{"id":139651,"date":"2021-02-20T10:18:05","date_gmt":"2021-02-20T10:18:05","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=139651"},"modified":"2023-09-15T05:27:15","modified_gmt":"2023-09-15T05:27:15","slug":"ncert-solutions-for-7th-class-maths-chapter-2-fractions-and-decimals","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-2-fractions-and-decimals\/","title":{"rendered":"NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals"},"content":{"rendered":"\n

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

NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals<\/strong><\/h2>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No: 31<\/p>\n\n\n\n

Exercise 2.1<\/h4>\n\n\n\n

1. Solve:
(i) 2 – 3\/5
(ii) 4 + 7\/8
(iii) 3\/5 + 2\/7
(iv) 9\/11 \u2212 4\/15
(v) 7\/10 + 2\/5 + 3\/2
(vi) 2\u2154 + 3\u00bd
(vii) 8\u00bd \u2212 3\u215d<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2\u22123\/5 = (10\u22123)\/5 = 7\/5
(ii) 4+7\/8 = (32+7)\/8 = 39\/8
(iii) 3\/5 + 2\/7 = (21+10)\/35 = 31\/35
(iv) 9\/11 \u2212 4\/15 = (135\u221244)\/165 = 91\/165
(v) 7\/10 + 2\/5 + 3\/2 = (7+4+15)\/10 = 26\/10 = 13\/5
(vi) 2\u2154 + 3\u00bd = 8\/3 + 7\/2 = (16+21)\/6 = 37\/6 = 6.1\/6
(vii) 8\u00bd \u2212 3\u215d = 17\/2\u221229\/8 = (68\u221229)\/8 = 39\/8 = 4.7\/8<\/p>\n\n\n\n

2. Arrange the following in descending order:
(i) 2\/9,2\/3,8\/21
(ii) 1\/5,3\/7,7\/10<\/strong><\/p>\n\n\n\n


Answer<\/strong>
(i) 2\/9,2\/3,8\/21<\/p>\n\n\n\n


\u21d2 14\/63,42\/63,24\/63 [Converting into like fractions]
\u21d2 42\/63>24\/63>14\/63 [Arranging in descending order]
Therefore, 2\/3>8\/21>2\/9<\/p>\n\n\n\n

(ii) 1\/5,3\/7,7\/10 \u21d214\/70,30\/70,49\/70 [Converting into like fractions]
\u21d2 49\/70>30\/70>14\/70 [Arranging in descending order]
Therefore, 7\/10>3\/7>1\/5<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

3. In a \u201cmagic square\u201d, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?<\/strong><\/p>\n\n\n\n

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

(Along the first row 4\/11+9\/11+2\/11=15\/11)(Along the first row 4\/11+9\/11+2\/11=15\/11)<\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n


Sum of first row = 4\/11+9\/11+2\/11=15\/11 [Given]
Sum of second row = 3\/11+5\/11+7\/11=(3+5+7)\/11=15\/11
Sum of third row = 8\/11+1\/11+6\/11=(8+1+6)\/11=15\/11
Sum of first column = 4\/11+3\/11+8\/11=(4+3+8)\/11=15\/11
Sum of second column = 9\/11+5\/11+1\/11=(9+5+1)\/11=15\/11
Sum of third column = 2\/11+7\/11+6\/11=(2+7+6)\/11=15\/11
Sum of first diagonal (left to right) = 4\/11+5\/11+6\/11=(4+5+6)\/11=15\/11
Sum of second diagonal (left to right) = 2\/11+5\/11+8\/11=(2+5+8)\/11=15\/11
Since, the sum of fractions in each row, in each column and along the diagonals are same, therefore, it s a magic square.<\/p>\n\n\n\n


4. A rectangular sheet of paper is 12\u00bd cm long and 10\u2154 cm wide. Find its perimeter.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The sheet of paper is in rectangular form.
Length of sheet = 12\u00bd cm and Breadth of sheet = 10\u2154 cm
Perimeter of rectangle = 2 (length + breadth)
= 2(12\u00bd +10\u2154) = 2(25\/2+32\/3)
= 2{(25\u00d73+32\u00d72)\/6} = 2{(75+64)\/6}
= 2\u00d7139\/6 = 139\/3 = 46.(1\/3) cm.
Thus, the perimeter of the rectangular sheet is 46.1\/3 cm.<\/p>\n\n\n\n

5. Find the perimeter of (i) \u0394ABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) In \u0394ABE, AB = 5\/2 cm, BE = 2.(3\/4) cm, AE = 3.(3\/5) cm
The perimeter of \u0394ABE = AB + BE + AE
= 5\/2 + 2.(3\/4) + 3.(3\/5) = 5\/2 + 11\/4 + 18\/5
= (50+55+72)\/20 = 177\/20 = 8.(17\/20) cm
Thus, the perimeter of \u0394ABE is 8.(17\/20) cm.<\/p>\n\n\n\n

(ii) In rectangle BCDE, BE = 2.3\/4 cm, ED = 7\/6 cm
Perimeter of rectangle = 2 (length + breadth)
= 2(2.3\/4+7\/6) = 2(11\/4+7\/6)
= 2{(33+14)\/12} = 47\/6 = 7.(5\/6) cm
Thus, the perimeter of rectangle BCDE is 7.(5\/6) cm.
Comparing the perimeter of triangle and that of rectangle,
8.17\/20 cm > 7.5\/6 cm
Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

6. Salil wants to put a picture in a frame. The picture is 7.(3\/5) cm wide. To fit in the frame the picture cannot be more than 7.3\/10 cm wide. How much should the picture be trimmed?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: The width of the picture = 7.(3\/5) cm and the width of picture frame = 7.(3\/10) cm
Therefore, the picture should be trimmed = 7.(3\/5) \u2212 7.(3\/10) = 38\/5 = 73\/10
= (76\u221273)\/10 = 3\/10 cm
Thus, the picture should be trimmed by 3\/10 cm.<\/p>\n\n\n\n

7. Ritu ate 3\/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The part of an apple eaten by Ritu = 3\/5
The part of an apple eaten by Somu = 1\u22123\/5 = (5\u22123)\/5 = 2\/5
Comparing the parts of apple eaten by both Ritu and Somu 3\/5 > 2\/5
Larger share will be more by 3\/5\u22122\/5 = 1\/5 part.
Thus, Ritu\u2019s part is 1\/5 more than Somu\u2019s part.<\/p>\n\n\n\n

8. Michael finished colouring a picture in 7\/12 hour. Vaibhav finished colouring the same picture in 3\/4 hour. Who worked longer? By what fraction was it longer?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Time taken by Michael to colour the picture = 7\/12 hour
Time taken by Vaibhav to colour the picture = 3\/4 hour
Converting both fractions in like fractions, 7\/12 and (3\u00d73)\/(4\u00d73)=9\/12
Here, 7\/12 < 9\/12 \u21d2 7\/12<3\/4
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by 3\/4\u22127\/12=(9\u22127)\/12=2\/12=1\/6 hour.
Thus, Vaibhav took 1\/6 hour more than Michael.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 36<\/p>\n\n\n\n

