/p>\n\n\n\n

(ix) 7\/26 + 2 + -11\/13<\/strong><\/p>\n\n\n\n

(x) -1 + 2\/-3 + 5\/6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/3 + 3\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 3 and 4 is 12<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-2 \u00d7 4)\/ (3 \u00d7 4) + (3 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 9)\/ 12<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

(ii) 7\/-27 + 11\/18<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 27 and 18 is 54<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 2)\/ (-27 \u00d7 2) + (11 \u00d7 3)\/ (18 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 33)\/ 54<\/p>\n\n\n\n

= 19\/54<\/p>\n\n\n\n

(iii) -3\/8 + -5\/12<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 12 is 24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 \u00d7 3)\/ (8 \u00d7 3) + (-5 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-9 \u2013 10)\/ 24<\/p>\n\n\n\n

= -19\/24<\/p>\n\n\n\n

(iv) 9\/-16 + -5\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/-16 + 5\/12<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 16 and 12 is 48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 \u00d7 3)\/ (-16 \u00d7 3) + (5 \u00d7 4)\/ (12 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-27 + 20)\/48<\/p>\n\n\n\n

= -7\/48<\/p>\n\n\n\n

(v) -5\/9 + -7\/12 + 11\/18<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 9, 12 and 18 is 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u00d7 4)\/ (9 \u00d7 4) \u2013 (7 \u00d7 3)\/ (12 \u00d7 3) + (11 \u00d7 2)\/ (18 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-20 \u2013 21 + 22)\/ 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-41 + 22)\/ 36<\/p>\n\n\n\n

= -19\/36<\/p>\n\n\n\n

(vi) 7\/-26 + 16\/39<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 26 and 39 is 78<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-7 \u00d7 3)\/ (26 \u00d7 3) + (16 \u00d7 2)\/ (39 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-21 + 32)\/ 78<\/p>\n\n\n\n

= 11\/78<\/p>\n\n\n\n

(vii) -2\/3 \u2013 (-5\/7)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/3 + 5\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 7 is 21<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-2 \u00d7 7)\/ (3 \u00d7 7) + (5 \u00d7 3)\/ (7 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 15)\/ 21<\/p>\n\n\n\n

= 1\/21<\/p>\n\n\n\n

(viii) -5\/7 \u2013 (-3\/8)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/7 + 3\/8<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 7 and 8 is 56<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u00d7 8)\/ (7 \u00d7 8) + (3 \u00d7 7)\/ (8 \u00d7 7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-40 + 21)\/ 56<\/p>\n\n\n\n

= -19\/56<\/p>\n\n\n\n

(ix) 7\/26 + 2 + -11\/13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/26 + 2\/1 + -11\/13<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 26 and 13 is 26<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 1)\/ (26 \u00d7 1) + (2 \u00d7 26)\/ (1 \u00d7 26) \u2013 (11 \u00d7 2)\/ (13 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (7 + 52 \u2013 22)\/ 26<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (59 \u2013 22)\/ 26<\/p>\n\n\n\n

= 37\/26<\/p>\n\n\n\n

= 1 11\/26<\/p>\n\n\n\n

(x) -1 + 2\/-3 + 5\/6<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 6 is 6<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-1 \u00d7 6)\/ (1 \u00d7 6) \u2013 (2 \u00d7 2)\/ (3 \u00d7 2) + (5 \u00d7 1)\/ (6 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 \u2013 4 + 5)\/ 6<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= (-10 + 5)\/ 6<\/p>\n\n\n\n

= -5\/6<\/p>\n\n\n\n

2. The sum of two rational numbers is -3\/8. If one of them is 3\/16, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -3\/8<\/p>\n\n\n\n

One rational number = 3\/16<\/p>\n\n\n\n

Other rational number = -3\/8 \u2013 3\/16<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 16 is 16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 \u00d7 2)\/ (8 \u00d7 2) \u2013 (3 \u00d7 1)\/ (16 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 \u2013 3)\/ 16<\/p>\n\n\n\n

= -9\/16<\/p>\n\n\n\n

3. The sum of two rational numbers is -5. If one of them is -52\/25, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -5<\/p>\n\n\n\n

One rational number = -52\/25<\/p>\n\n\n\n

Other rational number = \u2013 5 \u2013 (-52\/25)<\/p>\n\n\n\n

Here LCM is 25<\/p>\n\n\n\n

= (-5 \u00d7 25)\/ (1 \u00d7 25) + (52 \u00d7 1)\/ (25 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-125 + 52)\/ 25<\/p>\n\n\n\n

= \u2013 73\/25<\/p>\n\n\n\n

4. What rational number should be added to -3\/16 to get 11\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = 11\/24<\/p>\n\n\n\n

One rational number = -3\/16<\/p>\n\n\n\n

Other number = 11\/24 \u2013 (-3\/16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 11\/24 + 3\/16<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 16 and 24 is 48<\/p>\n\n\n\n

= (11 \u00d7 2)\/ (24 \u00d7 2) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (22 + 9)\/ 48<\/p>\n\n\n\n

= 31\/48<\/p>\n\n\n\n

5. What rational number should be added to -3\/5 to get 2?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

So the required rational number = 2 \u2013 (-3\/5)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 2 + 3\/5<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

= (2 \u00d7 5)\/ (1 \u00d7 5) + (3 \u00d7 1)\/ (5 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (10 + 3)\/ 5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 13\/5<\/p>\n\n\n\n

= 2 3\/5<\/p>\n\n\n\n

6. What rational number should be subtracted from -5\/12 to get 5\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Required rational number = -5\/12 \u2013 5\/24<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 12 and 24 is 72<\/p>\n\n\n\n

= (-5 \u00d7 6)\/ (12 \u00d7 6) \u2013 (5 \u00d7 3)\/ (24 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-30 \u2013 15)\/ 72<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= \u2013 45\/72<\/p>\n\n\n\n

= -5\/8<\/p>\n\n\n\n

7. What rational number should be subtracted from 5\/8 to get 8\/5?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Required rational number = 5\/8 \u2013 8\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 5 is 40<\/p>\n\n\n\n

= (5 \u00d7 5)\/ (8 \u00d7 5) \u2013 (8 \u00d7 8)\/ (5 \u00d7 8)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (25 \u2013 64)\/ 40<\/p>\n\n\n\n

= -39\/40<\/p>\n\n\n\n

8. Evaluate:<\/strong><\/p>\n\n\n\n

(i) (7\/8 \u00d7 24\/21) + (-5\/9 \u00d7 6\/-25)<\/strong><\/p>\n\n\n\n

(ii) (8\/15 \u00d7 -25\/16) + (-18\/35 \u00d7 5\/6)<\/strong><\/p>\n\n\n\n

(iii) (18\/33 \u00d7 -22\/27) \u2013 (13\/25 \u00d7 -75\/26)<\/strong><\/p>\n\n\n\n

(iv) (-13\/7 \u00d7 -35\/39) \u2013 (-7\/45 \u00d7 9\/14)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (7\/8 \u00d7 24\/21) + (-5\/9 \u00d7 6\/-25)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (7 \u00d7 24)\/ (8 \u00d7 21) + (-5 \u00d7 6)\/ (9 \u00d7 -25)<\/p>\n\n\n\n

By further simplification<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (1 \u00d7 3) + (1 \u00d7 2)\/ (3 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3\/3 + 2\/15<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 15 is 15<\/p>\n\n\n\n

= (3 \u00d7 5)\/ (3 \u00d7 5) + (2 \u00d7 1)\/ (15 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (15 + 2)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n

(ii) (8\/15 \u00d7 -25\/16) + (-18\/35 \u00d7 5\/6)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (8 \u00d7 -25)\/ (15 \u00d7 16) + (-18 \u00d7 5)\/ (35 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1 \u00d7 -5)\/ (3 \u00d7 2) + (-3 \u00d7 1)\/ (7 \u00d7 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -5\/6 \u2013 3\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 6 and 7 is 42<\/p>\n\n\n\n

= (-5 \u00d7 7)\/ (6 \u00d7 7) \u2013 (3 \u00d7 6)\/ (7 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-35 \u2013 18)\/ 42<\/p>\n\n\n\n

= -53\/42<\/p>\n\n\n\n

(iii) (18\/33 \u00d7 -22\/27) \u2013 (13\/25 \u00d7 -75\/26)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (18 \u00d7 -22)\/ (33 \u00d7 27) \u2013 (13 \u00d7 -75)\/ (25 \u00d7 26)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (2 \u00d7 -2)\/ (3 \u00d7 3) \u2013 (1 \u00d7 -3)\/ (1 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -4\/9 \u2013 (-3\/2)<\/p>\n\n\n\n

= -4\/9 + 3\/2<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 9 and 2 is 18<\/p>\n\n\n\n

= (-4 \u00d7 2)\/ (9 \u00d7 2) + (3 \u00d7 9)\/ (2 \u00d7 9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 27)\/ 18<\/p>\n\n\n\n

= 19\/18<\/p>\n\n\n\n

= 1 1\/18<\/p>\n\n\n\n

(iv) (-13\/7 \u00d7 -35\/39) \u2013 (-7\/45 \u00d7 9\/14)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-13 \u00d7 -35)\/ (7 \u00d7 39) + (7 \u00d7 9)\/ (45 \u00d7 14)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -5)\/ (1 \u00d7 3) + (1 \u00d7 1)\/ (5 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 5\/3 + 1\/10<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 3 and 10 is 30<\/p>\n\n\n\n

= (5 \u00d7 10)\/ (3 \u00d7 10) + (1 \u00d7 3)\/ (10 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (50 + 3)\/ 30<\/p>\n\n\n\n

