{"id":547228,"date":"2021-10-09T10:01:17","date_gmt":"2021-10-09T10:01:17","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=547228"},"modified":"2021-10-11T06:33:05","modified_gmt":"2021-10-11T06:33:05","slug":"rd-sharma-solutions-for-class-6-maths-chapter-5-negative-numbers-and-integers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-6-maths-chapter-5-negative-numbers-and-integers\/","title":{"rendered":"RD Sharma Solutions for Class 6 Maths Chapter 5\u2013Negative Numbers and Integers"},"content":{"rendered":"\n

Class 6: Maths Chapter 5 solutions. Complete Class 6 Maths Chapter 5 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 5\u2013Negative Numbers and Integers<\/h2>\n\n\n\n

RD Sharma 6th Maths Chapter 5, Class 6 Maths Chapter 5 solutions<\/p>\n\n\n\n

Exercise 5.1 PAGE: 5.5<\/h4>\n\n\n\n

1. Write the opposite of each of the following:<\/strong><\/p>\n\n\n\n

(i) Increase in population<\/strong><\/p>\n\n\n\n

(ii) Depositing money in a bank<\/strong><\/p>\n\n\n\n

(iii) Earning money<\/strong><\/p>\n\n\n\n

(iv) Going North<\/strong><\/p>\n\n\n\n

(v) Gaining a weight of 4kg<\/strong><\/p>\n\n\n\n

(vi) A loss of Rs 1000<\/strong><\/p>\n\n\n\n

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

(viii) \u2013 15<\/strong><\/p>\n\n\n\n

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

(i) The opposite of Increase in population is Decrease in population.<\/p>\n\n\n\n

(ii) The opposite of Depositing money in a bank is Withdrawing money from a bank.<\/p>\n\n\n\n

(iii) The opposite of earning money is Spending money.<\/p>\n\n\n\n

(iv) The opposite of Going North is Going South.<\/p>\n\n\n\n

(v) The opposite of gaining a weight of 4kg is losing a weight of 4kg.<\/p>\n\n\n\n

(vi) The opposite of a loss of Rs 1000 is a gain of Rs 1000.<\/p>\n\n\n\n

(vii) The opposite of 25 is \u2013 25.<\/p>\n\n\n\n

(viii) The opposite of \u2013 15 is 15.<\/p>\n\n\n\n

2. Indicate the following by using integers:<\/strong><\/p>\n\n\n\n

(i) 25o<\/sup> above zero<\/strong><\/p>\n\n\n\n

(ii) 5o<\/sup> below zero<\/strong><\/p>\n\n\n\n

(iii) A profit of Rs 800<\/strong><\/p>\n\n\n\n

(iv) A deposit of Rs 2500<\/strong><\/p>\n\n\n\n

(v) 3km above sea level<\/strong><\/p>\n\n\n\n

(vi) 2km below level<\/strong><\/p>\n\n\n\n

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

(i) 25o<\/sup> above zero is + 25o<\/sup>.<\/p>\n\n\n\n

(ii) 5o<\/sup> below zero is \u2013 5o<\/sup>.<\/p>\n\n\n\n

(iii) A profit of Rs 800 is + 800.<\/p>\n\n\n\n

(iv) A deposit of Rs 2500 is + 2500.<\/p>\n\n\n\n

(v) 3km above sea level is + 3.<\/p>\n\n\n\n

(vi) 2km below level is \u2013 2.<\/p>\n\n\n\n

3. Mark the following integers on a number line:<\/strong><\/p>\n\n\n\n

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

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

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

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

The following integers are marked on a number line as given below:<\/p>\n\n\n\n

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

4. Which number in each of the following pairs is smaller?<\/strong><\/p>\n\n\n\n

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

(ii) -3 , 12<\/strong><\/p>\n\n\n\n

(iii) 8, 13<\/strong><\/p>\n\n\n\n

(iv) \u2013 15, -27<\/strong><\/p>\n\n\n\n

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

(i) 0 is greater than the negative integers<\/p>\n\n\n\n

So we get \u2013 4 < 0<\/p>\n\n\n\n

Therefore, \u2013 4 is smaller.<\/p>\n\n\n\n

(ii) 12 is greater than -3 on a number line<\/p>\n\n\n\n

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

-3 < 12<\/p>\n\n\n\n

Therefore, \u2013 3 is smaller.<\/p>\n\n\n\n

(iii) 13 is greater than 8 on a number line<\/p>\n\n\n\n

So we get 8 < 13<\/p>\n\n\n\n

Therefore, 8 is smaller.<\/p>\n\n\n\n

(iv) \u2013 15 is greater than \u2013 27 on a number line<\/p>\n\n\n\n

So we get \u2013 27 < \u2013 15<\/p>\n\n\n\n

Therefore, \u2013 27 is smaller.<\/p>\n\n\n\n

5. Which number in each of the following pairs is larger?<\/strong><\/p>\n\n\n\n

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

(ii) \u2013 12, \u2013 8<\/strong><\/p>\n\n\n\n

(iii) 0, 7<\/strong><\/p>\n\n\n\n

(iv) 12, \u2013 18<\/strong><\/p>\n\n\n\n

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

(i) We know that 3 is larger than \u2013 4 on a number line<\/p>\n\n\n\n

So we get 3 > \u2013 4<\/p>\n\n\n\n

Therefore, 3 is larger.<\/p>\n\n\n\n

(ii) We know that \u2013 8 is larger than \u2013 12 on a number line<\/p>\n\n\n\n