Exercise 2.2
<\/h4>\n\n\n\n

1. Which of the drawings (a) to (d) show:<\/strong><\/p>\n\n\n\n

(i) 2\u00d71\/5<\/strong><\/p>\n\n\n\n

(ii) 2\u00d71\/2<\/strong><\/p>\n\n\n\n

(iii) 3\u00d72\/3<\/strong><\/p>\n\n\n\n

(iv) 3\u00d71\/4<\/strong><\/p>\n\n\n\n

(a) <\/p>\n\n\n\n

(b) <\/p>\n\n\n\n

(c) <\/p>\n\n\n\n

(d) <\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) \u2013 (d) Since 2\u00d71\/5=1\/5+1\/5<\/p>\n\n\n\n

(ii) \u2013 (b) Since 2\u00d71\/2=1\/2+1\/2<\/p>\n\n\n\n

(iii) \u2013 (a) Since 3\u00d72\/3=2\/3+2\/3+2\/3<\/p>\n\n\n\n

(iv) \u2013 (c) Since 3\u00d71\/4=1\/4+1\/4+1\/4<\/p>\n\n\n\n

2. Some pictures (a) to (c) are given below. Tell which of them show:<\/strong><\/p>\n\n\n\n

(i) 3\u00d71\/5 = 3\/5<\/strong><\/p>\n\n\n\n

(ii) 2\u00d71\/3 = 2\/3 <\/strong><\/p>\n\n\n\n

(iii) 3\u00d73\/4 = 2.(1\/4)<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) \u2013 (c) Since 3\u00d71\/5=1\/5+1\/5+1\/5<\/p>\n\n\n\n

(ii) \u2013 (a) Since 2\u00d71\/3=1\/3+1\/3<\/p>\n\n\n\n

(iii) \u2013 (b) Since 3\u00d73\/4=3\/4+3\/4+3\/4<\/p>\n\n\n\n

3. Multiply and reduce to lowest form and convert into a mixed fraction:<\/strong><\/p>\n\n\n\n

(i) 7 \u00d7 3\/5<\/strong><\/p>\n\n\n\n

(ii) 4 \u00d7 1\/3<\/strong><\/p>\n\n\n\n

(iii) 2 \u00d7 6\/7<\/strong><\/p>\n\n\n\n

(iv) 5 \u00d7 2\/9<\/strong><\/p>\n\n\n\n

(v) 2\/3 \u00d7 4<\/strong><\/p>\n\n\n\n

(vi) 5\/2 \u00d7 6<\/strong><\/p>\n\n\n\n

(vii) 11 \u00d7 4\/7<\/strong><\/p>\n\n\n\n

(viii) 20 \u00d7 4\/5<\/strong><\/p>\n\n\n\n

(ix) 13 \u00d7 1\/3<\/strong><\/p>\n\n\n\n

(x) 15 \u00d7 3\/5<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 7\u00d7 3\/5
= (7\u00d73)\/5
= 21\/5
=4.(1\/5)<\/p>\n\n\n\n

(ii) 4\u00d7 1\/3
= (4\u00d71)\/3
= 4\/3
=1.(1\/3)<\/p>\n\n\n\n

(iii) 2\u00d7 6\/7
= (2\u00d76)\/7
= 12\/7
=1.(5\/7)<\/p>\n\n\n\n

(iv) 5\u00d7 2\/9
= (5\u00d72)\/9
= 10\/9
=1.(1\/9)<\/p>\n\n\n\n

(v) 2\/3 \u00d7 4
= (2\u00d74)\/3
= 8\/3
= 2.(2\/3)<\/p>\n\n\n\n

(vii) 11\u00d7 4\/7
= (11\u00d74)\/7
= 44\/7
= 6.(2\/7)<\/p>\n\n\n\n

(viii) 20\u00d7 4\/5
= 4\u00d74
= 16<\/p>\n\n\n\n

(ix) 13\u00d7 1\/3
= (13\u00d71)\/3
= 13\/3
=4.(1\/3)<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

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

(i) 2\/3 of the triangles in box (a) <\/strong><\/p>\n\n\n\n

(ii) 3\/5 of the squares in box (b)<\/strong><\/p>\n\n\n\n

(i) 1\/2 of the circles in box (c)<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) 2\/3 of 9 triangles = 2\/3 \u00d79 = 2\u00d73 = 6 triangles<\/p>\n\n\n\n

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

(ii) 3\/5 of 15 squares = 3\/5 \u00d715 = 3 \u00d7 3 = 9 squares<\/p>\n\n\n\n

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

(iii) 1\/2 of 12 circles = 1\/2 \u00d712 = 6 circles<\/p>\n\n\n\n


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

(a) 1\/2 of (i) 24 (ii) 46<\/p>\n\n\n\n

(b) 2\/3 of (i) 18 (ii) 27<\/p>\n\n\n\n

(c) 3\/4 of (i) 16 (ii) 36<\/p>\n\n\n\n

(d) 4\/5 of (i) 20 (ii) 35<\/p>\n\n\n\n

(i) 1\/2 of 24 = 12<\/p>\n\n\n\n

(ii) 1\/2 of 46 = 23<\/p>\n\n\n\n

(i) 2\/3 of 18 = 2\/3\u00d718 = 2 \u00d7 6 = 12<\/p>\n\n\n\n

(ii) 2\/3 of 27 = 2\/3\u00d727 = 2 \u00d7 9 = 18<\/p>\n\n\n\n

(i) 3\/4 of 16 = 3\/4 \u00d7 16 = 3 \u00d7 4 = 12<\/p>\n\n\n\n

(ii) 3\/4 of 36 = 3\/4 \u00d7 36 = 3 \u00d7 9 = 27<\/p>\n\n\n\n

(i) 4\/5 of 20 = 4\/5 \u00d7 20 = 4 \u00d7 4 = 16<\/p>\n\n\n\n

(ii) 4\/5 of 35 = 4\/5 \u00d7 35 = 4 \u00d7 7 = 28<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n


6. Multiply and express as a mixed fraction:<\/strong><\/p>\n\n\n\n

(a) 3\u00d7 5.(1\/5)<\/p>\n\n\n\n

(b) 5\u00d7 6.(3\/4)<\/p>\n\n\n\n

(c) 7\u00d7 2.(1\/4)<\/p>\n\n\n\n

(d) 4\u00d7 6.(1\/3)<\/p>\n\n\n\n

(e) 3.(1\/4) \u00d76<\/p>\n\n\n\n

(f) 3.(2\/5) \u00d78<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 3\u00d7 5.1\/5
= 3\u00d7 26\/5
= (3\u00d726)\/5
= 78\/5
= 15.(3\/5)<\/p>\n\n\n\n

