RS Aggarwal 2019,2020 solution class 7 chapter 3 Decimals Test Paper 3

Test Paper 3

Page-53

Question 1:

If the cost of a pen is Rs 32.50, find the cost of 24 such pens.

Answer 1:

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

Question 2:

A bus can cover 64.5 km in an hour. How much distance can it cover in 18 hours?

Answer 2:

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

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

Question 3:

Find the product 0.68 × 6.5 × 0.04.

Answer 3:

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

∴ 0.68 $\times$ 6.5 $\times$ 0.04 = 0.17680
                                   = 0.1768

Question 4:

Each bag of cement weighs 48.5 kg. How many such bags will weigh 2231 kg?

Answer 4:

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

Number of bags = $\frac{\mathrm{Total} \mathrm{weight}}{\mathrm{Weight} \mathrm{of} \mathrm{each} \mathrm{bag}}$

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

                           = $\left(\frac{2231\times 10}{48.5\times 10}\right)=\frac{22310}{485}= 46$

Hence, 46 bags of cement will weigh 2231 kg.

Question 5:

Divide:
 
(i) 0.196 by 1.4
(ii) 39.168 by 1.2
(iii) 0.228 by 0.38

Answer 5:

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

    

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

      

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

      

Question 6:

The product of two decimals is 1.824. If one of them is 0.64, find the other.

Answer 6:

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

Hence, the other decimal is 2.85.

Question 7:

How many pieces of plywood, each of 0.45 cm thick, are required to make a pile 2.43 m high?

Answer 7:

Thickness of the pile of plywoods = 2.43 m = 2.43 $\times$ 100 cm = 243 cm
Thickness of one piece of plywood = 0.45 cm
∴ Required number of pieces of plywood = $\frac{243}{0.45}=\frac{243\times 100}{0.45\times 100}=\frac{24300}{45}=540$
   
   
Hence, the required number of pieces of plywood is 540.

Question 8:

Each side of a polygon is 3.8 cm in length and its perimeter is 22.8 cm. How many sides does the polygon have?

Answer 8:

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 $\times$ n) cm
But it is given that its perimeter is 22.8 cm.
∴ (3.8 $\times$ n) cm = 22.8 cm
⇒ n = $\left(\frac{22.8}{3.8}\right)$ = $\frac{228}{38}$ = 6

Hence, the given polygon has six sides.

Question 9:

 Mark (✓) against the correct answer
$2\frac{1}{25}=?$

(a) 2.4
(b) 2.04
(c) 2.004
(d) none of these

Answer 9:

(b) 2.04

$2\frac{1}{25}=\frac{51}{25} = 2.04$

Question 10:

 Mark (✓) against the correct answer
1.008 = ?

(a) $1\frac{2}{25}$
(b) $1\frac{1}{125}$
(c) $1\frac{2}{125}$
(d) none of these

Answer 10:

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

1.008 = $\frac{1.008 \times 1000}{1 \times 1000}=\frac{1008}{1000}=\frac{1008 ÷ 4}{1000 ÷ 4}=\frac{126}{125}=1\frac{1}{125}$

Question 11:

 Mark (✓) against the correct answer
2 kg 5 g = ?

(a) 2.5 kg
(b) 2.05 kg
(c) 2.005 kg
(d) none of these

Answer 11:

(c) 2.005 kg

2 kg 5 g = (2 $\times$ 1000) g + 5 g = (2005) g

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

Question 12:

 Mark (✓) against the correct answer
0.12 ÷ .15 = ?

(a) 0.8
(b) 0.08
(c) 0.008
(d) none of these

Answer 12:

(b) 0.08

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

Question 13:

 Mark (✓) against the correct answer
1.1 × .1 × .01 = ?

(a) .11
(b) .011
(c) .0011
(d) none of these

Answer 13:

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

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

Question 14:

Mark (✓) against the correct answer
4.669 ÷ 2.3 = ?

(a) 2.3
(b) 2.03
(c) 2.003
(d) none of these

Answer 14:

(b) 2.03

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

Question 15:

Mark (✓) against the correct answer
What should be added to 2.06 to get 3.1?

(a) 1.4
(b) 1.24
(c) 1.04
(d) none of these

Answer 15:

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.

Question 16:

Mark (✓) against the correct answer
What should be subtracted from .1 to get .04?

(a) 0.6
(b) 0.06
(c) 0.006
(d) none of these

Answer 16:

(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:

Fill in the blanks.

(i) 1.001 ÷ 14 = ......
(ii) 204 ÷ 0.17 = ......
(iii) 0.47 × 5.3 = ......
(iv) 0.023 × 0.03 = ......
(v) (0.7)² = ......
(vi) (0.05)³ = ......

Answer 17:

(i) 1.001 ÷ 14 = 0.0715

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

(ii) 204 ÷ 0.17 = 1200

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

(iii) 0.47 × 5.3 = 2.491

       Explanation: First, we will multiply 47 by 53.
                              
        ∴ 47 $\times$ 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
     

Question 18:

Write 'T' for true and 'F' for false

(i) 0.5 × 0.05 = 0.25
(ii) 0.25 × 0.8 = 0.2
(iii) 0.35 ÷ 0.7 = 0.5
(iv) .4 × .4 × .4 = 0.64
(v) 6 cm = 0.06 m

Answer 18:

(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 = $\frac{0.35}{0.7}=\frac{0.35\times 10}{0.7\times 10}=\frac{3.5}{7}=0.5$

(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