= 53\/30<\/p>\n\n\n\n

= 1 23\/30<\/p>\n\n\n\n

9. The product of two rational numbers is 24. If one of them is -36\/11, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Product of two rational numbers = 24<\/p>\n\n\n\n

One rational number = -36\/11<\/p>\n\n\n\n

Other rational number = 24 \u00f7 (-36\/ 11)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 24 \u00d7 (-11\/36)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2 \u00d7 (-11\/3)<\/p>\n\n\n\n

= -22\/3<\/p>\n\n\n\n

10. By what rational number should we multiply 20\/-9, so that the product may be -5\/9?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here the required rational number = -5\/9 \u00f7 (20\/-9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -5\/9 \u00d7 (-9\/20)<\/p>\n\n\n\n

= 1\/4<\/p>\n\n\n\n

Exercise 2B<\/h3>\n\n\n\n

1. Mark the following pairs of rational numbers on the separate number lines:<\/strong><\/p>\n\n\n\n

(i) 3\/4 and -1\/4<\/strong><\/p>\n\n\n\n

(ii) 2\/5 and -3\/5<\/strong><\/p>\n\n\n\n

(iii) 5\/6 and -2\/3<\/strong><\/p>\n\n\n\n

(iv) 2\/5 and -4\/5<\/strong><\/p>\n\n\n\n

(v) 1\/4 and -5\/4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/4 and -1\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) 2\/5 and -3\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) 5\/6 and -2\/3<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iv) 2\/5 and -4\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(v) 1\/4 and -5\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

2. Compare:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 5\/7<\/strong><\/p>\n\n\n\n

(ii) -7\/2 and 5\/2<\/strong><\/p>\n\n\n\n

(iii) -3 and 2 \u00be<\/strong><\/p>\n\n\n\n

(iv) \u2013 1 \u00bd and 0<\/strong><\/p>\n\n\n\n

(v) 0 and 3\/4<\/strong><\/p>\n\n\n\n

(vi) 3 and -1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 5\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

5\/7 is on the right side of the number line.<\/p>\n\n\n\n

Hence, 3\/5 < 5\/7.<\/p>\n\n\n\n

(ii) -7\/2 and 5\/2<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -7\/2 < 5\/2.<\/p>\n\n\n\n

(iii) -3 and 2 \u00be<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -3 < 11\/4 or \u2013 3 < 2 \u00be.<\/p>\n\n\n\n

(iv) \u2013 1 \u00bd and 0<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, -3\/2 < 0 or \u2013 1 \u00bd < 0.<\/p>\n\n\n\n

(v) 0 and \u00be<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, 0 < 3\/4.<\/p>\n\n\n\n

(vi) 3 and -1<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

P is on the right of Q<\/p>\n\n\n\n

Hence, 3 > \u2013 1.<\/p>\n\n\n\n

3. Compare:<\/strong><\/p>\n\n\n\n

(i) -1\/4 and 0<\/strong><\/p>\n\n\n\n

(ii) 1\/4 and 0<\/strong><\/p>\n\n\n\n

(iii) -3\/8 and 2\/5<\/strong><\/p>\n\n\n\n

(iv) -5\/8 and 7\/-12<\/strong><\/p>\n\n\n\n

(v) 5\/-9 and -5\/-9<\/strong><\/p>\n\n\n\n

(vi) -7\/8 and 5\/-6<\/strong><\/p>\n\n\n\n

(vii) 2\/7 and -3\/-8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -1\/4 and 0<\/p>\n\n\n\n

-1\/4 is a negative rational number which is always less than 0.<\/p>\n\n\n\n

Hence, -1\/4 < 0.<\/p>\n\n\n\n

(ii) 1\/4 and 0<\/p>\n\n\n\n

1\/4 is a positive rational number which is always greater than 0.<\/p>\n\n\n\n

Hence, 1\/4 > 0.<\/p>\n\n\n\n

(iii) -3\/8 and 2\/5<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 3 \u00d7 5 and 2 \u00d7 8<\/p>\n\n\n\n

\u2013 15 < 16<\/p>\n\n\n\n

Hence, -3\/8 < 2\/5.<\/p>\n\n\n\n

(iv) -5\/8 and 7\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-5\/ 8 and -7\/12<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 5 \u00d7 12 and \u2013 7 \u00d7 8<\/p>\n\n\n\n

-60 < \u2013 56<\/p>\n\n\n\n

Hence, -5\/8 < 7\/-12.<\/p>\n\n\n\n

(v) 5\/-9 and -5\/-9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

5 \u00d7 -9 and -5 \u00d7 -9<\/p>\n\n\n\n

\u2013 45 < 45<\/p>\n\n\n\n

Hence, 5\/-9 < -5\/-9.<\/p>\n\n\n\n

(vi) -7\/8 and 5\/-6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-7\/ 8 and -5\/6<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2013 7 \u00d7 6 and -5 \u00d7 8<\/p>\n\n\n\n

-42 < -40<\/p>\n\n\n\n

Hence, -7\/8 < 5\/-6.<\/p>\n\n\n\n

(vii) 2\/7 and -3\/-8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

2\/7 and 3\/8<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

a\/b and c\/d = a \u00d7 d and b \u00d7 c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a \u00d7 d < b \u00d7 c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

2 \u00d7 8 and 7 \u00d7 3<\/p>\n\n\n\n

16 < 21<\/p>\n\n\n\n

Hence, 2\/7 < \u2013 3\/ -8.<\/p>\n\n\n\n

4. Arrange the given rational numbers in ascending order:<\/strong><\/p>\n\n\n\n

(i) 7\/10, -11\/-30 and 5\/-15<\/strong><\/p>\n\n\n\n

(ii) 4\/-9, -5\/12 and 2\/-3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/10, -11\/-30 and 5\/-15<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

= 7\/10, -11\/-30 and -5\/-5<\/p>\n\n\n\n

LCM of 10, 30 and 15 = 30<\/p>\n\n\n\n

= (7 \u00d7 3)\/ (10 \u00d7 3), 11\/30 and (-5 \u00d7 2)\/ (15 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 21\/30, 11\/30 and -10\/30<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -10 < 11 < 21<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-10\/ 30 < 11\/30 < 21\/30<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

5\/ -15 < -11\/ -30 < 7\/10<\/p>\n\n\n\n

(ii) 4\/-9, -5\/12 and 2\/-3<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

= -4\/9, -5\/12 and -2\/3<\/p>\n\n\n\n

LCM of 9, 12 and 3 is 36<\/p>\n\n\n\n

= (-4 \u00d7 4)\/ (9 \u00d7 4), (-5 \u00d7 3)\/ (12 \u00d7 3) and (-2 \u00d7 12)\/ (3 \u00d7 12)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -16\/36, -15\/36 and -24\/36<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -24 < -16 < -15<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-24\/36 < -16\/36 < -15\/36<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\/ -3 < 4\/ -9 < -5\/12<\/p>\n\n\n\n

5. Arrange the given rational numbers in descending order:<\/strong><\/p>\n\n\n\n

(i) 5\/8, 13\/-16 and -7\/12<\/strong><\/p>\n\n\n\n

(ii) 3\/-10, -13\/30 and 8\/-20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/8, 13\/-16 and -7\/12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/8, -13\/16 and -7\/12<\/p>\n\n\n\n

LCM of 8, 16 and 12 is 48<\/p>\n\n\n\n

= (5 \u00d7 6)\/ (8 \u00d7 6), (-13 \u00d7 3)\/ (16 \u00d7 3) and (-7 \u00d7 4)\/ (12 \u00d7 4)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 30\/48, -39\/48 and -28\/48<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, 30 > \u2013 28 > \u2013 39<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

30\/48 > -28\/48 > -39\/48<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

5\/8 > -7\/12 > 13\/-16<\/p>\n\n\n\n

(ii) 3\/-10, -13\/30 and 8\/-20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/10, -13\/30 and -8\/20<\/p>\n\n\n\n

LCM of 10, 20 and 30 is 60<\/p>\n\n\n\n

= (- 3 \u00d7 6)\/ (10 \u00d7 6), (-13 \u00d7 2)\/ (30 \u00d7 2) and (-8 \u00d7 3)\/ (20 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -18\/60, -26\/60 and -24\/60<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here, -18 > -24 > -26<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

-18\/60 > -24\/60 > -26\/60<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3\/ -10 > 8\/-20 > -13\/30<\/p>\n\n\n\n

6. Fill in the blanks:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 3\/10 are on the \u2026\u2026. side of zero.<\/strong><\/p>\n\n\n\n

(ii) -5\/8 and 3\/10 are on the \u2026\u2026.. sides of zero.<\/strong><\/p>\n\n\n\n

(iii) -5\/8 and -3\/10 are on the \u2026\u2026. side of zero.<\/strong><\/p>\n\n\n\n

(iv) 5\/8 and -3\/10 are on the \u2026\u2026.. sides of zero.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 3\/10 are on the same side of zero.<\/p>\n\n\n\n

(ii) -5\/8 and 3\/10 are on the opposite sides of zero.<\/p>\n\n\n\n

(iii) -5\/8 and -3\/10 are on the same side of zero.<\/p>\n\n\n\n

(iv) 5\/8 and -3\/10 are on the opposite sides of zero.<\/p>\n\n\n\n

Exercise 2C<\/h3>\n\n\n\n

1. Add:<\/strong><\/p>\n\n\n\n

(i) 7\/5 and 2\/5<\/strong><\/p>\n\n\n\n

(ii) -4\/9 and 2\/9<\/strong><\/p>\n\n\n\n

(iii) 5\/-12 and 1\/12<\/strong><\/p>\n\n\n\n

(iv) 4\/-15 and -7\/-15<\/strong><\/p>\n\n\n\n

(v) -7\/25 and 9\/-25<\/strong><\/p>\n\n\n\n

(vi) -7\/26 and 7\/-26<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/5 and 2\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/5 + 2\/5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (7 + 2)\/ 5<\/p>\n\n\n\n