(b) 5\u00d7 6.(3\/4)
= 5\u00d7 27\/4
= (5\u00d727)\/4
= 135\/4
= 33.(3\/4)<\/p>\n\n\n\n

(c) 7\u00d7 2.(1\/4)
= 7\u00d7 9\/4
= (7\u00d79)\/4
= 63\/4
= 15.(3\/4)<\/p>\n\n\n\n

(d) 4\u00d76.1\/3
= 4\u00d7 19\/3
= (4\u00d719)\/3
= 76\/3
= 25.1\/3<\/p>\n\n\n\n

(e) 3.(1\/4) \u00d76
= 13\/4 \u00d76
= (13\u00d73)\/2
= 39\/2
= 19.(1\/2)<\/p>\n\n\n\n

(f) 3.2\/5\u00d78
= 17\/5 \u00d78
= (17\u00d78)\/5
= 136\/5
= 27.(1\/5)<\/p>\n\n\n\n

7. Find:<\/strong><\/p>\n\n\n\n

(a) 1\/2 of (i) 2.(3\/4) (ii) 4.(2\/9)<\/p>\n\n\n\n

(b) 5\/8 of (i) 3.(5\/6) (ii) 9.(2\/3)<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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

(i) 1\/2 of 2.(3\/4) = 1\/2\u00d711\/4 = 11\/8 = 1.(3\/8)<\/p>\n\n\n\n

(ii) 1\/2 of 4.(2\/9) = 1\/2 \u00d7 38\/9 = 19\/9 = 2.(1\/9)<\/p>\n\n\n\n

(i) 5\/8 of 3.(5\/6) = 5\/8 \u00d7 23\/6 = 115\/48 = 2.(19\/48) <\/p>\n\n\n\n

(ii) 5\/8 of 9.(2\/3) = 5\/8 \u00d7 29\/3 = 145\/24 = 6.(1\/24)<\/p>\n\n\n\n

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2\/5 of the water. Pratap consumed the remaining water.<\/strong><\/p>\n\n\n\n

(i) How much water did Vidya drink?<\/p>\n\n\n\n

(ii) What fraction of the total quantity of water did Pratap drink?<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Total quantity of water in bottle = 5 litres<\/p>\n\n\n\n

(i) Vidya consumed = 2\/5 of 5 litres = 2\/5\u00d75 = 2 litres<\/p>\n\n\n\n

Thus, Vidya drank 2 litres water from the bottle.<\/p>\n\n\n\n

(ii) Pratap consumed = (1\u22122\/5) part of bottle = 5\u22122\/5=3\/5 part of bottle<\/p>\n\n\n\n

Pratap consumed 3\/5 of 5 litres water = 3\/5\u00d75 = 3 litres<\/p>\n\n\n\n

Thus, Pratap drank 3\/5 part of the total quantity of water.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 41<\/p>\n\n\n\n

Exercise 2.3<\/h4>\n\n\n\n

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

(i) 1\/4 of<\/p>\n\n\n\n

(a) 1\/4<\/p>\n\n\n\n

(b) 3\/4<\/p>\n\n\n\n

(c) 4\/3<\/p>\n\n\n\n

(ii)1\/7 of<\/p>\n\n\n\n

(a) 2\/9<\/p>\n\n\n\n

(b) 6\/5<\/p>\n\n\n\n

(c) 3\/10<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) (a) 1\/4 of 1\/4
= 1\/4\u00d71\/4
=(1\u00d71)\/(4\u00d74)
=1\/16<\/p>\n\n\n\n

(b) 1\/4 of 3\/4
= 1\/4\u00d73\/4
=(1\u00d73)\/(4\u00d74)
= 3\/16<\/p>\n\n\n\n

(c) 1\/4 of 4\/3
= 1\/4\u00d74\/3
=(1\u00d74)\/(4\u00d73)
=1\/3<\/p>\n\n\n\n

(ii) (a) 1\/7 of 2\/9
= 1\/7\u00d72\/9
=(1\u00d72)\/(7\u00d79)
=2\/63<\/p>\n\n\n\n

(b) 1\/7 of 2\/9
= 1\/7\u00d76\/5
=(1\u00d76)\/(7\u00d75)
=6\/35<\/p>\n\n\n\n

(c) 1\/7 of 3\/10
= 1\/7\u00d73\/10
=(1\u00d73)\/(7\u00d710)
=3\/70<\/p>\n\n\n\n


2. Multiply and reduce to lowest form (if possible):<\/strong><\/p>\n\n\n\n

(i) 2\/3 \u00d7 2.(2\/3)<\/p>\n\n\n\n

(ii) 2\/7 \u00d7 7\/9<\/p>\n\n\n\n

(iii) 3\/8 \u00d7 6\/4<\/p>\n\n\n\n

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

(v) 1\/3 \u00d7 15\/8<\/p>\n\n\n\n

(vi) 11\/2 \u00d7 3\/10<\/p>\n\n\n\n

(vii) 4\/5 \u00d7 12\/7<\/p>\n\n\n\n

Answer<\/strong>
<\/p>\n\n\n\n

(i) 2\/3 \u00d7 2.2\/3
= 2\/3 \u00d7 8\/3
= (2\u00d78)\/(3\u00d73)
= 16\/9
= 1.(7\/9)<\/p>\n\n\n\n

(ii) 2\/7 \u00d7 7\/9
= (2\u00d77)\/(7\u00d79)
= 2\/9<\/p>\n\n\n\n

(iii) 3\/8 \u00d7 6\/4
= (3\u00d76)\/(8\u00d74)
= (3\u00d73)\/(8\u00d72)
= 9\/16<\/p>\n\n\n\n

(iv) 9\/5 \u00d7 3\/5
= (9\u00d73)\/(5\u00d75)
= 27\/25
= 1.(2\/25)<\/p>\n\n\n\n

(v) 1\/3 \u00d7 15\/8
= (1\u00d715)\/(3\u00d78)
= (1\u00d75)\/(1\u00d78)
= 5\/8<\/p>\n\n\n\n

(vi) 11\/2 \u00d7 3\/10
=(11\u00d73)\/(2\u00d710)
=33\/20
=1.(3\/20)<\/p>\n\n\n\n

(vii) 4\/5 \u00d7 12\/7
=(4\u00d712)\/(5\u00d77)
=48\/35
=1.(13\/35)<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

