**NCERT Exemplar Class 6 Maths Chapter 4 Fractions and Decimals** are part of NCERT Exemplar Class 6 Maths. Here we have given NCERT Exemplar Class 6 Maths Solutions Chapter 4 Fractions and Decimals.

## NCERT Exemplar Class 6 Maths Solutions Chapter 4 Fractions and Decimals

Directions: In questions 1 to 20, out of the four options, only one answer is correct. Choose the correct answer.

Question 1.

The fraction which is not equal to \(\frac{4}{5}\) is

(A) \(\frac{40}{50}\)

(B)\(\frac{12}{15}\)

(C) \(\frac{16}{20}\)

(D) \(\frac{9}{15}\)

Solution:

(D) : Since, \(\frac{9}{15}=\frac{9 \div 3}{15 \div 3}=\frac{3}{5}\), which is not equal to \(\frac{4}{5}\).

∴ \(\frac{9}{15}\) is the required fraction.

Remaining cloth = 20.05 m – 4.50 m = 15.55 m

Therefore, 15.55 m of cloth is left with Tina.

Question 2.

The two consecutive integers between which the fraction \(\frac{5}{7}\) lies are

(A) 5 and 6

(B) 0and1

(C) 5 and 7

(D) 6 and 7

Solution:

(B) : Since, we know that a proper fraction whose numerator is less than its denominator always lies between 0 and 1.

∴ \(\frac{5}{7}\), which is also a proper fraction, must lie between 0 and 1.

Question 3.

When \(\frac{1}{4}\) is written with denominator as 12, its numerator is

(A) 3

(B) 8

(C) 24

(D) 12

Solution:

(A): Since, \(\frac{1}{4}=\frac{1}{4} \times \frac{3}{3}\)

[By multiplying the numerator and denominator by 3 to get 12 as the denominator]

\(=\frac{3}{12}\), which shows that the required numerator is 3.

Question 4.

Which of the following is not in the lowest form?

(A) \(\frac{7}{5}\)

(B) \(\frac{15}{20}\)

(C) \(\frac{13}{33}\)

(D) \(\frac{27}{28}\)

Solution:

(B) : In case of \(\frac{7}{5}\), the common factor of 7 and 5 is 1.

i.e., already given in the simplest form.

Similarly,

In \(\frac{15}{20}\), the common factor of 15 and 20 is 4

In \(\frac{13}{33}\) the common factor of 13 and 33 is 1

i.e., already given in the simplest form

In \(\frac{27}{28}\), the common factor of 27 and 28 is 1

i.e., already given in the simplest form.

So, the above cases show that \(\frac{15}{20}\) is not in the lowest form.

Question 5.

If \(\frac{5}{8}=\frac{20}{p}\), then value of p is

(A) 23

(B) 2

(C) 32

(D) 16

Solution:

(C) : we have given, \(\frac{5}{8}=\frac{20}{p}\)

To find the value of p, we must multiply the numerator and denominator of \(\frac{5}{8}\) by 4

∴ The value of p is 32.

Question 6.

Which of the following is not equal to the others?

(A) \(\frac{6}{8}\)

(B) \(\frac{12}{16}\)

(C) \(\frac{15}{25}\)

(D) \(\frac{18}{24}\)

Solution:

(C) : Firstly, we will simplify all the fractions into their lowest form, we get,

Thus, on comparing the above cases, we observe that \(\frac{3}{5}=\frac{15}{25}\) is not equal to the above given fractions.

Question 7.

Which of the following fractions is the greatest?

(A) \(\frac{5}{7}\)

(B) \(\frac{5}{6}\)

(C) \(\frac{5}{9}\)

(D) \(\frac{5}{8}\)

Solution:

(B): We have given, \(\frac{5}{7}, \frac{5}{6}, \frac{5}{9} \text { and } \frac{5}{8}\)

Since we know that, among all the fractions with the same numerator, the one with smaller denominator will be greatest.

i.e., \(\frac{5}{6}\) is the required fraction.

Question 8.

Which of the following fractions is the smallest ?

(A) \(\frac{7}{8}\)

(B) \(\frac{9}{8}\)

(C) \(\frac{3}{8}\)

(D) \(\frac{5}{8}\)

Solution:

We have given, \(\frac{7}{8}, \frac{9}{8}, \frac{3}{8} \text { and } \frac{5}{8}\)

since we know that, among all the fractions with the same denominator, the one with smaller numerator will be smallest i.e., \(\frac{3}{8}\) is the required fraction

Question 9.

Sum of \(\frac{4}{17}\) and \(\frac{15}{17}\) is

(A) \(\frac{19}{17}\)

(B) \(\frac{11}{17}\)

(C) \(\frac{19}{34}\)

(D) \(\frac{2}{17}\)

Solution:

(A) : we have \(\frac{4}{12}+\frac{15}{17}\)

Question 10.

On subtracting \(\frac{5}{9}\) from \(\frac{19}{9}\) the result is

(A) \(\frac{24}{9}\)

(B) \(\frac{14}{9}\)

(C) \(\frac{14}{18}\)

(D) \(\frac{14}{0}\)

Solution:

(B) According to question, \(\frac{19}{9}-\frac{5}{9}\)

Question 11.