So we get \u2013 8 > \u2013 12<\/p>\n\n\n\n

Therefore, \u2013 8 is larger.<\/p>\n\n\n\n

(iii) We know that 7 is larger than 0 on a number line<\/p>\n\n\n\n

So we get 7 > 0<\/p>\n\n\n\n

Therefore, 7 is larger.<\/p>\n\n\n\n

(iv) We know that 12 is larger than \u2013 18 on a number line<\/p>\n\n\n\n

So we get 12 > \u2013 18<\/p>\n\n\n\n

Therefore, 12 is larger.<\/p>\n\n\n\n

6. Write all integers between:<\/strong><\/p>\n\n\n\n

(i) \u2013 7 and 3<\/strong><\/p>\n\n\n\n

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

(iii) \u2013 4 and 0<\/strong><\/p>\n\n\n\n

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

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

(i) The integers between \u2013 7 and 3 are<\/p>\n\n\n\n

\u2013 6, \u2013 5, \u2013 4, \u2013 3, \u2013 2, \u2013 1, 0, 1, 2<\/p>\n\n\n\n

(ii) The integers between \u2013 2 and 2 are<\/p>\n\n\n\n

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

(iii) The integers between \u2013 4 and 0 are<\/p>\n\n\n\n

-3, -2, -1<\/p>\n\n\n\n

(iv) The integers between 0 and 3 are<\/p>\n\n\n\n

1, 2.<\/p>\n\n\n\n

7. How many integers are between?<\/strong><\/p>\n\n\n\n

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

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

(iii) \u2013 9 and \u2013 2<\/strong><\/p>\n\n\n\n

(iv) 0 and 5<\/strong><\/p>\n\n\n\n

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

(i) The integers between \u2013 4 and 3 are<\/p>\n\n\n\n

-3, -2, -1, 0, 1, 2<\/p>\n\n\n\n

Therefore, number of integers between \u2013 4 and 3 are 6.<\/p>\n\n\n\n

(ii) The integers between 5 and 12 are<\/p>\n\n\n\n

6, 7, 8, 9, 10, 11<\/p>\n\n\n\n

Therefore, number of integers between 5 and 12 are 6.<\/p>\n\n\n\n

(iii) The integers between \u2013 9 and \u2013 2 are<\/p>\n\n\n\n

-8, -7, -6, -5, -4, -3<\/p>\n\n\n\n

Therefore, number of integers between -9 and -2 are 6.<\/p>\n\n\n\n

(iv) The integers between 0 and 5 are<\/p>\n\n\n\n

1, 2, 3, 4<\/p>\n\n\n\n

Therefore, number of integers between 0 and 5 are 4.<\/p>\n\n\n\n

8. Replace * in each of the following by < or > so that the statement is true:<\/strong><\/p>\n\n\n\n

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

(ii) 0 * 3<\/strong><\/p>\n\n\n\n

(iii) 0 * \u2013 7<\/strong><\/p>\n\n\n\n

(iv) \u2013 18 * 15<\/strong><\/p>\n\n\n\n

(v) \u2013 235 * \u2013 532<\/strong><\/p>\n\n\n\n

(vi) \u2013 20 * 20<\/strong><\/p>\n\n\n\n

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

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

(ii) 0 < 3<\/p>\n\n\n\n

(iii) 0 > \u2013 7<\/p>\n\n\n\n

(iv) \u2013 18 < 15<\/p>\n\n\n\n

(v) \u2013 235 > \u2013 532<\/p>\n\n\n\n

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

9. Write the following integers in increasing order:<\/strong><\/p>\n\n\n\n

(i) \u2013 8, 5, 0, -12, 1, -9, 15<\/strong><\/p>\n\n\n\n

(ii) \u2013 106, 107, \u2013 320, \u2013 7, 185<\/strong><\/p>\n\n\n\n

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

(i) \u2013 8, 5, 0, -12, 1, -9, 15 can be written in increasing order as<\/p>\n\n\n\n

\u2013 12, \u2013 9, \u2013 8, 0, 1, 5, 15<\/p>\n\n\n\n

(ii) \u2013 106, 107, \u2013 320, \u2013 7, 185 can be written in increasing order as<\/p>\n\n\n\n

-320, \u2013 106, \u2013 7, 107, 185.<\/p>\n\n\n\n

10. Write the following integers in decreasing order:<\/strong><\/p>\n\n\n\n

(i) \u2013 15, 0, -2, -9, 7, 6, -5, 8<\/strong><\/p>\n\n\n\n

(ii) -154, 123, -205, -89, -74<\/strong><\/p>\n\n\n\n

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

(i) \u2013 15, 0, -2, -9, 7, 6, -5, 8 can be written in decreasing order as<\/p>\n\n\n\n

8, 7, 6, 0, -2, -5, -9, -15<\/p>\n\n\n\n

(ii) -154, 123, -205, -89, -74 can be written in decreasing order as<\/p>\n\n\n\n

123, \u2013 74, \u2013 89, \u2013 154, \u2013 205<\/p>\n\n\n\n

11. Using the number line, write the integer which is:<\/strong><\/p>\n\n\n\n

(i) 2 more than 3<\/strong><\/p>\n\n\n\n

(ii) 5 less than 3<\/strong><\/p>\n\n\n\n

(iii) 4 more than \u2013 9<\/strong><\/p>\n\n\n\n

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

(i) 2 more than 3<\/p>\n\n\n\n

In order to get the integer 2 more than 3<\/p>\n\n\n\n

We draw a number line from 2 and proceed 3 units to the right to obtain 5<\/p>\n\n\n\n

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

Therefore, 2 more than 3 is 5.<\/p>\n\n\n\n

(ii) 5 less than 3<\/p>\n\n\n\n

In order to get the integer 5 less than 3<\/p>\n\n\n\n

We draw a number line from 3 and proceed 5 units to the left to obtain \u2013 2<\/p>\n\n\n\n

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

Therefore, 5 less than 3 is \u2013 2.<\/p>\n\n\n\n

(iii) 4 more than \u2013 9<\/p>\n\n\n\n

In order to get the integer 4 more than \u2013 9<\/p>\n\n\n\n

We draw a number line from \u2013 9 and proceed 4 units to the right to obtain -5<\/p>\n\n\n\n

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

Therefore, 4 more than \u2013 9 is \u2013 5.<\/p>\n\n\n\n

12. Write the absolute value of each of the following:<\/strong><\/p>\n\n\n\n

(i) 14<\/strong><\/p>\n\n\n\n

(ii) \u2013 25<\/strong><\/p>\n\n\n\n

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

(iv) \u2013 125<\/strong><\/p>\n\n\n\n

(v) \u2013 248<\/strong><\/p>\n\n\n\n

(vi) a \u2013 7, if a is greater than 7<\/strong><\/p>\n\n\n\n

(vii) a \u2013 7, if a \u2013 2 is less than 7<\/strong><\/p>\n\n\n\n

(viii) a + 4, if a is greater than -4<\/strong><\/p>\n\n\n\n

(ix) a + 4 if a is less than \u2013 4<\/strong><\/p>\n\n\n\n

(x) |-3|<\/strong><\/p>\n\n\n\n

(xi) -|-5|<\/strong><\/p>\n\n\n\n

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

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

(i) The absolute value of 14 is<\/p>\n\n\n\n

|14| = 14<\/p>\n\n\n\n

(ii) The absolute value of \u2013 25 is<\/p>\n\n\n\n

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

(iii) The absolute value of 0 is<\/p>\n\n\n\n

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

(iv) The absolute value of \u2013 125 is<\/p>\n\n\n\n

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

(v) The absolute value of \u2013 248 is<\/p>\n\n\n\n

|-248| = 248<\/p>\n\n\n\n

(vi) The absolute value of a \u2013 7, if a is greater than 7 is<\/p>\n\n\n\n

|a \u2013 7| = a \u2013 7 where a > 7<\/p>\n\n\n\n

(vii) The absolute value of a \u2013 7, if a \u2013 2 is less than 7 is<\/p>\n\n\n\n

|a \u2013 7| = \u2013 (a \u2013 7) where a \u2013 2 < 7<\/p>\n\n\n\n

(viii) The absolute value of a + 4, if a is greater than -4 is<\/p>\n\n\n\n

|a + 4| = a + 4 where a > \u2013 4<\/p>\n\n\n\n

(ix) The absolute value of a + 4 if a is less than \u2013 4 is<\/p>\n\n\n\n

|a + 4| = \u2013 (a + 4) where a < -4<\/p>\n\n\n\n

(x) The absolute value of |-3| is<\/p>\n\n\n\n

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

(xi) The absolute value of -|-5| is<\/p>\n\n\n\n

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

(xii) The absolute value of |12 \u2013 5| is<\/p>\n\n\n\n

|12 \u2013 5| = 7<\/p>\n\n\n\n

13. (i) Write 4 negative integers less than \u2013 10.<\/strong><\/p>\n\n\n\n

(ii) Write 6 negative integers just greater than \u2013 12.<\/strong><\/p>\n\n\n\n

