CBSE NCERT Solutions for Class 7 Mathematics …...Class- VII-CBSE-Mathematics Fractions and...
Transcript of CBSE NCERT Solutions for Class 7 Mathematics …...Class- VII-CBSE-Mathematics Fractions and...
Class- VII-CBSE-Mathematics Fractions and Decimals
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CBSE NCERT Solutions for Class 7 Mathematics Chapter 2 Back of Chapter Questions
Exercise: 2.1
1. Solve:
(i) 2 − 35
(ii) 4 + 78
(iii) 3 5
+ 27
(iv) 911− 4
15
(v) 710
+ 25
+ 32
(vi) 2 23
+ 3 12
(vii) 8 12− 3 5
8
Solution:
(i) 2 − 35
21
× 55
= 105
(since the LCM of the denominators 1 and 5 is 5)
105−
35
=10 − 3
5
=75
= 1 25 (7÷5 gives quotient 1 and remainder 2)
(ii) 4 + 78
41
× 88
= 328
(since the LCM of the denominators 1 and 8 is 8)
328
+78
=32 + 7
8
=398
= 4 78 (39 ÷ 8 gives quotient 4 and remainder 7)
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(iii) 3 5
+ 27
35
× 77
= 2135
and 27
× 55
= 1035
(since the LCM of the denominators 5 and 7 is 35)
2135
+1035
=21 + 10
35
=3135
(iv) 911− 4
15
911
× 1515
= 135165
and 415
× 1111
= 44165
(since the LCM of the denominators 11 and 15 is 165)
135165
−44
165=
135 − 44165
=91
165
(v) 710
+ 25
+ 32
= 7+4+1510
(since the LCM of the denominators 10, 5 and 2 is 10)
=2610
=135
= 2 35
(13 ÷ 5 gives quotient 2 and remainder 3)
(vi) 2 23
+ 3 12
= 83
+ 72
= 16+216
(since the LCM of the denominators 3 and 2 is 6)
=376
= 6 16 (37 ÷ 6 gives quotient 6 and remainder 1)
(vii) 8 12− 3 5
8= 17
2− 29
8
= 68− 298
(since the LCM of the denominators 2 and 8 is 8)
=398
= 4 78 (39 ÷ 8 gives quotient 4 and remainder 7)
2. Arrange the following in descending order:
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(i) 29 , 23 , 821
(ii) 15 , 37 , 710
Solution:
(i) 29 , 23 , 821
The above fractions are unlike fractions (i.e., the denominators are not the same)
The LCM of the denominators 9, 3 and 21 is 63
29
×77
=1463
23
×2121
=4263
821
×33
=2463
∴ the descending order of the fractions is 4263
> 2463
> 1463
i.e., 23 > 8
21 > 2
9
(ii) 15 , 37 , 710
The above fractions are unlike fractions (i.e., the denominators are not the same)
The LCM of the denominators 5, 7 and 10 is 70
15
×1414
=1470
37
×1010
=3070
710
×77
=4970
∴ the descending order of the fractions is 4970
> 3070
> 1470
i.e., 710
> 37 > 1
5
3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square? Along the first row
4
11+
911
+2
11=
1511
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411
9
11
211
311
5
11
711
8
11
111
6
11
Solution:
Sum of the remaining rows:
3
11+
511
+7
11=
1511
8
11+
111
+6
11=
1511
Sum of the three columns
4
11+
311
+8
11=
1511
9
11+
511
+1
11=
1511
2
11+
711
+6
11=
1511
Sum of the diagonals
4
11+
511
+6
11=
1511
2
11+
511
+8
11=
1511
The sum of the fractions row wise, column wise and diagonal wise is the same i.e., 15
11
Hence, it is a magic square
4. A rectangular sheet of paper is 12 12 cm long and 10 2
3 cm wide. Find its perimeter.
