## RS Aggarwal Class 7 Solutions Chapter 3 Decimals CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper.

**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:**

Cost of 1 pen = ₹ 32.50

Cost of 24 such pens = ₹ (32.50 x 24) = ₹ 80

Hence, the cost of 24 pens is ₹ 780.

**Question 2.**

**Solution:**

Distance covered by the bus in 1 hour = 64.5 km

Distance covered in 18h = (64.5 x 18) km = 1161 km

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

**Question 3.**

**Solution:**

First, we will find the product 68 x 65 x 4

Now, 68 x 65 x 4 = 4420 x 4 = 17680

Sum of decimal places in the given decimals = (2 + 1 + 2) = 5

So, the product have five decimal places.

0.68 x 6.5 x 0.04 = 0.17680 = 0.1768

**Question 4.**

**Solution:**

Total weight of all the bags = 2231 kg

Weight of each bag = 48.5 kg

**Question 5.**

**Solution:**

**Question 6.**

**Solution:**

Product of the given decimals = 1.824

One decimal = 0.64

The other decimal = 1.824 ÷ 0.64

Hence, the other decimal is 2.85.

**Question 7.**

**Solution:**

Thickness of the pile of plywoods = 2.43 m = 2.43 x 100 cm = 243 cm

Thickness of one piece of plywood = 0.45 cm

Required no. of pieces of plywood

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

**Question 8.**

**Solution:**

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 x n) cm

But it is given that its perimeter is 22.8 cm.

Hence, the given polygon has six sides.

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

**Question 9.**

**Solution:**

(b) 2.04

**Question 10.**

**Solution:**

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

**Question 11.**

**Solution:**

(c) 2.005 kg

2 kg 5 g = (2 x 1000) g + 5 g = (2005)g

= \(\frac { 2005 }{ 1000 }\) kg = 2.005 kg

**Question 12.**

**Solution:**

(b) 0.08

We have :

**Question 13.**

**Solution:**

(c) 0.011

First, we will find the product 11 x 1 x 1

i.e. 11 x 1 x 1 = 11 x 1 = 11

Sum of decimal places in the given decimals = (1 + 1 + 2) = 4

1.1 x 0.1 x 0.01 = 0.0011 [4 places of decimal]

**Question 14.**

**Solution:**

(b) 2.03

**Question 15.**

**Solution:**

(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.

**Question 16.**

**Solution:**

(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.

**Question 17.**

**Solution:**

(i) 1.001 ÷ 14 = 0.0715

47 x 53 = 2491

Sum of decimal places in the given decimals = (2 + 1) = 3

0.47 x 5.3 = 2.491

(iv) 0.023 x 0.03 = 0.69

Explanation first, we will multiply 23 by 3

23 x 3 = 69

Sum of decimal places in the given decimals = (3 + 2) = 5

0.023 x 0.03 =0.00069

(v) (0.7)^{2} = 0.69

Explanation : (0.7)^{2} = 0.7 x 0.7

First, we will find the product 0.7 x 0.7

Now, 7 x 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 x 0.7 = 0.49

(vi) (0.05)^{3} = 0.000125

Explanation : First, we will find the

product 0.05 x 0.05 x 0.05

Now, 5 x 5 x 5 = 125

Sum of decimal places in the given decimals = (2 + 2 + 2) = 6

So, the product must have six decimal places.

(0.05)^{2} = 0.05 x 0.05 x 0.05 = 0.000125

**Question 18.**

**Solution:**

(i) False

We have :

0.5 x 0.05 Now, 5 x 5 = 25

Sum of decimal places in the given decimals = (1 + 2) = 3

0.5 x 0.5 = 0.025

(ii) True

We have :

0.25 x 0.8

Now, 25 x 8 = 200

Sum of decimal places in the given decimals = (2 + 1) = 3

0.25 x 0.8 = 0.200 = 0.2

(iii) True

We have :

0.35 ÷ 0.7

(iv) False We have :

0.4 x 0.4 x 0.4

Now, 4 x 4 x 4 = 64

Sum of decimal places in the given decimals = (1 + 1 + 1) = 3

0.4 x 0.4 x 0.4 = 0.064

(v) True

6 cm = \(\frac { 6 }{ 100 }\) m = 0.06 m

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper are helpful to complete your math homework.

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