3. Multiply the following fractions:<\/strong><\/p>\n\n\n\n

(i) 2\/5 \u00d7 5.(1\/4)<\/p>\n\n\n\n

(ii) 6.(2\/5) \u00d7 7\/9<\/p>\n\n\n\n

(iii) 3\/2 \u00d7 5.(1\/3)<\/p>\n\n\n\n

(iv) 5\/6 \u00d7 2.(3\/7)<\/p>\n\n\n\n

(v) 3.(2\/5) \u00d7 4\/7<\/p>\n\n\n\n

(vi) 2.(3\/5) \u00d7 3<\/p>\n\n\n\n

(vii) 3.(4\/7) \u00d7 3\/5<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2\/5 \u00d7 5.(1\/4)
= 2\/5 \u00d7 21\/4
= (2\u00d721)\/(5\u00d74)
= (1\u00d721)\/(5\u00d72)
= 21\/10
= 2.(1\/10)<\/p>\n\n\n\n

(ii) 6.(2\/5) \u00d7 7\/9
= 32\/5 \u00d7 7\/9
= (32\u00d77)\/(5\u00d79)
= 224\/45
= 4.(44\/45)<\/p>\n\n\n\n

(iii) 3\/2\u00d75.(1\/3)
= 3\/2 \u00d7 16\/3
= 48\/6
= 8<\/p>\n\n\n\n

(iv) 5\/6\u00d72.(3\/7)
= 5\/6 \u00d7 17\/7
= 85\/42
= 2.(1\/42)<\/p>\n\n\n\n

(v) 3.(2\/5) \u00d7 4\/7
= 17\/7 \u00d7 4\/7
= 68\/35
= 1.(33\/35)<\/p>\n\n\n\n

(vi) 2.(3\/5) \u00d7 3
= 13\/5 \u00d7 3\/1
= (13\u00d73)\/(5\u00d71)
= 39\/5
= 7.(4\/5)<\/p>\n\n\n\n

(vii) 3.(4\/7) \u00d7 3\/5
= 25\/7 \u00d7 3\/5
= (5\u00d73)\/(7\u00d71)
= 15\/7
= 2.(1\/7)<\/p>\n\n\n\n

4. Which is greater:<\/strong><\/p>\n\n\n\n

(i) 2\/7 of 3\/4 or 3\/5 of 5\/8<\/p>\n\n\n\n

(ii) 1\/2 of 6\/7 or 2\/3 of 3\/7<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2\/7 of 3\/4 or 3\/5 of 5\/8
\u21d2 2\/7 \u00d7 3\/4 or 3\/5 \u00d7 5\/8<\/p>\n\n\n\n

\u21d2 3\/14 or 3\/8
\u21d2 3\/14<3\/8<\/p>\n\n\n\n

Thus, 3\/5 of 5\/8 is greater.<\/p>\n\n\n\n

(ii) 1\/2 of 6\/7 or 2\/3 of 3\/7
\u21d2 1\/2 \u00d7 6\/7 or 2\/3 \u00d7 3\/7<\/p>\n\n\n\n

\u21d2 3\/7 or 2\/7
\u21d2 3\/7 > 2\/7<\/p>\n\n\n\n

Thus, 1\/2 of 6\/7 is greater.<\/p>\n\n\n\n

5. Saili plants 4 saplings in a row in her garden. The distance between two adjacent saplings is 3\/4 m. Find the distance between the first and the last sapling.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
<\/p>\n\n\n\n

The distance between two adjacent saplings = 3\/4 m<\/p>\n\n\n\n

Saili planted 4 saplings in a row, then number of gap in saplings<\/p>\n\n\n\n

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

Therefore, the distance between the first and the last saplings = 3\u00d73\/4 = 9\/4m = 2.(1\/4) m<\/p>\n\n\n\n

Thus the distance between the first and the last saplings is 2.(1\/4) m.<\/p>\n\n\n\n

6. Lipika reads a book for 1.3\/4 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Time taken by Lipika to read a book = 1.(3\/4) hours.<\/p>\n\n\n\n

She reads entire book in 6 days.<\/p>\n\n\n\n

Now, total hours taken by her to read the entire book = 1.3\/4\u00d76 = 7\/4\u00d76=21\/2=10.1\/2 hours<\/p>\n\n\n\n

Thus, 10 hours were required by her to read the book.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n


7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2.3\/4 litres of petrol?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

In 1 litre of pertrol, car covers the distance = 16 km<\/p>\n\n\n\n

In 2.3\/4 litres of petrol, car covers the distance = 2.3\/4 of 16 km = 11\/4\u00d716 = 44 km<\/p>\n\n\n\n

Thus, car will cover 44 km distance.<\/p>\n\n\n\n

8. (a) (i) Provide the number in the box ___ , such that 2\/3\u00d7 ___ = 10\/30<\/strong><\/p>\n\n\n\n

(ii) The simplest form of the number obtained in ___ is _____.<\/strong><\/p>\n\n\n\n

(b) (i) Provide the number in the box ____ such that 3\/5\u00d7 ____ = 24\/75<\/strong><\/p>\n\n\n\n

(ii) The simplest form of the number obtained in ___ is ____.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) (i) 2\/3\u00d75\/10=10\/30<\/p>\n\n\n\n

(ii) The simplest form of 5\/10 is 1\/2<\/p>\n\n\n\n

(b) (i) 3\/5\u00d78\/15=24\/75<\/p>\n\n\n\n

(ii) The simplest form of 8\/15 is 8\/15<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 46<\/p>\n\n\n\n

Exercise 2.4<\/h4>\n\n\n\n

1. Find:
(i) 12 \u00f7 3\/4
(ii) 14 \u00f7 5\/6
(iii) 8 \u00f7 7\/3
(iv) 4 \u00f7 8\/3
(v) 3 \u00f7 2.(1\/3)
(vi) 5 \u00f7 3.(4\/7)<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 12\u00f73\/4 = 12 \u00d7 4\/3 = 16
(ii) 14 \u00f7 5\/6 = 14 \u00d7 6\/5 = 84\/5 = 16.(4\/5)
(iii) 8 \u00f7 7\/3 = 8\u00d7 3\/7 = 24\/7 = 3.(3\/7)
(iv) 4\u00f7 8\/3 = 4\u00d7 3\/8 = 3\/2 = 1.(1\/2)
(v) 3\u00f7 2.(1\/3) = 3\u00f7 7\/3 = 3\u00d7 3\/7 = 9\/7 = 1.(2\/7)
(vi) 5\u00f7 3.(4\/7) = 5\u00f7 25\/7 = 5\u00d7 7\/25 = 7\/5 = 1.(2\/5)<\/p>\n\n\n\n