= 9\/5<\/p>\n\n\n\n

= 1 4\/5<\/p>\n\n\n\n

(ii) -4\/9 and 2\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -4\/9 + 2\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 + 2)\/ 9<\/p>\n\n\n\n

= -2\/9<\/p>\n\n\n\n

(iii) 5\/-12 and 1\/12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/12 + 1\/12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 + 1)\/ 12<\/p>\n\n\n\n

= -4\/ 12<\/p>\n\n\n\n

= -1\/3<\/p>\n\n\n\n

(iv) 4\/-15 and -7\/-15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 4\/15 + 7\/15<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 + 7)\/ 15<\/p>\n\n\n\n

= 3\/15<\/p>\n\n\n\n

= 1\/5<\/p>\n\n\n\n

(v) -7\/25 and 9\/-25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/25 + -9\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= [(-7) + (-9)]\/ 25<\/p>\n\n\n\n

= -16\/25<\/p>\n\n\n\n

(vi) -7\/26 and 7\/-26<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 7\/26 + -7\/26<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= [(-7) + (-7)]\/ 26<\/p>\n\n\n\n

= \u2013 14\/26<\/p>\n\n\n\n

= -7\/13<\/p>\n\n\n\n

2. Add:<\/strong><\/p>\n\n\n\n

(i) -2\/5 and 3\/7<\/strong><\/p>\n\n\n\n

(ii) -5\/6 and 4\/9<\/strong><\/p>\n\n\n\n

(iii) \u2013 3 and 2\/3<\/strong><\/p>\n\n\n\n

(iv) -5\/9 and 7\/18<\/strong><\/p>\n\n\n\n

(v) -7\/24 and -5\/48<\/strong><\/p>\n\n\n\n

(vi) 1\/-18 and 5\/-27<\/strong><\/p>\n\n\n\n

(vii) -9\/25 and 1\/-75<\/strong><\/p>\n\n\n\n

(viii) 13\/-16 and -11\/24<\/strong><\/p>\n\n\n\n

(ix) -9\/-16 and -11\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/5 and 3\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 \u00d7 7)\/ (5 \u00d7 7) + (3 \u00d7 5)\/ (7 \u00d7 5)<\/p>\n\n\n\n

LCM of 5 and 7 is 35<\/p>\n\n\n\n

= -14\/35 + 15\/35<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 15)\/35<\/p>\n\n\n\n

= 1\/35<\/p>\n\n\n\n

(ii) -5\/6 and 4\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/6 + 4\/9<\/p>\n\n\n\n

LCM of 6 and 9 is 36<\/p>\n\n\n\n

= (-5 \u00d7 6)\/ (6 \u00d7 6) + (4 \u00d7 4)\/ (9 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -30\/36 + 16\/36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-30 + 16)\/ 36<\/p>\n\n\n\n

= -14\/36<\/p>\n\n\n\n

= \u2013 7\/18<\/p>\n\n\n\n

(iii) \u2013 3 and 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/1 + 2\/3<\/p>\n\n\n\n

LCM of 1 and 3 is 3<\/p>\n\n\n\n

= (- 3 \u00d7 3)\/ (1 \u00d7 3) + (2 \u00d7 1)\/ (3 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -9\/3 + 2\/3<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-9 + 2)\/ 3<\/p>\n\n\n\n

= -7\/3<\/p>\n\n\n\n

(iv) -5\/9 and 7\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/9 + 7\/18<\/p>\n\n\n\n

LCM of 9 and 18 is 18<\/p>\n\n\n\n

= (-5 \u00d7 2)\/ (9 \u00d7 2) + (7 \u00d7 1)\/ (18 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -10\/18 + 7\/18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-10 + 7)\/ 18<\/p>\n\n\n\n

= \u2013 3\/18<\/p>\n\n\n\n

= -1\/6<\/p>\n\n\n\n

(v) -7\/24 and -5\/48<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/24 + -5\/48<\/p>\n\n\n\n

LCM of 24 and 48 is 48<\/p>\n\n\n\n

= (-7 \u00d7 2)\/ (24 \u00d7 2) + (-5 \u00d7 1)\/ (48 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -14\/48 + -5\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-14 \u2013 5)\/ 48<\/p>\n\n\n\n

= \u2013 19\/48<\/p>\n\n\n\n

(vi) 1\/-18 and 5\/-27<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 1\/18 + -5\/27<\/p>\n\n\n\n

LCM of 18 and 27 is 54<\/p>\n\n\n\n

= (-1 \u00d7 3)\/ (18 \u00d7 3) + (-5 \u00d7 2)\/ (27 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -3\/54 + -10\/54<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (- 3 \u2013 10)\/ 54<\/p>\n\n\n\n

= -13\/54<\/p>\n\n\n\n

(vii) -9\/25 and 1\/-75<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -9\/25 + -1\/75<\/p>\n\n\n\n

LCM of 24 and 75 is 75<\/p>\n\n\n\n

= (-9 \u00d7 3)\/ (25 \u00d7 3) + (-1 \u00d7 1)\/ (75 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -27\/75 + -1\/75<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-27 \u2013 1)\/ 75<\/p>\n\n\n\n

= -28\/75<\/p>\n\n\n\n

(viii) 13\/-16 and -11\/24<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -13\/16 + -11\/24<\/p>\n\n\n\n

LCM of 16 and 24 is 48<\/p>\n\n\n\n

= (-13 \u00d7 3)\/ (16 \u00d7 3) + (-11 \u00d7 2)\/ (24 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -39\/48 + -22\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-39 \u2013 22)\/ 48<\/p>\n\n\n\n

= -61\/48<\/p>\n\n\n\n

(ix) -9\/-16 and -11\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/16 + -11\/8<\/p>\n\n\n\n

LCM of 16 and 8 is 16<\/p>\n\n\n\n

= (9 \u00d7 1)\/ (16 \u00d7 1) + (-11 \u00d7 2)\/ (8 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 9\/16 + -22\/16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 \u2013 22)\/ 16<\/p>\n\n\n\n

= -13\/16<\/p>\n\n\n\n

3. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -2\/5 + 3\/5 + -1\/5<\/strong><\/p>\n\n\n\n

(ii) -8\/9 + 4\/9 + -2\/9<\/strong><\/p>\n\n\n\n

(iii) 5\/-24 + -1\/8 + 3\/16<\/strong><\/p>\n\n\n\n

(iv) -7\/6 + 4\/-15 + -4\/-30<\/strong><\/p>\n\n\n\n

(v) -2 + 2\/5 + -2\/15<\/strong><\/p>\n\n\n\n

(vi) -11\/12 + 5\/16 + -3\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/5 + 3\/5 + -1\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 + 3 \u2013 1)\/ 5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 0\/5<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

(ii) -8\/9 + 4\/9 + -2\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-8 + 4 \u2013 2)\/ 9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-10 + 4)\/ 9<\/p>\n\n\n\n

= -6\/9<\/p>\n\n\n\n

= -2\/3<\/p>\n\n\n\n

(iii) 5\/-24 + -1\/8 + 3\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/24 + -1\/8 + 3\/16<\/p>\n\n\n\n

LCM of 8, 16 and 24 is 48<\/p>\n\n\n\n

= (-5 \u00d7 2)\/ (24 \u00d7 2) + (-1 \u00d7 6)\/ (8 \u00d7 6) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -10\/ 48 + -6\/48 + 9\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-10 \u2013 6 + 9)\/ 48<\/p>\n\n\n\n

= (-16 + 9)\/ 48<\/p>\n\n\n\n

= -7\/48<\/p>\n\n\n\n

(iv) -7\/6 + 4\/-15 + -4\/-30<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7\/6 + -4\/15 + 4\/30<\/p>\n\n\n\n

LCM of 6, 15 and 30 is 30<\/p>\n\n\n\n

= (-7 \u00d7 5)\/ (6 \u00d7 5) + (-4 \u00d7 2)\/ (15 \u00d7 2) + (4 \u00d7 1)\/ (30 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -35\/30 + -8\/30 + 4\/30<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-35 \u2013 8 + 4)\/ 30<\/p>\n\n\n\n

= (-43 + 4)\/ 30<\/p>\n\n\n\n

= \u2013 39\/30<\/p>\n\n\n\n

= -13\/10<\/p>\n\n\n\n

(v) -2 + 2\/5 + -2\/15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/1 + 2\/5 + -2\/15<\/p>\n\n\n\n

LCM of 1, 5 and 15 is 15<\/p>\n\n\n\n

= (-2 \u00d7 15)\/ (1 \u00d7 15) + (2 \u00d7 3)\/ (5 \u00d7 3) + (-2 \u00d7 1)\/ (15 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -30\/15 + 6\/15 + -2\/15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-30 + 6 \u2013 2)\/ 15<\/p>\n\n\n\n

= (-32 + 6)\/ 15<\/p>\n\n\n\n

= -26\/15<\/p>\n\n\n\n

(vi) -11\/12 + 5\/16 + -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -11\/12 + 5\/16 + -3\/8<\/p>\n\n\n\n

LCM of 12, 16 and 8 is 48<\/p>\n\n\n\n

= (-11 \u00d7 4)\/ (12 \u00d7 4) + (5 \u00d7 3)\/ (16 \u00d7 3) + (-3 \u00d7 6)\/ (8 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -44\/48 + 15\/48 + -18\/48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-44 + 15 \u2013 18)\/ 48<\/p>\n\n\n\n

