Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 3 Decimals are provided here with simple step-by-step explanations. These solutions for Decimals are extremely popular among Class 7 students for Maths Decimals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 39:

We have:

(i) $0.8=\frac{8}{10}=\frac{8÷2}{10÷2}=\frac{4}{5}$

(ii)

(iii) $0.06=\frac{6}{100}=\frac{6÷2}{100÷2}=\frac{3}{50}$

(iv) $0.285=\frac{285}{1000}=\frac{285÷5}{1000÷5}=\frac{57}{200}$

#### Page No 39:

We have:

(i) $5.6=\frac{56}{10}=\frac{56÷2}{10÷2}=\frac{28}{5}=5\frac{3}{5}$

(ii) $12.25=\frac{1225}{100}=\frac{1225÷25}{100÷25}=\frac{49}{4}=12\frac{1}{4}$

(iii) $6.004=\frac{6004}{1000}=\frac{6004÷4}{1000÷4}=\frac{1501}{250}=6\frac{1}{250}$

(iv) $4.625=\frac{4625}{1000}=\frac{4625÷125}{1000÷125}=\frac{37}{8}=4\frac{5}{8}$

#### Page No 39:

(i) $\frac{47}{10}$
On dividing, we get:

$\frac{47}{10}$ = 4.7

(ii)  $\frac{156}{100}$

On dividing, we get:

$\frac{156}{100}$ = 1.56

(iii) $\frac{2516}{100}$

On dividing, we get:

$\frac{2516}{100}$ = 25.16

(iv) $\frac{3524}{1000}$

On dividing, we get:

$\frac{3524}{1000}$ = 3.524

(v) $\frac{25}{8}$
On dividing, we get:

$\frac{25}{8}$ = 3.125

(vi) $3\frac{2}{5}=\frac{17}{5}$
On dividing, we get:

$\frac{17}{5}$ = 3.4

(vii) $2\frac{2}{25}=\frac{52}{25}$
On dividing, we get:

$\frac{52}{25}$ = 2.08

(viii) $\frac{17}{20}$

On dividing, we get:

∴ $\frac{17}{20}$ = 0.85

#### Page No 39:

Converting the given decimals into like decimals, we have:
(i) 6.500, 16.030, 0.274 and 119.400
(ii) 3.50, 0.67, 15.60 and 4.00

#### Page No 39:

We have,

(i) Comparing the whole number part, 78 > 69.
Thus, 78.23  > 69.85

(ii) Converting the decimals into like decimals, we get 3.406  and 3.460.
Comparing the whole number parts, 3 = 3
Comparing the tenths digit, 4 = 4
Comparing the hundredths digit, 6 > 0
Thus, 3.406  < 3.46

(iii)  Comparing the whole number parts, 5 = 5
Comparing the tenths digit, 6 <  8
Thus, 5.68  < 5.86

(iv) Converting the decimals into like decimals, we get 14.050 and 14.005.
Comparing the whole number parts, 14 = 14
Comparing the tenths digit, 0 = 0
Comparing the hundredths digit, 5 > 0
Thus,  14.05  > 14.005

(v) Converting the decimals into like decimals, we get 1.850 and 1.805.
Comparing the whole number parts, 1 = 1
Comparing the tenths digit, 8 = 8
Comparing the hundredths digit, 5 > 0
Thus, 1.85 > 1.805

(vi)  Comparing the whole number parts, 0 < 1
Thus, 0.98 < 1.07

#### Page No 39:

(i) Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06
Clearly, 4.06 < 4.58 < 4.60 < 7.32 < 7.40
Hence, the given decimals in ascending order are 4.06, 4.58, 4.6, 7.32 and 7.4.

(ii) Converting the given decimals into like decimals, we get:
0.50, 5.50, 5.05, 0.05, 5.55
Clearly, 0.05 < 0.50 < 5.05< 5.50 < 5.55
Hence, the given decimals in ascending order are 0.05, 0.5, 5.05, 5.5 and 5.55.

(iii) Converting the given decimals into like decimals, we get:
6.84, 6.48, 6.80, 6.40, 6.08
Clearly, 6.08 < 6.40 < 6.48 < 6.80< 6.84
Hence, the given decimals in ascending order are 6.08, 6.4, 6.48, 6.8 and 6.84.

(iv)  Converting the given decimals into like decimals, we get:
2.200, 2.202, 2.020, 22.200, 2.002
Clearly, 2.002 < 2.020 < 2.200 < 2.202 < 22.200
Hence, the given decimals in ascending order are 2.002, 2.02, 2.2, 2.202 and 22.2.

#### Page No 39:

(i) Converting the given decimals into like decimals, we get:
7.40, 8.34, 74.40, 7.44, 0.74
Clearly, 74.40 > 8.34 > 7.44 > 7.40 > 0.74
Hence, the given decimals in descending order are 74.4, 8.34, 7.44, 7.4 and 0.74.

(ii) Converting the given decimals into like decimals, we get:
2.600, 2.260, 2.060, 2.007, 2.300
Clearly, 2.600 > 2.300 > 2.260 > 2.060 > 2.007
Hence, the given decimals in descending order are 2.6, 2.3, 2.26, 2.06 and 2.007.

#### Page No 39:

45 mm = $\frac{45}{10}$ cm = 4.5 cm

= 4.5 cm = = 0.045 m

= 0.045 m = = 0.000045 km

∴ 45 mm = 4.5 cm = 0.045 m = 0.000045 km

#### Page No 39:

We have:
(i) 8 paise = Rs $\frac{8}{100}$ = Rs 0.08

(ii) 9 rupees 75 paise = Rs = Rs 9.75

(iii) 8 rupees 5 paise  = = Rs 8.05

#### Page No 39:

We have:

(i) 65 m = $\frac{65}{1000}$ km = 0.065 km
∴ 65 m = 0.065 km

(ii) 284 m = $\frac{284}{1000}$km = 0.284 km

(iii) 3 km 5 m =

#### Page No 41:

Converting the given decimals into like decimals, we get:
16.00, 8.70, 0.94, 6.80 and 7.77
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 40.21

#### Page No 41:

Converting the given decimals into like decimals, we get:
18.600, 206.370, 8.008, 26.400 and 6.900
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 266.278.