2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fraction, improper fractions and whole numbers.
(i) 3\/7
(ii) 5\/8
(iii) 9\/7
(iv) 6\/5
(v) 12\/7
(vi) 1\/8
(vii) 1\/11<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Reciprocal of 3\/7 = 7\/3  \u2192 Improper fraction
(ii) Reciprocal of 5\/8 = 8\/5 \u2192 Improper fraction
(iii) Reciprocal of 9\/7 = 7\/9 \u2192 Proper fraction
(iv) Reciprocal of 6\/5 = 5\/6 \u2192Proper fraction
(v)Reciprocal of 12\/7 = 7\/12 \u2192 Proper fraction
(vi) Reciprocal of 1\/8 = 8 \u2192 Whole number
(vi) Reciprocal of 1\/11 = 11 \u2192 Whole number<\/p>\n\n\n\n

3. Find:
(i) 7\/3 \u00f72
(ii) 4\/9 \u00f75
(iii) 6\/13 \u00f77
(iv) 4.(1\/3) \u00f73
(v) 3.(1\/2) \u00f74
(vi) 4.(3\/7) \u00f77<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 7\/3 \u00f7 2 = 7\/3 \u00d7 \u00bd = (7\u00d71)\/(3\u00d72) = 7\/6 = 1.(1\/6)
(ii) 4\/9 \u00f7 5 = 4\/9 \u00d7 1\/5 = (4\u00d71)\/(9\u00d75) = 4\/45
(iii) 6\/13 \u00f7 7 = 6\/13 \u00d7 1\/7 = (6\u00d71)\/(13\u00d77) = 6\/91
(iv) 4.(1\/3) \u00f7 3 = 13\/3 \u00f7 3 = 13\/3 \u00d7 1\/3 = 13\/9 = 1.(4\/9)
(v) 3.(1\/2) \u00f7 4 = 7\/2 \u00f7 4 = 7\/2 \u00d7 \u00bc = 7\/8
(vi) 4.(3\/7) \u00f7 7 = 31\/7 \u00f7 7 = 31\/7 \u00d7 1\/7 = 31\/49<\/p>\n\n\n\n

4. Find:
(i) 2\/5 \u00f7 1\/2
(ii) 4\/9 \u00f7 2\/3
(iii) 3\/7 \u00f7 8\/7
(iv) 2.1\/3 \u00f7 3\/5
(v) 3.1\/2 \u00f7 8\/3
(vi) 2\/5 \u00f7 1.1\/2
(vii) 3.1\/5 \u00f7 1.2\/3
(viii) 2.1\/5\u00f71.1\/5<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2\/5 \u00f7 \u00bd
= 2\/5 \u00d7 2\/1 = (2\u00d72)\/(5\u00d71) =4\/5<\/p>\n\n\n\n

(ii) 4\/9 \u00f7 2\/3
= 4\/9 \u00d7 3\/2 = 2\/3<\/p>\n\n\n\n

(iii) 3\/7 \u00f7 8\/7
= 3\/7 \u00d7 7\/8 = 3\/8<\/p>\n\n\n\n

(iv) 2.(1\/3) \u00f7 3\/5
= 7\/3 \u00f7 3\/5 = 7\/3 \u00d7 5\/3 = 35\/9 = 3.(8\/9)<\/p>\n\n\n\n

(v) 3.(1\/2) \u00f7 8\/3
= 7\/2 \u00f7 8\/3 = 7\/2 \u00d7 3\/8 = (7\u00d73)\/(2\u00d78) = 21\/16 = 1.(5\/16)<\/p>\n\n\n\n

(vi) 2\/5 \u00f7 1.(1\/2)
= 2\/5 \u00f7 3\/2 = 2\/5 \u00d7 2\/3 = (2\u00d72)\/(5\u00d73) = 4\/15<\/p>\n\n\n\n

(vii) 3.(1\/5) \u00f7 1.(2\/3)
= 16\/5 \u00f7 5\/3 = 16\/5 \u00d7 3\/5 = (16\u00d73)\/(5\u00d75) = 48\/25 = 1.(23\/25)<\/p>\n\n\n\n

(viii) 2.(1\/5) \u00f7 1.(1\/5)<\/p>\n\n\n\n

= 11\/5 \u00f7 6\/5 = 11\/5 \u00d7 5\/6 = 11\/6 = 1.(5\/6)<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 47<\/p>\n\n\n\n

Exercise 2.5<\/h4>\n\n\n\n

1. Which is greater:<\/strong><\/p>\n\n\n\n

(i) 0.5 or 0.05<\/p>\n\n\n\n

(ii) 0.7 or 0.5<\/p>\n\n\n\n

(iii) 7 or 0.7<\/p>\n\n\n\n

(iv) 1.37 or 1.49<\/p>\n\n\n\n

(v) 2.03 or 2.30<\/p>\n\n\n\n

(vi) 0.8 or 0.88<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 0.50 > 0.05<\/p>\n\n\n\n

(ii) 0.7 > 0.5<\/p>\n\n\n\n

(iii) 7.0 > 0.7<\/p>\n\n\n\n

(iv) 1.37 < 1.49<\/p>\n\n\n\n

(v) 2.03 < 2.30<\/p>\n\n\n\n

(vi) 0.80 < 0.88<\/p>\n\n\n\n

2. Express as rupees using decimals:<\/strong><\/p>\n\n\n\n

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

(ii) 7 rupees 7 paise<\/p>\n\n\n\n

(iii) 77 rupees 77 paise<\/p>\n\n\n\n

(iv) 50 paise<\/p>\n\n\n\n

(v) 235 paise<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\u2235 100 paise = Re. 1<\/p>\n\n\n\n

\u2234 1 paisa = Re. 1\/100<\/p>\n\n\n\n

7 paise = Re. 7\/100 = Re. 0.07<\/p>\n\n\n\n

7 rupees 7 paise = Rs. 7 + Re. 7\/100 = Rs. 7 + Re. 0.07 = Rs. 7.07<\/p>\n\n\n\n

77 rupees 77 paise = Rs. 77 + Re. 77\/100 = Rs. 77 + Re. 0.77 = Rs. 77.77<\/p>\n\n\n\n

50 paise = Re. 50\/100 = Re. 0.50<\/p>\n\n\n\n

235 paise = Re. 235\/100 = Rs. 2.35<\/p>\n\n\n\n

3. (i) Express 5 cm in metre and kilometer.<\/strong><\/p>\n\n\n\n

(ii) Express 35 mm in cm, m and km.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Express 5 cm in meter and kilometer.<\/p>\n\n\n\n

