Important Questions for CBSE Class 7 Maths Chapter 2 - Fractions and Decimals

1 Mark Questions:

1. Convert the given fraction into mixed fraction \[\dfrac{{22}}{7}\].

Ans. A fraction can be written in the form of mixed fraction in the following way:

\[Q\dfrac{R}{D}\] , where Q is the quotient, R is the remainder and D is the divisor of the fraction.

So, \[\dfrac{{22}}{7}\] in mixed fraction form will be \[3\dfrac{1}{7}\] .

2. Write an equivalent fraction of \[\dfrac{9}{{15}}\].

Ans. To find an equivalent fraction, we simply multiply the numerator and denominator of the given fraction with the same number.

A fraction equivalent to \[\dfrac{9}{{15}}\] will be \[\dfrac{9}{{15}} \times \dfrac{2}{2} = \dfrac{{18}}{{30}}\] .

3. Find the value of \[3\dfrac{4}{7} \div 7\].

Ans. We solve the expression \[3\dfrac{4}{7} \div 7\] as follows:

\[ \Rightarrow \dfrac{{25}}{7} \div 7\]

\[ \Rightarrow \dfrac{{25}}{7} \times \dfrac{1}{7}\]

\[ \Rightarrow \dfrac{{25}}{{49}}\]

4. Express \[8\] rupee \[5\] paise in decimal.

Ans. \[8\] rupees and \[5\] paise in decimal form can be written as Rs. \[8.05\].

5. Write the place value of \[5\] in \[498.05\].

Ans. The place value of \[5\] in \[498.05\] is hundredths.

Refer to page 2 - 6 for 2 Mark Questions

6. Find the value of \[\dfrac{5}{6}\] of:

(i) \[30\]

Ans. The value of \[\dfrac{5}{6}\] of \[30\] is:

\[\dfrac{5}{6} \times 30 = 25\]

(ii) \[54\]

Ans. The value of \[\dfrac{5}{6}\] of \[54\] is:

\[\dfrac{5}{6} \times 54 = 45\]

7. Multiply and reduce to lowest form:

(i) \[\dfrac{3}{8} \times \dfrac{4}{9}\]

Ans. Multiplying and simplifying \[\dfrac{3}{8} \times \dfrac{4}{9}\] :

\[ \Rightarrow \dfrac{{12}}{{72}}\]

\[ \Rightarrow \dfrac{1}{6}\]

(ii)  \[\dfrac{{11}}{{10}} \times \dfrac{2}{5}\]

Ans. Multiplying and simplifying \[\dfrac{{11}}{{10}} \times \dfrac{2}{5}\] :

\[ \Rightarrow \dfrac{{22}}{{50}}\]

\[ \Rightarrow \dfrac{{11}}{{25}}\]

8. Multiply and express as mixed fractions:

(i) \[4 \times 6\dfrac{2}{3}\]

Ans. Solving the expression:

\[4 \times 6\dfrac{2}{3}\] 

\[ \Rightarrow 4 \times \dfrac{{20}}{3}\]

\[ \Rightarrow \dfrac{{80}}{3}\]

\[ \Rightarrow 26\dfrac{2}{3}\]

(ii) \[3\dfrac{2}{3} \times 5\]

Ans. Solving the expression:

\[3\dfrac{2}{3} \times 5\] 

\[ \Rightarrow \dfrac{{11}}{3} \times 5\]

\[ \Rightarrow \dfrac{{55}}{3}\]

\[ \Rightarrow 18\dfrac{1}{3}\]

9. Shade:

(i) \[\dfrac{1}{3}\] of the ice creams in box:

Ans. There are total \[9\] ice creams in the box. We have to shade \[\dfrac{1}{3}\] , that is, \[\dfrac{1}{3} \times 9 = 3\] ice creams. 

(ii) \[\dfrac{3}{4}\] of the apples in box:

Ans. There are a total of \[16\] apples in the box. We have to shade \[\dfrac{3}{4}\] , that is, \[\dfrac{3}{4} \times 16 = 12\] apples. 

10. Sarah and Darshan went for a picnic. Their mother gave them a juice bottle of \[3\] litres.

Sarah consumed \[{\dfrac{1}{3}^{rd}}\] of the juice. Darshan consumed the rest.