Solution:
Length, l = 12 12
= 252
cm Breadth, b = 10 23
= 323
cm
Perimeter of the sheet = 2 (l + b)
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= 2 �252
+323�
= 2 �75+646
� (since the LCM of the denominators 2 and 3 is 6)
= 2 ×139
6
=139
3
= 46 13
cm (139 ÷ 3 gives quotient 46 and remainder 1)
Therefore, the perimeter of the rectangular sheet is 46 13 cm
5. Find the perimeters of (i) Δ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution:
Given that AB = 52 cm, BE = 2 3
4= 11
4 cm AE = 3 3
5= 18
5 cm
Perimeter of a triangle = sum of the three sides
Perimeter of the Δ ABE = AB + BE + AE
=52
+114
+185
= 50+55+7220
(since the LCM of the denominators 2, 4 and 5 is 20)
=17720
= 8 1720
cm (177 ÷ 20 gives quotient 8 and remainder 17)
Therefore, the perimeter of the Δ ABE is 8 1720
cm
(ii) Length (BE = CD) l = 2 34
= 114
cm Breadth (BC = ED) b = 76 cm
Perimeter of a rectangle BCDE = 2 ( l + b)
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= 2 �114
+76�
= 2 �33+1412
� (since the LCM of the denominators 4 and 6 is 12)
= 2 ×4712
=476
= 7 56 cm (47 ÷ 6 gives quotient 7 and remainder 5)
Now, 8 1720
cm > 7 56
cm
Therefore, the perimeter of the triangle ABE is greater than the rectangle BCDE is 7 5
6 cm
6. Salil wants to put a picture in a frame. The picture is 7 35 cm wide. To fit in the
frame the picture cannot be more than 7 310
cm wide. How much should the picture be trimmed?
Solution:
The size of the picture = 7 35
= 385
cm
The size of the picture that will fit in the frame = 7 310
= 7310
cm
The picture should be trimmed by = 385− 73
10
= 76 − 7310
(since the LCM of the denominators 5 and 10 is 10)
= 310
cm
7. Ritu ate 35 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?
Solution:
Part of the apple eaten by Ritu = 35
Part of the apple eaten by Somu = 1 − 35
= 5 − 35
(since the LCM of the denominators 1 and 5 is 5)
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=25
⇒ 35
> 25
∴ Ritu has eaten the larger share by=35− 2
5
=15
8. Michael finished colouring a picture in 712
hour. Vaibhav finished colouring the
same picture in 34 hour. Who worked longer? By what fraction was it longer?
Solution:
Time taken by Michael to colour the picture = 712
hour
Time taken by Vaibhav to colour the picture = 34 hour
712
× 11
= 712
(since the LCM of the denominators 12 and 4 is 12)
34
× 33
= 912
(since the LCM of the denominators 12 and 4 is 12)
912
> 712
i.e., 34 > 7
12
∴ Vaibhav has taken longer time than Michael
⇒ 9
12−
712
=2
12=
16
Therefore, Vaibhav has taken 16 hour longer than Michael to colour the picture
Exercise: 2.2
1. Which of the drawings (a) to (d) show:
(i) 2 × 15
(ii) 2 × 12
(iii) 3 × 23
(iv) 3 × 14
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(a)
(b)
(c)
(d)
Solution:
2 ×15
= (d) ⇒ 2 ×15
=15
+15
2 ×12
= (b) ⇒ 2 ×12
=12
+12
3 ×23
= (a) ⇒ 3 ×23
=23
+23
+23
3 ×14
= (c) ⇒ 14
+14
+14
2. Some pictures (a) to (c)are given below. Tell which of them show:
(i) 3 × 15
= 35
(ii) 2 × 13
= 23
(iii) 3 × 34
= 2 14
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(a)
(b)
(c)
Solution:
3 ×15
=35
= (c) ⇒ 3 ×15
=15
+15
+15
2 ×13
=23
= (a) ⇒ 2 ×13
=13
+13
3 ×34
= 214
= (b) ⇒ 3 ×34
= 214
=34
+34
+34
3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 × 35
(ii) 4 × 13
(iii) 2 × 67
(iv) 5 × 29
(v) 23
× 4
(vi) 52
× 6
(vii) 11 × 47
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(viii) 20 × 45
(ix) 13 × 13
(x) 15 × 35
Solution:
7 ×35
=215
= 415
4 ×13
=43
= 113
2 ×67
=127
= 157
5 ×29
=109
= 119
23
× 4 =83
= 223
52
× 6 = 15
11 ×47
=447
= 627
20 ×45
= 16
13 ×13
=133
= 413
15 ×35
= 9
4. Shade: (i) 12 of the circles in box (a) (ii) 2
3 of the triangles in box (b) (iii) 3
5 of the
squares in box (c)
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Solution:
Shade 12 of the circles in box (a)
Shade 23 of the triangles in box (b)
Shade 35 of the squares in box (c)
5. Find
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(a) 12 of (i) 24 (ii) 46
(b) 23 of (i) 18 (ii) 27
(c) 34 of (i) 16(ii) 36
(d) 45 of (i) 20 (ii) 35
Solution:
(i) 12 of 24 = 1
2× 24
= 12
(ii) 12 of 46 = 1
2× 46
= 23
(i) 23 of 18 = 2
3× 18
= 12
(ii) 23 of 27 = 2
3× 27
= 18
(i) 34 of 16 = 3
4× 16
= 12
(ii) 34 of 36 = 3
4× 36
= 27
(i) 45 of 20 = 4
5× 20
= 16
(ii) 45 of 35 = 4
5× 35
=28
6. Multiply and express as a mixed fraction:
(a) 3 × 5 15
(b) 5 × 6 34
(c) 7 × 2 14
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(d) 4 × 6 13
(e) 3 14
× 6
(f) 3 25
× 8
Solution:
(a) 3 × 5 15
= 3 × 265
=785
= 15 35(since 78 ÷ 5 gives 15 as quotient and 3 as remainder)
(b) 5 × 6 34
= 5 × 274
=135
4
= 33 34(since 135 ÷ 4 gives 33 as quotient and 3 as remainder)
(c) 7 × 2 14
= 7 × 94
=634
= 15 34(since 63 ÷ 4 gives 15 as quotient and 3 as remainder)
(d) 4 × 6 13
= 4 × 193
=763
=25 13(since 76 ÷3 gives 25 as quotient and 1 as remainder)
(e) 3 14×6=13
4×6
=392
= 19 12(since 39 ÷ 2 gives 19 as quotient and 1 as remainder)
(f) 3 25×8=17
5×8
=136
5
= 27 15(since 136 ÷ 5 gives 27 as quotient and 1 as remainder)
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7. Find:
(a) 12of (i) 2 3
4 (ii) 4 2
9
(b) 58of (i) 3 5
6 (ii) 9 2
3
Solution:
(a) (i) 12 of 2 3
4= 1
2× 11
4
=118
= 1 38(since 11 ÷ 8 gives 1 as quotient and 3 as remainder)
(ii) 12 of 4 2
9= 1
2× 38
9
=199
= 2 19(since 19 ÷ 9 gives 2 as quotient and 1 as remainder)
(b) (i) 58 of 3 5
6= 5
8× 23
6
=11548
= 2 1948
(since 115 ÷ 48 gives 2 as quotient and 19 as remainder)
(ii) 58 of 9 2
3= 5
8× 29
3
=14524
= 6 124
(since 145 ÷ 24 gives 6 as quotient and 1 as remainder)
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.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:
Amount of water in the bottle = 5 litres
water consumed by Vidya = 25
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Therefore, amount of water Vidya drank = 25
× 5
= 2 litres
Fraction of water in the bottle = 1 (whole)