#### Page No 41:

Converting the given decimals into like decimals, we get:
63.50, 9.70, 0.80, 26.66 and 12.17
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 112.83.

#### Page No 41:

Converting the given decimals into like decimals, we get:
17.400, 86.390, 9.435, 8.800 and 0.060
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 122.085.

#### Page No 41:

Converting the given decimals into like decimals, we get:
26.900, 19.740, 231.769 and 0.048
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 278.457.

#### Page No 41:

Converting the given decimals into like decimals, we get:
23.800, 8.940, 0.078 and 214.600
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 247.418.

#### Page No 41:

Converting the given decimals into like decimals, we get:
6.606, 66.600, 666.000, 0.066 and 0.660
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 739.932.

#### Page No 41:

Converting the given decimals into like decimals, we get:
9.090, 0.909, 99.900, 9.990 and 0.099
Writing these decimals in column form and adding, we get:

Hence, the sum of the given decimals is 119.988.

#### Page No 41:

The given decimals are like decimals. Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (72.43 − 14.79) = 57.64

#### Page No 41:

Converting the given decimals into like decimals, we get:
36.74 and 52.60
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (52.60 − 36.74) = 15.86

#### Page No 41:

Converting the given decimals into like decimals, we get:
13.876 and 22.000
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (22.000 − 13.876) = 8.124

#### Page No 41:

Converting the given decimals into like decimals, we get:
15.079 and 24.160
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (24.160 − 15.079) = 9.081

#### Page No 41:

Converting the given decimals into like decimals, we get:
0.680 and 1.007
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (1.007 − 0.680) = 0.327

#### Page No 41:

Converting the given decimals into like decimals, we get:
0.4678 and 5.0500
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (5.0500 − 0.4678) = 4.5822

#### Page No 41:

Converting the given decimals into like decimals, we get:
2.5307 and 8.0000
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (8.0000 − 2.5307) = 5.4693

#### Page No 41:

Writing the given like decimals in column form with the larger one at the top and subtracting them, we get:

∴ (9.001 − 6.732) = 2.269

#### Page No 41:

Converting the given decimals into like decimals, we get:
5.746 and 9.100
Writing them in column form with the larger one at the top and subtracting them, we get:

∴ (9.100 − 5.746) = 3.354

#### Page No 41:

Converting the given decimals into like decimals, we get:
63.58 and 92.00
Thus, required number = (92.00 − 63.58) = 28.42
Hence, 28.42 should be added to 63.58 to get 92.

#### Page No 41:

Converting the given decimals into like decimals, we get:
8.100 and 0.813
Thus, required number = (8.100 − 0.813) = 7.287
Hence, 7.287 should be subtracted from 8.1 to get 0.813.

#### Page No 41:

Converting the given decimals into like decimals, we get:
32.67 and 60.10
Thus, required number = (60.10 − 32.67) = 27.43
Hence, 32.67 should be increased by 27.43 to get 60.1.

#### Page No 41:

Converting the given decimals into like decimals, we get:
74.30 and 26.87
Thus, required number = (74.30 − 26.87) = 47.43
Hence, 74.3 should be decreased by 47.43 to get 26.87.

#### Page No 41:

Total amount spent by Rohit on purchasing of the given articles = Rs (23.75 + 2.85 + 15.90)
= Rs 42.50
Money given to the shopkeeper = Rs 50
∴ Money returned by the shopkeeper = Rs (50 − 42.50)
= Rs 7.50
Thus, amount received by Rohit = Rs 7.50

#### Page No 43:

We have the following:

(i) 73.92 × 10 = 739.2                 [Shifting the decimal point to the right by 1 place]
(ii) 7.54 × 10 = 75.4                    [Shifting the decimal point to the right by 1 place]
(iii) 84.003 × 10 = 840.03           [Shifting the decimal point to the right by 1 place]
(iv) 0.83 × 10 = 8.3                     [Shifting the decimal point to the right by 1 place]
(v) 0.7 × 10 = 7                           [Shifting the decimal point to the right by 1 place]
(vi) 0.032 × 10 = 0.32                 [Shifting the decimal point to the right by 1 place]

#### Page No 44:

We have the following:

(i) 2.397 × 100 = 239.7           [Shifting the decimal point to the right by 2 places]
(ii) 6.83 × 100 = 683               [Shifting the decimal point to the right by 2 places]
(iii) 2.9 × 100 = 290                [Shifting the decimal point to the right by 2 places]
(iv) 0.08 ×100 = 8                   [Shifting the decimal point to the right by 2 places]
(v) 0.6 × 100 = 60                   [Shifting the decimal point to the right by 2 places]
(vi) 0.003 × 100 = 0.3             [Shifting the decimal point to the right by 2 places]

#### Page No 44:

We have:
(i) 6.7314 × 1000 = 6731.4            [Shifting the decimal point to the right by 3 places]
(ii) 0.182 × 1000 = 182                  [Shifting the decimal point to the right by 3 places]
(iii) 0.076 × 1000 = 76                   [Shifting the decimal point to the right by 3 places]
(iv) 6.25 × 1000 = 6250                  [Shifting decimal point to the right by 3 places]
(v) 4.8 × 1000 = 4800                     [Shifting the decimal point to the right by 3 places]
(vi) 0.06 × 1000 = 60                      [Shifting the decimal point to the right by 3 places]

#### Page No 44:

We have the following:

(i)      54 $×$ 16 = 864
∴ 5.4 $×$ 16 = 86.4                            [1 place of decimal]

(ii)    365 $×$ 19 = 6935
∴ 3.65 $×$ 19 = 69.35                      [2 places of decimal]

(iii)  854 $×$ 12 = 10248
∴ 0.854 $×$ 12 = 10.248                 [3 places of decimal]

(iv)  3673 $×$ 48 = 176304
∴ 36.78 $×$ 48 = 1763.04                [2 places of decimal]

(v) 4125 $×$ 86 = 354750
∴ 4.125 $×$ 86 = 354.750                [3 places of decimal]
= 354.75

(vi) 10406 $×$ 75 = 780450
∴ 104.06 $×$ 75 = 7804.50               [2 places of decimal]
= 7804.5

(vii)  6032 $×$ 124 = 747968
∴ 6.032 $×$ 124 = 747.968             [3 places of decimal]

(viii) 146 $×$ 69 = 10074
∴ 0.0146 $×$ 69 = 1.0074              [4 places of decimal]

(ix) 125 $×$ 327 = 40875
∴ 0.00125 $×$ 327 = 0.40875           [5 places of decimal]

#### Page No 44:

(i) First, we will multiply 76 by 24.