= (-62 + 15)\/ 48<\/p>\n\n\n\n

= -47\/ 48<\/p>\n\n\n\n

4. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -11\/18 + -3\/9 + 2\/-3<\/strong><\/p>\n\n\n\n

(ii) -9\/4 + 13\/3 +25\/6<\/strong><\/p>\n\n\n\n

(iii) \u2013 5 + 5\/-8 + -5\/-12<\/strong><\/p>\n\n\n\n

(iv) -2\/3 + 5\/2 + 2<\/strong><\/p>\n\n\n\n

(v) 5 + -3\/4 + -5\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -11\/18 + -3\/9 + 2\/-3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= \u2013 11\/18 + -3\/9 + -2\/3<\/p>\n\n\n\n

LCM of 3, 9 and 18 is 18<\/p>\n\n\n\n

= (-11 \u00d7 1)\/ (18 \u00d7 1) + (-3 \u00d7 2)\/ (9 \u00d7 2) + (-2 \u00d7 6)\/ (3 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -11\/18 + -6\/18 + -12\/18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-11 \u2013 6 \u2013 12)\/18<\/p>\n\n\n\n

= -29\/18<\/p>\n\n\n\n

(ii) -9\/4 + 13\/3 +25\/6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -9\/4 + 13\/3 + 25\/6<\/p>\n\n\n\n

LCM of 4, 3 and 6 is 24<\/p>\n\n\n\n

= (-9 \u00d7 6)\/ (4 \u00d7 6) + (13 \u00d7 8)\/ (3 \u00d7 8) + (25 \u00d7 4)\/ (6 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -54\/24 + 104\/24 + 100\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-54 + 104 + 100)\/ 24<\/p>\n\n\n\n

= 150\/24<\/p>\n\n\n\n

= 25\/4<\/p>\n\n\n\n

= 6 1\/4<\/p>\n\n\n\n

(iii) \u2013 5 + 5\/-8 + -5\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/1 + -5\/8 + 5\/12<\/p>\n\n\n\n

LCM of 1, 8 and 12 is 24<\/p>\n\n\n\n

= (-5 \u00d7 24)\/ (1 \u00d7 24) + (-5 \u00d7 3)\/ (8 \u00d7 3) + (5 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -120\/24 + -15\/24 + 10\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-120 \u2013 15 + 10)\/24<\/p>\n\n\n\n

= -125\/24<\/p>\n\n\n\n

(iv) -2\/3 + 5\/2 + 2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/3 + 5\/2 + 2\/1<\/p>\n\n\n\n

LCM of 3, 2 and 1 is 6<\/p>\n\n\n\n

= (-2 \u00d7 2)\/ (3 \u00d7 2) + (5 \u00d7 3)\/ (2 \u00d7 3) + (2 \u00d7 6)\/ (1 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -4\/6 + 15\/6 + 12\/6<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-4 + 15 + 12)\/ 6<\/p>\n\n\n\n

= 23\/6<\/p>\n\n\n\n

= 3 5\/6<\/p>\n\n\n\n

(v) 5 + -3\/4 + -5\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/1 + -3\/4 + -5\/8<\/p>\n\n\n\n

LCM of 1, 4 and 8 is 8<\/p>\n\n\n\n

= (5 \u00d7 8)\/ (1 \u00d7 8) + (-3 \u00d7 2)\/ (4 \u00d7 2) + (-5 \u00d7 1)\/ (8 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 40\/8 + -6\/8 + -5\/8<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (40 \u2013 6 \u2013 5)\/ 8<\/p>\n\n\n\n

= (40 \u2013 11)\/8<\/p>\n\n\n\n

= 29\/8<\/p>\n\n\n\n

= 3 5\/8<\/p>\n\n\n\n

5. Subtract:<\/strong><\/p>\n\n\n\n

(i) 2\/9 from 5\/9<\/strong><\/p>\n\n\n\n

(ii) -6\/11 from -3\/-11<\/strong><\/p>\n\n\n\n

(iii) -2\/15 from -8\/15<\/strong><\/p>\n\n\n\n

(iv) 11\/18 from -5\/18<\/strong><\/p>\n\n\n\n

(v) -4\/11 from -2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 2\/9 from 5\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 5\/9 \u2013 2\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (5 \u2013 2)\/ 9<\/p>\n\n\n\n

= 3\/9<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

(ii) -6\/11 from -3\/-11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/ 11 \u2013 (-6\/11)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 3\/11 + 6\/11<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (3 + 6)\/ 11<\/p>\n\n\n\n

= 9\/11<\/p>\n\n\n\n

(iii) -2\/15 from -8\/15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -8\/15 \u2013 (-2\/15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -8\/15 + 2\/15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-8 + 2)\/ 15<\/p>\n\n\n\n

= -6\/ 15<\/p>\n\n\n\n

= -2\/5<\/p>\n\n\n\n

(iv) 11\/18 from -5\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/18 \u2013 11\/18<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u2013 11)\/ 18<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -16\/18<\/p>\n\n\n\n

= -8\/9<\/p>\n\n\n\n

(v) -4\/11 from -2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/1 \u2013 (-4\/11)<\/p>\n\n\n\n

LCM of 1 and 11 is 11<\/p>\n\n\n\n

= (- 2 \u00d7 11)\/ (1 \u00d7 11) + (4 \u00d7 1)\/ (11 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -22\/11 + 4\/11<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-22 + 4)\/ 11<\/p>\n\n\n\n

= -18\/11<\/p>\n\n\n\n

6. Subtract:<\/strong><\/p>\n\n\n\n

(i) -3\/10 from 1\/5<\/strong><\/p>\n\n\n\n

(ii) -6\/25 from -8\/5<\/strong><\/p>\n\n\n\n

(iii) -7\/4 from -2<\/strong><\/p>\n\n\n\n

(iv) -16\/21 from 1<\/strong><\/p>\n\n\n\n

(v) -8\/15 from 0<\/strong><\/p>\n\n\n\n

(vi) 0 from -3\/8<\/strong><\/p>\n\n\n\n

(vii) -2 from -3\/10<\/strong><\/p>\n\n\n\n

(viii) 5\/8 from -5\/16<\/strong><\/p>\n\n\n\n

(ix) 4 from -3\/13<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -3\/10 from 1\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1\/5 \u2013 (-3\/10)<\/p>\n\n\n\n

LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (1 \u00d7 2)\/ (5 \u00d7 2) + 3\/10<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/10 + 3\/10<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (2 + 3)\/ 10<\/p>\n\n\n\n

= 5\/10<\/p>\n\n\n\n

= 1\/2<\/p>\n\n\n\n

(ii) -6\/25 from -8\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -8\/5 \u2013 (-6\/25)<\/p>\n\n\n\n

LCM of 5 and 25 is 25<\/p>\n\n\n\n

= (-8 \u00d7 5)\/ (5 \u00d7 5) + 6\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -40\/25 + 6\/25<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-40 + 6)\/ 25<\/p>\n\n\n\n

= -34\/25<\/p>\n\n\n\n

(iii) -7\/4 from -2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2\/1) \u2013 (-7\/4)<\/p>\n\n\n\n

LCM of 1 and 4 is 4<\/p>\n\n\n\n

= (-2 \u00d7 4)\/ (1 \u00d7 4) + 7\/4<\/p>\n\n\n\n

= -8\/4 + 7\/4<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 7)\/ 4<\/p>\n\n\n\n

= -1\/4<\/p>\n\n\n\n

(iv) -16\/21 from 1<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1\/1 \u2013 (-16\/21)<\/p>\n\n\n\n

= 1\/1 + 16\/21<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (21 + 16)\/ 21<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (21 + 16)\/21<\/p>\n\n\n\n

= 37\/21<\/p>\n\n\n\n

= 1 16\/21<\/p>\n\n\n\n

(v) -8\/15 from 0<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 0 \u2013 (-8\/15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 0 + 8\/15<\/p>\n\n\n\n

= 8\/15<\/p>\n\n\n\n

(vi) 0 from -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/8 \u2013 0<\/p>\n\n\n\n

= -3\/8<\/p>\n\n\n\n

(vii) -2 from -3\/10<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -3\/10 \u2013 (-2\/1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -3\/10 + 2\/1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 + 2 \u00d7 10)\/10<\/p>\n\n\n\n

= 17\/10<\/p>\n\n\n\n

= 1 7\/10<\/p>\n\n\n\n

(viii) 5\/8 from -5\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/16 \u2013 5\/8<\/p>\n\n\n\n

LCM of 8 and 16 is 16<\/p>\n\n\n\n

= -5\/16 \u2013 (5 \u00d7 2)\/ (8 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -5\/16 \u2013 10\/16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u2013 10)\/16<\/p>\n\n\n\n

= \u2013 15\/16<\/p>\n\n\n\n

(ix) 4 from -3\/13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/13 \u2013 4\/1<\/p>\n\n\n\n

LCM of 13 and 1 is 13<\/p>\n\n\n\n

= (- 3 \u2013 4 \u00d7 13)\/ 13<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u2013 52)\/13<\/p>\n\n\n\n

= -55\/13<\/p>\n\n\n\n

7. The sum of two rational numbers is 11\/24. If one of them is 3\/8, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = 11\/24<\/p>\n\n\n\n

One of the rational number = 3\/8<\/p>\n\n\n\n

Other rational number = 11\/24 \u2013 3\/8<\/p>\n\n\n\n

LCM of 24 and 8 is 24<\/p>\n\n\n\n

= 11\/24 \u2013 (3 \u00d7 3)\/ (8 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 11\/24 \u2013 9\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (11 \u2013 9)\/ 24<\/p>\n\n\n\n

= 2\/24<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

8. The sum of two rational numbers is -7\/12. If one of them is 13\/24, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -7\/12<\/p>\n\n\n\n