water consumed by Vidya = 25
Therefore, fraction of water consumed by Pratap = 1 − 25
= 5 − 25
(LCM of 1 and 5 is 5)
=35
Exercise: 2.3
1. Find: 12 of (a) 1
4 (b) 3
5 (c) 4
3
17 of (a) 2
9 (b) 6
5 (c) 3
10
Solution:
(a) 12 of 1
4= 1
2× 1
4
=18
(b) 12 of 3
5= 1
2× 3
5
=3
10
(c) 12 of 4
3= 1
2× 4
3
=23
(a) 17 of 2
9= 1
7× 2
9
=2
63
(b) 17 of 6
5= 1
7× 6
5
=6
35
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(c) 17 of 3
10= 1
7× 3
10
=3
70
2. Multiply and reduce to lowest form (if possible)
(i) 23
× 2 23
(ii) 27
× 79
(iii) 38
× 64
(iv) 95
× 35
(v) 13
× 158
(vi) 112
× 310
(vii) 45
× 127
Solution:
(i) 23
× 2 23
= 23
× 83
=169
= 179
(ii) 27
× 79
= 29
(iii) 38
× 64
= 916
(iv) 95
× 35
= 2725
= 12
25
(v) 13
× 158
= 58
(vi) 112
× 310
= 3320
= 11320
(vii) 45
× 127
= 4835
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= 11335
3. Multiply the following fractions:
(i) 25
× 5 14
(ii) 6 25
× 79
(iii) 32
× 5 13
(iv) 56
× 2 37
(v) 3 25
× 47
(vi) 2 35
× 3
(vii) 3 47
× 35
Solution:
(i) 25
× 5 14
= 25
× 214
=2110
= 21
10
(ii) 6 25
× 79
= 325
× 79
=22445
= 44445
(iii) 32
× 5 13
= 32
× 163
= 8
(iv) 56
× 2 37
= 56
× 177
=8542
= 21
42
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(v) 3 25
× 47
= 175
× 47
=6835
= 13335
(vi) 2 35
× 3 = 135
× 3
=395
= 745
(vii) 3 47
× 35
= 257
× 35
=157
= 217
4. Which is greater: 27 of 3
4 or 3
5 of 5
8
12 of 6
7 or 2
3 of 3
7
Solution: 27 of 3
4 or 3
5 of 5
8
27 of 3
4= 2
7× 3
4
=3
14
35 of 5
8= 3
5× 5
8
=38
⇒ 38
> 314
(when the numerator is equal, the fraction with smaller denominator is greater)
Therefore, 38 is greater than 3
14
12 of 6
7 or 2
3 of 3
7
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12 of 6
7= 1
2× 6
7
=37
23 of 3
7= 2
3× 3
7
=27
⇒37
>27
Therefore, 37 is greater than 2
7
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.
Solution:
No of saplings in a row in the garden = 4
The distance between two adjacent saplings = 34
m
Therefore, the distance between the first and the last sapling = 3 × 34
=94
= 214
m
6. Lipika reads a book for 1 34 hours every day. She reads the entire book in 6 days.
How many hours in all were required by her to read the book?
Solution:
Time spent every day by Lipika in reading the book = 1 34
= 74 hours
No of days she takes to finish the entire book = 6 days
Therefore, total hours required by her to read the book = 74
× 6
=212
= 1012
hours
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7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 34 litres of petrol?
Solution:
The distance covered by the car using 1 litre of petrol = 16 kms
Therefore, distance covered using 2 34 litres = 2 3
4× 16
=114
× 16
= 44 kms
(a) (i) Provide the number in the box � , such that 23
�= 1030
(ii) The simplest form of the number obtained in � is _____.
(b) (i) Provide the number in the box � , such that 35
�= 2475
(ii) The simplest form of the number obtained in � is _____.