0.7499 lies between

(A) 0.7 and 0.74

(B) 0.75 and 0.79

(C) 0.749 and 0.75

(D) 0.74992 and 0.75

Solution:

(C) On observing the given options, we notice that 0.7499 lies between 0.749 and 0.75.

[As 0.7499 is successor of 0.749 and the predecessor of 0.75]

Question 12.

0.023 lies between

(A) 0.2 and 0.3

(B) 0.02 and 0.03

(C) 0.03 and 0.029

(D) 0.026 and 0.024

Solution:

(B) : On observing the given options, we notice that 0.023 lies between 0.02 and 0.03

Question 13.

\(\frac{11}{7}\) can be expressed in the form

(A) \(7 \frac{1}{4}\)

(B) \(4 \frac{1}{7}\)

(C) \(1 \frac{4}{7}\)

(D) \(11 \frac{1}{7}\)

Solution:

(C) : Dividing 11 by 7, we get the quotient = 1 and the remainder = 4

Question 14.

The mixed fraction \(5 \frac{4}{7}\) can be expressed as

(A) \(\frac{33}{7}\)

(B) \(\frac{39}{7}\)

(C) \(\frac{33}{4}\)

(D) \(\frac{39}{4}\)

Solution:

(B) :

we have, \(5 \frac{4}{7}=\frac{(5 \times 7)+4}{7}=\frac{35+4}{7}=\frac{39}{7}\)

Question 15.

0.07 + 0.008 is equal to

(A) 0.15

(B) 0.015

(C) 0.078

(D) 0.78

Solution:

we have to find the value of 0.07 + 0.008

First converting the given decimals into like decimals, we get 0.070 and 0.008

Writing these decimals in column form and adding, we get

Question 16.

Which of the following decimals is the greatest ?

(A) 0.182

(B) 0.0925

(C) 0.29

(D) 0.038

Solution:

(C) : Firstly converting the given decimals in like decimals, we get 0.1820, 0.0925, 0.2900 and 0.0380

Thus, on comparing the above, 0.2900 is the greatest among the given decimals.

Question 17.

Which of the following decimals is the smallest ?

(A) 0.27

(B) 1.5

(C) 0.082

(D) 0.103

Solution:

(C) : Firstly converting the given decimals in like decimals, we get 0.270, 1.500, 0.082 and 0.103

Thus, on comparing the above, 0.082 is the smallest among the given decimals.

Question 18.

13.572 correct to the tenths place is

(A) 10

(B) 13.57

(C) 14.5

(D) 13.6

Solution:

(D) : When 13.572 is estimated correct to the tenths place, we get 13.6

Question 19.

15.8 – 6.73 is equal to

(A) 8.07

(B) 9.07

(C) 9.13

(D) 9.25

Solution:

(B) : We have given, 15.8 – 6.73.

Firstly converting the given decimals into like fractions, then writing the decimals in column form and subtracting, we get

Question 20.

The decimals 0.238 is equal to the fraction

(A) \(\frac{119}{500}\)

(B) \(\frac{238}{25}\)

(C) \(\frac{119}{25}\)

(D) \(\frac{119}{50}\)

Solutions:

(A) : We have given 0.238, which can be written in the form of

Directions : In questions 21 to 44, fill in the blanks to make the statement true

Question 21.

A number representing a part of a ____ is called a fraction.

Solution:

Whole

Question 22.

A fraction with denominator greater than the numerator is called a ____ fraction.

Solution:

Proper

Question 23.

Fractions with the same denominator are called _____ fractions

Solution:

Like

Question 24.

\(13 \frac{5}{18}\) is a ____ fraction

Solution:

Mixed

Question 25.

\(\frac{18}{5}\) is an ____ fraction

Solution:

Improper

Question 26.

\(\frac{7}{19}\) is a ___ fraction.

Solution:

Proper

Question 27.

\(\frac{5}{8} \text { and } \frac{3}{8}\) are ____ proper fractions.

Solution:

Like

Question 28.

\(\frac{6}{11} \text { and } \frac{6}{13}\) are ___ proper fractions

Solution:

Unlike

Question 29.

The fraction \(\frac{6}{15}\) in simplest form is ____.

Solution:

\(\frac{2}{5} : \text { We have, } \frac{6}{15}=\frac{6 \div 3}{15 \div 3}=\frac{2}{5}\)

Question 30.

The fraction \(\frac{17}{34}\) in simplest form is

Solution:

\(\frac{1}{2} : \text { We have, } \frac{17}{34}=\frac{17 \div 17}{34 \div 17}=\frac{1}{2}\)

Question 31.

\(\frac{18}{135} \text { and } \frac{90}{675}\) are proper, unlike and ____ fractions.

Solution:

Equivalent

Question 32.

\(8 \frac{2}{7}\) is equal to the improper fraction ___.

Solution:

\(\frac{58}{7} : 8 \frac{2}{7}=\frac{(8 \times 7)+2}{7}=\frac{56+2}{7}=\frac{58}{7}\)

Question 33.