\u2235 100 cm = 1 meter<\/p>\n\n\n\n

\u2234 1 cm = 1\/100 meter \u21d2 5 cm = 5\/100 = 0.05 meter.<\/p>\n\n\n\n

Now, \u2235 1000 meters = 1 kilometers<\/p>\n\n\n\n

\u2234 1 meter = 1\/1000 kilometer<\/p>\n\n\n\n

\u21d2 0.05 meter = 0.05\/1000 = 0.00005 kilometer<\/p>\n\n\n\n

(ii) Express 35 mm in cm, m and km.<\/p>\n\n\n\n

\u2235 10 mm = 1 cm<\/p>\n\n\n\n

\u2234 1 mm = 1\/10 cm \u21d2 35 mm = 35\/10 = 3.5 cm<\/p>\n\n\n\n

Now, \u2235 100 cm = 1 meter<\/p>\n\n\n\n

\u2234 1 cm = 1\/100 meter \u21d23.5 cm = 3.5\/100 = 0.035 meter<\/p>\n\n\n\n

Again, \u2235 1000 meters = 1 kilometers<\/p>\n\n\n\n

\u2234 1 meter = 1\/1000 kilometer<\/p>\n\n\n\n

\u21d2 0.035 meter = 0.035\/1000 = 0.000035 kilometer<\/p>\n\n\n\n

4. Express in kg.:<\/strong><\/p>\n\n\n\n

(i) 200 g<\/p>\n\n\n\n

(ii) 3470 g<\/p>\n\n\n\n

(iii)4 kg 8 g<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

1000 g = 1 kg \u21d2 1 g = 1\/1000 kg<\/p>\n\n\n\n

(i) 200 g = (200\u00d7 1\/1000) kg = 0.2 kg<\/p>\n\n\n\n

(ii) 3470 g = (3470\u00d7 1\/1000) kg = 3.470 kg<\/p>\n\n\n\n

(iii) 4 kg 8 g = 4 kg + (8\u00d7 1\/1000) kg = 4 kg + 0.008 kg = 4.008 kg<\/p>\n\n\n\n

5. Write the following decimal numbers in the expanded form:<\/strong><\/p>\n\n\n\n

(i) 20.03<\/p>\n\n\n\n

(ii) 2.03<\/p>\n\n\n\n

(iii) 200.03<\/p>\n\n\n\n

(iv) 2.034<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 20.03 = 2\u00d710 + 0\u00d71 + 0\u00d71\/10 + 3\u00d71\/100<\/p>\n\n\n\n

(ii) 2.03 = 2\u00d71+0\u00d71\/10+3\u00d71\/100<\/p>\n\n\n\n

(iii) 200.03 = 2\u00d7100+0\u00d710+0\u00d71+0\u00d71\/10+3\u00d71\/100<\/p>\n\n\n\n

(iv) 2.034 = 2\u00d71+0\u00d71\/10+3\u00d71\/100+4\u00d71\/1000<\/p>\n\n\n\n

6. Write the place value of 2 in the following decimal numbers:<\/strong><\/p>\n\n\n\n

(i) 2.56<\/p>\n\n\n\n

(ii) 21.37<\/p>\n\n\n\n

(iii) 10.25<\/p>\n\n\n\n

(iv) 9.42<\/p>\n\n\n\n

(v) 63.352<\/p>\n\n\n\n

Answer<\/strong>
<\/p>\n\n\n\n

(i) Place value of 2 in 2.56 = 2 \u00d7 1 = 2 ones<\/p>\n\n\n\n

(ii) Place value of 2 in 21.37 = 2 \u00d7 10 = 2 tens<\/p>\n\n\n\n

(iii) Place value of 2 in 10.25 = 2\u00d7 1\/10 = 2 tenths<\/p>\n\n\n\n

(iv) Place value of 2 in 9.42 = 2\u00d7 1\/100 = 2 hundredth<\/p>\n\n\n\n

(v) Place value of 2 in 63.352 = 2\u00d7 1\/1000 = 2 thousandth<\/p>\n\n\n\n

7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km.<\/p>\n\n\n\n

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

Total distance covered by Dinesh = AB + BC<\/p>\n\n\n\n

= 7.5 + 12.7 = 20.2 km<\/p>\n\n\n\n

Total distance covered by Ayub = AD + DC<\/p>\n\n\n\n

= 9.3 + 11.8 = 21.1 km<\/p>\n\n\n\n

On comparing the total distance of Ayub and Dinesh,<\/p>\n\n\n\n

21.1 km > 20.2 km<\/p>\n\n\n\n

Therefore, Ayub covered more distance by 21.1 \u2013 20.2 = 0.9 km = 900 m<\/p>\n\n\n\n

8. Shyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g<\/p>\n\n\n\n

Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g<\/p>\n\n\n\n

On comparing the quantity of fruits,<\/p>\n\n\n\n

8 kg 550 g < 8 kg 950 g<\/p>\n\n\n\n

Therefore, Sarala bought more fruits.<\/p>\n\n\n\n

9. How much less is 28 km than 42.6 km?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We have to find the difference of 42.6 km and 28 km.<\/p>\n\n\n\n

42.6 \u2013 28.0 = 14.6 km<\/p>\n\n\n\n

Therefore 14.6 km less is 28 km than 42.6 km<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 52<\/p>\n\n\n\n

Exercise 2.6<\/h4>\n\n\n\n

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

(i) 0.2 \u00d7 6<\/p>\n\n\n\n

(ii) 8 \u00d7 4.6<\/p>\n\n\n\n

(iii) 2.71 \u00d7 5<\/p>\n\n\n\n

(iv) 20.1 \u00d7 4<\/p>\n\n\n\n

(v) 0.05 \u00d7 7<\/p>\n\n\n\n

(vi) 211.02 \u00d7 4<\/p>\n\n\n\n

(vii) 2 \u00d7 0.86<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 0.2 \u00d7 6 = 1.2<\/p>\n\n\n\n

(ii) 8 \u00d7 4.6 = 36.8<\/p>\n\n\n\n

(iii) 2.71 \u00d7 5 = 13.55<\/p>\n\n\n\n

(iv) 20.1 \u00d7 4 = 80.4<\/p>\n\n\n\n

(v) 0.05 \u00d7 7 = 0.35<\/p>\n\n\n\n

(vi) 211.02 \u00d7 4 = 844.08<\/p>\n\n\n\n

(vii) 2 \u00d7 0.86 = 1.72<\/p>\n\n\n\n

2. Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm ?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: Length of rectangle = 5.7 cm and Breadth of rectangle = 3 cm<\/p>\n\n\n\n