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

(i) The 4 negative integers less than \u2013 10 are<\/p>\n\n\n\n

\u2013 11, \u2013 12, \u2013 13, \u2013 14<\/p>\n\n\n\n

(ii) The 6 negative integers just greater than \u2013 12 are<\/p>\n\n\n\n

-11, \u2013 10, \u2013 9, \u2013 8, \u2013 7, \u2013 6<\/p>\n\n\n\n

14. Which of the following statements are true?<\/strong><\/p>\n\n\n\n

(i) The smallest integer is zero.<\/strong><\/p>\n\n\n\n

(ii) The opposite of zero is zero.<\/strong><\/p>\n\n\n\n

(iii) Zero is not an integer.<\/strong><\/p>\n\n\n\n

(iv) 0 is larger than every negative integer.<\/strong><\/p>\n\n\n\n

(v) The absolute value of an integer is greater than the integer.<\/strong><\/p>\n\n\n\n

(vi) A positive integer is greater than its opposite.<\/strong><\/p>\n\n\n\n

(vii) Every negative integer is less than every natural number.<\/strong><\/p>\n\n\n\n

(viii) 0 is the smallest positive integer.<\/strong><\/p>\n\n\n\n

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

(i) False. The smallest integer is 1.<\/p>\n\n\n\n

(ii) True. 0 is neither positive nor negative so the opposite is 0.<\/p>\n\n\n\n

(iii) False. Zero is an integer which is neither positive nor negative.<\/p>\n\n\n\n

(iv) True. 0 is larger than \u2013 1.<\/p>\n\n\n\n

(v) False. The absolute value of an integer is the numerical value.<\/p>\n\n\n\n

(vi) True. 3 is greater than \u2013 3.<\/p>\n\n\n\n

(vii) True. \u2013 3 is less than 1.<\/p>\n\n\n\n

(viii) False. 1 is the smallest positive integer.<\/p>\n\n\n\n

\n\n\n\n

Exercise 5.2 PAGE: 5.9<\/h4>\n\n\n\n

1. Draw a number line and represent each of the following on it:<\/strong><\/p>\n\n\n\n

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

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

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

(iv) 6 + (-6)<\/strong><\/p>\n\n\n\n

(v) (-1) + (-2) + 2<\/strong><\/p>\n\n\n\n

(vi) (-2) + 7 + (-9)<\/strong><\/p>\n\n\n\n

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

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

From 0 move towards right of first five units to obtain + 5<\/p>\n\n\n\n

So the second number is \u2013 2 so move 2 units towards left of + 5 we get + 3<\/p>\n\n\n\n

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

Therefore, 5 + (-2) = 3.<\/p>\n\n\n\n

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

From 0 move towards left of nine units to obtain \u2013 9<\/p>\n\n\n\n

So the second number is 4 so move 4 units towards right of \u2013 9 we get \u2013 5<\/p>\n\n\n\n

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

Therefore, (-9) + 4 = \u2013 5.<\/p>\n\n\n\n

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

From 0 move towards left of three units to obtain \u2013 3<\/p>\n\n\n\n

So the second number is \u2013 5 so move 5 units towards left of \u2013 3 we get \u2013 8<\/p>\n\n\n\n

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

Therefore, (-3) + (-5) = \u2013 8.<\/p>\n\n\n\n

(iv) 6 + (-6)<\/p>\n\n\n\n

From zero move towards right of six units to obtain 6<\/p>\n\n\n\n

So the second number is \u2013 6 so move 6 units towards left of 6 we get 0<\/p>\n\n\n\n

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

Therefore, 6 + (-6) = 0.<\/p>\n\n\n\n

(v) (-1) + (-2) + 2<\/p>\n\n\n\n

From zero move towards left of one unit to obtain \u2013 1<\/p>\n\n\n\n

So the second number is \u2013 2 so move 2 units towards left of \u2013 1 we get \u2013 3<\/p>\n\n\n\n

The third number is 2 so move 2 units towards right of \u2013 3 we get \u2013 1<\/p>\n\n\n\n

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

Therefore, (-1) + (-2) + 2 = \u2013 1.<\/p>\n\n\n\n

(vi) (-2) + 7 + (-9)<\/p>\n\n\n\n

From zero move towards left of two units to obtain \u2013 2<\/p>\n\n\n\n

So the second number is 7 so move 7 units towards right of \u2013 2 we get 5<\/p>\n\n\n\n

The third number is \u2013 9 so move 9 units towards left of 5 we get \u2013 4<\/p>\n\n\n\n