\(\frac{87}{7}\) is equal to the mixed fraction ___.

Solution:

\(12 \frac{3}{7} :\)

Question 34.

\(9+\frac{2}{10}+\frac{6}{100}\) is equal to the decimal number ____.

Solution:

9.26:

Question 35.

Decimal 16.25 is equal to the fraction ___ .

Solution:

\(\frac{65}{4} : \text { We have, } 16.25=\frac{1625}{100}=\frac{1625 \div 25}{100 \div 25}\)

\(=\frac{65}{4} \text { or } 16 \frac{1}{4}\)

Question 36.

Fraction \(\frac{7}{25}\) is equal to the decimal number ____.

Solution:

\(0.28 : \text { We have, } \frac{7}{25}=\frac{7}{25} \times \frac{4}{4}=\frac{28}{100}=0.28\)

Question 37.

\(\frac{17}{9}+\frac{41}{9}=\) ____

Solution:

\(\frac{58}{9} : \frac{17}{9}+\frac{41}{9}=\frac{17+41}{9}=\frac{58}{9}\)

Question 38.

\(\frac{67}{14}-\frac{24}{14}=\) _____

Solution:

\(\frac{43}{14} : \frac{67}{14}-\frac{24}{14}=\frac{67-24}{14}=\frac{43}{14}\)

Question 39.

\(\frac{17}{2}+3 \frac{1}{2}=\) ____.

Solution:

\(12 : \frac{17}{2}+3 \frac{1}{2}=\frac{17}{2}+\frac{7}{2}=\frac{17+7}{2}=\frac{24}{2}=12\)

Question 40.

\(9 \frac{1}{4}-\frac{5}{4}=\) ____.

Solution:

\(8 : 9 \frac{1}{4}-\frac{5}{4}=\frac{37}{4}-\frac{5}{4}=\frac{37-5}{4}=\frac{32}{4}=8\)

Question 41.

4.55 + 9.73 = ____.

Solution:

14.28 : 4.55 + 9.73 = 14.28

Question 42.

8.76 – 2.68 = ____ .

Solution:

6.08 : 8.76 – 2.68 = 6.08

Question 43.

The value of 50 coins of 50 paisa = ₹ ____

Solution:

25 : The value of 50 coins of 50 paisa

= 50 × 50 paisa

\(=2500 \text { paisa }=\mathrm{Rs} . \frac{2500}{100}\)

= Rs. 25

Question 44.

3 Hundredths + 3 tenths = _____.

Solution:

0.33 : 3 hundredths + 3 tenths = 0.33

Directions: In each of the questions 45 to 65, state whether the statement is true or false.

Question 45.

Fractions with same numerator are called like fractions.

Solution:

False

Because, fractions with same denominators are called like fractions.

Question 46.

Fraction \(\frac{18}{39}\) is in its lowest form

Solution:

False

Since, the common factor of 18 and 39 is 3 and its simplest form is

\(\frac{18 \div 3}{39 \div 3}=\frac{6}{13}\)

Question 47.

Fractions \(\frac{15}{39} \text { and } \frac{45}{117}\) are equivalent fractions.

Solution:

True

Since, \(\frac{45 \div 3}{117 \div 3}=\frac{15}{39}\)

∴ Both are equivalent

Question 48.

The sum of two fractions is always a fraction.

Solution:

True

Let \(\frac{2}{3} \text { and } \frac{4}{3}\) are two fractions.

When we add them, we get \(\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}\)

\(=\frac{6}{3}=\frac{2}{1},\) which is a fraction.

Question 49.

The result obtained by subtracting a fraction from another fraction is necessarily a fraction.

Solution:

False

Let \(\frac{1}{2} \text { and } \frac{2}{3}\) are two fractions.

When we subtract \(\frac{2}{3} \text { from } \frac{1}{2}\), we get

\(\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=\frac{-1}{6}\)

Which is not a fraction.

Question 50.

If a whole or an object is divided into a number of equal parts, then each part represents a fraction.

Solution:

True

Question 51.

The place value of a digit at the tenths place is 10 times the same digit at the ones place.

Solution:

False

Since, the place value of a digit at the tenths place is \(\frac{1}{10}\) times the same digit at the ones place.

i.e., \(\frac{x}{10}=10 x\), which is not possible

Question 52.

The place value of a digit at the hundredths place is \(\frac{1}{10}\) times the same digit at the tenths place.

Solution:

True

Question 53.

The decimal 3.725 is equal to 3.72 correct to two decimal places.

Solution:

False

Since the digit at thousandths place of 3.725 is 5

3.725 can be written as 3.73 (after rounding off at hundredths place)

Question 54.

In the decimal form, fraction \(\frac{25}{8}=3.125\).

Solution:

True

[On multiplying the numerator and denominator by 125]

\(=\frac{3125}{1000}=3.125\)

Question 55.

The decimal \(23.2=23 \frac{2}{5}\)

Solution:

False

we have, \(23.2=\frac{232}{10}=\frac{116}{5}\)

\(=23 \frac{1}{5}\)

Question 56.

The fraction represented by the shaded portion in the adjoining figure is \(\frac{3}{8}\).