Solution:
(a) (i) 23
�=1030
=1030
×32
=5
10
(ii) The simplest form of the number = 12
(b) (i) 35
�= 2475
=2475
× 53
=8
15
(ii) The simplest form of the number = 815
Exercise: 2.4
1. Find:
(i) 12 ÷ 34
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(ii) 14 ÷ 56
(iii) 8 ÷ 73
(iv) 4 ÷ 83
(v) 3 ÷ 2 13
(vi) 5 ÷ 3 47
Solution:
(i) 12 ÷ 34
= 12 × 43
= 16
(ii) 14 ÷ 56
= 14 × 65
=845
= 1645
(iii) 8 ÷ 73
= 8 × 37
=247
= 337
(iv) 4 ÷ 83
= 4 × 38
=32
= 112
(v) 3 ÷ 2 13
= 3 × 37
=97
= 127
(vi) 5 ÷ 3 47
= 5 × 725
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=75
= 125
2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 37
(ii) 58
(iii) 97
(iv) 65
(v) 127
(vi) 18
(vii) 111
Solution:
(i) 37
Reciprocal of 37 is 7
3 and 7
3 is an improper fraction
(ii) 58
Reciprocal of 58 is 8
5 and 8
5 is an improper fraction
(iii) 97
Reciprocal of 97 is 7
9 and 7
9 is a proper fraction
(iv) 65
Reciprocal of 65 is 5
6 and 5
6 is a proper fraction
(v) 127
Reciprocal of 127
is 712
and 712
is a proper fraction
(vi) 18
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Reciprocal of 18 is 8
1=8 and 8 is a whole number
(vii) 111
Reciprocal of 111
is 111
=11 and 11 is a whole fraction
3. Find:
(i) 73
÷ 2
(ii) 49
÷ 5
(iii) 613
÷ 7
(iv) 4 13
÷ 3
(v) 3 12
÷ 4
(vi) 4 37
÷ 7
Solution:
(i) 73
÷ 2 = 73
× 12
=76
(ii) 49
÷ 5 = 49
× 15
=4
45
(iii) 613
÷ 7 = 613
× 17
=6
91
(iv) 4 13
÷ 3 = 133
÷ 3
=133
×13
=139
(v) 3 12
÷ 4 = 72
÷ 4
=72
×14
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=78
(vi) 4 37
÷ 7 = 317
÷ 7
=317
×17
=3149
4. Find:
(i) 25
÷ 12
(ii) 49
÷ 23
(iii) 37
÷ 87
(iv) 2 13
÷ 35
(v) 3 12
÷ 83
(vi) 25
÷ 1 12
(vii) 3 15
÷ 1 23
(viii) 2 15
÷ 1 15
Solution:
(i) 25
÷ 12
= 25
× 21
=45
(ii) 49
÷ 23
= 49
× 32
=23
(iii) 37
÷ 87
= 37
× 78
=38
(iv) 2 13
÷ 35
= 73
÷ 35
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=73
×53
=359
= 3 89
(v) 3 12
÷ 83
= 72
÷ 83
=72
×38
=2116
= 15
16
(vi) 25
÷ 1 12
= 25
÷ 32
=25
×23
=4
15
(vii) 3 15
÷ 1 23
= 165
÷ 53
=165
×35
=4825
= 12325
(viii) 2 15
÷ 1 15
= 115
÷ 65
=115
×56
=116
= 156
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Exercise: 2.5
1. Which is greater?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88
Solution:
(i) 0.5 or 0.05
0.5 =5
10
0.05 =5
100
⇒ 510
> 5100
(when the numerator is equal, the fraction with the smaller denominator is greater)
Therefore, 0.5 is greater than 0.05
(ii) 0.7 or 0.5
0.7 =7
10
0.5 =5
10
⇒ 710
> 510
[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]
Therefore, 0.7 is greater than 0.5
(iii) 7 or 0.7
7 =71
0.7 =7
10
⇒ 71
> 710
(when the numerator is equal, the fraction with the smaller denominator is greater)
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Therefore, 7 is greater than 0.7
(iv) 1.37 or 1.49
1. 37 =137100
1.49 =149100
⇒ 149100
> 137100
[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]
Therefore, 1.49 is greater than 1.37
(v) 2.03 or 2.30
2.03 =203100
2.30 =230100
⇒ 230100
> 203100
[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]
Therefore, 2.30 is greater than 2.03
(vi) 0.8 or 0.88
0.8 =80
100
0.88 =88
100
⇒ 80100
> 88100
[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]
Therefore, 0.88 is greater than 0.8
2. Express as rupees using decimals:
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise
Solution:
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(i) 7 paise
Rs. 1 = 100 paise
7 paise = 7100
= Rs 0.07
(ii) 7 rupees 7 paise
Rs. 1 = 100 paise
7 rupees 7 paise = 7 + 7100
= 7 + 0.07
= Rs 7.07