∴ 76 $×$ 24 = 1824
Sum of decimal places in the given numbers = (1 + 1) = 2
∴ 7.6 $×$ 2.4 = 18.24     [2 places of decimal]

(ii) First, we will multiply 345 by 63.

∴ 345 $×$ 63 = 21735
Sum of decimal places in the given numbers = (2 + 1) = 3
∴ 3.45 $×$ 6.3 = 21.735     [3 places of decimal]

(iii) First, we will multiply 54 by 27.

∴ 54 $×$ 27 = 1458
Sum of decimal places in the given numbers = (2 + 2) = 4
∴ 0.54 $×$ 0.27 = 0.1458    [4 places of decimal]

(iv) First, we will multiply 568 by 49.

∴ 568 $×$ 49 = 27832
Sum of decimal places in the given numbers = (3 + 1) = 4
∴ 0.568 $×$ 4.9 = 2.7832    [4 places of decimal]

(v) First, we multiply 654 by 9.

∴ 654 $×$ 9 = 5886
Sum of decimal places in the given numbers = (2 + 2) = 4
∴ 6.54 $×$ 0.09 = 0.5886     [4 places of decimal]

(vi) First, we will multiply 387 by 125.

∴ 387 $×$ 125 = 48375
Sum of decimal places in the given numbers = (2 + 2) = 4
∴ 3.87 $×$ 1.25 = 4.8375    [4 places of decimal]

(vii) First, we will multiply 38 by 6.

∴ 38 $×$ 6 = 228
Sum of decimal places in the given numbers = (2 + 2) = 4
∴ 0.06 $×$ 0.38 = 0.0228    [4 places of decimal]

(viii) First, we will multiply 623 by 75.

∴ 623 $×$ 75 = 46725
Sum of decimal places in the given numbers = (3 + 2) = 5
∴ 0.623 $×$ 0.75 = 0.46725    [5 places of decimal]

(ix) First, we will multiply 14 by 46.

∴ 14 $×$ 46 = 644
Sum of decimal places in the given numbers = (3 + 2) = 5
∴ 0.014 $×$ 0.46 = 0.00644    [5 places of decimal]

(x) First, we will multiply 545 by 176.

∴ 545 $×$ 176 = 95920
Sum of decimal places in the given numbers = (1 + 2) = 3
∴ 54.5 $×$ 1.76 = 95.920  [3 places of decimal]
= 95.92

(xi) First, we will multiply 45 by 24.

∴ 45 $×$ 24 = 1080
Sum of decimal places in the given numbers = (3 + 1) = 4
∴ 0.045 $×$ 2.4 = 0.1080    [4 places of decimal]
= 0.108

(xii) First, we will multiply 1245 by 64.

∴ 1245 $×$ 64 = 79680
Sum of decimal places in the given numbers = (3 + 1) = 4
∴ 1.245 $×$ 6.4 = 7.9680   [4 places of decimal]
= 7.968

#### Page No 44:

(i) First, we will find the product 13 ⨯ 1.3 ⨯ 0.13.
Now, 13 ⨯ 13 ⨯ 13 = 169 x 13
= 2197

Sum of decimal places in the given numbers = (1 + 2) = 3
So, the product must have three decimal places.

∴ 13 ⨯ 1.3 ⨯ 0.13 = 2.197

(ii) First, we will find the product 2.4 ⨯ 1.5 ⨯ 2.5.
Now, 24 ⨯ 15 ⨯ 25 = 360 x 25
= 9000

Sum of decimal places in the given numbers = (1 + 1 + 1) = 3
So, the product must have three decimal places.

∴ 2.4 ⨯ 1.5 ⨯ 2.5 = 9.000
= 9

(iii) First, we will find the product 0.8 ⨯ 3.5 ⨯ 0.05.
Now, 8 ⨯ 35 ⨯ 5 = 280 ⨯ 5
= 1400

Sum of decimal places in the given numbers = (1 + 1 + 2) = 4
So, the product must have four decimal places.

∴ 0.8 ⨯ 3.5 ⨯ 0.05 = 0.1400
= 0.14

(iv) First, we will find the product 0.2 ⨯ 0.02 ⨯ 0.002.
Now, 2 ⨯ 2 ⨯ 2 = 4 ⨯ 2
= 8
Sum of decimal places in the given numbers = (1 + 2 + 3) = 6
So, the product must have six decimal places.

∴ 0.2 ⨯ 0.02 ⨯ 0.002 = 0.000008

(v) First, we will find the product 11.1 ⨯ 1.1 ⨯ 0.11.
Now, 111 ⨯ 11 ⨯ 11 = 1221 ⨯ 11
= 13431

Sum of decimal places in the given numbers = (1 + 1 + 2) = 4
So, the product must have four decimal places.

∴ 11.1 ⨯ 1.1 ⨯ 0.11 = 1.3431

(vi) First, we will find the product 2.1 ⨯ 0.21 ⨯ 0.021.
Now, 21 ⨯ 21 ⨯ 21 = 441 ⨯ 21
= 9261

Sum of decimal places in the given numbers = ( 1 + 2 + 3) = 6
So, the product must have six decimal places.

∴ 2.1 ⨯ 0.21 ⨯ 0.021 = 0.009261

#### Page No 44:

(i) (1.2)2 = 1.2 $×$ 1.2
First, we will find the product 1.2 $×$ 1.2.
Now, 12 $×$ 12 = 144
Sum of decimal places in the given numbers = (1 + 1) = 2
So, the product must have two decimal places.