Solution:

True

Total number of parts in a given figure is 8 and shaded parts is 3.

∴ The required fraction is \(\frac{3}{8}\).

Question 57.

The fraction represented by the unshaded portion in the adjoining figure is \(\frac{5}{9}\).

Solution:

False

Total number of parts in a given figure is 9 from which unshaded parts is 4.

4

∴ The required fraction is \(\frac{4}{9}\).

Question 58.

\(\frac{25}{19}+\frac{6}{19}=\frac{31}{38}\)

Solution:

False

\(\frac{25}{19}+\frac{6}{19}=\frac{25+6}{19}=\frac{31}{19} \neq \frac{31}{38}\)

∴ The above condition is false.

Question 59.

\(\frac{8}{18}-\frac{8}{15}=\frac{8}{3}\)

Solution:

False

∵ The L.C.M. of 18 and 15 is 2 × 3 × 3 × 5 = 90

∴ The above condition is false

Question 60.

\(\frac{7}{12}+\frac{11}{12}=\frac{3}{2}\)

Solution:

True

\(\frac{7}{12}+\frac{11}{12}=\frac{7+11}{12}=\frac{18}{12}=\frac{18 \div 6}{12 \div 6}=\frac{3}{2}\)

Question 61.

3.03 + 0.016 = 3.019

Solution:

False

Firstly convert 3.03 and 0.016 into like fractions, writing the decimals in column form and finally by adding we get,

Question 62.

42.28 – 3.19 = 39.09

Solution:

True

Question 63.

\(\frac{16}{25}>\frac{13}{25}\)

Solution:

True

The given fractions are like fractions. On comparing the numerators, we get

\(\frac{16}{25}>\frac{13}{25}\)

Question 64.

19.25 < 19.053

Solution:

False

Since, the digit at tenth place of 19.25 is 2 and the digit at tenth place of 19.053 is 0, where 2 > 0.

∴ 19.25 >19.053

Question 65.

13.730 = 13.73

Solution:

True

Since, after converting the given decimals in like decimals we get, 13.730 = 13.73.

Directions: In each of the questions 66 to 71, fill in the blanks using ’<’, ‘>’ or ‘=’

Question 66.

\(\frac{11}{16} \dots \frac{14}{15}\)

Solution:

< : The L.C.M. of 16 and 15 is 2 × 2 × 2 × 2 × 3 × 5 = 240

Thus, \(\frac{11}{16}=\frac{11 \times 15}{16 \times 15}=\frac{165}{240}\)

and \(\frac{14}{15}=\frac{14 \times 16}{15 \times 16}=\frac{224}{240}\)

On comparing, we observe that

\(\frac{224}{240}>\frac{165}{240}, i . e ., \frac{11}{16}<\frac{14}{15}\)

Question 67.

\(\frac{8}{15} \dots \frac{95}{14}\)

Solution:

< : The L.C.M of 15 and 14 is

2 × 3 × 5 × 7 = 120

Question 68.

\(\frac{12}{75} \dots \frac{32}{200}\)

Solution:

= : The L.C.M of 75 and 200 is 2 × 2 × 2 × 3 × 5 × 5 = 600

Question 69.

3.25 … 3.4

Solution:

< : Converting the given decimals into like decimals, they become 3.25 and 3.40. The whole number part of these is same. On comparing, we get their tenths digits 2 < 4

∴ 3.25 < 3.4

Question 70.

\(\frac{18}{15} \ldots 1.3\)

Solution:

< : \(1.3=\frac{13}{10}\)

∵ The L.C.M. of 15 and 10 is 2 × 3 × 5 = 30

Now, \(\frac{18}{15}=\frac{18 \times 2}{15 \times 2}=\frac{36}{30} \text { and } \frac{13}{10}=\frac{13 \times 3}{10 \times 3}=\frac{39}{30}\)

Thus, \(\frac{18}{15}<1.3\)

Question 71.

\(6.25 \dots \frac{25}{4}\)

Solution:

=: ∵ \(6.25=\frac{625}{100}=\frac{625 \div 25}{100 \div 25}=\frac{25}{4}\)

Question 72.

Write the fraction represented by the shaded portion of the adjoining figure:

Solution:

In the given figure, total parts in which figure has been divided is 8 and out of which 7 parts are shaded.

∴ The required fraction is \(\frac{7}{8}\).

Question 73.

Write the fraction represented by the unshaded portion of the adjoining figure:

Solution:

In the given figure, total parts in which figure has been divided is 15 and out of which 4 parts are unshaded.

∴ The required fraction is \(\frac{4}{15}\).

Question 74.

Ali divided one fruit cake equally among six persons. What part of the cake he gave to each person?

Solution:

Since, Ali has to divide one fruit cake equally among 6 persons

∴ Each person will get \(\frac{1}{6}\) part.

Question 75.

Arrange 12.142, 12.124, 12.104, 12.401 and 12.214 in ascending order.

Solution:

∵ The digits are already given in the form of like decimals.

Clearly,

12.104 < 12.124 < 12.142 < 12.214 < 12.401

Question 76.

Write the largest four digit decimal number less than 1 using the digits 1,5,3 and 8 once.

