## 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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