One of the rational number = 13\/24<\/p>\n\n\n\n

Other rational number = -7\/12 \u2013 13\/24<\/p>\n\n\n\n

LCM of 12 and 24 is 24<\/p>\n\n\n\n

= (-7 \u00d7 2)\/ (12 \u00d7 2) \u2013 13\/24<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -14\/24 \u2013 13\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-14 \u2013 13)\/ 24<\/p>\n\n\n\n

= -27\/24<\/p>\n\n\n\n

= -9\/8<\/p>\n\n\n\n

9. The sum of two rational numbers is -4. If one of them is -13\/12, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -4<\/p>\n\n\n\n

One of the rational number = -13\/12<\/p>\n\n\n\n

Other rational number = \u2013 4 \u2013 (-13\/12)<\/p>\n\n\n\n

LCM of 1 and 12 is 12<\/p>\n\n\n\n

= \u2013 4 + 13\/12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-4 \u00d7 12 + 13)\/ 12<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-48 + 13)\/ 12<\/p>\n\n\n\n

= -35\/12<\/p>\n\n\n\n

10. What should be added to -3\/16 to get 11\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -3\/16<\/p>\n\n\n\n

Sum of two numbers = 11\/24<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-3\/16 + x = 11\/24<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 11\/24 + 3\/16<\/p>\n\n\n\n

LCM of 16 and 24 is 48<\/p>\n\n\n\n

x = (11 \u00d7 2)\/ (24 \u00d7 2) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 22\/48 + 9\/48<\/p>\n\n\n\n

x = (22 + 9)\/ 48 = 31\/48<\/p>\n\n\n\n

11. What should be added to -3\/5 to get 2?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -3\/5<\/p>\n\n\n\n

Here the sum of two numbers is 2<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-3\/5 + x = 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 2 + 3\/5<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

x = (2 \u00d7 5 + 3)\/ 5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (10 + 3)\/ 5<\/p>\n\n\n\n

= 13\/5<\/p>\n\n\n\n

= 2 3\/5<\/p>\n\n\n\n

12. What should be subtracted from -4\/5 to get 1?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other number = -4\/5<\/p>\n\n\n\n

Here the difference between two numbers is 1<\/p>\n\n\n\n

From the question<\/p>\n\n\n\n

-4\/5 \u2013 x = 1<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

-4\/5 \u2013 1 = x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = (-4 \u2013 1 \u00d7 5)\/ 5<\/p>\n\n\n\n

x = (-4 \u2013 5)\/ 5 = -9\/5<\/p>\n\n\n\n

13. The sum of two numbers is -6\/5. If one of them is -2, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two numbers = -6\/5<\/p>\n\n\n\n

One of the numbers = -2<\/p>\n\n\n\n

Other number = -6\/5 \u2013 (-2\/1)<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

= \u2013 6\/5 \u2013 (2 \u00d7 5)\/ (1 \u00d7 5)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 + 10)\/ 5<\/p>\n\n\n\n

= 4\/5<\/p>\n\n\n\n

14. What should be added to -7\/12 to get 3\/8?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the required rational number<\/p>\n\n\n\n

Other rational number = -7\/12<\/p>\n\n\n\n

Sum of two numbers = 3\/8<\/p>\n\n\n\n

Using the question<\/p>\n\n\n\n

-7\/12 + x = 3\/8<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 3\/8 \u2013 (-7\/12)<\/p>\n\n\n\n

LCM of 8 and 12 is 24<\/p>\n\n\n\n

x = (3 \u00d7 3)\/ (8 \u00d7 3) + (7 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 9\/24 + 14\/24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 + 14)\/ 24 = 23\/34<\/p>\n\n\n\n

15. What should be subtracted from 5\/9 to get 9\/5?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider x as the first number<\/p>\n\n\n\n

Other number is 5\/9<\/p>\n\n\n\n

Here the difference between two numbers is 9\/5<\/p>\n\n\n\n

Using the question<\/p>\n\n\n\n

5\/9 \u2013 x = 9\/5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 5\/9 \u2013 9\/5<\/p>\n\n\n\n

LCM of 9 and 5 is 45<\/p>\n\n\n\n

x = (5 \u00d7 5)\/ (9 \u00d7 5) \u2013 (9 \u00d7 9)\/ (5 \u00d7 9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 25\/45 \u2013 81\/45<\/p>\n\n\n\n

x = (25 \u2013 81)\/ 45 = \u2013 56\/45<\/p>\n\n\n\n

Exercise 2D<\/h3>\n\n\n\n

1. Evaluate:<\/strong><\/p>\n\n\n\n

(i) 5\/4 \u00d7 3\/7<\/strong><\/p>\n\n\n\n

(ii) 2\/3 \u00d7 -6\/7<\/strong><\/p>\n\n\n\n

(iii) (-12\/5) \u00d7 (10\/-3)<\/strong><\/p>\n\n\n\n

(iv) -45\/39 \u00d7 -13\/ 15<\/strong><\/p>\n\n\n\n

(v) 3 1\/8 \u00d7 (-2 2\/5)<\/strong><\/p>\n\n\n\n

(vi) 2 14\/25 \u00d7 (-5\/16)<\/strong><\/p>\n\n\n\n

(vii) (-8\/9) \u00d7 (-3\/ 16)<\/strong><\/p>\n\n\n\n

(viii) (5\/-27) \u00d7 (-9\/ 20)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 5\/4 \u00d7 3\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5 \u00d7 3)\/ (4 \u00d7 7)<\/p>\n\n\n\n

= 15\/28<\/p>\n\n\n\n

(ii) 2\/3 \u00d7 -6\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 -6)\/ (3 \u00d7 7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (2 \u00d7 -2)\/ 7<\/p>\n\n\n\n

= -4\/7<\/p>\n\n\n\n

(iii) (-12\/5) \u00d7 (10\/-3)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-12 \u00d7 10)\/ (5 \u00d7 -3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 4 \u00d7 2<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(iv) -45\/39 \u00d7 -13\/ 15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-45 \u00d7 -13)\/ (39 \u00d7 15)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u00d7 -1)\/ (3 \u00d7 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3\/3<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

(v) 3 1\/8 \u00d7 (-2 2\/5)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (3 \u00d7 8 + 1)\/ 8 \u00d7 (-2 \u00d7 5 + 2)\/ 5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 25\/8 \u00d7 (-12\/5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (25 \u00d7 -12)\/ (8 \u00d7 5)<\/p>\n\n\n\n

On further simplification<\/p>\n\n\n\n

= (5 \u00d7 -3)\/ (2 \u00d7 1)<\/p>\n\n\n\n

= -15\/2<\/p>\n\n\n\n

(vi) 2 14\/25 \u00d7 (-5\/16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 25 + 14)\/ 25 \u00d7 (-5\/16)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 64\/25 \u00d7 (-5\/16)<\/p>\n\n\n\n

= (64 \u00d7 -5)\/ (25 \u00d7 16)<\/p>\n\n\n\n

On further simplification<\/p>\n\n\n\n

= (4 \u00d7 -1)\/ (5 \u00d7 1)<\/p>\n\n\n\n

= -4\/5<\/p>\n\n\n\n

(vii) (-8\/9) \u00d7 (-3\/ 16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-8 \u00d7 -3)\/ (9 \u00d7 16)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -1)\/ (3 \u00d7 2)<\/p>\n\n\n\n

= 1\/6<\/p>\n\n\n\n

(viii) (5\/-27) \u00d7 (-9\/ 20)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5 \u00d7 -9)\/ (-27 \u00d7 20)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1 \u00d7 1)\/ (3 \u00d7 4)<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

2. Multiply:<\/strong><\/p>\n\n\n\n

(i) 3\/25 and 4\/5<\/strong><\/p>\n\n\n\n

(ii) 1 1\/8 and 10 2\/3<\/strong><\/p>\n\n\n\n

(iii) 6 2\/3 and -3\/8<\/strong><\/p>\n\n\n\n

(iv) -13\/15 and -25\/26<\/strong><\/p>\n\n\n\n

(v) 1 1\/6 and 18<\/strong><\/p>\n\n\n\n

(vi) 2 1\/14 and -7<\/strong><\/p>\n\n\n\n

(vii) 5 1\/8 and -16<\/strong><\/p>\n\n\n\n

(viii) 35 and -18\/25<\/strong><\/p>\n\n\n\n

(ix) 6 2\/3 and -3\/8<\/strong><\/p>\n\n\n\n

(x) 3 3\/5 and -10<\/strong><\/p>\n\n\n\n

(xi) 27\/28 and -14<\/strong><\/p>\n\n\n\n

(xii) -24 and 5\/16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3\/25 and 4\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 3\/25 \u00d7 4\/5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (3 \u00d7 4)\/ (25 \u00d7 5)<\/p>\n\n\n\n

= 12\/125<\/p>\n\n\n\n

(ii) 1 1\/8 and 10 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/8 \u00d7 32\/2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (9 \u00d7 32)\/ (8 \u00d7 3)<\/p>\n\n\n\n

= 3 \u00d7 4<\/p>\n\n\n\n

= 12<\/p>\n\n\n\n

(iii) 6 2\/3 and -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 20\/3 \u00d7 -3\/8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (20 \u00d7 -3)\/ (3 \u00d7 8)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 -1)\/ (1 \u00d7 2)<\/p>\n\n\n\n

= -5\/2<\/p>\n\n\n\n

(iv) -13\/15 and -25\/26<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-13 \u00d7 -25)\/ (15 \u00d7 26)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -5)\/ (3 \u00d7 2)<\/p>\n\n\n\n

= 5\/6<\/p>\n\n\n\n

(v) 1 1\/6 and 18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/6 \u00d7 18<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 7 \u00d7 3<\/p>\n\n\n\n