Solution:

The required number is 0.8531, which is the largest four digit decimal number less than 1.

Question 77.

Using the digits 2, 4, 5 and 3 once, write the smallest four digit decimal number.

Solution:

The required number is 0.2345, which is the smallest four digit decimal number.

Question 78.

Express \(\frac{11}{20}\) as a decimal.

Solution:

We have, \(\frac{11}{20}=\frac{11 \times 5}{20 \times 5}=\frac{55}{100}=0.55\)

Question 79.

Express \(6 \frac{2}{3}\) as an improper fraction.

Solution:

We have, \(6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Question 80.

Express \(3 \frac{2}{5}\) as a decimal.

Solution:

We have, \(3 \frac{2}{5}=\frac{(3 \times 5)+2}{5}=\frac{15+2}{5}=\frac{17}{5}\)

Now, \(\frac{17}{5}=\frac{17 \times 2}{5 \times 2}=\frac{34}{10}=3.4\)

Question 81.

Express 0.041 as a fraction.

Solution:

We have, \(0.041=\frac{41}{1000}\)

Question 82.

Express 6.03 as a mixed fraction.

Solution:

We have, \(6.03=\frac{603}{100}\)

\(=6 \frac{3}{100}\)

Question 83.

Convert 5201 g to kg.

Solution:

We have, \(5201 \mathrm{g}=\frac{5201}{1000} \mathrm{kg}\)

[∵ 1 kg = 1000 g]

= 5.201 kg

Question 84.

Convert 2009 paise to rupees and express the result as a mixed fraction.

Solution:

We have,

\(2009 \text { paise }=\mathrm{Rs} . \frac{2009}{100}=\mathrm{Rs} .20 .09\)

\(\quad \mathrm{Rs} .20 \frac{9}{100}\)

Question 85.

Convert 1537 cm to m and express the result as an improper fraction.

Solution:

We have, \(1537 \mathrm{cm}=\frac{1537}{100} \mathrm{m}\)

[∵ 1 m = 100 cm]

= 15.37 m

Question 86.

Convert 2435 m to km and express the result as mixed fraction.

Solution:

We have, \(2435 \mathrm{m}=\frac{2435}{1000} \mathrm{km}\)

[ ∵ 1 km = 1000 m]

= 2.435 km

Firstly, convert the fraction \(\frac{2435}{1000}\) into the simplest form, for this dividing the numerator and denominator by 5, we get

\(\frac{2435 \div 5}{1000 \div 5} \mathrm{km}=\frac{487}{200} \mathrm{km}\)

i.e., \(\quad 2 \frac{87}{200} \mathrm{km}\)

Question 87.

Arrange the fractions \(\frac{2}{3}, \frac{3}{4}, \frac{1}{2} \text { and } \frac{5}{6}\) in ascending order.

Solution:

We have given, \(\frac{2}{3}, \frac{3}{4}, \frac{1}{2} \text { and } \frac{5}{6}\)

Firstly find the L.C.M. of 3, 4, 2 and 6.

The L.C.M. of 3, 4, 2 and 6 is 2 × 2 × 3 = 12

Question 88.

Arrange the fractions \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4}\) in descending order.

Solution:

We have given, \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4}\).

Question 89.

Write \(\frac{3}{4}\) as a fraction with denominator 44.

Solution:

Let \(\frac{3}{4}=\frac{?}{44}\)

Then, we have to find the missing numeral.

To get 44 in the denominator, we multiply 4 by 11.

So, we multiply the numerator and denominator by 11.

Hence, \(\frac{3}{4} \text { and } \frac{33}{44}\) are equivalent fractions.

Question 90.

Write \(\frac{5}{6}\) as a fraction with numerator 60.

Solution:

Let \(\frac{5}{6}=\frac{60}{?}\)

Then, we have to find the missing numeral. To get 60 in the numerator, we multiply 5 by 12.

So, we multiply the numerator and denominator by 12.

Hence, \(\frac{5}{6} \text { and } \frac{60}{72}\) are equivalent fractions.

Question 91.

Write \(\frac{129}{8}\) as a mixed fraction.

Solution:

We have, \(\frac{129}{8}=16 \frac{1}{8}\)

Question 92.

Round off 20.83 to nearest tenths.

Solution:

The estimated value of 20.83 to the nearest tenths is 20.8

Question 93.

Round off 75.195 to nearest hundredths.

Solution:

The estimated value of 75.195 to the nearest hundredths is 75.20

Question 94.

Round off 27.981 to nearest tenths.

Solution:

The estimated value of 27.981 to the nearest tenths is 28.0

Question 95.

Add the fractions \(\frac{3}{8} \text { and } \frac{2}{3}\).

Solution:

L.C.M. of 8 and 3 is 2 × 2 × 2 × 3 = 24

Now, \(\frac{3}{8}+\frac{2}{3}=\frac{3 \times 3}{8 \times 3}+\frac{2 \times 8}{3 \times 8}\)

\(=\frac{9}{24}+\frac{16}{24}=\frac{9+16}{24}=\frac{25}{24}\)

Question 96.

