CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:
We know that by multiplying by 10, the decimal point is shifted one place to its right side.
(i) 73.92 x 10 = 739.2
(ii) 7.54 x 10 = 75.4
(iii) 84.003 x 10 = 840.03
(iv) 0.83 x 10 = 8.3
(v) 0.7 x 10 = 7.0
(vi) 0.032 x 10 = 0.32

Question 2.
Solution:
We know that by multiplying a decimal by 100, two decimal points are shifted to it right side
(i) 2.397 x 100 = 239.7
(ii) 6.83 x 100 = 683.0
(iii) 2.9 x 100 = 290
(iv) 0.08 x 100 = 8
(v) 0.6 x 100 = 60
(vi) 0.003 x 100 = 0.3

Question 3.
Solution:
We know that by multiplying a decimal by 1000, three places of decimal are shifted to its right.
(i) 6.7314 x 1000 = 6731.4
(ii) 0.182 x 1000 = 182
(iii) 0.076 x 1000 = 76
(iv) 6.25 x 1000 = 6250
(v) 4.8 x 1000=4800
(vi) 0.06 x 1000 = 60

Question 4.
Solution:
(i) 5.4 x 16 = 86.4 (One place of decimal)
(ii) 3.65 x 19 = 69.35 (Two place of decimal)
(iii) 0.854 x 12 = 10.2468 (Three place of decimal)
(iv) 36.73 x 48 = 1763.04 (Two places of decimal)

(v) 4.125 x 86=354.750 (Three places of decimal)
= 354.75

Question 5.
Solution:
(i) 7.6 x 2.4= 18.24
{Sum of decimal places = 1 + 1 = 2}

Question 6.
Solution:
(i) 13 x 1.3 x 0.13 = 2.197
{Sum of decimal places = 1 + 2 = 3}

(ii) 2.4 x 1.5 x 2.5 = 9.000 = 9
{Sum of decimal places = 1 + 1 + 1 = 3}
(iii) 0.8 x 3.5 x 0.05 = 0.1400 = 0.14
{Sum of decimal places = 1 + 1 + 2 = 4}
(iv) 0.2 x 0.02 x 0.002 = 0.000008
{Sum of decimal places = 1 + 2 + 3 = 6}
(v) 11.1 x 1.1 x 0.11 = 1.3431
{Sum of decimal places = 1 + 1 + 2 = 4}

(vi) 2.1 x 0.21 x 0.021 = 0.00926
21 x 21 = 441
441 x 21 = 9261
{Sum of decimal places = 1 + 2 + 3 = 6}

Question 7.
Solution:
(i) (1.2)²= 1.2 x 1.2 = 1.44
{Sum of decimal places = 1 + 1 = 2}
(ii) (0.7)² = 0.7 x 0.7 = 0.49
{Sum of decimal places = 1 + 1 = 2}
(iii) (0.04)² = 0.04 x 0.04 = 0.0016
{Sum of decimal places = 2 + 2 = 4}
(iv) (0.11)² = 0.11 x 0.11 =0.0121
{Sum of decimal places = 2 + 2 = 4}

Question 8.
Solution:
(i) (0.3)³ = 0.3 x 0.3 x 0.3 = 0.027
{Sum of decimal places = 1 + 1 + 1 = 3}
(ii) (0.05)³= 0.05 x 0.05 x 0.05 = 0.000125
{Sum of decimal places = 2 + 2 + 2 = 6}
(iii) (1.5)³ = 1.5 x 1.5 x 1.5 = 3.375
{Sum of decimal places = 1 + 1 + 1 = 3}

Question 9.
Solution:
Distance covered in one hour = 62.5 km
Distance covered in 18 hours = 62.5 x 18 km = 1125.0 km

Question 10.
Solution:
Weight of one tin of oil = 16.8 kg
Weight of 45 tins = 16.8 x 45 kg = 756.0 kg = 756 kg

Question 11.
Solution:
Weight of wheat in one bag = 97.8 kg
weight of wheat in 500 bags = 97.8 x 500 kg = 48900.0 kg = 48900 kg

Question 12.
Solution:
Weight of one bag = 48.450 kg
Weight of 16 bags = 48.450 x 16 = 775.200 kg

Question 13.
Solution:
Quantity of sauce in one bottle = 0.845 kg
quantity of sauce in 72 bottles = 0.845 x 72 kg = 60.840 kg

Question 14.
Solution:
Quantity of jam in one bottle = 925 .
Quantity of jam in 25 bottles = 925 x 25 g = 23135 g = 23.125 kg

Question 15.
Solution:
Oil in one drum = 16.850 litres
Oil in 48 drums = 16.850 x 48 = 808.800 = 808.800 litres

Question 16.
Solution:
Cost of 1 kg rice = Rs 56.80
Cost of 16.25 kg of rice = Rs 56.80 x 16.25 = Rs 923.0000 = Rs 923

Question 17.
Solution:
Cost of one metre of cloth = Rs 108.5 0
Costof 18.5 metres of cloth = Rs 108.50 x 18.5 = Rs 2007.250 = Rs 2007.25