(iii) 77 rupees 77 paise
Rs. 1 = 100 paise
77 rupees 77 paise = 77 + 77100
= 77 + 0.77
= Rs 77.77
(iv) 50 paise
Rs. 1 = 100 paise
50 paise = 50100
= Rs 0.50
(v) 235 paise
Rs. 1 = 100 paise
235 paise = 235100
= Rs 2.35
(i) Express 5 cm in metre and kilometer
(ii) Express 35 mm in cm, m and km
Solution:
3. Express 5 cm in metre and kilometer
Solution:
100cm = 1m
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∴ 5cm =5
100
= 0.05 m
100cm = 1m
1000m = 1km
⇒ 100000cm = 1km
100000cm = 1 km
∴ 5cm =5
100000
= 0.00005 km
4. Express 35 mm in cm, m and km
Solution:
10mm = 1cm
∴ 35mm =3510
= 3.5 cm
10mm = 1cm
100cm = 1m
⇒ 1000mm = 1m
∴ 35mm =35
1000
= 0.035 m
10mm = 1cm
100cm = 1m
⇒ 1000000mm = 1km
∴ 35mm =35
1000000
= 0.000035 km
5. Express in kg:
(i) 200 g
(ii) 3470 g
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(iii) 4 kg 8 g
Solution:
200 g
1000 g = 1kg
200 g =200
1000
= 0.2Kg
3470 g
1000 g = 1kg
3470g =34701000
= 3.47 Kg
4 kg 8 g
1000 g = 1kg
4 kg 8 g = 4 kg +8
1000
= 4 kg + 0.008
= 4.008 Kg
6. Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:
20.03 = (2 × 10) + (0 × 1) + �0 ×1
10� + �3 ×
1100
�
2.03 = (2 × 10) + �0 ×1
10� + �3 ×
1100
�
200.03 = (2 × 100) + (0 × 10) + (0 × 1) + �0 ×1
10� + �3 ×
1100
�
2.034 = (2 × 1) + �0 ×1
10� + �3 ×
1100
� + �4 ×1
1000�
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7. Write the place value of 2 in the following decimal numbers:
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352
Solution:
(i) 2.56
Place value of 2 is ones
(ii) 21.37
Place value of 2 is tens
(iii) 10.25
Place value of 2 is one tenths
(iv) 9.42
Place value of 2 is one hundredths
(v) 63.352
Place value of 2 is one thousandths
8. 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?
Solution:
Distance from A to B = 7.5 km
Distance from B to C = 12.7 km
Distance travelled by Dinesh = A to B + B to C
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= 7.5 + 12.7
= 20.2 km
Distance from A to D = 9.3 km
Distance from D to C = 11.8 km
Distance travelled by Ayub = A to D + D to C
= 9.3 + 11.8
= 21.1 km
Difference between the distance travelled = 21.1 − 20.2
= 0.9 km
Therefore, Ayub travelled 0.9 kms more than Dinesh
9. Shyama 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?
Solution:
Quantity of fruits bought by Shyama:
Apples = 5.300 Kg
Mangoes = 3.250 kg
Total fruits bought by Shyama = 8.550 kg
Quantity of fruits bought by Sarala:
Oranges = 4.800 Kg
Bananas = 4.150 kg
Total fruits bought by Shyama = 8.950 kg
Difference in the quantity = 8.950 – 8.550
= 0.4 kg
Therefore, Sarala has bought 0.4 kg or 400 g more fruits than Shyama
10. How much less is 28 km than 42.6 km?
Solution:
42.6 km – 28 Km = 14.6 Km
Therefore, 28 km is less than 42.6 km by 14.6 km
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Exercise: 2.6
1. Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Solution:
(i) 0.2 × 6 = 1.2
(ii) 8 × 4.6 = 36.8
(iii) 2.71 × 5 = 13.55
(iv) 20.1 × 4 = 80.4
(v) 0.05 × 7 = 0.35
(vi) 211.02 × 4 = 844.08
(vii) 2 × 0.86 = 1.72
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length of the rectangle, l = 5.7cm
Breadth of the rectangle, b = 3cm
∴ Area of the rectangle = l × b
= 5.7 × 3
= 17.1cm2
2. Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
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(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 311
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30
3. 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?