= 21<\/p>\n\n\n\n

(vi) 2 1\/14 and -7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (2 \u00d7 14 + 1)\/ 14 \u00d7 (-7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 29\/4 \u00d7 (-7)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (29 \u00d7 -1)\/ 2<\/p>\n\n\n\n

= -29\/2<\/p>\n\n\n\n

(vii) 5 1\/8 and -16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 41\/8 \u00d7 -16<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 41 \u00d7 -2<\/p>\n\n\n\n

= -82<\/p>\n\n\n\n

(viii) 35 and -18\/25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 35 \u00d7 -18\/25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (35 \u00d7 -18)\/ 25<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 -18)\/ 5<\/p>\n\n\n\n

= -126\/5<\/p>\n\n\n\n

(ix) 6 2\/3 and -3\/8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 20\/3 \u00d7 -3\/8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (20 \u00d7 -3)\/ (3 \u00d7 8)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 -1)\/ (1 \u00d7 2)<\/p>\n\n\n\n

= -5\/2<\/p>\n\n\n\n

(x) 3 3\/5 and -10<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (3 \u00d7 5 + 3)\/ 5 \u00d7 -10<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 18\/5 \u00d7 -10<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 18 \u00d7 -2<\/p>\n\n\n\n

= -36<\/p>\n\n\n\n

(xi) 27\/28 and -14<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 27\/28 \u00d7 -14<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (27 \u00d7 -1)\/ 2<\/p>\n\n\n\n

= -27\/2<\/p>\n\n\n\n

(xii) -24 and 5\/16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-24 \u00d7 5)\/ 16<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-3 \u00d7 5)\/ 2<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -15\/2<\/p>\n\n\n\n

3. Evaluate:<\/strong><\/p>\n\n\n\n

(i) (6 \u00d7 5\/18) \u2013 (- 4 2\/9)<\/strong><\/p>\n\n\n\n

(ii) (7\/8 \u00d7 8\/7) + (-5\/9) \u00d7 (6\/-25)<\/strong><\/p>\n\n\n\n

(iii) (11\/-9 \u00d7 21\/44) + (-5\/9) \u00d7 (63\/ -100)<\/strong><\/p>\n\n\n\n

(iv) (-5\/9 \u00d7 6\/-25) + (24\/21 \u00d7 7\/8)
(v) (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/strong><\/p>\n\n\n\n

(vi) (-4\/5 \u00d7 3\/2) + (9\/-5 \u00d7 10\/3) \u2013 (-3\/2 \u00d7 -1\/4)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (6 \u00d7 5\/18) \u2013 (- 4 2\/9)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-1 \u00d7 5\/3) \u2013 [- (4 \u00d7 9 + 2)\/ 9]<\/p>\n\n\n\n

LCM of 3 and 9 is 9<\/p>\n\n\n\n

= -5\/3 \u2013 (-38\/9)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -5\/3 + 38\/9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u00d7 3)\/ (3 \u00d7 3) + (38 \u00d7 1)\/ (9 \u00d7 1)<\/p>\n\n\n\n

= (-15 + 38)\/ 9<\/p>\n\n\n\n

= 23\/9<\/p>\n\n\n\n

= 2 5\/9<\/p>\n\n\n\n

(ii) (7\/8 \u00d7 8\/7) + (-5\/9) \u00d7 (6\/-25)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (7\/8 \u00d7 8\/7) + (-5\/9 \u00d7 6\/-25)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 1\/1 + (1 \u00d7 2)\/ (3 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 1\/1 + 2\/15<\/p>\n\n\n\n

= (15 + 2)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n

(iii) (11\/-9 \u00d7 21\/44) + (-5\/9) \u00d7 (63\/ -100)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (11\/-9 \u00d7 21\/44) + (5\/9 \u00d7 63\/100)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 7)\/ (3 \u00d7 4) + (1 \u00d7 7)\/ (1 \u00d7 20)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -7\/12 + 7\/20<\/p>\n\n\n\n

LCM of 12 and 20 is 60<\/p>\n\n\n\n

= (-7 \u00d7 5)\/ (12 \u00d7 5) + (7 \u00d7 3)\/ (20 \u00d7 3)<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

= -35\/60 + 211\/60<\/p>\n\n\n\n

= (-35 + 21)\/ 60<\/p>\n\n\n\n

= -14\/60<\/p>\n\n\n\n

= -7\/30<\/p>\n\n\n\n

(iv) (-5\/9 \u00d7 6\/-25) + (24\/21 \u00d7 7\/8)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (5\/9 \u00d7 6\/25) + (24\/21 \u00d7 7\/8)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/ (3 \u00d7 5) + 1<\/p>\n\n\n\n

= 2\/15 + 1<\/p>\n\n\n\n

LCM of 15 and 1 is 15<\/p>\n\n\n\n

= (2 + 15)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n


(v) (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-35\/39 \u00d7 -13\/7) \u2013 (7\/90 \u00d7 -18\/14)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-5 \u00d7 -1)\/ (3 \u00d7 1) \u2013 (1 \u00d7 -1)\/ (5 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 5\/3 \u2013 (-1\/10)<\/p>\n\n\n\n

LCM of 3 and 10 is 30<\/p>\n\n\n\n

= (5 \u00d7 10)\/ (3 \u00d7 10) + 1\/ (10 \u00d7 3)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= (50 + 3)\/ 30<\/p>\n\n\n\n

= 53\/30<\/p>\n\n\n\n

= 1 23\/30<\/p>\n\n\n\n

(vi) (-4\/5 \u00d7 3\/2) + (9\/-5 \u00d7 10\/3) \u2013 (-3\/2 \u00d7 -1\/4)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-2 \u00d7 3)\/ (5 \u00d7 1) + (3 \u00d7 2)\/ (-1 \u00d7 1) \u2013 (-3 \u00d7 -1)\/ (2 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -6\/5 + -6\/1 \u2013 3\/8<\/p>\n\n\n\n

LCM of 5, 1 and 8 is 40<\/p>\n\n\n\n

= = (-6 \u00d7 8)\/ (5 \u00d7 8) \u2013 (6 \u00d7 40)\/ (1 \u00d7 40) \u2013 (3 \u00d7 5)\/ (8 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-48 \u2013 240 \u2013 15)\/ 40<\/p>\n\n\n\n

= \u2013 303\/40<\/p>\n\n\n\n

4. Find the cost of 3 \u00bd m cloth, if one metre cloth costs \u20b9 325 \u00bd.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that cost of one metre cloth = \u20b9 325 \u00bd<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 325 + 1)\/ 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (650 + 1)\/ 2<\/p>\n\n\n\n

= \u20b9 651\/2<\/p>\n\n\n\n

Cost of 3 \u00bd m cloth<\/p>\n\n\n\n

(2 \u00d7 3 + 1)\/ 2 = 7\/2 m<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 651\/2 \u00d7 7\/2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (651 \u00d7 7)\/ (2 \u00d7 2)<\/p>\n\n\n\n

= 4557\/4<\/p>\n\n\n\n

= \u20b9 1139 \u00bc<\/p>\n\n\n\n

5. A bus is moving with a speed of 65 \u00bd km per hour. How much distance will it cover in 1 1\/3 hours.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Speed of bus = 65 \u00bd km per hour<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 65 + 1)\/ 2<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (130 + 1)\/ 2<\/p>\n\n\n\n

= 131\/ 2 km<\/p>\n\n\n\n

Distance covered in 1 1\/3 hour = 4\/3 hour can be written as<\/p>\n\n\n\n

= 131\/2 \u00d7 4\/3<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 131\/1 \u00d7 2\/3<\/p>\n\n\n\n

We know that distance covered = speed \u00d7 time<\/p>\n\n\n\n

= 131\/2 \u00d7 4\/3<\/p>\n\n\n\n

= (131 \u00d7 2)\/ (1 \u00d7 3)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 262\/3<\/p>\n\n\n\n

= 87 1\/3 km<\/p>\n\n\n\n

6. Divide:<\/strong><\/p>\n\n\n\n

(i) 15\/28 by \u00be<\/strong><\/p>\n\n\n\n

(ii) -20\/9 by -5\/9<\/strong><\/p>\n\n\n\n

(iii) 16\/-5 by -8\/7<\/strong><\/p>\n\n\n\n

(iv) -7 by -14\/5<\/strong><\/p>\n\n\n\n

(v) -14 by 7\/-2<\/strong><\/p>\n\n\n\n

(vi) -22\/9 by 11\/18<\/strong><\/p>\n\n\n\n

(vii) 35 by -7\/9<\/strong><\/p>\n\n\n\n

(viii) 21\/44 by -11\/9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 15\/28 by \u00be<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 15\/28 \u00f7 3\/4<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 15\/28 \u00d7 4\/3<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 5\/7 \u00d7 1\/1<\/p>\n\n\n\n

= 5\/7<\/p>\n\n\n\n

(ii) -20\/9 by -5\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -20\/9 \u00f7 -5\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -20\/9 \u00d7 9\/-5<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -4\/-1<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

(iii) 16\/-5 by -8\/7<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 16\/-5 \u00f7 -8\/7<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 16\/-5 \u00d7 7\/-8<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/-5 \u00d7 7\/-1<\/p>\n\n\n\n

= (2 \u00d7 7)\/ (-5 \u00d7 -1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 14\/5<\/p>\n\n\n\n

= 2 4\/5<\/p>\n\n\n\n

(iv) -7 by -14\/5<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -7 \u00f7 \u2013 14\/5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -7 \u00d7 5\/-14<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 1 \u00d7 5\/2<\/p>\n\n\n\n