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

Therefore, (-2) + 7 + (-9) = \u2013 4.<\/p>\n\n\n\n

2. Find the sum of<\/strong><\/p>\n\n\n\n

(i) -557 and 488<\/strong><\/p>\n\n\n\n

(ii) -522 and -160<\/strong><\/p>\n\n\n\n

(iii) 2567 and \u2013 325<\/strong><\/p>\n\n\n\n

(iv) -10025 and 139<\/strong><\/p>\n\n\n\n

(v) 2547 and -2548<\/strong><\/p>\n\n\n\n

(vi) 2884 and -2884<\/strong><\/p>\n\n\n\n

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

(i) -557 and 488<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

-557 + 488<\/p>\n\n\n\n

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

|-557| \u2013 |488| = 557 \u2013 488 = 69.<\/p>\n\n\n\n

(ii) -522 and -160<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

-522 + (-160)<\/p>\n\n\n\n

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

-522 \u2013 160 = \u2013 682<\/p>\n\n\n\n

(iii) 2567 and \u2013 325<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

2567 + (-325)<\/p>\n\n\n\n

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

2567 \u2013 325 = 2242<\/p>\n\n\n\n

(iv) -10025 and 139<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

-10025 + 139<\/p>\n\n\n\n

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

-10025 + 139 = -9886<\/p>\n\n\n\n

(v) 2547 and -2548<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

2547 + (-2548)<\/p>\n\n\n\n

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

2547 \u2013 2548 = -1<\/p>\n\n\n\n

(vi) 2884 and -2884<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

2884 + (-2884)<\/p>\n\n\n\n

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

2884 \u2013 2884 = 0<\/p>\n\n\n\n

\n\n\n\n

Exercise 5.3 page: 5.11<\/h4>\n\n\n\n

1. Find the additive inverse of each of the following integers:<\/strong><\/p>\n\n\n\n

(i) 52<\/strong><\/p>\n\n\n\n

(ii) \u2013 176<\/strong><\/p>\n\n\n\n

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

(iv) 1<\/strong><\/p>\n\n\n\n

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

(i) The additive inverse of 52 is \u2013 52.<\/p>\n\n\n\n

(ii) The additive inverse of \u2013 176 is 176.<\/p>\n\n\n\n

(iii) The additive inverse of 0 is 0.<\/p>\n\n\n\n

(iv) The additive inverse of 1 is \u2013 1.<\/p>\n\n\n\n

2. Find the successor of each of the following integers:<\/strong><\/p>\n\n\n\n

(i) \u2013 42<\/strong><\/p>\n\n\n\n

(ii) -1<\/strong><\/p>\n\n\n\n

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

(iv) \u2013 200<\/strong><\/p>\n\n\n\n

(v) -99<\/strong><\/p>\n\n\n\n

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

(i) The successor of \u2013 42 is<\/p>\n\n\n\n

\u2013 42 + 1 = \u2013 41<\/p>\n\n\n\n

(ii) The successor of \u2013 1 is<\/p>\n\n\n\n

-1 + 1 = 0<\/p>\n\n\n\n

(iii) The successor of 0 is<\/p>\n\n\n\n

0 + 1 = 1<\/p>\n\n\n\n

(iv) The successor of \u2013 200 is<\/p>\n\n\n\n

-200 + 1 = \u2013 199<\/p>\n\n\n\n

(v) The successor of \u2013 99 is<\/p>\n\n\n\n

\u2013 99 + 1 = \u2013 98<\/p>\n\n\n\n

3. Find the predecessor of each of the following integers:<\/strong><\/p>\n\n\n\n

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

(ii) 1<\/strong><\/p>\n\n\n\n

(iii) \u2013 1<\/strong><\/p>\n\n\n\n

(iv) \u2013 125<\/strong><\/p>\n\n\n\n

(v) 1000<\/strong><\/p>\n\n\n\n

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

(i) The predecessor of 0 is<\/p>\n\n\n\n

0 \u2013 1 = \u2013 1<\/p>\n\n\n\n

(ii) The predecessor of 1 is<\/p>\n\n\n\n

1 \u2013 1 = 0<\/p>\n\n\n\n

(iii) The predecessor of -1 is<\/p>\n\n\n\n

-1 \u2013 1 = -2<\/p>\n\n\n\n

(iv) The predecessor of \u2013 125 is<\/p>\n\n\n\n

-125 \u2013 1 = \u2013 126<\/p>\n\n\n\n

(v) The predecessor of 1000 is<\/p>\n\n\n\n

1000 \u2013 1 = 999<\/p>\n\n\n\n

4. Which of the following statements are true?<\/strong><\/p>\n\n\n\n

(i) The sum of a number and its opposite is zero.<\/strong><\/p>\n\n\n\n

(ii) The sum of two negative integers is a positive integer.<\/strong><\/p>\n\n\n\n

(iii) The sum of a negative integer and a positive integer is always a negative integer.<\/strong><\/p>\n\n\n\n

(iv) The successor of \u2013 1 is 1.<\/strong><\/p>\n\n\n\n

(v) The sum of three different integers can never be zero.<\/strong><\/p>\n\n\n\n

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

(i) True. 1 \u2013 1 = 0<\/p>\n\n\n\n

(ii) False. -1 \u2013 1 = -2<\/p>\n\n\n\n

(iii) False. \u2013 2 + 3 = 1<\/p>\n\n\n\n

(iv) False. The successor of \u2013 1 is 0.<\/p>\n\n\n\n

(v) False. 1 + 2 \u2013 3 = 0<\/p>\n\n\n\n

5. Write all integers whose absolute values are less than 5.<\/strong><\/p>\n\n\n\n

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

The integers whose absolute values are less than 5 are<\/p>\n\n\n\n

-4, \u2013 3, \u2013 2, \u2013 1, 0, 1, 2, 3, 4<\/p>\n\n\n\n

6. Which of the following is false:<\/strong><\/p>\n\n\n\n

(i) |4 + 2| = |4| + |2|<\/strong><\/p>\n\n\n\n

(ii) |2 \u2013 4| = |2| + |4|<\/strong><\/p>\n\n\n\n

(iii) |4 \u2013 2| = |4| \u2013 |2|<\/strong><\/p>\n\n\n\n

(iv) |(-2) + (-4)| = |-2| + |-4|<\/strong><\/p>\n\n\n\n

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

(i) True.<\/p>\n\n\n\n

(ii) False.<\/p>\n\n\n\n

(iii) True.<\/p>\n\n\n\n

(iv) True.<\/p>\n\n\n\n

7. Complete the following table:<\/strong><\/p>\n\n\n\n

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

From the above table:<\/strong><\/p>\n\n\n\n

(i) Write all the pairs of integers whose sum is 0.<\/strong><\/p>\n\n\n\n

(ii) Is (-4) + (-2) = (-2) + (-4)?<\/strong><\/p>\n\n\n\n

(iii) Is 0 + (-6) = -6?<\/strong><\/p>\n\n\n\n

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

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

(i) The pairs of integers whose sum is 0 are<\/p>\n\n\n\n

(6, -6), (4, \u2013 4), (2, \u2013 2), (0, 0)<\/p>\n\n\n\n

(ii) Yes. By using commutativity of addition (-4) + (-2) = (-2) + (-4)<\/p>\n\n\n\n

(iii) Yes. By using additive identity 0 + (-6) = -6.<\/p>\n\n\n\n

8. Find an integer x such that<\/strong><\/p>\n\n\n\n

(i) x + 1 = 0<\/strong><\/p>\n\n\n\n

(ii) x + 5 = 0<\/strong><\/p>\n\n\n\n

(iii) \u2013 3 + x = 0<\/strong><\/p>\n\n\n\n

(iv) x + (-8) = 0<\/strong><\/p>\n\n\n\n

(v) 7 + x = 0<\/strong><\/p>\n\n\n\n

(vi) x + 0 = 0<\/strong><\/p>\n\n\n\n

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

(i) x + 1 = 0<\/p>\n\n\n\n

Subtracting 1 on both sides<\/p>\n\n\n\n

x + 1 \u2013 1 = 0 \u2013 1<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