∴ (1.2)2 = 1.2$×$ 1.2 = 1.44

(ii) (0.7)2 = 0.7 $×$ 0.7
First, we will find the product 0.7 $×$ 0.7.
Now, 7 $×$ 7 = 49
Sum of decimal places in the given numbers = (1 + 1) = 2
So, the product must have two decimal places.
∴ (0.7)2 = 0.7 $×$ 0.7 = 0.49

(iii) (0.04)2 = 0.04 $×$ 0.04
First, we will find the product 0.04 $×$ 0.04.
Now, 4$×$ 4 = 16
Sum of decimal places in the given numbers = (2 + 2) = 4
So, the product must have four decimal places.
∴ (0.04)2 = 0.04$×$ 0.04 = 0.0016

(iv) (0.11)2 = 0.11 $×$ 0.11
First, we will find the product 0.11 $×$ 0.11.
Now, 11 $×$ 11 = 121
Sum of decimal places in the given numbers = ( 2 + 2) = 4
So, the product must have four decimal places.
∴ (0.11)2 = 0.11 $×$ 0.11 = 0.0121

#### Page No 44:

(i) (0.3)3 = 0.3 $×$ 0.3 $×$ 0.3
First, we will find the product 3 $×$ 3 $×$ 3.
Now, 3 $×$ 3 $×$ 3 = 27
Sum of decimal places in the given numbers = (1 + 1 + 1) = 3
So, the product must have three places of decimal.
∴ (0.3)3 = 0.3 $×$ 0.3 $×$ 0.3 = 0.027

(ii) (0.05)3 = 0.05 $×$ 0.05 $×$ 0.05
First, we will find the product 5 $×$ 5 $×$ 5.
Now, 5 $×$5 $×$ 5 = 125
Sum of decimal places in the given numbers = (2 + 2 + 2) = 6
So, the product must have six decimal places.
∴ (0.05)3 = 0.05 $×$ 0.05 $×$ 0.05 = 0.000125

(iii) (1.5)3 = 1.5 $×$1.5 $×$1.5
First, we will find the product 15 $×$15 $×$ 15.
Now, 15 $×$15 $×$ 15 = 225 $×$ 15 = 3375

Sum of decimal places in the given numbers = (1 + 1 + 1) = 3
So, the product must have three decimal places.
∴ (1.5)3 = 1.5 $×$ 1.5 $×$ 1.5 = 3.375

#### Page No 44:

Distance covered by the bus in 1 hour = 62.5 km
∴ Distance covered in 18 hours = (62.5 $×$ 18) km
= 1125 km
Hence, the bus can cover a distance of 1125 km in 18 hours.

#### Page No 44:

Weight of 1 tin of oil = 16.8 kg
∴ Weight of 45 such tins = (16.8 $×$ 45) kg
= 756 kg
Hence, the weight of 45 tins of oil is 756 kg.

#### Page No 44:

Weight of 1 bag of wheat = 97.8 kg
∴ Weight of 500 such bags = (97.8 x 500) kg
= 48900 kg
Hence, the weight of 500 bags of wheat is 48900 kg.

#### Page No 44:

Weight of 1 bag of sugar = 48.450 kg
∴ Weight of 16 bags of sugar = (48.450 ⨯ 16) kg
= 775.2 kg

Hence, the weight of 16 bags of sugar is 775.2 kg.

#### Page No 44:

Capacity of 1 sauce bottle = 0.845 kg
∴ Capacity of 72 such bottles = (0.845 $×$ 72) kg
= 60.84 kg

Hence, the capacity of 72 bottles of sauce will be 60.84 kg.

#### Page No 44:

Weight of 1 bottle of jam = 925 g =0.925 kg
∴ Weight of 25 such bottles = (0.925 $×$ 25) kg
= 23.125 kg

∴ The weight of 25 bottles of jam will be 23.125 kg.

#### Page No 44:

Capacity of 1 drum of oil = 16.850 litres
∴ Capacity of 48 such drums = (16.850 x 48) litres
= 808.800 litres

Hence, the capacity of 48 drums of oil is 808.800 litres.

#### Page No 44:

Cost of 1 kg of rice =Rs 56.80
∴ Cost of 16.25 kg of rice = Rs (56.80 $×$ 16.25)
= Rs 923

Hence, the cost of 16.25 kg of rice is Rs 923.

#### Page No 44:

Cost of 1 m of cloth = Rs 108.50
∴ Cost of 18.5 m of cloth = Rs (108.50 x 18.5)
= Rs 2007.25

Hence, the cost of 18.5 m of cloth is Rs 2007.25.

#### Page No 44:

Distance covered by the car with 1 litre of petrol = 8.6 km
∴ Distance covered with 36.5 litres of petrol = (8.6 $×$ 36.5) km
= 313.900 km
Hence, the distance covered by the car with 36.5 litres of petrol is 313.900 km.

#### Page No 44:

Charges for 1 km = Rs 9.80
∴ Charges for 106.5 km = Rs (9.80 $×$ 106.5)
= Rs 1043.70
Hence, the taxi driver will charge Rs 1043.70 for a journey of 106.5 km.

#### Page No 49:

We have the following:

(i) 131.6 ÷ 10 = $\frac{131.6}{10}=13.16$                      [Shift the decimal point to the left by 1 place]

(ii) 32.56 ÷ 10 = $\frac{32.56}{10}=3.256$                    [Shift the decimal point to the left by 1 place]

(iii) 4.38 ÷ 10 = $\frac{4.38}{10}=0.438$                       [Shift the decimal point to the left by 1 place]

(iv) 0.34 ÷ 10 = $\frac{0.34}{10}=0.034$                       [Shift the decimal point to the left by 1 place]

(v) 0.08 ÷ 10 = $\frac{0.08}{10}=0.008$                        [Shift the decimal point to the left by 1 place]

(vi) 0.062 ÷ 10 = $\frac{0.062}{10}=0.0062$                 [Shift the decimal point to the left by 1 place]

#### Page No 49:

We have the following:

(i) 137.2 ÷ 100 = $\frac{137.2}{100}=1.372$                          [Shifting the decimal point to the left by 2 places]