Area of rectangle = Length \u00d7 Breadth<\/p>\n\n\n\n

= 5.7 \u00d7 3 = 17.1 cm2<\/sup><\/p>\n\n\n\n

Thus, the area of rectangle is 17.1 cm2<\/sup><\/p>\n\n\n\n

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

(i) 1.3 \u00d7 10<\/p>\n\n\n\n

(ii) 36.8 \u00d7 10<\/p>\n\n\n\n

(iii) 153.7 \u00d7 10<\/p>\n\n\n\n

(iv) 168.07 \u00d7 10<\/p>\n\n\n\n

(v) 31.1 \u00d7 100<\/p>\n\n\n\n

(vi) 156.1 \u00d7 100<\/p>\n\n\n\n

(vii) 3.62 \u00d7 100<\/p>\n\n\n\n

(viii) 3.07 \u00d7 100<\/p>\n\n\n\n

(ix) 0.5 \u00d7 10<\/p>\n\n\n\n

(x) 0.08 \u00d7 10<\/p>\n\n\n\n

(xi) 0.9 \u00d7 100<\/p>\n\n\n\n

(xii) 0.03 \u00d7 1000<\/p>\n\n\n\n

(i) 1.3 \u00d7 10 = 13.0<\/p>\n\n\n\n

(ii) 36.8 \u00d7 10 = 368.0<\/p>\n\n\n\n

(iii) 153.7 \u00d7 10 = 1537.0<\/p>\n\n\n\n

(iv) 168.07 \u00d7 10 = 1680.7<\/p>\n\n\n\n

(v) 31.1 \u00d7 100 = 3110.0<\/p>\n\n\n\n

(vi) 156.1 \u00d7 100 = 15610.0<\/p>\n\n\n\n

(vii) 3.62 \u00d7 100 = 362.0<\/p>\n\n\n\n

(viii) 43.07 \u00d7 100 = 4307.0<\/p>\n\n\n\n

(ix) 0.5 \u00d7 10 = 5.0<\/p>\n\n\n\n

(x) 0.08 \u00d7 10 = 0.80<\/p>\n\n\n\n

(xi) 0.9 \u00d7 100 = 90.0<\/p>\n\n\n\n

(xii) 0.03 \u00d7 1000 = 30.0<\/p>\n\n\n\n

4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\u2235 In one litre, a two-wheeler covers a distance = 55.3 km<\/p>\n\n\n\n

\u2234 In 10 litrs, a two- wheeler covers a distance = 55.3 \u00d7 10 = 553.0 km<\/p>\n\n\n\n

Thus, 553 km distance will be covered by it in 10 litres of petrol.<\/p>\n\n\n\n

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

(i) 2.5 \u00d7 0.3<\/p>\n\n\n\n

(ii) 0.1 \u00d7 51.7<\/p>\n\n\n\n

(iii) 0.2 \u00d7 316.8<\/p>\n\n\n\n

(iv) 1.3 \u00d7 3.1<\/p>\n\n\n\n

(v) 0.5 \u00d7 0.05<\/p>\n\n\n\n

(vi) 11.2 \u00d7 0.15<\/p>\n\n\n\n

(vii) 1.07 \u00d7 0.02<\/p>\n\n\n\n

(viii) 10.05 \u00d7 1.05<\/p>\n\n\n\n

(ix) 101.01 \u00d7 0.01<\/p>\n\n\n\n

(x) 100.01 \u00d7 1.1<\/p>\n\n\n\n

(i) 2.5 \u00d7 0.3 = 0.75<\/p>\n\n\n\n

(ii) 0.1 \u00d7 51.7 = 5.17<\/p>\n\n\n\n

(iii) 0.2 \u00d7 316.8 = 63.36<\/p>\n\n\n\n

(iv) 1.3 \u00d7 3.1 = 4.03<\/p>\n\n\n\n

(v) 0.5 \u00d7 0.05 = 0.025<\/p>\n\n\n\n

(vi) 11.2 \u00d7 0.15 = 1.680<\/p>\n\n\n\n

(vii) 1.07 \u00d7 0.02 = 0.0214<\/p>\n\n\n\n

(viii) 10.05 \u00d7 1.05 = 10.5525<\/p>\n\n\n\n

(ix) 101.01 \u00d7 0.01 = 1.0101<\/p>\n\n\n\n

(x) 100.01 \u00d7 1.1 = 110.11<\/p>\n\n\n\n

NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No. 55<\/p>\n\n\n\n

Exercise 2.7<\/h4>\n\n\n\n

1. Find:
(i) 0.4 \u00f7 2      
(ii) 0.35 \u00f7 5      
(iii) 2.48 \u00f7 4      
(iv) 65.4 \u00f7 6
(v) 651.2 \u00f7 4  
(v) 14.49 \u00f7 7  
(vii) 3.96 \u00f7 4    
(viii) 0.80 \u00f7 5<\/strong><\/p>\n\n\n\n

Answer <\/strong><\/p>\n\n\n\n

(i) 0.4 \u00f7 2
= 4\/10 \u00d7 1\/ 2= 2\/10 = 0.2<\/p>\n\n\n\n

(ii) 0.35 \u00f7 5
= 35\/100 \u00d7 1\/5 = 7\/100 = 0.07<\/p>\n\n\n\n

(iii)2.48 \u00f7 4
= 248\/100 \u00d7 1\/4 = 62\/100 = 0.62<\/p>\n\n\n\n

(iv) 65.4 \u00f7 6
= 654\/10 \u00d7 1\/6 = 109\/10 = 10.9<\/p>\n\n\n\n

(v) 651.2 \u00f7 4
= 6512\/10 \u00d7 1\/4 = 1628\/10 = 162.8<\/p>\n\n\n\n

(vi) 14.49 \u00f7 7
= 1449\/100 \u00d7 1\/7 = 207\/100 = 2.07<\/p>\n\n\n\n

(vii) 3.96 \u00f7 4
= 396\/100 \u00d7 1\/4 = 99\/100 = 0.99<\/p>\n\n\n\n

(viii) 0.80 \u00f7 5
= 80\/100 \u00d7 1\/5 = 16\/100 = 0.16<\/p>\n\n\n\n

2. Find:
(i) 4.8 \u00f7 10
(ii) 52.5 \u00f7 10
(iii) 0.7 \u00f7 10
(iv) 33.1 \u00f7 10
(v) 272.23 \u00f7 10
(vi) 0.56 \u00f7 10
(vii) 3.97 \u00f7 10<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 4.8 \u00f7 10
= 4.8\/10 = 0.48<\/p>\n\n\n\n

(ii) 52.5 \u00f7 10
= 52.5\/10 = 5.25<\/p>\n\n\n\n

(iii) 0.7 \u00f7 10
= 0.7\/10 = 0.07<\/p>\n\n\n\n

(iv) 33.1 \u00f7 10
= 33.1\/10 = 3.31<\/p>\n\n\n\n

(v) 272.23 \u00f7 10
= 272.23\/10 = 27.223<\/p>\n\n\n\n

(vi) 0.56 \u00f7 10
= 0.56\/10 = 0.056<\/p>\n\n\n\n

(vii) 3.97 \u00f7 10
= 3.97\/10 = 0.397<\/p>\n\n\n\n

3. Find:
(i) 2.7 \u00f7 100
(ii) 0.3 \u00f7 100
(iii) 0.78 \u00f7 100
(iv) 432.6 \u00f7 100
(v) 23.6 \u00f7 100
(vi) 98.53 \u00f7 100<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2.7 \u00f7 100
= 27\/10 \u00d7 1\/100 = 27\/1000 = 0.027<\/p>\n\n\n\n