(a) How much did Sarah drink?

Ans. Total quantity of juice in the bottle is \[3\] litres.

Sarah consumed \[{\dfrac{1}{3}^{rd}}\] of the juice, that is, \[\dfrac{1}{3} \times 3 = 1\] litre.

(b) What fraction of the total quantity did Darshan drink?

Ans. Darshan consumed \[1 - \dfrac{1}{3} = \dfrac{2}{3}\] of the total juice.

11. Find:

(i) \[\dfrac{2}{9} \div 4\]

Ans. Solving:

\[\dfrac{2}{9} \div 4\]

\[ \Rightarrow \dfrac{2}{9} \times \dfrac{1}{4}\]

\[ \Rightarrow \dfrac{2}{{36}}\]

\[ \Rightarrow \dfrac{1}{{18}}\]

(ii) \[\dfrac{{11}}{7} \div 2\]

Ans. Solving:

\[\dfrac{{11}}{7} \div 2\]

\[ \Rightarrow \dfrac{{11}}{7} \times \dfrac{1}{2}\]

\[ \Rightarrow \dfrac{{11}}{{14}}\]

12. Find:

(i) \[\dfrac{{12}}{7} \div \dfrac{3}{{14}}\]

Ans. Solving:

\[\dfrac{{12}}{7} \div \dfrac{3}{{14}}\]

\[ \Rightarrow \dfrac{{12}}{7} \times \dfrac{{14}}{3}\]

\[ \Rightarrow \dfrac{{168}}{{21}}\]

\[ \Rightarrow 8\]

(ii) \[\dfrac{2}{5} \div \dfrac{4}{5}\]

Ans. Solving:

\[\dfrac{2}{5} \div \dfrac{4}{5}\]

\[ \Rightarrow \dfrac{2}{5} \times \dfrac{5}{4}\]

\[ \Rightarrow \dfrac{2}{4}\]

\[ \Rightarrow \dfrac{1}{2}\]

13. Which is greater:

(i) \[{\bf{0}}.{\bf{02}}\] or \[{\bf{0}}.{\bf{2}}\]

Ans. We convert the decimals into equivalent fractions:

\[0.02 = \dfrac{2}{{100}}\] and \[0.2 = \dfrac{{20}}{{100}}\]

On comparing, we conclude that \[\dfrac{{20}}{{100}} > \dfrac{2}{{100}}\] .

Hence, \[0.2\] is greater.

(ii) \[1.98\] or \[1.98\]

Ans. We convert the decimals into equivalent fractions:

\[1.98 = \dfrac{{198}}{{100}}\] and \[1.89 = \dfrac{{189}}{{100}}\]

On comparing, we conclude that \[\dfrac{{198}}{{100}} > \dfrac{{189}}{{100}}\] .

Hence, \[1.98\] is greater.

14. How much \[5.6\] kg is less than \[9.4\] kg?

Ans. Calculating the difference:

\[9.4 - 5.6 = 3.8\] kg

Hence, \[5.6\] kg is \[3.8\] kg less than \[9.4\] kg.

15. Find:

(i) \[1.08 \times 0.3\]

Ans. Converting into fractions and solving:

\[1.08 \times 0.3\]

\[ \Rightarrow \dfrac{{108}}{{100}} \times \dfrac{3}{{10}}\]

\[ \Rightarrow \dfrac{{324}}{{1000}}\]

\[ \Rightarrow 0.324\]

(ii) \[158.3 \times 2.9\]

Ans. Converting into fractions and solving:

\[158.3 \times 2.9\]

\[ \Rightarrow \dfrac{{1583}}{{10}} \times \dfrac{{29}}{{10}}\]

\[ \Rightarrow \dfrac{{45907}}{{100}}\]

\[ \Rightarrow 459.07\]

3 Mark Questions

16. Arrange in ascending order \[\dfrac{3}{5},\dfrac{4}{7},\dfrac{3}{{10}},\dfrac{4}{5}\] .

Ans. To arrange given fractions in ascending order, we first make their denominators equivalent.