(ii) 23.4 ÷ 100 =$\frac{23.4}{100}=0.234$                              [Shifting the decimal point to the left by 2 places]

(iii) 4.7 ÷ 100 = $\frac{4.7}{100}=0.047$                               [Shifting the decimal point to the left by 2 places]

(iv) 0.3 ÷ 100 = $\frac{0.3}{100}=0.003$                               [Shifting the decimal point to the left by 2 places]

(v) 0.58 ÷ 100 = $\frac{0.58}{100}=0.0058$                           [Shifting the decimal point to the left by 2 places]

(vi) 0.02 ÷ 100 = $\frac{0.02}{100}=0.0002$                          [Shifting the decimal point to the left by 2 places]

#### Page No 49:

We have the following:

(i) 1286.5 ÷ 1000 = $\frac{1286.5}{1000}=1.2865$                 [Shift the decimal point to the left by 3 places]

(ii) 354.16 ÷ 1000 = $\frac{354.16}{1000}=0.35416$              [Shift the decimal point to the left by 3 places]

(iii) 38.9 ÷ 1000 = $\frac{38.9}{1000}=0.0389$                      [Shift the decimal point to the left by 3 places]

(iv) 4.6 ÷ 1000 = $\frac{4.6}{1000}=0.0046$                        [Shift the decimal point to the left by 3 places]

(v) 0.8 ÷ 1000 = $\frac{0.8}{1000}=0.0008$                         [Shift the decimal point to the left by 3 places]

(vi) 2 ÷ 1000 = $\frac{2}{1000}=0.002$                           [Shift the decimal point to the left by 3 places]

#### Page No 49:

(i) 12 ÷ 8 = $\frac{12}{8}=\frac{3}{2}$

∴ 12 ÷ 8 = 1.5

(ii) 63 ÷ 15 = $\frac{63}{15}=\frac{21}{5}$

∴ 63 ÷ 15 = 4.2

(iii) 47 ÷ 20 = $\frac{47}{20}$

∴ 47 ÷ 20 = 2.35

(iv) 101 ÷ 25 = $\frac{101}{25}$

∴ 101 ÷ 25 = 4.04

(v ) 31 ÷ 40

∴ 31 ÷ 40 = 0.775

(vi) 11 ÷ 16 = $\frac{11}{16}$

∴ 11 ÷ 16 = 0.6875

#### Page No 49:

(i) We have:
43.2 ÷ 6

∴ 43.2 ÷ 6 = 7.2
(ii) We have:
60.48 ÷ 12

∴ 60.48 ÷ 12 = 5.04

(iii) We have:
117.6 ÷ 21

∴ 117.6 ÷ 21 = 5.6

(iv) We have:
217.44 ÷ 18

∴ 217.44 ÷ 18 = 12.08

(v) We have:
2.575 ÷ 25

∴ 2.575 ÷ 25 = 0.103

(vi) We have:
6.08 ÷ 8

∴ 6.08 ÷ 8 = 0.76

(vii) We have:
0.765 ÷ 9

∴ 0.765 ÷ 9 = 0.085

(viii) We have:
0.768 ÷ 16

∴ 0.768 ÷ 16 = 0.048
(ix) We have:
0.175 ÷ 25

$=\frac{0.175}{25}\phantom{\rule{0ex}{0ex}}=\frac{0.175×1000}{25×1000}\phantom{\rule{0ex}{0ex}}=\frac{175}{25×1000}\phantom{\rule{0ex}{0ex}}=\frac{7}{1000}\phantom{\rule{0ex}{0ex}}=0.007$

(x) We have:
0.3322 ÷ 11

∴ 0.3322 ÷ 11 = 0.0302

(xi) We have:
2.13 ÷ 15

∴ 2.13 ÷ 15 = 0.142

(xii) We have:
6.54 ÷ 12

∴ 6.54 ÷ 12 = 0.545

(xiii) We have:
5.52 ÷ 16

∴ 5.52 ÷ 16 = 0.345

(xiv) We have:
1.001 ÷ 14

∴ 1.001 ÷ 14 = 0.0715

(xv) We have:
0.477 ÷ 18

∴ 0.477 ÷ 18 =  0.0265

#### Page No 49:

(i) 16.46 ÷ 20 = $\frac{16.46}{20}=\frac{16.46×100}{20×100}=\frac{1646}{2×1000}=\frac{823}{1000}=0.823$

(ii) 403.8 ÷ 30 = $\frac{403.8}{30}=\frac{403.8×10}{30×10}=\frac{4038}{3×100}=\frac{1346}{100}=13.46$

(iii) 19.2 ÷ 80 = $\frac{19.2}{80}=\frac{19.2×10}{80×10}=\frac{192}{800}=\frac{192}{8×100}=\frac{24}{100}=0.24$

(iv) 156.8 ÷ 200 = $\frac{156.8}{200}=\frac{156.8×10}{200×10}=\frac{1568}{2000}=\frac{784}{1000}=0.784$

(v) 12.8 ÷ 500 = $\frac{12.8}{500}=\frac{12.8×10}{500×10}=\frac{128}{5000}=\frac{25.6}{1000}=0.0256$

(vi) 18.08 ÷ 400 = $\frac{18.08}{400}=\frac{18.08×100}{400×100}=\frac{1808}{40000}=\frac{452}{10000}=0.0452$

#### Page No 49:

(i) 3.28 ÷ 0.8 = $\frac{3.28}{0.8}=\frac{3.28×10}{0.8×10}=\frac{32.8}{8}$
Now, we have:

∴ $\frac{3.28}{0.8}=\frac{32.8}{8}=4.1$

(ii)  0.288 ÷ 0.9 = $\frac{0.288}{0.9}=\frac{0.288×10}{0.9×10}=\frac{2.88}{9}$
Now, we have:

∴ $\frac{0.288}{0.9}=\frac{2.88}{9}=0.32$

(iii)  25.395 ÷ 1.5 = $\frac{25.395}{1.5}=\frac{25.395×10}{1.5×10}=\frac{253.95}{15}$
Now, we have:

∴ $\frac{25.395}{1.5}=\frac{253.95}{15}=16.93$

(iv) 2.0484 ÷ 0.18 = $\frac{2.0484}{0.18}=\frac{2.0484×100}{0.18×100}=\frac{204.84}{18}$
Now, we have:

∴ $\frac{2.0484}{0.18}=\frac{204.84}{18}=11.38$

(v)  0.228 ÷ 0.38 = $\frac{0.228}{0.38}=\frac{0.228×100}{0.38×100}=\frac{22.8}{38}$
Now, we have:

∴ $\frac{0.228}{0.38}=\frac{22.8}{38}=0.6$

(vi) 0.8085 ÷ 0.35 = $\frac{0.8085}{0.35}=\frac{0.8085×100}{0.35×100}=\frac{80.85}{35}$
Now, we have:

∴$\frac{0.8085}{0.35}=\frac{80.85}{35}=2.31$

(vii) 21.976 ÷ 1.64 = $\frac{21.976}{1.64}=\frac{21.976×100}{1.64×100}=\frac{2197.6}{164}$
Now, we have:

∴ $\frac{21.976}{1.64}=\frac{2197.6}{164}=13.4$

(viii) 11.04 ÷ 1.6 = $\frac{11.04}{1.6}=\frac{11.04×10}{1.6×10}=\frac{110.4}{16}$
Now, we have:

∴$\frac{11.04}{1.6}=\frac{110.4}{16}=6.9$

(ix)  6.612 ÷ 11.6 = $\frac{6.612}{11.6}=\frac{6.612×10}{11.6×10}=\frac{66.12}{116}$
Now, we have:

∴ $\frac{6.612}{11.6}=\frac{66.12}{116}=0.57$

(x) 0.076 ÷  0.19 = $\frac{0.076}{0.19}=\frac{0.076×100}{0.19×100}=\frac{7.6}{19}$
Now, we have:

∴ $\frac{0.076}{0.19}=\frac{7.6}{19}=0.4$

(xi) 48 ÷ 0.074

$=\frac{148}{0.074}\phantom{\rule{0ex}{0ex}}=\frac{148×1000}{0.074×1000}\phantom{\rule{0ex}{0ex}}=\frac{148000}{74}\phantom{\rule{0ex}{0ex}}=2×1000\phantom{\rule{0ex}{0ex}}=2000$

(xii)  16.578 ÷ 5.4 = $\frac{16.578}{5.4}=\frac{16.578×10}{5.4×10}=\frac{165.78}{54}$
Now, we have:

∴ $\frac{16.578}{5.4}=\frac{165.78}{54}=3.07$

(xiii)  28 ÷ 0.56

$=\frac{28}{0.56}\phantom{\rule{0ex}{0ex}}=\frac{28×100}{0.56×100}\phantom{\rule{0ex}{0ex}}=\frac{2800}{56}\phantom{\rule{0ex}{0ex}}=\frac{1×100}{2}\phantom{\rule{0ex}{0ex}}=50$

(xiv) 204 ÷ 0.17

$=\frac{204}{0.17}\phantom{\rule{0ex}{0ex}}=\frac{204×100}{0.17×100}\phantom{\rule{0ex}{0ex}}=\frac{20400}{17}\phantom{\rule{0ex}{0ex}}=12×100\phantom{\rule{0ex}{0ex}}=1200$

(xv) 3 ÷ 80 = $\frac{3}{80}$

Now, we have:

∴ $\frac{3}{80}$ = 0.0375

#### Page No 50:

Cost of 24 chairs = Rs 9255.60
∴ Cost of one chair = Rs $\left(\frac{9255.60}{24}\right)$
= Rs $\left(\frac{9255.60×10}{24×10}\right)$
= Rs $\left(\frac{92556}{240}\right)$
= Rs 385.65

Hence, the cost of one chair is Rs 385.65.

#### Page No 50:

Cloth required for 1 shirt = 1.8 m
∴ Number of shirts that can be made from 45 m of cloth =  $\frac{45}{1.8}$ = $\frac{15}{0.6}$ =$\frac{5}{0.2}$ = $\frac{50}{2}$= 25

Hence, 25 shirts can be made from a piece of cloth of length 45 m.

#### Page No 50:

Distance covered by the car with 2.4 litres of petrol = 22.8 km
∴ Distance covered with 1 litre of petrol = $\left(\frac{22.8}{2.4}\right)$ km
= $\left(\frac{228}{24}\right)$ km = $\left(\frac{228÷12}{24÷12}\right)$ km = $\left(\frac{19}{2}\right)$ km = $9\frac{1}{2}$ km

Hence, the distance covered by the car with 1 litre of petrol is $9\frac{1}{2}$ km.

#### Page No 50:

Capacity of 1 tin of oil = 16.5 litres
∴ Number of tins required to hold 478.5 litres of oil = $\left(\frac{478.5}{16.5}\right)=\left(\frac{4785}{165}\right)=\left(\frac{4785÷15}{165÷15}\right)=\frac{319}{11}=29$
Hence, 29 oil tins will be required to hold 478.5 litres of oil.

#### Page No 50:

Weight of 37 bags of sugar = 3644.5 kg
∴ Weight of 1 bag of sugar = $\left(\frac{3644.5}{37}\right)$ = 98.5 kg

Hence, each bag of sugar weighs 98.5 kg.

#### Page No 50:

Capacity of 69 buckets of water = 586.5 litres
∴ Capacity of one such bucket = $\left(\frac{586.5}{69}\right)$ litres = 8.5 litres.

Hence, the capacity of each water bucket is 8.5 litres.

#### Page No 50:

Length of one piece of cloth = 1.15 m
∴ Number of pieces she gets from 46 m of cloth = $\left(\frac{46}{1.15}\right)$
= $\left(\frac{46×100}{1.15×100}\right)$$\left(\frac{4600}{115}\right)$ = 40

Hence, Monica has 40 pieces of cloth each of length 1.15 m.

#### Page No 50:

Total weight of all the bags of cement = 1792.8 kg
Weight of each bag = 49.8 kg
Number of bags =
= $\left(\frac{1792.8}{49.8}\right)=\left(\frac{17928}{498}\right)=36$

Hence, Mr. Soni bought 36 bags of cement.