Question 18.
Solution:
Distance covered in one litre = 8.6 km
Distance covered in 36.5 litres = 8.6 x 36.5 km = 313.90 km = 313.9 km

Question 19.
Solution:
Charges for 1 km = Rs 9.80
Charges for 106.5 km = Rs 9.80 x 106.5 = Rs 1043.700 = Rs 1043.70

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Add:
Question 1.
Solution:
Converting them into like decimals 16.00, 8.70, 0.94, 6.80 and 7.77
Now, adding them,
16.0 + 8.70 + 0.94 + 6.80 + 7.77 = 40.21

Question 2.
Solution:
Converting them into like decimals 18.600, 206.370, 8.008, 26.400, 6.900
Adding we get
18.600 + 206.370 + 8.008 + 26.400 + 6.900 = 266.278

Question 3.
Solution:
Converting them into like decimals, 63.50, 9.70, 0.80, 26.66, 12.17
Adding we get:
63.50 + 9.70 + 0.80 + 26.66 + 12.17 = 112.83

Question 4.
Solution:
Converting them into like decimals 17.400, 86.390, 9.435, 8.800, 0.060
Adding we get:
17.400 + 86.390 + 9.435 + 8.800 + 0.060 = 122.085

Question 5.
Solution:
Converting them into like decimals 26.900, 19.740, 231.769, 0.048
Now adding we get:
26.900 + 19.740 + 231.769 + 0.048 = 278.457

Question 6.
Solution:
Converting them into like decimals 23.800, 8.940, 0.078 and 214.600
Now adding we get:
23.800 + 8.940 + 0.078 + 214.600 = 247.418

Question 7.
Solution:
Converting them into like decimals.
6.606, 66.600, 666.000,0.066, 0.660
Now adding we get:
6.606 + 66.600 + 666.000 + 0.066 + 0,660 = 739.932

Question 8.
Solution:
9.090, 0.909, 99.900, 9.990, 0.099
Now adding we get:
9.090 + 0.909 + 99.900 + 9.990 + 0.099 = 119.988

Subtract:
Question 9.
Solution:
14.79 from 72.43
72.43 – 14.79 = 57.64

Question 10.
Solution:
Converting them into like decimals, We get
36.74 and 52.60
Now 52.60 – 36.74 = 15.86

Question 11.
Solution:
Converting them into like decimals, We get
13.876 and 22.000
22.000 – 13.876 = 8.124

Question 12.
Solution:
Converting them into like decimals, We get
15.079 and 24.160
24.160 – 15.079 = 9.081

Question 13.
Solution:
Converting them into like decimals We get
0.680 and 1.007
1.007 – 0.680 = 0.327

Question 14.
Solution:
Converting them into like decimals,
We get 0.4678 and 5.0500
5.0500 – 0.4678 = 4.5822

Question 15.
Solution:
Converting them into like decimals,
We get 2.5307 and 8.0000
8.0 – 2.5307 = 5.4693

Question 16.
Solution:
There are like decimals
9.1 – 6.732 = 2.269

Question 17.
Solution:
Converting them into like decimals,
We get 5.746 and 9.100
9.100 – 5.746 = 3.354

Question 18.
Solution:
Converting into like decimals, we get,
63.59 and 92.00
Required number = 92.00 – 63.58 = 28.42

Question 19.
Solution:
Converting into like decimals, we get:
8.100 and 0.813
Required number = 8.100 – 0.813 = 7.287

Question 20.
Solution:
Converting them into like decimals, we get: 32.67 and 60.10
Required number = 60.10 – 32.67 = 27.43

Question 21.
Solution:
Converting into like decimals, we get 74.3 and 26.87
Required number = 74.30 – 26.87 = 47.43

Question 22.
Solution:
Cost of notebook = Rs. 23.75
Cost ofpencil = Rs. 2.85
Costofpen =Rs. 15.90
Total cost = Rs. 42.50
Amount gave to the shop keeper = 50 rupees
Balance amount got = Rs 50.00 – Rs 42.50 = 7.50

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
(i) 6.5, 16.03, 0.274, 119.4
In these decimals, the greatest places of decimal is 3
6.5 = 6.500
16.03 = 16.030
0. 274 = 0.274
119.4 = 119.400 are like decimals.
(ii) 3.5, 0.67, 15.6, 4
In these decimal, the greatest place of decimal is 2
3.5 = 3.50
0.67 = 0.67
15.6 = 15.60
4 = 4.00 are the like decimals

Question 5.
Solution:
(i) Among 78.23 and 69.85,
78.23 is greater than 69.85 (78 > 69)
78.23 > 69.85
(ii) Among 3.406 and 3.46,
3.406 is less than 3.46 (40 < 46)
3.406 < 3.46
(iii) Among 5.68 and 5.86,
5.68 is less than 5.86 (68 < 86)
5.68 < 5.86
(iv) Among 14.05 and 14.005
14.5 is greater than 14.005 (05 > 00)
14.5 >14.005
(v) Among 1.85 and 1.805,
1.85 is greater than 1.805 (85 > 80)
1.85 > 1.805
(vi) Among 0.98 and 1.07,
0.98 is less than 1.07 (0 < 1)
0.98 < 1.07

