Free Fractions and Decimals 03 Practice Test - 7th grade

Question 1

$(0.89345 \times 1000) + 8.9345 - 893.45 =$

$89.345$

$8.9345$

$893.45$

$8934.50$

SOLUTION

Solution : B

Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
$0.89345 \times 1000 = 893.45$

$∴ (0.89345 \times 1000) + 8.9345 - 893.45$
$= 893.45 + 8.9345 - 893.45$
$= 0 + 8.9345$
$= 8.9345$

Question 2

$(12.4 + 9.6) \times (0.1 + 0.001)$ _____ $(1.24 \times 10.96) \times 0.2$

$>$

$<$

$=$

$can't say$

SOLUTION

Solution : B

$(12.4 + 9.6) \times (0.1 + 0.001)$

$= 22.0 \times 0.101$

$= 2.222$

And, $(1.24 \times 10.96) \times 0.2$

$= 13.59 \times 0.2$

$= 2.718$

$2.718$ is greater than $2.222$

Therefore, $(12.4 + 9.6) \times (0.1 + 0.001)$ < $(1.24 \times 10.96) \times 0.2$

Question 3

The length of a cuboid is $0.3 m$ . Its breadth and height is $0.03 m$ and $0.003 m$ respectively. The volume of the cuboid is given as the product of its length, breadth and height. The volume of cuboid is:

{0.0027 m}^{3}

2.7 \times 10^{5} m^{3}

2.7 \times 10^{4} m^{3}

0.000027 m^{3}

SOLUTION

Solution : B and D

Given, length = $0.3 m$

breadth = $0.03 m$

height = $0.003 m$

$Volume = Length \times B r e a d t h \times H e i g h t$

$Volume = 0.3 \times 0.03 \times 0.003 m^{3}$

$= 0.000027 m^{3}$

Question 4

$(60 \times 0.06 \times 6) - (0.0006 \times 600 \times 0.6)$ is equal to:

$21.384$

$213.84$

$2.1834$

$20.484$

SOLUTION

Solution : A

$(60 \times 0.06 \times 6) - (0.0006 \times 600 \times 0.6)$

$= (21.6) - (0.216)$

$= 21.384$

Question 5

A bag contains 25 kg of rice at ₹ 20.04 per kg and 20 kg of dal at ₹ 15.28 per kg. What is the total cost of 5 such bags?

$₹ 4033$

$₹ 4038.36$

$₹ 4038.92$

$₹ 4040.40$

SOLUTION

Solution : A

$Cost of rice in one bag = 25 \times 20.04$

$= ₹ 501$

$Cost of dal in one bag = 20 \times 15.28$

$= ₹ 305.6$

$Total cost of one bag = 501 + 305.6$

$= ₹ 806.6$

$Cost of 5 such bags = 5 \times 806.6$

$= ₹ 4033$

Question 6

Convert the following fractions into decimals and find the sum of these decimals that rounded up to $2$ decimal places.

$\frac{1}{2}$ , $\frac{5}{2}$ , $\frac{4}{9}$ , $\frac{5}{20}$ , $\frac{3}{10}$ , $\frac{7}{100}$ , $\frac{100}{2}$ , $\frac{100000}{10000000}$ , $\frac{12}{48}$ , $\frac{9}{99}$ , $\frac{5}{25}$
Ans: ___

SOLUTION

Solution :
$\frac{1}{2} = 0.5$

$\frac{5}{2} = 2.5$

$\frac{4}{9} = 0.4444$

$\frac{5}{20} = 0.25$

$\frac{3}{10} = 0.3$

$\frac{7}{100} = 0.07$

$\frac{100}{2} = 50$

$\frac{100000}{10000000} = 0.01$

$\frac{12}{48} = 0.25$

$\frac{9}{99} = 0.09090909$

$\frac{5}{25} = 0.20$

∴ $Sum = 54.615$

When rounded up to two decimal place,

Then $Sum = 54.62$

Question 7

$\frac{0.01}{0.001} + \frac{0.001}{0.01}$ equals ______.

$10.101$

$100.1$

$10.01$

$10.1$

SOLUTION

Solution : D

$\frac{0.01}{0.001} = 0.01 \times \frac{1}{0.001} = 0.01 \times \frac{1000}{1}$

$= 10$

$\frac{0.001}{0.01} = 0.001 \times \frac{1}{0.01} = 0.001 \times \frac{100}{1}$

$=$ $0.1$

$⟹$ $\frac{0.01}{0.001} + \frac{0.001}{0.01}$ $= 10 + 0.1 = 10.1$

Question 8

On multiplying a decimal number by $100$ , we shift the decimal point (i) ______ places to the (ii) ______ and on dividing a decimal number by $100$ , we shift the decimal point, (iii)_______ places to the (iv) _______.

i) two ii) left iii) two iv) right

i) one ii) right iii) one iv) left

i) two ii) right iii) one iv) left

i) two ii) right iii) two iv) left

SOLUTION

Solution : D

Let's take an example and visualize.

$2.48 \times 100 = 248.0$ [Decimal point shifted two places to the right]

$2480 \div 100 = \frac{2480}{100} = 24.80$ [Decimal point shifted two places to the left]

Question 9

Find the value of $\frac{7}{22}$ correct up to $4$ decimal places.

$0.3183$

$0.3182$

$0.3180$

$0.1830$

SOLUTION

Solution : B

Now, the $5^{t h}$ decimal place is greater than $5$ . So, we add $1$ to the previous digit, which gives $0.3182$

Hence, the required answer is $0.3182$ .

Question 10

4 kg of fruits are shared equally among 5 children. How many grams of fruits does each child get?

700 g

800 g

80 g

70 g

SOLUTION

Solution : B

Total weight of fruits = 4 kg
Total number of children among whom the fruits are to be shared equally = 5
Thus, the total quantity of fruits each child would get = $\frac{4}{5} kg$

= $\frac{4}{5}$ $\times 1000 grams$

= $\frac{4000}{5} g$

= 800 g

Free Fractions and Decimals 03 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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