#### Page No 50:

Thickness of the pile of plywood pieces = 1.89 m = 189 cm
Thickness of one piece of plywood = 0.35 cm
∴ Required number of plywood pieces = $\left(\frac{189}{0.35}\right)=\left(\frac{189×100}{0.35×100}\right)=\left(\frac{18900}{35}\right)=540$

Hence, 540 pieces of plywood are required to make a pile of height 1.89 m.

#### Page No 50:

Product of the given decimals = 261.36
One decimal = 17.6
The other decimal = 261.36 ÷ 17.6
= $\left(\frac{261.36}{17.6}\right)=\left(\frac{261.36×10}{17.6×10}\right)=\left(\frac{2613.6}{176}\right)$
= 14.85

Hence, the other decimal is 14.85.

#### Page No 50:

(b) $\frac{3}{50}$

0.06 = $\frac{6}{100}=\frac{3}{50}$

#### Page No 50:

(c) $1\frac{1}{25}$

1.04 = $\frac{104}{100}=\frac{26}{25}=1\frac{1}{25}$

#### Page No 50:

(b) 2.08

$2\frac{2}{25}=\frac{52}{25}$

On dividing, we get:

$2\frac{2}{25}=\frac{52}{25}$ = 2.08

#### Page No 50:

(c) 0.00006 km

6 cm = $\frac{6}{100}$m = 0.06 m

0.06 m = $\frac{0.06}{1000}$ km = 0.00006 km

∴ 6 cm = 0.00006 km

#### Page No 50:

(b) 0.07 kg

70 g =$\frac{70}{1000}$ kg = $\frac{7}{100}$ kg

= 0.07 kg

∴ 70 g = 0.07 kg

#### Page No 50:

(c) 5.006 kg

5 kg 6 g = (5 $×$ 1000) g + 6 g = 5006 g

= $\frac{5006}{1000}$ kg = 5.006 kg

∴ 5 kg 6 g = 5.006 kg

#### Page No 50:

(c) 2.005 km

2 km 5 m = (2 $×$ 1000) m + 5 m = 2005 m

$\frac{2005}{1000}$ km = 2.005 km

∴ 2 km 5 m = 2.005 km

#### Page No 50:

(c) 0.307

Converting the given decimals into like decimals, we get:
1.007 and 0.700
Writing them in column form with the larger one at the top and subtracting, we get:

Hence, the required number is 0.307.

#### Page No 50:

(b) .07

We have:
0.1 − x = 0.03
⇒ x = 0.1 − 0.03

Converting the given decimals into like decimals, we get:
0.10 and 0.03

Writing them in column form with the larger one at the top and subtracting, we get:

∴ x = 0.07
Hence, the required number is 0.07.

#### Page No 51:

(c) .43
We have:
3.07 + x = 3.5
x = 3.5 − 3.07
Converting the given decimals into like decimals, we get:
3.07 and 3.50
Writing them in column form with the larger one at the top and subtracting, we get:

∴ x  = 0.43

Hence, 0.43 should be added to 3.07 to get 3.5.

#### Page No 51:

(c) 0.069
First, we will multiply 23 by 3.
i.e., 23 $×$ 3 = 69
Sum of decimal places in the given decimals  = (2 + 1) = 3
∴ 0.23 $×$ 0.3 = 0.069     ( 3 places of decimal)

#### Page No 51:

(b) 0.6
We have:
2 $×$ 30 = 60
∴ 0.02 $×$ 30 = 0.60            (2 places of decimal)
= 0.6

#### Page No 51:

(b) 0.2
First, we will multiply 25 by 8.
∴ 25 $×$ 8 = 200
Sum of decimal places in the given decimals = (2 + 1) = 3
∴ 0.25 $×$ 0.8 = 0.200    [3 places of decimal]
= 0.2

#### Page No 51:

(c) .064

First, we will find the product 4 $×$ 4 $×$ 4 = 64
Sum of decimal places in the given decimals = (1 + 1 + 1) = 3
∴ 0.4 × 0.4 × 0.4 = 0.064   ( 3 places of decimal)

#### Page No 51:

(b) .0011

First, we will find the product 11 $×$ 1 $×$ 1.
Sum of decimal places in the given decimals  = (1 + 1 + 2) = 4
∴ 1.1 $×$  0.1 $×$ 0.01 = 0.0011   (4 places of decimal)

#### Page No 51:

(a) 13

2.08 ÷ 0.16 = $\frac{2.08}{0.16}$ = $\frac{2.08×100}{0.16×100}$ = $\frac{208}{16}=13$

#### Page No 51:

(b) 0.17

1.02 ÷ 6 = $\frac{1.02}{6}=\frac{1.02×100}{6×100}=\frac{102}{6×100}=\frac{17}{100}=0.17$

#### Page No 51:

(a) 44.2

30.94 ÷ 0.7 = $\frac{30.94}{0.7}=\frac{30.94×100}{0.7×100}=\frac{3094}{70}=44.2$

(b) 2.1

2.73 ÷ 1.3 =

(a) 40.5

89.1 ÷ 2.2 =

#### Page No 51:

(c) 0.025
First, we will multiply 5 by 5.
i.e., 5 $×$ 5 = 25
Sum of decimal places in the given decimals = (1 + 2) = 3
∴ 0.5 × 0.05 = 0.025   (3 places of decimal)

#### Page No 53:

Cost of 1 pen = Rs 32.50
∴ Cost of 24 such pens = Rs (32.50 $×$ 24)
= Rs 780
Hence, the cost of 24 pens is Rs 780.

#### Page No 53:

Distance covered by the bus in 1 h = 64.5 km
∴ Distance covered in 18 h = (64.5 $×$ 18) km
= 1161 km

Hence, the bus can cover a distance of 1161 km in 18 h.

#### Page No 53:

First, we will find the product 68 $×$ 65 $×$ 4.
Now, 68 $×$ 65 $×$ 4 = 4420 $×$ 4 = 17680
Sum of decimal places in the given decimals = (2 + 1 + 2) = 5
So, the product have five decimal places.