Add the fractions \(\frac{3}{8} \text { and } 6 \frac{3}{4}\).

Solution:

L.C.M of 8 and 4 is 2 × 2 × 2 = 8

Question 97.

Subtract \(\frac{1}{6} \mathrm{from} \frac{1}{2}\).

Solution:

The L.C.M of 6 and 2 = 6

Now, \(\frac{1}{2}-\frac{1}{6}=\frac{1 \times 3}{2 \times 3}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}\)

\(=\frac{3-1}{6}=\frac{2}{6}=\frac{2 \div 2}{6 \div 2}\)

[∵ H.C.F of 2 and 6 is 2]

\(=\frac{1}{3}\)

Question 98.

Subtract \(8 \frac{1}{3} \text { from } \frac{100}{9}\).

Solution:

The L.C.M of 3 and 9 = 9

Now, \(\frac{100}{9}-8 \frac{1}{3}=\frac{100}{9}-\frac{25}{3}\)

\(=\frac{100}{9}-\frac{25 \times 3}{3 \times 3}=\frac{100}{9}-\frac{75}{9}=\frac{100-75}{9}=\frac{25}{9}\)

\(=2 \frac{7}{9}\)

Question 99.

Subtract \(1 \frac{1}{4} \text { from } 6 \frac{1}{2}\).

Solution:

The L.C.M of 4 and 2 = 4

Now, \(6 \frac{1}{2}-1 \frac{1}{4}=\frac{13}{2}-\frac{5}{4}\)

\(=\frac{13 \times 2}{2 \times 2}-\frac{5}{4}=\frac{26}{4}-\frac{5}{4}=\frac{26-5}{4}\)

\(=\frac{21}{4}=5 \frac{1}{4}\)

Question 100.

Add \(1 \frac{1}{4} \text { and } 6 \frac{1}{2}\).

Solution:

The L.C.M. of 4 and 2 = 4

Now, \(\frac{1}{2}-\frac{1}{6}=\frac{1 \times 3}{2 \times 3}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}\)

Question 101.

Katrina rode her bicycle \(6 \frac{1}{2} \mathrm{km}\) in the morning and \(8 \frac{3}{4} \mathrm{km}\) in the evening. Find the distance travelled by her altogether on that day.

Solution:

The distance travelled by Katrina in the morning \(=6 \frac{1}{2} \mathrm{km}=\frac{13}{2} \mathrm{km}\)

The distance travelled by Katrina in the evening \(=8 \frac{3}{4} \mathrm{km}=\frac{35}{4} \mathrm{km}\)

∴ Total distance travelled by her

Question 102.

A rectangle is divided into certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\), find the number of parts in which the rectangle has been divided.

Solution:

Let the number of parts in which the rectangle has been divided be x.

According to question, \(\frac{1}{4}=\frac{16}{x}\)

By cross-multiplication, x = 16 × 4 = 64

∴ The required number of parts is 64.

Question 103.

Grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{cm}\). Express the size as an improper fraction.

Solution:

We have given, a grip size of a tennis racquet \(=11 \frac{9}{80} \mathrm{cm}=\frac{889}{80} \mathrm{cm}\), which is the required improper fraction.

Question 104.

On an average \(\frac{1}{10}\) of the food eaten is turned into organism’s own body and is available for the next level of consumer in a food chain. What fraction of the food eaten is not available for the next level?

Solution:

We have given, \(\frac{1}{10}\) of the food eaten is turned into organism’s own body.

∴ The required fraction of the food eaten not available for the next level is \(1-\frac{1}{10}=\frac{10-1}{10}\)

[∵ L.C.M. of 1 and 10 is 10]

\(=\frac{9}{10}\)

Question 105.

Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years. What fraction of his age till retirement was he in the job?

Solution:

Mr. Rajan got a job at the age of 24 years.

He got retired at the age of 60 years.

He worked for (60 – 24) years = 36 years

∴ The required fraction \(=\frac{36}{60}=\frac{36 \div 12}{60 \div 12}=\frac{3}{5}\)

[ ∵ H.C.F. of 36 and 60 is 12]

Question 106.

The food we eat remains in the stomach for a maximum of 4 hours. For what fraction of a day, does it remain there?

Solution:

The food we eat remains in the stomach for a maximum of 4 hours.

Total number of hours in a day = 24 hours

∴ The required fraction \(=\frac{4}{24}=\frac{4 \div 4}{24 \div 4}=\frac{1}{6}\)

[∵ H.C.F of 4 and 24 is 4]

Question 107.

What should be added to 25.5 to get 50?

Solution:

Question 108.

Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.

Solution:

Alok purchased,

Potatoes = 1 kg 200 g = 1.200 kg

Dhania = 250 g = 0.250 kg

Onion = 5 kg 300 g = 5.300 kg

Palak = 500 g = 0.500 kg

Tomatoes = 2 kg 600 g = 2.600 kg

∴ The total weight of the above purchases

Question 109.

Arrange in ascending order:

0.011, 1.001, 0.101, 0.110

Solution:

Since, all the decimals are already given in like fractions, i.e., 0.011, 1.001, 0.101, 0.110

∴ Arranging them in ascending order, we get

0.011, 0.101, 0.110, 1.001

Question 110.