L.C.M. of all the denominators \[5,7,10\] is \[70\] .

\[\dfrac{3}{5} \times \dfrac{{14}}{{14}} = \dfrac{{42}}{{70}}\]

\[\dfrac{4}{7} \times \dfrac{{10}}{{10}} = \dfrac{{40}}{{70}}\]

\[\dfrac{3}{{10}} \times \dfrac{7}{7} = \dfrac{{21}}{{70}}\]

\[\dfrac{4}{5} \times \dfrac{{14}}{{14}} = \dfrac{{56}}{{70}}\]

Now, \[\dfrac{{21}}{{70}} < \dfrac{{40}}{{70}} < \dfrac{{42}}{{70}} < \dfrac{{56}}{{70}}\] .

Hence, \[\dfrac{3}{{10}} < \dfrac{4}{7} < \dfrac{3}{5} < \dfrac{4}{5}\] .

17. Find the perimeter of the rectangle whose length is \[7\dfrac{3}{{10}}\] cm and breadth is \[\dfrac{3}{5}\] cm.

Ans. We know that the perimeter of a rectangle is twice the sum of its length and breadth, that is, \[2(l + b)\] .

\[ \Rightarrow 2(7\dfrac{3}{{10}} + \dfrac{3}{5})\]

\[ \Rightarrow 2(\dfrac{{73}}{{10}} + \dfrac{6}{{10}})\]

\[ \Rightarrow 2(\dfrac{{79}}{{10}})\]

\[ \Rightarrow \dfrac{{79}}{5}\]

\[ \Rightarrow 15\dfrac{4}{5}\] cm

18. Find \[\dfrac{3}{7}\] of:

(i) \[3\dfrac{5}{8}\]

Ans. The value of \[\dfrac{3}{7}\] of \[3\dfrac{5}{8}\] is:

\[\dfrac{3}{7} \times 3\dfrac{5}{8} = \dfrac{3}{7} \times \dfrac{{29}}{8}\]

\[ \Rightarrow \dfrac{{87}}{{56}}\]

\[ \Rightarrow 1\dfrac{{31}}{{56}}\]

(ii) \[4\dfrac{3}{9}\]

Ans. The value of \[\dfrac{3}{7}\] of \[4\dfrac{3}{9}\] is:

\[\dfrac{3}{7} \times 4\dfrac{3}{9} = \dfrac{3}{7} \times \dfrac{{39}}{9}\]

\[ \Rightarrow \dfrac{{13}}{7}\]

\[ \Rightarrow 1\dfrac{6}{7}\]

19. Find:

(i) \[3\dfrac{1}{8} \div 2\dfrac{1}{4}\]

Ans. Solving:

\[3\dfrac{1}{8} \div 2\dfrac{1}{4}\]

\[ \Rightarrow \dfrac{{25}}{8} \times \dfrac{4}{9}\]

\[ \Rightarrow \dfrac{{25}}{{18}}\]

\[ \Rightarrow 1\dfrac{7}{{18}}\]

(ii) \[4\dfrac{4}{3} \div 6\dfrac{1}{2}\]

Ans. Solving:

\[4\dfrac{4}{3} \div 6\dfrac{1}{2}\]

\[ \Rightarrow \dfrac{{16}}{3} \times \dfrac{2}{{13}}\]

\[ \Rightarrow \dfrac{{32}}{{39}}\]

20. Write the following decimal number in expanded form:

(i) \[208.183\]

Ans. In expanded form, the given decimal can be written as:

\[(2 \times 100) + (0 \times 10) + (8 \times 1) + (1 \times \dfrac{1}{{10}}) + (8 \times \dfrac{1}{{100}}) + (3 \times \dfrac{1}{{1000}})\]

(ii) \[5.018\]

Ans. In expanded form, the given decimal can be written as:

\[(5 \times 1) + (0 \times \dfrac{1}{{10}}) + (1 \times \dfrac{1}{{100}}) + (8 \times \dfrac{1}{{1000}})\]

(iii) \[360.05\]

Ans. In expanded form, the given decimal can be written as:

\[(6 \times 100) + (3 \times 10) + (0 \times 1) + (0 \times \dfrac{1}{{10}}) + (5 \times \dfrac{1}{{100}})\]

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