= (1 \u00d7 5)\/ 2<\/p>\n\n\n\n

= 5\/2<\/p>\n\n\n\n

= 2 \u00bd<\/p>\n\n\n\n

(v) -14 by 7\/-2<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= -14 \u00f7 7\/-2<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -14 \u00d7 -2\/7<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-2 \u00d7 -2)\/ (1 \u00d7 1)<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

(vi) -22\/9 by 11\/18<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= \u2013 22\/9 \u00f7 11\/18<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -22\/9 \u00d7 18\/11<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -2\/1 \u00d7 2\/1<\/p>\n\n\n\n

= (-2 \u00d7 2)\/ (1 \u00d7 1)<\/p>\n\n\n\n

= \u2013 4\/1<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

(vii) 35 by -7\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 35 \u00f7 -7\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 35 \u00d7 9\/-7<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 5 \u00d7 9\/-1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 9)\/ -1<\/p>\n\n\n\n

= 45\/-1<\/p>\n\n\n\n

= -45<\/p>\n\n\n\n

(viii) 21\/44 by -11\/9<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= 21\/44 \u00f7 -11\/9<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 21\/44 \u00d7 -9\/11<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (21 \u00d7 -9)\/ (44 \u00d7 11)<\/p>\n\n\n\n

= -189\/484<\/p>\n\n\n\n

7. Evaluate:<\/strong><\/p>\n\n\n\n

(i) 3 5\/12 + 1 2\/3<\/strong><\/p>\n\n\n\n

(ii) 3 5\/12 \u2013 1 2\/3<\/strong><\/p>\n\n\n\n

(iii) (3 5\/12 + 1 2\/3) \u00f7 (3 5\/12 \u2013 1 2\/3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3 5\/12 + 1 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (12 \u00d7 3 + 5)\/ 12 + (3 \u00d7 1 + 2)\/ 3<\/p>\n\n\n\n

= 41\/12 + 5\/3<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 \u00d7 1)\/ (12 \u00d7 1) + (5 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 41\/12 + 20\/12<\/p>\n\n\n\n

= (41 + 20)\/ 12<\/p>\n\n\n\n

= 61\/12<\/p>\n\n\n\n

= 5 1\/12<\/p>\n\n\n\n

(ii) 3 5\/12 \u2013 1 2\/3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (12 \u00d7 3 + 5)\/ 12 \u2013 (3 \u00d7 1 + 2)\/ 3<\/p>\n\n\n\n

= 41\/12 \u2013 5\/3<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 \u00d7 1)\/ (12 \u00d7 1) \u2013 (5 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (41 \u2013 20)\/ 12<\/p>\n\n\n\n

= 21\/12<\/p>\n\n\n\n

= 2\/4<\/p>\n\n\n\n

= 1 \u00be<\/p>\n\n\n\n

(iii) (3 5\/12 + 1 2\/3) \u00f7 (3 5\/12 \u2013 1 2\/3)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= [(12 \u00d7 3 + 5)\/ 12 + (3 \u00d7 1 + 2)\/ 3] \u00f7 [(12 \u00d7 3 + 5)\/ 12 \u2013 (3 \u00d7 1 + 2)\/ 3]<\/p>\n\n\n\n

= (41\/12 + 5\/3) \u00f7 (41\/12 \u2013 5\/3)<\/p>\n\n\n\n

LCM of 12 and 3 is 12<\/p>\n\n\n\n

= (41 + 20)\/ 12 \u00f7 (41 \u2013 20)\/ 12<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 61\/12 \u00f7 21\/12<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 61\/12 \u00d7 12\/21<\/p>\n\n\n\n

= 61\/21<\/p>\n\n\n\n

= 2 19\/21<\/p>\n\n\n\n

8. The product of two numbers is 14. If one of the numbers is -8\/7, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Product of two numbers = 14<\/p>\n\n\n\n

One of the number = -8\/7<\/p>\n\n\n\n

Other number = 14 \u00f7 -8\/7<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 14 \u00d7 -7\/8<\/p>\n\n\n\n

= -98\/8<\/p>\n\n\n\n

= \u2013 49\/4<\/p>\n\n\n\n

9. The cost of 11 pens is \u20b9 24 \u00be. Find the cost of one pen.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Cost of 11 pens = \u20b9 24 \u00be<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (24 \u00d7 4 + 3)\/ 4<\/p>\n\n\n\n

= \u20b9 99\/4<\/p>\n\n\n\n

So the cost of one pen = 99\/4 \u00f7 11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 99\/4 \u00d7 1\/11<\/p>\n\n\n\n

= \u20b9 9\/4<\/p>\n\n\n\n

= \u20b9 2 \u00bc<\/p>\n\n\n\n

10. If 6 identical articles can be bought for \u20b9 2 6\/17. Find the cost of each article.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Cost of 6 articles = \u20b9 2 6\/17<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= (2 \u00d7 17 + 6)\/ 17<\/p>\n\n\n\n

= \u20b9 40\/17<\/p>\n\n\n\n

So the cost of each article = 40\/17 \u00f7 6<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 40\/17 \u00d7 1\/6<\/p>\n\n\n\n

= \u20b9 20\/51<\/p>\n\n\n\n

11. By what number should -3\/8 be multiplied so that the product is -9\/16?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Number = -3\/8 \u00f7 (-9\/16)<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= -3\/8 \u00d7 16\/-9<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2\/3<\/p>\n\n\n\n

= 1 \u00bd<\/p>\n\n\n\n

12. By what number should -5\/7 be divided so that the result is -15\/28?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the number as x<\/p>\n\n\n\n

-5\/7 \u00f7 x = -15\/28<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

-5\/7 \u00d7 1\/x = -15\/28<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

-5\/7x = -15\/28<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 5\/7 \u00d7 28\/15 = 4\/3<\/p>\n\n\n\n

x = 1 1\/3<\/p>\n\n\n\n

13. Evaluate: (32\/15 + 8\/5) \u00f7 (32\/15 \u2013 8\/5).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

(32\/15 + 8\/5) \u00f7 (32\/15 \u2013 8\/5)<\/p>\n\n\n\n

LCM of 15 and 5 is 15<\/p>\n\n\n\n

= [(32 \u00d7 1)\/ (15 \u00d7 1) + (8 \u00d7 3)\/ (5 \u00d7 3)] \u00f7 [(32 \u00d7 1)\/ (15 \u00d7 1) \u2013 (8 \u00d7 1)\/ (5 \u00d7 1)]<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (32 + 24)\/ 15 \u00f7 (32 \u2013 24)\/ 15<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 56\/15 \u00f7 8\/15<\/p>\n\n\n\n

= 56\/15 \u00d7 15\/8<\/p>\n\n\n\n

= 7<\/p>\n\n\n\n

14. Seven equal pieces are made out of a rope of 21 5\/7 m. Find the length of each piece.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Length of 7 pieces of rope = 21 5\/7 m<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (21 \u00d7 7 + 5)\/ 7<\/p>\n\n\n\n

= 152\/7<\/p>\n\n\n\n

So the length of each piece = 152\/7 \u00f7 7<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 152\/7 \u00d7 1\/7<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 152\/49<\/p>\n\n\n\n

= 3 5\/49 m<\/p>\n\n\n\n

Exercise 2E<\/h3>\n\n\n\n

1. Evaluate:<\/strong><\/p>\n\n\n\n

(i) -2\/3 + 3\/4<\/strong><\/p>\n\n\n\n

(ii) 7\/-27 + 11\/18<\/strong><\/p>\n\n\n\n

(iii) -3\/8 + -5\/12<\/strong><\/p>\n\n\n\n

(iv) 9\/-16 + -5\/-12<\/strong><\/p>\n\n\n\n

(v) -5\/9 + -7\/12 + 11\/18<\/strong><\/p>\n\n\n\n

(vi) 7\/-26 + 16\/39<\/strong><\/p>\n\n\n\n

(vii) -2\/3 \u2013 (-5\/7)<\/strong><\/p>\n\n\n\n

(viii) -5\/7 \u2013 (-3\/8)<\/strong><\/p>\n\n\n\n

(ix) 7\/26 + 2 + -11\/13<\/strong><\/p>\n\n\n\n

(x) -1 + 2\/-3 + 5\/6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) -2\/3 + 3\/4<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 3 and 4 is 12<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-2 \u00d7 4)\/ (3 \u00d7 4) + (3 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 9)\/ 12<\/p>\n\n\n\n

= 1\/12<\/p>\n\n\n\n

(ii) 7\/-27 + 11\/18<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 27 and 18 is 54<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 2)\/ (-27 \u00d7 2) + (11 \u00d7 3)\/ (18 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 33)\/ 54<\/p>\n\n\n\n

= 19\/54<\/p>\n\n\n\n

(iii) -3\/8 + -5\/12<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 12 is 24<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 \u00d7 3)\/ (8 \u00d7 3) + (-5 \u00d7 2)\/ (12 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-9 \u2013 10)\/ 24<\/p>\n\n\n\n

= -19\/24<\/p>\n\n\n\n

(iv) 9\/-16 + -5\/-12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 9\/-16 + 5\/12<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 16 and 12 is 48<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (9 \u00d7 3)\/ (-16 \u00d7 3) + (5 \u00d7 4)\/ (12 \u00d7 4)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-27 + 20)\/48<\/p>\n\n\n\n

= -7\/48<\/p>\n\n\n\n

(v) -5\/9 + -7\/12 + 11\/18<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 9, 12 and 18 is 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u00d7 4)\/ (9 \u00d7 4) \u2013 (7 \u00d7 3)\/ (12 \u00d7 3) + (11 \u00d7 2)\/ (18 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-20 \u2013 21 + 22)\/ 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-41 + 22)\/ 36<\/p>\n\n\n\n