Add the following:

20.02 and 2.002

Solution:

We have, 20.02 and 2.002

To add the above decimals, we must convert them into like decimals first.

Writing 20.020 and 2.002 in a column

So,

which is the required sum.

Question 111.

It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel and 21000 tonnes of petrol was saved by the end of year 2007. Find the fraction of:

(i) the quantity of diesel saved to the quantity of petrol saved.

(ii) the quantity of diesel saved to the quantity of CNG saved.

Solution:

By the end of year 2007,

The quantity of CNG saved 33000 tonnes,

The quantity of diesel saved 3300 tonnes and The quantity of petrol saved 21000 tonnes

Question 112.

Energy content of different foods are as follows:

Food | Energy content per kg. |

Wheat | 3.2 Joules |

Rice | 5.3 Joules |

Potatoes(Cooked) | 3.7 Joules |

Milk | 3.0 Joules |

Which food provides the least energy and which provides the maximum?

Express the least energy as a fraction of the maximum energy.

Solution:

Milk provides the least energy and rice provides the maximum energy.

∴ The required fraction \(=\frac{3.0}{5.3}=\frac{3.0 \times 10}{5.3 \times 10}\)

\(=\frac{30}{53}\)

Question 113.

A cup is \(\frac{1}{3}\) full of milk. What part of the cup is still to be filled by milk to make it full?

Solution:

A cup is \(\frac{1}{3}\) full of milk.

∴ The remaining part of the cup which is still to be filled by milk \(=1-\frac{1}{3}\)

\(=\frac{3-1}{3}\)

[∵ L.C.M of 1 and 3 is 3]

\(=\frac{2}{3}\)

Question 114.

Mary bought \(3 \frac{1}{2} m\) of lace. She used \(1 \frac{3}{4} m\) of lace for her new dress. How much lace is left with her ?

Solution:

Mary bought the lace \(=3 \frac{1}{2} \mathrm{m}=\frac{7}{2} \mathrm{m}\)

Lace used by Mary \(=1 \frac{3}{4} \mathrm{m}=\frac{7}{4} \mathrm{m}\)

∴ She is left with \(\frac{7}{2} m-\frac{7}{4} m\)

\(=\frac{7 \times 2}{2 \times 2} \mathrm{m}-\frac{7}{4} \mathrm{m}=\frac{14 \mathrm{m}-7 \mathrm{m}}{4}\)

[ ∵ L.C.M. of 2 and 4 is 4]

\(=\frac{7}{4} \mathrm{m}=1 \frac{3}{4} \mathrm{m}\) of lace

Question 115.

When Sunita weighed herself on Monday, she found that she had gained \(1 \frac{1}{4}\) kg. Earlier her weight was \(46 \frac{3}{8}\) kg. What was her weight on Monday?

Solution:

Sunita had gained \(=1 \frac{1}{4} \mathrm{kg}=\frac{5}{4} \mathrm{kg}\)

Earlier her weight was \(46 \frac{3}{8} k g=\frac{371}{8} k g\)

∴ Her total weight on Monday

Question 116.

Sunil purchased \(12 \frac{1}{2}\) litres of juice on Monday and \(14 \frac{3}{4}\) litres of juice on Tuesday. How many litres of juice did he purchase together in two days?

Solution:

Sunil purchased juice on Monday

\(=12 \frac{1}{2}\) litres \(=\frac{25}{2}\) litres

∴ Total quantity of juice Sunil purchased in two days \(=\left(\frac{25}{2}+\frac{59}{4}\right)\) litres

Question 117.

Nazima gave \(2 \frac{3}{4}\) litres out of the \(5 \frac{1}{2}\) litres of juice she purchased to her friends. How many litres of juice is left with her?

Solution:

Total quantity of juice = \(5 \frac{1}{2}\) litres

= \(\frac{11}{2}\) litres

Nazima gave to her friends = \(2 \frac{3}{4}\) litres

= \(\frac{11}{4}\) litres

∴ The required quantity of juice she is left

Question 118.

Roma gave a wooden board of length \(150 \frac{1}{4}\) cm to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5}\) cm from it. What is the length of the remaining piece?

Solution:

Total length of a wooden board

\(=150 \frac{1}{4} \mathrm{cm}=\frac{601}{4} \mathrm{cm}\)

The carpenter sawed off a piece of length

\(=40 \frac{1}{5} \mathrm{cm}=\frac{201}{5} \mathrm{cm}\)

∴ The length of the remaining piece

Question 119.

Nasir travelled \(3 \frac{1}{2}\) km in a bus and then walked \(1 \frac{1}{8}\) km to reach a town. How much did he travel to reach the town?

Solution:

Nasir travelled by a bus \(=3 \frac{1}{2}\) km

\(=\frac{7}{2} \mathrm{km}\)

Nasir walked = \(1 \frac{1}{8}\) km = \(\frac{9}{8}\) km

∴ Total distance travelled by him

Question 120.