(ii) 0.3 \u00f7 100
= 3\/10 \u00d7 1\/100 =3\/1000 = 0.003<\/p>\n\n\n\n

(iii) 0.78 \u00f7 100
= 78\/100 \u00d7 1\/100 = 78\/10000 = 0.0078<\/p>\n\n\n\n

(iv) 432.6 \u00f7 100
= 4326\/10 \u00d7 1\/100 = 4326\/1000 = 4.326<\/p>\n\n\n\n

(v) 23.6 \u00f7 100
= 236\/10 \u00d7 1\/100 = 236\/1000 = 0.236<\/p>\n\n\n\n

(vi) 98.53 \u00f7 100
= 9853\/100 \u00d7 1\/100 = 9853\/10000 = 0.9853<\/p>\n\n\n\n

4. Find:
(i) 7.9 \u00f7 1000
(ii) 26.3 \u00f7 1000
(iii) 38.53 \u00f7 1000
(iv) 128.9 \u00f7 1000
(v) 0.5 \u00f7 1000<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 7.9 \u00f7 1000
= 79\/10 \u00d7 1\/1000 = 79\/10000 = 0.0079<\/p>\n\n\n\n

(ii) 26.3 \u00f7 1000
= 263\/10 \u00d7 1\/1000 = 263\/10000 = 0.0263<\/p>\n\n\n\n

(iii) 38.53 \u00f7 1000
= 3853\/100 \u00d7 1\/1000 = 3853\/100000 = 0.03853<\/p>\n\n\n\n

(iv) 128.9 \u00f7 1000
= 1289\/10 \u00d7 1\/1000 = 1289\/10000 = 0.1289<\/p>\n\n\n\n

(v) 0.5 \u00f7 1000
= 5\/10 \u00d7 1\/1000 = 5\/10000 = 0.0005<\/p>\n\n\n\n

5. Find:
(i) 7 \u00f7 3.5
(ii) 36 \u00f7 0.2
(iii) 3.25 \u00f7 0.5
(iv) 30.94 \u00f7 0.7
(v) 0.5 \u00f7 0.25
(vi) 7.75 \u00f7 0.25
(vii) 76.5 \u00f7 0.15
(viii) 37.8 \u00f7 1.4
(ix) 2.73 \u00f7 1.3<\/strong><\/p>\n\n\n\n

Answer <\/strong><\/p>\n\n\n\n

(i) 7 \u00f7 3.5
= 7 \u00f7 35\/10 = 7 \u00d7 10\/35 = 10\/5 = 2<\/p>\n\n\n\n

(ii) 36 \u00f7 0.2
= 36 \u00f7 2\/10 = 36 \u00d7 10\/2 = 18 \u00d7 10 = 180<\/p>\n\n\n\n

(iii) 3.25 \u00f7 0.5
= 325\/100 \u00f7 5\/10 = 325\/100 \u00d7 10\/5 = 65\/10 = 6.5<\/p>\n\n\n\n

(iv) 30.94 \u00f7 0.7
= 3094\/100 \u00f7 7\/10 = 3094\/100 \u00d7 10\/7 = 442\/10 = 44.2<\/p>\n\n\n\n

(v) 0.5 \u00f7 0.25
= 5\/10 \u00f7 25\/100 = 5\/10 \u00d7 100\/25 = 10\/5 = 2<\/p>\n\n\n\n

(vi) 7.75 \u00f7 0.25
= 775\/100 \u00f7 25\/100 = 775\/100 \u00d7 100\/25 = 31<\/p>\n\n\n\n

(vii) 76.5 \u00f7 0.15
= 765\/10 \u00f7 15\/100 = 765\/10 \u00d7 100\/15 = 51\u00d710 = 510<\/p>\n\n\n\n

(viii) 37.8 \u00f7 1.4
= 378\/10 \u00f7 14\/10 = 378\/10 \u00d7 10\/14 = 27<\/p>\n\n\n\n

(ix) 2.73 \u00f7 1.3
= 273\/100 \u00f7 13\/10 = 273\/100 \u00d7 10\/13 = 21\/10 = 2.1<\/p>\n\n\n\n

6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre petrol?<\/strong><\/p>\n\n\n\n

Answer <\/strong><\/p>\n\n\n\n

\u2235 In 2.4 litres of petrol, distance covered by the vehicle = 43.2 km
\u2234 In 1 litre of petrol, distance covered by the vehicle = 43.2 \u00f7 2.4
= 432\/10 \u00f7 24\/10 = 432\/10 \u00d7 10\/24
= 18 km
Thus, it covered 18 km distance in one litre of petrol.<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals PDF<\/a><\/p>\n\n\n\n

Chapter-wise NCERT Solutions Class 7 Maths<\/h2>\n\n\n\n
Chapter 1 Integers<\/a><\/td><\/tr>
Chapter 2 Fractions and Decimals<\/a><\/td><\/tr>
Chapter 3 Data Handling<\/a><\/td><\/tr>
Chapter 4 Simple Equations<\/a><\/td><\/tr>
Chapter 5 Lines and Angles<\/a><\/td><\/tr>
Chapter 6 The Triangle and its Properties<\/a><\/td><\/tr>
Chapter 7 Congruence of Triangles<\/a><\/td><\/tr>
Chapter 8 Comparing Quantities<\/a><\/td><\/tr>
Chapter 9 Rational Numbers<\/a><\/td><\/tr>
Chapter 10 Practical Geometry<\/a><\/td><\/tr>
Chapter 11 Perimeter and Area<\/a><\/td><\/tr>
Chapter 12 Algebraic Expressions<\/a><\/td><\/tr>
Chapter 13 Exponents and Powers<\/a><\/td><\/tr>
Chapter 14 Symmetry<\/a><\/td><\/tr>
Chapter 15 Visualising Solid Shapes<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Class 7: Maths Chapter 2 solutions. Complete Class 7 Maths Chapter 2 Notes. NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals NCERT 7th Maths Chapter 2, class 7 Maths Chapter 2 solutions Page No: 31 Exercise 2.1 1. Solve:(i) 2 – 3\/5(ii) 4 + 7\/8(iii) 3\/5 + 2\/7(iv) 9\/11 \u2212 4\/15(v) 7\/10 + […]<\/p>\n","protected":false},"author":302,"featured_media":627309,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1550],"boards":[1180],"class_list":["post-139651","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-ncert-maths-class-7","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions Class 7, Maths Chapter 2- IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals | Browse all Class 7 Maths Solutions of NCERT books- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-2-fractions-and-decimals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 2 solutions. 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