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

(ii) x + 5 = 0<\/p>\n\n\n\n

By subtracting 5 on both sides<\/p>\n\n\n\n

x + 5 \u2013 5 = 0 \u2013 5<\/p>\n\n\n\n

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

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

(iii) \u2013 3 + x = 0<\/p>\n\n\n\n

By adding 3 on both sides<\/p>\n\n\n\n

-3 + x + 3 = 0 + 3<\/p>\n\n\n\n

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

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

(iv) x + (-8) = 0<\/p>\n\n\n\n

By adding 8 on both sides<\/p>\n\n\n\n

x \u2013 8 + 8 = 0 + 8<\/p>\n\n\n\n

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

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

(v) 7 + x = 0<\/p>\n\n\n\n

By subtracting 7 on both sides<\/p>\n\n\n\n

7 + x \u2013 7 = 0 \u2013 7<\/p>\n\n\n\n

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

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

(vi) x + 0 = 0<\/p>\n\n\n\n

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

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

\n\n\n\n

Exercise 5.4 page: 5.17<\/h4>\n\n\n\n

1. Subtract the first integer from the second in each of the following:<\/strong><\/p>\n\n\n\n

(i) 12, -5<\/strong><\/p>\n\n\n\n

(ii) \u2013 12, 8<\/strong><\/p>\n\n\n\n

(iii) \u2013 225, \u2013 135<\/strong><\/p>\n\n\n\n

(iv) 1001, 101<\/strong><\/p>\n\n\n\n

(v) \u2013 812, 3126<\/strong><\/p>\n\n\n\n

(vi) 7560, \u2013 8<\/strong><\/p>\n\n\n\n

(vii) \u2013 3978, \u2013 4109<\/strong><\/p>\n\n\n\n

(viii) 0, \u2013 1005<\/strong><\/p>\n\n\n\n

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

(i) 12, -5<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

-5 \u2013 12 = \u2013 17<\/p>\n\n\n\n

(ii) \u2013 12, 8<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

8 \u2013 (-12) = 8 + 12 = 20<\/p>\n\n\n\n

(iii) \u2013 225, \u2013 135<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

-135 \u2013 (-225) = 225 \u2013 135 = 90<\/p>\n\n\n\n

(iv) 1001, 101<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

101 \u2013 1001 = \u2013 900<\/p>\n\n\n\n

(v) \u2013 812, 3126<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

3126 \u2013 (-812) = 3126 + 812 = 3938<\/p>\n\n\n\n

(vi) 7560, \u2013 8<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

-8 \u2013 7560 = \u2013 7568<\/p>\n\n\n\n

(vii) \u2013 3978, \u2013 4109<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

-4109 \u2013 (-3978) = \u2013 4109 + 3978 = -131<\/p>\n\n\n\n

(viii) 0, \u2013 1005<\/p>\n\n\n\n

So by subtracting the first integer from the second<\/p>\n\n\n\n

-1005 \u2013 0 = \u2013 1005<\/p>\n\n\n\n

2. Find the value of:<\/strong><\/p>\n\n\n\n

(i) \u2013 27 \u2013 (- 23)<\/strong><\/p>\n\n\n\n

(ii) \u2013 17 \u2013 18 \u2013 (-35)<\/strong><\/p>\n\n\n\n

(iii) \u2013 12 \u2013 (-5) \u2013 (-125) + 270<\/strong><\/p>\n\n\n\n

(iv) 373 + (-245) + (-373) + 145 + 3000<\/strong><\/p>\n\n\n\n

(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900<\/strong><\/p>\n\n\n\n

(vi) (-1) + (-304) + 304 + 304 + (-304) + 1<\/strong><\/p>\n\n\n\n

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

(i) \u2013 27 \u2013 (- 23)<\/p>\n\n\n\n

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

= \u2013 27 + 23<\/p>\n\n\n\n

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

= 23 \u2013 27<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

(ii) \u2013 17 \u2013 18 \u2013 (-35)<\/p>\n\n\n\n

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

= \u2013 35 + 35<\/p>\n\n\n\n

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

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

(iii) \u2013 12 \u2013 (-5) \u2013 (-125) + 270<\/p>\n\n\n\n

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

= \u2013 12 + 5 + 125 + 270<\/p>\n\n\n\n

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

= 400 \u2013 12<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 388<\/p>\n\n\n\n

(iv) 373 + (-245) + (-373) + 145 + 3000<\/p>\n\n\n\n

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

= 373 \u2013 245 \u2013 373 + 145 + 3000<\/p>\n\n\n\n

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

= 3145 + 373 \u2013 373 \u2013 245<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= 3145 \u2013 245<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

= 2900<\/p>\n\n\n\n

(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900<\/p>\n\n\n\n

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

= 1 \u2013 950 \u2013 950 + 1900<\/p>\n\n\n\n

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

= 1900 + 1 \u2013 1900<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

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

(vi) (-1) + (-304) + 304 + 304 + (-304) + 1<\/p>\n\n\n\n

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

= \u2013 1 + 1 \u2013 304 + 304 \u2013 304 + 304<\/p>\n\n\n\n

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

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

3. Subtract the sum of \u2013 5020 and 2320 from \u2013 709.<\/strong><\/p>\n\n\n\n

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

We know that the sum of -5020 and 2320 is<\/p>\n\n\n\n

-5020 + 2320<\/p>\n\n\n\n

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

= 2320 \u2013 5020<\/p>\n\n\n\n

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

= \u2013 2700<\/p>\n\n\n\n

Subtracting from \u2013 709 we get<\/p>\n\n\n\n

= \u2013 709 \u2013 (-2700)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= \u2013 709 + 2700<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

= 1991<\/p>\n\n\n\n

4. Subtract the sum of \u2013 1250 and 1138 from the sum of 1136 and \u2013 1272.<\/strong><\/p>\n\n\n\n

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

We know that the sum of \u2013 1250 and 1138 is<\/p>\n\n\n\n

-1250 + 1138<\/p>\n\n\n\n

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

= 1138 \u2013 1250<\/p>\n\n\n\n

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

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

We know that the sum of 1136 and \u2013 1272 is<\/p>\n\n\n\n

1136 \u2013 1272 = \u2013 136<\/p>\n\n\n\n

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

-136 \u2013 (-112) = \u2013 136 + 112 = -24<\/p>\n\n\n\n

5. From the sum of 233 and \u2013 147, subtract \u2013 284.<\/strong><\/p>\n\n\n\n

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

We know that the sum of 233 and \u2013 147 is<\/p>\n\n\n\n

233 \u2013 147 = 86<\/p>\n\n\n\n

Subtracting \u2013 284 we get<\/p>\n\n\n\n

86 \u2013 (-284) = 86 + 284 = 370<\/p>\n\n\n\n

6. The sum of two integers is 238. If one of the integers is \u2013 122, determine 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 integers = 238<\/p>\n\n\n\n

One of the integers = \u2013 122<\/p>\n\n\n\n

So the other integer = \u2013 (-122) + 238<\/p>\n\n\n\n

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

Other integer = 238 + 122 = 360<\/p>\n\n\n\n

7. The sum of two integers is \u2013 223. If one of the integers is 172, 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 integers = \u2013 223<\/p>\n\n\n\n

One of the integers = 172<\/p>\n\n\n\n

So the other integer = \u2013 223 \u2013 172 = \u2013 395<\/p>\n\n\n\n

8. Evaluate the following:<\/strong><\/p>\n\n\n\n

(i) \u2013 8 \u2013 24 + 31 \u2013 26 \u2013 28 + 7 + 19 \u2013 18 \u2013 8 + 33<\/strong><\/p>\n\n\n\n