Solution:
Distance covered in one litre of petrol = 55.3 km
∴ Distance covered in 10 litres = 55.3 × 10
= 553 km
4. Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
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(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Solution:
(i) 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7 = 5.17
(iii) 0.2 × 316.8 = 63.36
(iv) 1.3 × 3.1 = 4.03
(v) 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15 = 1.68
(vii) 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1 = 110.011
Exercise: 2.7
1. Find:
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
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(viii) 0.80 ÷ 5
Solution:
(i) 0.4 ÷ 2 = 410
÷ 2
=4
10×
12
=2
10
= 0.2
(ii) 0.35 ÷ 5 = 35100
÷ 5
=35
100×
15
=7
100
= 0.07
(iii) 2.48 ÷ 4 = 248100
÷ 4
=248100
×14
=62
100
= 0.62
(iii) 65.4 ÷ 6 = 65410
÷ 6
=65410
×16
=10910
= 10.9
(iv) 651.2 ÷ 4 = 651210
÷ 4
=6512
10×
14
=1628
10
= 162.8
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(v) 14.49 ÷ 7 = 1449100
÷ 7
=1449100
×17
=207100
= 2.07
(vi) 3.96 ÷ 4 = 396100
÷ 4
=396100
×14
=99
100
= 0.99
(vii) 0.80 ÷ 5 = 80100
÷ 5
=80
100×
15
=16
100
= 0.16
2. Find:
(i) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(v) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Solution:
(i) 4.8 ÷ 10 = 0.48
(ii) 52.5 ÷ 10 = 5.25
(iii) 0.7 ÷ 10 = 0.07
(iv) 33.1 ÷ 10 = 3.31
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(v) 272.23 ÷ 10 = 27.223
(vi) 0.56 ÷ 10 = 0.056
(vii) 3.97 ÷ 10 = 0.397
3. Find:
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Solution:
(i) 2.7 ÷ 100 = 0.027
(ii) 0.3 ÷ 100 = 0.003
(iii) 0.78 ÷ 100 = 0.0078
(iv) 432.6 ÷ 100 = 4.326
(v) 23.6 ÷ 100 = 0.236
(vi) 98.53 ÷ 100 = 0.9853
4. Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Solution:
(i) 7.9 ÷ 1000 = 0.0079
(ii) 26.3 ÷ 1000 = 0.0263
(iii) 38.53 ÷ 1000 = 0.03853
(iv) 128.9 ÷ 1000 = 0.1289
(v) 0.5 ÷ 1000 = 0.0005
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5. Find:
(i) 7 ÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Solution:
(i) 7 ÷ 3.5 = 7 ÷ 3510
= 7 ×1035
= 2
(ii) 36 ÷ 0.2 = 36 ÷ 210
= 36 ×102
= 180
(iii) 3.25 ÷ 0.5 = 325100
÷ 510
=325100
× 105
=6510
= 6.5
(iv) 30.94 ÷ 0.7 = 3094100
÷ 710
=3094100
× 107
=44210
= 44.2
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(v) 0.5 ÷ 0.25 = 510
÷ 25100
=5
10×
10025
=2010
= 2
(vi) 7.75 ÷ 0.25 = 775100
÷ 25100
=775100
× 10025
=77525
= 31
(vii) 76.5 ÷ 0.15 = 76510
÷ 15100
=76510
× 10015
= 255 × 2
= 510
(viii) 37.8 ÷ 1.4 = 37810
÷ 1410
=37810
× 1014
=37814
= 27
(ix) 2.73 ÷ 1.3 = 273100
÷ 1310
=273100
× 1013
=2110
= 2.1
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 of petrol?
Solution:
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Distance covered by the vehicle = 43.2 km
Quantity of petrol used = 2.4 litres
Therefore, distance that can be covered in 1 litre = 43.22.4
× 1010
=43224
= 18 kms