The fish caught by Neetu was of weight \(3 \frac{3}{4}\) kg and the fish caught by Narendra was of weight \(2 \frac{1}{2}\) kg. How much more did Neetu’s fish weigh than that of Narendra?

Solution:

The weight of fish caught by Neetu

\(=3 \frac{3}{4} \mathrm{kg}=\frac{15}{4} \mathrm{kg}\)

The weight of fish caught by Narendra

\(=2 \frac{1}{2} \mathrm{kg}=\frac{5}{2} \mathrm{kg}\)

∴ Neetu’s fish weigh more than that of Narendra by

Question 121.

Neelam’s father needs \(1 \frac{3}{4}\)m of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2}\) m for the scarf. How much cloth must he buy in all?

Solution:

Neelam’s father purchased the length of cloth for the skirt \(=1 \frac{3}{4} \mathrm{m}=\frac{7}{4} \mathrm{m}\) and for the scarf \(=\frac{1}{2} \mathrm{m}\)

∴ Total length he buys in all \(=\frac{7}{4} \mathrm{m}+\frac{1}{2} \mathrm{m}\)

Question 122.

What is wrong in the following additions?

Solution:

(a) Equal denominators are added.

(b) Numerators and denominators are added.

Question 123.

Which one is greater?

1 metre 40 centimetres + 60 centimetres or 2.6 metres.

Solution:

1 metre 40 centimetres + 60 centimetres = 1.40 metres + 0.60 metres

[ 100 centimetres = 1 metre]

= 2.00 metres

Since, 2.6 > 2.00

∴ 2.6 metres is greater.

Question 124.

Match the fractions of Column I with the shaded or marked portion of figures of Column II:

Solution:

(i) ➝ (D) ; (ii) ➝ (A) ; (iii) ➝ (E) ; (iv) ➝ (B)

Marked point in (A) \(=\frac{6}{10}\)

Shaded fraction in (B) \(=\frac{6}{16}\)

Shaded fraction in (C) \(=\frac{6}{7}\)

Shaded fraction in (D) \(=\frac{4}{4}+\frac{2}{4}=\frac{6}{4}\)

Shaded fraction in (E) \(=\frac{6}{6}\)

Question 125.

Find the fraction that represents the number of natural numbers to total numbers in the collection 0, 1, 2, 3, 4, 5. What fraction will it be for whole numbers?

Solution:

Out of 0, 1, 2, 3, 4, 5 ➝ 1, 2, 3, 4 and 5 are the natural numbers.

∴ The fraction that represents the number of natural numbers to the total numbers \(=\frac{5}{6}\) and the whole numbers are 0,1, 2, 3, 4 and 5.

∴ The fraction that represents the number of whole numbers to the total numbers \(=\frac{6}{6}\)

Question 126.

Write the fraction representing the total number of natural numbers in the collection of numbers -3, -2, -1,0,1,2, 3. What fraction will it be for whole numbers? What fraction will it be for integers?

Solution:

Out of -3, -2, -1, 0, 1, 2, 3 -> 1, 2 and 3 are the natural numbers, 0,1, 2 and 3 are the whole numbers and -3, -2, -1, 0, 1, 2, 3 are integers.

∴ The fraction representing the natural numbers to the total numbers \(=\frac{3}{7}\)

The fraction representing the whole numbers to the total numbers \(=\frac{4}{7}\)

And the required fraction representing the integers to the total numbers \(=\frac{7}{7}\)

Question 127.

Write a pair of fractions whose sum is \(\frac{7}{11}\) and difference is \(\frac{2}{11}\).

Solution:

Let one fraction be x.

Another fraction be \(\frac{7}{11}-x\)

Now, according to question,

Thus, one fraction is \(\frac{9}{22}\) and another fraction is

\(\frac{7}{11}-\frac{9}{22}=\frac{7 \times 2}{11 \times 2}-\frac{9}{22}=\frac{14-9}{22}=\frac{5}{22}\)

Question 128.

What fraction of a straight angle is a right angle?

Solution:

Since, we know that the measurement of a straight angle is 180° and a right angle is 90°.

∴ The required fraction is \(\frac{90^{\circ}}{180^{\circ}}=\frac{1}{2}\).

Question 129.

Put the right card in the right bag.

Solution:

We know that if numerator is less than the denominator, then the fraction is less than 1.

If numerator is equal to the denominator, then the fraction is equal to 1 and if numerator is greater than the denominator, then the fraction is greater than 1.

Cards in Bag I are

(i) \(\frac{3}{7}\),

(iv) \(\frac{8}{9}\)

(v) \(\frac{5}{6}\)

(vi) \(\frac{6}{11}\)

(viii) \(\frac{19}{25}\)

(ix) \(\frac{2}{3}\) and

(x) \(\frac{13}{17}\)

Cards in Bag II are (ii) \(\frac{4}{4}\) , (vii) \(\frac{18}{18}\)

And cards in Bag III are (iii) \(\frac{9}{8}\)

We hope the NCERT Exemplar Class 6 Maths Chapter 4 Fractions and Decimals will help you. If you have any query regarding NCERT Exemplar Class 6 Maths Solutions Chapter 4 Fractions and Decimals, drop a comment below and we will get back to you at the earliest.