(ii) \u2013 26 \u2013 20 + 33 \u2013 (-33) + 21 + 24 \u2013 (-25) \u2013 26 \u2013 14 \u2013 34<\/strong><\/p>\n\n\n\n

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

(i) \u2013 8 \u2013 24 + 31 \u2013 26 \u2013 28 + 7 + 19 \u2013 18 \u2013 8 + 33<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= \u2013 8 \u2013 24 \u2013 26 \u2013 28 \u2013 18 \u2013 8 + 31 + 7 + 19 + 33<\/p>\n\n\n\n

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

= \u2013 32 \u2013 26 \u2013 28 \u2013 26 + 38 + 19 + 33<\/p>\n\n\n\n

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

= 38 \u2013 32 \u2013 26 \u2013 28 + 33 \u2013 26 + 19<\/p>\n\n\n\n

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

= 6 \u2013 26 \u2013 28 + 7 + 19<\/p>\n\n\n\n

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

= 6 \u2013 28 \u2013 26 + 26<\/p>\n\n\n\n

= 6 \u2013 28<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

= \u2013 22<\/p>\n\n\n\n

(ii) \u2013 26 \u2013 20 + 33 \u2013 (-33) + 21 + 24 \u2013 (-25) \u2013 26 \u2013 14 \u2013 34<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= \u2013 46 + 33 + 33 + 21 + 24 + 25 \u2013 26 \u2013 14 \u2013 34<\/p>\n\n\n\n

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

= \u2013 46 + 66 + 21 + 24 + 25 + (-74)<\/p>\n\n\n\n

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

= \u2013 46 + 66 + 70 \u2013 74<\/p>\n\n\n\n

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

= \u2013 46 \u2013 4 + 66<\/p>\n\n\n\n

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

= \u2013 50 + 66<\/p>\n\n\n\n

= 66 \u2013 50<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

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

9. Calculate<\/strong><\/p>\n\n\n\n

1 \u2013 2 + 3 \u2013 4 + 5 \u2013 6 + \u2026\u2026\u2026 + 15 \u2013 16<\/strong><\/p>\n\n\n\n

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

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

1 \u2013 2 + 3 \u2013 4 + 5 \u2013 6 + 7 \u2013 8 + 9 \u2013 10 + 11 \u2013 12 + 13 \u2013 14 + 15 \u2013 16<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

= \u2013 1 \u2013 1 \u2013 1 \u2013 1 \u2013 1 \u2013 1 \u2013 1 \u2013 1<\/p>\n\n\n\n

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

= \u2013 8<\/p>\n\n\n\n

10. Calculate the sum:<\/strong><\/p>\n\n\n\n

5 + (-5) + 5 + (-5) + \u2026..<\/strong><\/p>\n\n\n\n

(i) if the number of terms is 10.<\/strong><\/p>\n\n\n\n

(ii) if the number of terms is 11.<\/strong><\/p>\n\n\n\n

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

(i) if the number of terms is 10<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5)<\/p>\n\n\n\n

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

= 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 = 0<\/p>\n\n\n\n

(ii) if the number of terms is 11<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5<\/p>\n\n\n\n

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

= 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 + 5 \u2013 5 + 5 = 5<\/p>\n\n\n\n

11. Replace * by < or > in each of the following to make the statement true:<\/strong><\/p>\n\n\n\n

(i) (-6) + (-9) * (-6) \u2013 (-9)<\/strong><\/p>\n\n\n\n

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

(iii) (-20) \u2013 (-20) * 20 \u2013 (65)<\/strong><\/p>\n\n\n\n

(iv) 28 \u2013 (-10) * (-16) \u2013 (-76)<\/strong><\/p>\n\n\n\n

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

(i) (-6) + (-9) < (-6) \u2013 (-9)<\/p>\n\n\n\n

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

(iii) (-20) \u2013 (-20) > 20 \u2013 (65)<\/p>\n\n\n\n

(iv) 28 \u2013 (-10) < (-16) \u2013 (-76)<\/p>\n\n\n\n

12. If \u25b3 is an operation on integers such that a \u25b3 b = \u2013 a + b \u2013 (-2) for all integers a, b. Find the value of<\/strong><\/p>\n\n\n\n

(i) 4 \u25b3 3<\/strong><\/p>\n\n\n\n

(ii) (-2) \u25b3 (-3)<\/strong><\/p>\n\n\n\n

(iii) 6 \u25b3 (-5)<\/strong><\/p>\n\n\n\n

(iv) (-5) \u25b3 6<\/strong><\/p>\n\n\n\n

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

(i) 4 \u25b3 3<\/p>\n\n\n\n

By substituting values in a \u25b3 b = \u2013 a + b \u2013 (-2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

4 \u25b3 3 = \u2013 4 + 3 \u2013 (-2) = 1<\/p>\n\n\n\n

(ii) (-2) \u25b3 (-3)<\/p>\n\n\n\n

By substituting values in a \u25b3 b = \u2013 a + b \u2013 (-2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

(-2) \u25b3 (-3) = \u2013 (-2) + (-3) \u2013 (-2) = 1<\/p>\n\n\n\n

(iii) 6 \u25b3 (-5)<\/p>\n\n\n\n

By substituting values in a \u25b3 b = \u2013 a + b \u2013 (-2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

6 \u25b3 (-5) = \u2013 6 + (-5) \u2013 (-2) = \u2013 9<\/p>\n\n\n\n

(iv) (-5) \u25b3 6<\/p>\n\n\n\n

By substituting values in a \u25b3 b = \u2013 a + b \u2013 (-2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

(-5) \u25b3 6 = \u2013 (-5) + 6 \u2013 (-2) = 13<\/p>\n\n\n\n

13. If a and b are two integers such that a is the predecessor of b. Find the value of a \u2013 b.<\/strong><\/p>\n\n\n\n

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

It is given that a is the predecessor of b<\/p>\n\n\n\n

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

a + 1 = b<\/p>\n\n\n\n

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

a \u2013 b = \u2013 1<\/p>\n\n\n\n

14. If a and b are two integers such that a is the successor of b. Find the value of a \u2013 b.<\/strong><\/p>\n\n\n\n

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

It is given that a is the successor of b<\/p>\n\n\n\n

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

a \u2013 1 = b<\/p>\n\n\n\n

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

a \u2013 b = 1<\/p>\n\n\n\n

15. Which of the following statements are true:<\/strong><\/p>\n\n\n\n

(i) \u2013 13 > \u2013 8 \u2013 (-2)<\/strong><\/p>\n\n\n\n

(ii) \u2013 4 + (-2) < 2<\/strong><\/p>\n\n\n\n

(iii) The negative of a negative integer is positive.<\/strong><\/p>\n\n\n\n

(iv) If a and b are two integers such that a > b, then a \u2013 b is always a positive integer.<\/strong><\/p>\n\n\n\n

(v) The difference of two integers is an integer.<\/strong><\/p>\n\n\n\n

(vi) Additive inverse of a negative integer is negative.<\/strong><\/p>\n\n\n\n

(vii) Additive inverse of a positive integer is negative.<\/strong><\/p>\n\n\n\n

(viii) Additive inverse of a negative integer is positive.<\/strong><\/p>\n\n\n\n