= -19\/36<\/p>\n\n\n\n

(vi) 7\/-26 + 16\/39<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 26 and 39 is 78<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-7 \u00d7 3)\/ (26 \u00d7 3) + (16 \u00d7 2)\/ (39 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-21 + 32)\/ 78<\/p>\n\n\n\n

= 11\/78<\/p>\n\n\n\n

(vii) -2\/3 \u2013 (-5\/7)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -2\/3 + 5\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 7 is 21<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-2 \u00d7 7)\/ (3 \u00d7 7) + (5 \u00d7 3)\/ (7 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-14 + 15)\/ 21<\/p>\n\n\n\n

= 1\/21<\/p>\n\n\n\n

(viii) -5\/7 \u2013 (-3\/8)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= -5\/7 + 3\/8<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 7 and 8 is 56<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-5 \u00d7 8)\/ (7 \u00d7 8) + (3 \u00d7 7)\/ (8 \u00d7 7)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-40 + 21)\/ 56<\/p>\n\n\n\n

= -19\/56<\/p>\n\n\n\n

(ix) 7\/26 + 2 + -11\/13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 7\/26 + 2\/1 + -11\/13<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 26 and 13 is 26<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (7 \u00d7 1)\/ (26 \u00d7 1) + (2 \u00d7 26)\/ (1 \u00d7 26) \u2013 (11 \u00d7 2)\/ (13 \u00d7 2)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (7 + 52 \u2013 22)\/ 26<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (59 \u2013 22)\/ 26<\/p>\n\n\n\n

= 37\/26<\/p>\n\n\n\n

= 1 11\/26<\/p>\n\n\n\n

(x) -1 + 2\/-3 + 5\/6<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 6 is 6<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-1 \u00d7 6)\/ (1 \u00d7 6) \u2013 (2 \u00d7 2)\/ (3 \u00d7 2) + (5 \u00d7 1)\/ (6 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 \u2013 4 + 5)\/ 6<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= (-10 + 5)\/ 6<\/p>\n\n\n\n

= -5\/6<\/p>\n\n\n\n

2. The sum of two rational numbers is -3\/8. If one of them is 3\/16, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -3\/8<\/p>\n\n\n\n

One rational number = 3\/16<\/p>\n\n\n\n

Other rational number = -3\/8 \u2013 3\/16<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 16 is 16<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= (-3 \u00d7 2)\/ (8 \u00d7 2) \u2013 (3 \u00d7 1)\/ (16 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-6 \u2013 3)\/ 16<\/p>\n\n\n\n

= -9\/16<\/p>\n\n\n\n

3. The sum of two rational numbers is -5. If one of them is -52\/25, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = -5<\/p>\n\n\n\n

One rational number = -52\/25<\/p>\n\n\n\n

Other rational number = \u2013 5 \u2013 (-52\/25)<\/p>\n\n\n\n

Here LCM is 25<\/p>\n\n\n\n

= (-5 \u00d7 25)\/ (1 \u00d7 25) + (52 \u00d7 1)\/ (25 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-125 + 52)\/ 25<\/p>\n\n\n\n

= \u2013 73\/25<\/p>\n\n\n\n

4. What rational number should be added to -3\/16 to get 11\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Sum of two rational numbers = 11\/24<\/p>\n\n\n\n

One rational number = -3\/16<\/p>\n\n\n\n

Other number = 11\/24 \u2013 (-3\/16)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 11\/24 + 3\/16<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 16 and 24 is 48<\/p>\n\n\n\n

= (11 \u00d7 2)\/ (24 \u00d7 2) + (3 \u00d7 3)\/ (16 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (22 + 9)\/ 48<\/p>\n\n\n\n

= 31\/48<\/p>\n\n\n\n

5. What rational number should be added to -3\/5 to get 2?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

So the required rational number = 2 \u2013 (-3\/5)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 2 + 3\/5<\/p>\n\n\n\n

LCM of 1 and 5 is 5<\/p>\n\n\n\n

= (2 \u00d7 5)\/ (1 \u00d7 5) + (3 \u00d7 1)\/ (5 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (10 + 3)\/ 5<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 13\/5<\/p>\n\n\n\n

= 2 3\/5<\/p>\n\n\n\n

6. What rational number should be subtracted from -5\/12 to get 5\/24?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Required rational number = -5\/12 \u2013 5\/24<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 12 and 24 is 72<\/p>\n\n\n\n

= (-5 \u00d7 6)\/ (12 \u00d7 6) \u2013 (5 \u00d7 3)\/ (24 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-30 \u2013 15)\/ 72<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= \u2013 45\/72<\/p>\n\n\n\n

= -5\/8<\/p>\n\n\n\n

7. What rational number should be subtracted from 5\/8 to get 8\/5?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Required rational number = 5\/8 \u2013 8\/5<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 8 and 5 is 40<\/p>\n\n\n\n

= (5 \u00d7 5)\/ (8 \u00d7 5) \u2013 (8 \u00d7 8)\/ (5 \u00d7 8)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (25 \u2013 64)\/ 40<\/p>\n\n\n\n

= -39\/40<\/p>\n\n\n\n

8. Evaluate:<\/strong><\/p>\n\n\n\n

(i) (7\/8 \u00d7 24\/21) + (-5\/9 \u00d7 6\/-25)<\/strong><\/p>\n\n\n\n

(ii) (8\/15 \u00d7 -25\/16) + (-18\/35 \u00d7 5\/6)<\/strong><\/p>\n\n\n\n

(iii) (18\/33 \u00d7 -22\/27) \u2013 (13\/25 \u00d7 -75\/26)<\/strong><\/p>\n\n\n\n

(iv) (-13\/7 \u00d7 -35\/39) \u2013 (-7\/45 \u00d7 9\/14)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (7\/8 \u00d7 24\/21) + (-5\/9 \u00d7 6\/-25)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (7 \u00d7 24)\/ (8 \u00d7 21) + (-5 \u00d7 6)\/ (9 \u00d7 -25)<\/p>\n\n\n\n

By further simplification<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (1 \u00d7 3) + (1 \u00d7 2)\/ (3 \u00d7 5)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3\/3 + 2\/15<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 3 and 15 is 15<\/p>\n\n\n\n

= (3 \u00d7 5)\/ (3 \u00d7 5) + (2 \u00d7 1)\/ (15 \u00d7 1)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (15 + 2)\/ 15<\/p>\n\n\n\n

= 17\/15<\/p>\n\n\n\n

= 1 2\/15<\/p>\n\n\n\n

(ii) (8\/15 \u00d7 -25\/16) + (-18\/35 \u00d7 5\/6)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (8 \u00d7 -25)\/ (15 \u00d7 16) + (-18 \u00d7 5)\/ (35 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1 \u00d7 -5)\/ (3 \u00d7 2) + (-3 \u00d7 1)\/ (7 \u00d7 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -5\/6 \u2013 3\/7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 6 and 7 is 42<\/p>\n\n\n\n

= (-5 \u00d7 7)\/ (6 \u00d7 7) \u2013 (3 \u00d7 6)\/ (7 \u00d7 6)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-35 \u2013 18)\/ 42<\/p>\n\n\n\n

= -53\/42<\/p>\n\n\n\n

(iii) (18\/33 \u00d7 -22\/27) \u2013 (13\/25 \u00d7 -75\/26)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (18 \u00d7 -22)\/ (33 \u00d7 27) \u2013 (13 \u00d7 -75)\/ (25 \u00d7 26)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (2 \u00d7 -2)\/ (3 \u00d7 3) \u2013 (1 \u00d7 -3)\/ (1 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= -4\/9 \u2013 (-3\/2)<\/p>\n\n\n\n

= -4\/9 + 3\/2<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here LCM of 9 and 2 is 18<\/p>\n\n\n\n

= (-4 \u00d7 2)\/ (9 \u00d7 2) + (3 \u00d7 9)\/ (2 \u00d7 9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-8 + 27)\/ 18<\/p>\n\n\n\n

= 19\/18<\/p>\n\n\n\n

= 1 1\/18<\/p>\n\n\n\n

(iv) (-13\/7 \u00d7 -35\/39) \u2013 (-7\/45 \u00d7 9\/14)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= (-13 \u00d7 -35)\/ (7 \u00d7 39) + (7 \u00d7 9)\/ (45 \u00d7 14)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (-1 \u00d7 -5)\/ (1 \u00d7 3) + (1 \u00d7 1)\/ (5 \u00d7 2)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 5\/3 + 1\/10<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here the LCM of 3 and 10 is 30<\/p>\n\n\n\n

= (5 \u00d7 10)\/ (3 \u00d7 10) + (1 \u00d7 3)\/ (10 \u00d7 3)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (50 + 3)\/ 30<\/p>\n\n\n\n

= 53\/30<\/p>\n\n\n\n

= 1 23\/30<\/p>\n\n\n\n

9. The product of two rational numbers is 24. If one of them is -36\/11, find the other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Product of two rational numbers = 24<\/p>\n\n\n\n

One rational number = -36\/11<\/p>\n\n\n\n

Other rational number = 24 \u00f7 (-36\/ 11)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 24 \u00d7 (-11\/36)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2 \u00d7 (-11\/3)<\/p>\n\n\n\n

= -22\/3<\/p>\n\n\n\n

10. By what rational number should we multiply 20\/-9, so that the product may be -5\/9?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Here the required rational number = -5\/9 \u00f7 (20\/-9)<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= -5\/9 \u00d7 (-9\/20)<\/p>\n\n\n\n

= 1\/4<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 2- Rational Numbers<\/p>\n\n\n\n

Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 2-\u00a0Rational Numbers PDF<\/a><\/p>\n\n\n\n

Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :<\/strong><\/h2>\n\n\n\n