∴ 0.68 $×$ 6.5 $×$ 0.04 = 0.17680
= 0.1768

#### Page No 53:

Total weight of all the bags = 2231 kg
Weight of each bag = 48.5 kg

Number of bags =

=$\left(\frac{2231}{48.5}\right)$kg

=

Hence, 46 bags of cement will weigh 2231 kg.

#### Page No 53:

(i) 0.196 ÷ 1.4 =$\frac{0.196}{1.4}=\frac{0.196×10}{1.4×10}=\frac{1.96}{14}=0.14$

(ii) 39.168 ÷ 1.2 =$\frac{39.168}{1.2}=\frac{39.168×10}{1.2×10}=\frac{391.68}{12}=32.64$

(iii) 0.228 ÷ 0.38 = $\frac{0.228}{0.38}=\frac{0.228×100}{0.38×100}=\frac{22.8}{38}=0.6$

#### Page No 53:

Product of the given decimals = 1.824
One decimal = 0.64
The other decimal = 1.824 ÷ 0.64
= $\frac{1.824}{0.64}=\left(\frac{1.824×100}{0.64×100}\right)=\frac{182.4}{64}=2.85$

Hence, the other decimal is 2.85.

#### Page No 53:

Thickness of the pile of plywoods = 2.43 m = 2.43 $×$ 100 cm = 243 cm
Thickness of one piece of plywood = 0.45 cm
∴ Required number of pieces of plywood = $\frac{243}{0.45}=\frac{243×100}{0.45×100}=\frac{24300}{45}=540$

Hence, the required number of pieces of plywood is 540.

#### Page No 53:

Let the number of sides of the polygon be n.
Length of each side of the polygon = 3.8 cm
∴ Perimeter of the polygon = (3.8 $×$ n) cm
But it is given that its perimeter is 22.8 cm.
∴ (3.8 $×$ n) cm = 22.8 cm
n = $\left(\frac{22.8}{3.8}\right)$ = $\frac{228}{38}$ = 6

Hence, the given polygon has six sides.

(b) 2.04

#### Page No 53:

(b) $1\frac{1}{125}$

1.008 =

#### Page No 53:

(c) 2.005 kg

2 kg 5 g = (2 $×$ 1000) g + 5 g = (2005) g

= $\left(\frac{2005}{1000}\right)$ kg = 2.005 kg

#### Page No 53:

(b) 0.08

We have:
0.012 ÷ 0.15 = $\frac{0.012}{0.15}=\frac{0.012×100}{0.15×100}=\frac{1.2}{15}=0.08$

#### Page No 53:

(c) .0011
First, we will find the product 11 $×$ 1 $×$ 1.
i.e., 11 $×$ 1 $×$ 1 = 11 $×$ 1 = 11
Sum of decimal places in the given decimals = (1 + 1 + 2) = 4

∴ 1.1 $×$  0.1 $×$ 0.01 = 0.0011   [4 places of decimal]

#### Page No 53:

(b) 2.03

4.669 ÷ 2.3  = $\left(\frac{4.669}{2.3}\right)=\left(\frac{4.669×10}{2.3×10}\right)=\left(\frac{46.69}{23}\right)=2.03$

#### Page No 53:

Option (c) is correct.
Let the number added be x.
We have:
2.06 + x = 3.1
x = 3.1 − 2.06

Converting the given decimals into like decimals, we get:
2.06 and 3.10
Thus, required number = (3.10 − 2.06) = 1.04

Hence, 1.04 should be added to 2.06 to get 3.1.

#### Page No 53:

(b) 0.06
We have:
0.1 − x  = 0.04
x = 0.1 − 0.04
Converting the given decimals into like decimals, we get:
0.10 and 0.04
Thus, required number = (0.10 − 0.04) = 0.06
Hence, 0.06 should be subtracted from 0.1 to get 0.04.

#### Page No 53:

(i) 1.001 ÷ 14 = 0.0715

Explanation: $\frac{1.001}{14}=0.0715$

(ii) 204 ÷ 0.17 = 1200

Explanation: $\frac{204}{0.17}=\frac{204×100}{0.17×100}=\frac{204×100}{17}=12×100=1200$

(iii) 0.47 × 5.3 = 2.491

Explanation: First, we will multiply 47 by 53.

∴ 47 $×$ 53 = 2491
Sum of decimal places in the given decimals = (2 + 1) = 3
∴ 0.47 × 5.3 = 2.491

(iv) (0.7)2 = 0.49
Explanation: (0.7)2 = 0.7 × 0.7
First, we will find the product 0.7 × 0.7.
Now, 7 × 7 = 49
Sum of decimal places in the given decimals = (1 + 1) = 2
So, the product must have two decimal places.
∴ (0.7)2 = 0.7 × 0.7 = 0.49

(v) (0.05)3 = 0.000125
Explanation:  First, we will find the product 0.05 × 0.05 × 0.05.
Now, 5 × 5 × 5 = 125
Sum of decimal places in the given decimals = (2 + 2 + 2) = 6
So, the product must have six decimal places.
∴ (0.05)3 = 0.05 × 0.05 × 0.05 = 0.000125

#### Page No 53:

(i) T
We have:
0.5 × 0.05
Now, 5 × 5 = 25
Sum of decimal places in the given decimals = ( 1 + 2) = 3
∴ 0.5 × 0.05 = 0.025

(ii) T
We have:
0.25 × 0.8
Now, 25 × 8 = 200
Sum of decimal places in the given decimals = (2 +1) = 3
∴ 0.25 × 0.8 =0.200 = 0.2

(iii) T
We have:
0.35 ÷ 0.7 = $\frac{0.35}{0.7}=\frac{0.35×10}{0.7×10}=\frac{3.5}{7}=0.5$

(iv) F
We have:
0.4 × 0.4 × 0.4
Now, 4 × 4 × 4 = 64
Sum of decimal places in the given decimals = (1 +1 +1) = 3
∴ 0.4 × 0.4 × 0.4 = 0.064

(v) T
6 cm = $\left(\frac{6}{100}\right)$ m = 0.06 m

View NCERT Solutions for all chapters of Class 7