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

(i) False.<\/p>\n\n\n\n

(ii) True.<\/p>\n\n\n\n

(iii) True.<\/p>\n\n\n\n

(iv) True.<\/p>\n\n\n\n

(v) True.<\/p>\n\n\n\n

(vi) False.<\/p>\n\n\n\n

(vii) True.<\/p>\n\n\n\n

(viii) True.<\/p>\n\n\n\n

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

(i) \u2013 7 + \u2026.. = 0<\/strong><\/p>\n\n\n\n

(ii) 29 + \u2026.. = 0<\/strong><\/p>\n\n\n\n

(iii) 132 + (-132) = \u2026.<\/strong><\/p>\n\n\n\n

(iv) \u2013 14 + \u2026.. = 22<\/strong><\/p>\n\n\n\n

(v) \u2013 1256 + \u2026.. = \u2013 742<\/strong><\/p>\n\n\n\n

(vi) \u2026.. \u2013 1234 = \u2013 4539<\/strong><\/p>\n\n\n\n

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

(i) \u2013 7 + 7 = 0<\/p>\n\n\n\n

(ii) 29 + (-29) = 0<\/p>\n\n\n\n

(iii) 132 + (-132) = 0<\/p>\n\n\n\n

(iv) \u2013 14 + 36 = 22<\/p>\n\n\n\n

(v) \u2013 1256 + 514 = \u2013 742<\/p>\n\n\n\n

(vi) -3305 \u2013 1234 = \u2013 4539<\/p>\n\n\n\n

\n\n\n\n

Objective Type Questions page: 5.18<\/h4>\n\n\n\n

Mark the correct alternative in each of the following:<\/strong><\/p>\n\n\n\n

1. Which of the following statement is true?
(a) \u2212 7 > \u2212 5              (b) \u2212 7 < \u2212 5                 (c) (\u2212 7) + (\u2212 5) > 0                       (d) (\u2212 7) \u2212 (\u2212 5) > 0<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

In option (a)<\/p>\n\n\n\n

We know that \u2212 7 is to the left of \u2013 5<\/p>\n\n\n\n

Hence, \u2212 7 < \u2212 5.<\/p>\n\n\n\n

In option (c)<\/p>\n\n\n\n

We know that (\u2212 7) + (\u2212 5) = \u2212 (7 + 5) = \u2212 12.<\/p>\n\n\n\n

So \u2212 12 is to the left of 0<\/p>\n\n\n\n

Hence (\u2212 7) + (\u2212 5) < 0.<\/p>\n\n\n\n

In option (d)<\/p>\n\n\n\n

(\u2212 7) \u2212 (\u2212 5) = (\u2212 7) + (additive inverse of \u2212 5) = (\u2212 7) + (5) = \u2212 (7 \u2212 5) = \u2212 2<\/p>\n\n\n\n

We know that \u2212 2 is to the left of 0, so (\u2212 7) \u2212 (\u2212 5) < 0.<\/p>\n\n\n\n

2. 5 less than \u2212 2 is
(a) 3                            (b) \u2212 3                        (c) \u2212 7                                 (d) 7<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

We know that, 5 less than \u2212 2 = (\u2212 2) \u2212 (5) = \u2212 2 \u2212 5 = \u2212 7<\/p>\n\n\n\n

3. 6 more than \u2212 7 is
(a) 1                            (b) \u2212 1                        (c) 13                                 (d) \u2013 13<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that, 6 more than \u2212 7 = (\u2212 7) + 6 = \u2212 (7 \u2212 6) = \u2212 1<\/p>\n\n\n\n

4. If x is a positive integer, then
(a) x + |<\/em>x|<\/em> = 0              (b) x \u2212 |<\/em>x|<\/em> = 0                    (c) x + |<\/em>x|<\/em> = \u22122x <\/em>               (d) x = \u2212 <\/em>|x|<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that if x is positive integer, then |x| = x<\/p>\n\n\n\n

Hence, x + |x| = x + <\/em>x = 2x and x \u2212 |x| = x \u2212 x <\/em>= 0<\/p>\n\n\n\n

5. <\/em>If x is a negative integer, then
(a) x + |<\/em>x|<\/em> = 0              (b) x \u2212 |<\/em>x|<\/em> = 0                    (c) x + |<\/em>x| = <\/em>2x                (d) x \u2212 |<\/em>x|<\/em> = \u2212 2x<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

We know that x is negative integer, then |x| = \u2212x<\/p>\n\n\n\n

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

x + |x| = x \u2212 <\/em>x = 0 and x <\/em>\u2212 |x| = x \u2212 (\u2212 x) = x + x = 2x<\/p>\n\n\n\n

6. If x is greater than 2, then |2 \u2212 x| =
(a) 2 \u2212 x              (b) x \u2212 2                    (c) 2 + x                (d) \u2212 x \u2013 2<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that if a is negative integer, then |a| = \u2212 a<\/p>\n\n\n\n

It is given that x is greater than 2 where 2 \u2212 x is negative<\/p>\n\n\n\n

Hence, |2 \u2212 x| = \u2212 (2 \u2212 x) = \u2212 2 + x = x <\/em>\u2212 2.<\/p>\n\n\n\n

7. 9 + |\u2212 4| is equal to
(a) 5                         (b) \u2212 5                       (c) 13                            (d) \u221213<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

We know that, |\u2212 4| = 4<\/p>\n\n\n\n

Hence 9 + |\u2212 4| = 9 + 4 = 13<\/p>\n\n\n\n

8. (\u2212 35) + (\u2212 32) is equal to
(a) 67                         (b) \u2212 67                       (c) \u2212 3                            (d) 3<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It can be written as (\u2212 35) + (\u2212 32) = \u2212 (35 + 32) = \u2212 67<\/p>\n\n\n\n

9. (\u2212 29) + 5 is equal to
(a) 24                         (b) 34                       (c) \u2212 34                            (d) \u2013 24<\/strong><\/p>\n\n\n\n

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

The option (d) is correct answer.<\/p>\n\n\n\n

It can be written as (\u2212 29) + 5 = \u2212 (29 \u2212 5) = \u2212 24<\/p>\n\n\n\n

10. |\u2212 |\u2212 7| \u2212 3| is equal to
(a) \u2212 7                       (b) 7                              (c) 10                                    (d) \u2013 10<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

It can be written as |\u2212 |\u2212 7| \u2212 3| = |\u2212 7 \u2212 3| = |\u2212 10| = 10<\/p>\n\n\n\n

11. The successor of \u2212 22 is
(a) \u2212 23                       (b) \u2212 21                              (c) 23                                    (d) 21<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that if \u2018a\u2019 is an integer a + 1 is its successor.<\/p>\n\n\n\n

So the successor of \u2212 22 = \u2212 22 + 1 = \u2212 (22 \u2212 1) = \u2212 21<\/p>\n\n\n\n