Question 6.
Solution:
(i) 4.6, 7.4, 4.58, 7.32, 4.06
Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06.
We see that 4.06 < 4.58 < 4.60 < 7.32 < 7.40.
Writing in ascending order, 4.06, 4.58, 4.6, 7.32, 7.4
(ii) 0.5, 5.5, 5.05, 0.05, 5.55
Converting the given decimals into like decimals, we get:
0. 50, 5.50, 5.05, 0.05, 5.55
We see that 0.05 < 0.50 < 5.05 < 5.50 < 5.55.
Writing in ascending order, 0.05, 0.50, 5.05, 5.5, 5.55
(iii) 6.84, 6.84, 6.8, 6.4, 6.08
Converting the given decimals into like decimals
6.84, 6.48, 6.80, 6.40, 6.08
We see that 6.08 < 6.40 < 6.48 < 6.80 < 6.84
Writing in ascending order,
6.08, 6.4, 6.48, 6.8, 6.84
(iv) 2.2, 2.202, 2.02, 22.2, 2.002
Converting them into like decimals
2.200, 2.202, 2.020, 22.200, 2.002 we see that
2.002 < 2.020 < 2.200 < 2.202 < 22.200
Now writing in ascending order,
2.002, 2.020, 2.2, 2.202, 22.2

Question 7.
Solution:
(i) 7.4, 8.34, 74.4, 7.44, 0.74
Converting them into like decimals,
7.40, 8.34, 74.40, 7.44, 0.74
we see that
74.40 > 8.34 > 7.44 > 7.40 > 0.74
Writing in descending order,
74.4, 8.34, 7.44, 7.4, 0.74
(ii) 2.6, 2.26, 2.06, 2.007, 2.3
Converting them into like decimals,
2.600, 2.260, 2.060, 2.007, 2.300
We see that
2.600 > 2.300 > 2.260 > 2.060 > 2.007
Writing in descending order,
2.6, 2.3, 2.26, 2.06, 2.007

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:
(i) A number of the form \(\frac { a }{ b }\), where a and b are rational numbers, is called a natural number.
Here, a is the numerator and b is the denominator.
(a) \(\frac { 2 }{ 3 }\) is a fraction with 2 as the numerator and 3 as the denominator.
(b) \(\frac { 12 }{ 5 }\) is a fraction with 12 as the numerator and 5 as the denominator.
(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples : \(\frac { 2 }{ 5 }\) and \(\frac { 4 }{ 15 }\).
(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Example : \(\frac { 11 }{ 3 }\) and \(\frac { 41 }{ 35 }\)

Question 2.
Solution:
Required number to be added

Hence, the required number is 8\(\frac { 2 }{ 5 }\)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Mark (√) against the correct answer in each of the following:

Question 10.

Solution:
(d) \(\frac { 5 }{ 8 }\)
\(\frac { 5 }{ 8 }\) is a vulgar fraction, because its denominator is other than 10,100, 1000, etc.

Question 11.
Solution:
(c) \(\frac { 46 }{ 63 }\)
A fraction \(\frac { a }{ b }\) is said to be irreducible or in its lowest terms if the HCF of a and b is 1
46 = 2 x 23 x 1
63 = 3 x 3 x 21 x 1
Clearly, the HCF of 46 and 63 is 1.
Hence, \(\frac { 46 }{ 63 }\) is an irreducible fraction.

Question 12.
Solution:
(d) None of these
Reciprocal of 1\(\frac { 3 }{ 5 }\) = Reciprocal of \(\frac { 8 }{ 5 }\) = \(\frac { 5 }{ 8 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
(b) 33 km
Distance covered by the car on

Question 17.
Solution:

Question 18.
Solution:
(i) False.
By cross multiplication, we have:
9 x 24 = 216 and 13 x 16 = 208
However, 216 > 208
\(\frac { 9 }{ 16 }\) > \(\frac { 13 }{ 24 }\)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following:
Question 1.
Solution:
(c)
Denominator in (a) and (b) is 10
These are decimal fractions
But denominator of (c) is 3
\(\frac { 10 }{ 3 }\) is a vulgar fraction

Question 2.
Solution:
(c)
\(\frac { 7 }{ 10 }\) and \(\frac { 7 }{ 9 }\) are proper fractions as each of these have numerator less than its denominator
\(\frac { 9 }{ 7 }\) is improper fraction

Question 3.
Solution:
(a)
\(\frac { 105 }{ 112 }\) is reducible fraction because HCF 112 of 105 and 112 is 7

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)

Question 6.
Solution:
(d)

Question 7.
Solution:
(c)

Question 8.
Solution:
(d)

Question 9.
Solution:
(d)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(c)

Question 13.
Solution:
(d)

Question 14.
Solution:
(d)

Question 15.
Solution:
(b)
The correct statement will be

Question 16.
Solution:
(c)
A car runs in 1 litre of petrol = 16 km

Question 17.
Solution:
(a)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D are helpful to complete your math homework.

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