12. The predecessor of \u2013 14 is<\/strong><\/p>\n\n\n\n

(a) \u2013 15 (b) 15 (c) 13 (d) \u2013 13<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

The predecessor of \u2013 14 is \u2013 15.<\/p>\n\n\n\n

13. If the sum of two integers is \u2212 26 and one of them is 14, then the other integer is
(a) \u2212 12                       (b) 12                              (c) \u2212 40                               (d) 40<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

It is given that the sum of two integers = \u2212 26<\/p>\n\n\n\n

One of them = 14<\/p>\n\n\n\n

So the other integer = \u2212 26 \u2212 14 = \u2212 (26 + 14) = \u2212 40<\/p>\n\n\n\n

14. Which of the following pairs of integers have 5 as a difference?
(a) 10, 5                       (b) \u2212 10, \u2212 5                          (c) 15, \u2212 20                               (d) both (a) and (b)<\/strong><\/p>\n\n\n\n

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

The option (d) is correct answer.<\/p>\n\n\n\n

Consider option (a) 10 \u2212 5 = 5<\/p>\n\n\n\n

Consider option (b) (\u2212 5) \u2212 (\u2212 10) = \u2212 5 + 10 = 5<\/p>\n\n\n\n

Consider option (c) 15 \u2212 (\u2212 20) = 15 + 20 = 35<\/p>\n\n\n\n

15. If the product of two integers is 72 and one of them is \u2212 9, then the other integers is
(a) \u2212 8                       (b) 8                          (c) 81                               (d) 63<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

It is given that the product of two integers = 72<\/p>\n\n\n\n

One of them = \u2212 9<\/p>\n\n\n\n

Hence, the other integers = 72 \u00f7 (\u2212 9) = \u2212 8<\/p>\n\n\n\n

16. On subtracting \u2212 7 from \u2212 14, we get
(a) \u2212 12                       (b) \u2212 7                          (c) \u221214                               (d) 21<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

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

Required number = \u2212 14 \u2212 (\u2212 7) = \u2212 14 + 7 = \u2212 (14 \u2212 7) = \u2212 7<\/p>\n\n\n\n

17. The largest number that divides 64 and 72 and leave the remainders 12 and 7 respectively, is
(a) 17                             (b) 13                                       (c) 14                                      (d) 18<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

By subtracting 12 and 7 from 64 and 72<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

64 \u2212 12 = 52 and 72 \u2212 7 = 65<\/p>\n\n\n\n

So the required number is the HCF of 52 and 65.<\/p>\n\n\n\n

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

52 = 4 \u00d7 13 and 65 = 5 \u00d7 13<\/p>\n\n\n\n

HCF of 52 and 65 = 13<\/p>\n\n\n\n

Hence, the largest number that divides 64 and 72 and leave the remainders 12 and 7 respectively, is 13.<\/p>\n\n\n\n

18. The sum of two integers is \u2212 23. If one of them is 18, then the other is
(a) \u221214                             (b) 14                                    (c) 41                                    (d) \u221241<\/strong><\/p>\n\n\n\n

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

The option (d) is correct answer.<\/p>\n\n\n\n

It is given as the sum of two integers = \u2212 23<\/p>\n\n\n\n

One of them = 18<\/p>\n\n\n\n

So the other number = (\u2212 23) \u2212 (18) = \u2212 23 \u2212 18 = \u2212 (23 + 18) = \u2212 41<\/p>\n\n\n\n

Hence, the other number is \u2212 41.<\/p>\n\n\n\n

19. The sum of two integers is \u2212 35. If one of them is 40, then the other is
(a) 5                             (b) \u2212 75                                    (c) 75                                    (d) \u2013 5<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It is given that the sum of two integers = \u2212 35<\/p>\n\n\n\n

One of them = 40<\/p>\n\n\n\n

So the other number = (\u2212 35) \u2212 (40) = \u2212 35 \u2212 40 = \u2212 (35 + 40) = \u2212 75<\/p>\n\n\n\n

Hence, the other number is \u2212 75.<\/p>\n\n\n\n

20. On subtracting \u2212 5 from 0, we get
(a) \u2212 5                             (b) 5                                    (c) 50                                    (d) 0<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that, 0 \u2212 (\u2212 5) = 0 + 5 = 5<\/p>\n\n\n\n

Hence by subtracting \u2212 5 from 0, we obtain 5.<\/p>\n\n\n\n

21. (\u2212 16) + 14 \u2212 (\u2212 13) is equal to
(a) \u2212 11                             (b) 12                                    (c) 11                                    (d) \u2013 15<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

It can be written as (\u2212 16) + 14 \u2212 (\u2212 13) = (\u2212 16) + 14 + 13 = (\u2212 16) + 27 = 27 \u2212 16 = 11<\/p>\n\n\n\n

22. (\u2212 2) \u00d7 (\u2212 3) \u00d7 6 \u00d7 (\u2212 1) is equal to
(a) 36                             (b) \u2212 36                                (c) 6                                    (d) \u2013 6<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It can be written as (\u2212 2) \u00d7 (\u2212 3) \u00d7 6 \u00d7 (\u2212 1) = (2 \u00d7 3) \u00d7 6 \u00d7 (\u2212 1) = 6 \u00d7 6 \u00d7 (\u2212 1) = 36 \u00d7 (\u2212 1)<\/p>\n\n\n\n

So we get (\u2212 2) \u00d7 (\u2212 3) \u00d7 6 \u00d7 (\u2212 1) = \u2212 (36 \u00d7 1) = \u2212 36<\/p>\n\n\n\n

23. 86 + (- 28) + 12 + (- 34) is equal to
(a) 36                             (b) \u2212 36                                (c) 6                                    (d) \u2013 6<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

It can be written as 86 + (\u221228) + 12 + (\u221234) = 86 + (\u221228) \u2212 (34 \u2212 12) = 86 + (\u221228) \u2212 22<\/p>\n\n\n\n

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

86 + (\u221228) + 12 + (\u221234) = (86 \u2212 28) \u2212 (34 \u2212 12) = (86 \u2212 28) \u2212 22 = 58 \u2212 22 = 36<\/p>\n\n\n\n

24. (\u221212) \u00d7 (\u22129) \u2212 6 \u00d7 (\u22128) is equal to
(a) 156                             (b) 60                                (c) \u2212156                                    (d) \u2013 60<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

It can be written as (\u221212) \u00d7 (\u22129) \u2212 6 \u00d7 (\u22128) = (12 \u00d7 9) \u2212 6 \u00d7 (\u22128) = 108 \u2212 6 \u00d7 (\u22128)<\/p>\n\n\n\n

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

(\u221212) \u00d7 (\u22129) \u2212 6 \u00d7 (\u22128) = 108 + 6 \u00d7 8 = 108 + 48 = 156<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 5: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 5\u2013Negative Numbers and Integers<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 6 Maths Chapter 5\u2013Negative Numbers and Integers PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 6 Maths :<\/strong><\/h2>\n\n\n\n