# Fractions ICSE Class-6th Concise Selina Mathematics

**Fractions ICSE Class-6th Concise** Mathematics Selina Solutions Chapter-14. We provide step by step Solutions of Exercise / lesson-14 **Fraction** for **ICSE Class-6 Concise** Selina Mathematics. Our Solutions contain all type Questions of Exe-14 A, Exe-14 B, Exe-14 C, Exe-14 D, Exe-14 E and Exe-14 F to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 .

**Fractions ICSE Class-6th Concise** Mathematics Selina Solutions Chapter-14

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### Exercise – 14 A **Fractions ICSE Class-6th Concise** Mathematics

**Question -1.**

For each expression, given below, write a fraction :

(i) 2 out of 7 = ……….

(ii) 5 out of 17 = ……

(iii) three-fifths = ………

**Answer-1**

i) 2 out of 7 = .^{2}⁄_{7}

(ii) 5 out of 17 = ^{5}⁄_{7}

(iii) three-fifths = ^{3}⁄_{5}

**Question -2.**

Fill in the blanks :

**Answer-2**

(i) Proper

(ii) Improper

(iii) Improper

(iv) 1

(v) -1

(vi) Mixed

(vii) Like

(viii) Unlike fraction

(ix) Equal fraction

(x) Like

(xi) +2 = ^{41}⁄_{13}

(xii) – ^{(4×5+3)}⁄_{5}

= – ^{23}⁄_{5}

**Question-3**

From the following fractions, separate :

(i) Proper fractions

(ii) Improper fractions :

**Answer-3**

We know that proper fraction is a fraction whose numerator is less than its denominator and improper fraction is the fraction whose numerator is greater them its denominator

(i) Proper fractions:– ^{2}⁄_{9 , }^{7}⁄_{15 ,}^{11}⁄_{20 , }^{18}⁄_{23 and }^{27}⁄_{35}

(ii) Improper fractions :– ^{4}⁄_{3 , }^{20}⁄_{11}

**Question -4.**

Change the following mixed fractions to improper fractions :

**Answer -4**

**Question- 5.**

Change the following improper fractions to mixed fractions :

**Answer-5**

**Question -6.**

Change the following groups of fractions to like fractions :

**Answer-6**

= ^{32}⁄_{112 } ^{98}⁄_{112}, ^{40}⁄_{112}, ^{63}⁄_{112}

### Exercise – 14 B **Fractions Selina Solutions for ICSE Class-6th ** Mathematics

**Question -1.**

Reduce the given fractions to their lowest terms :

**Answer -1**

**Question -2.**

State, whether true or false?

**Answer-2**

**Question -3.**

Which fraction is greater?

**Answer-3**

**Question- 4.**

Which fraction is smaller ?

**Answer-4**

**Question- 5.**

Arrange the given fractions in descending order of magnitude :

**Answer-5**

(iii) …. ^{5}⁄_{7}….,….. ^{8}⁄_{8}…….. ^{9}⁄_{11}…

^{ LCM 0f Numerator 5 , 3, 9 = 45}

If numerator same then fraction with smallest denominator is the biggest fraction

hence

^{45}⁄_{55}….,….. ^{45}⁄_{63}……… ^{45}⁄_{120}….

ie.- ^{9}⁄_{11}….,….. ^{5}⁄_{7}……….. ^{3}⁄_{8}….

Note – (iii) part can be also solved by taking LCM of denominator as above 1st and 2nd part but due to Prime number of 11 and 7 the LCM is too big.

**Question -6.**

Arrange the given fractions in ascending order of magnitude :

**Answer-6**

**(i) ^{9}⁄_{16}….,….. ^{7}⁄_{12}……….. ^{1}⁄_{4}….**

LCM of 16,12,and 4 = 48

^{28}⁄

_{48}….,…..

^{28}⁄

_{48}………….

^{12}⁄

_{48}..

Arrange the fractions in ascending order

….^{12}⁄_{48}..<…..^{28}⁄_{48}….<…….^{28}⁄_{48}……

ie ..^{1}⁄_{4}..,<..^{9}⁄_{16}..<. ^{7}⁄_{12}…..

**(ii) ^{5}⁄_{6}….,….. ^{2}⁄_{7}…..^{8}⁄_{9}……. ^{1}⁄_{3}….**

LCM of 6, 7, 9 and 3 = 126

^{105}⁄_{126}….,^{36}⁄_{126}….. ^{112}⁄_{126}……..^{42}⁄_{126}..

Arrange the fractions in ascending order

….^{36}⁄_{126}..<…..^{42}⁄_{126}……<…….^{105}⁄_{126}…..^{112}⁄_{126}…

ie ..^{2}⁄_{7}…,<..^{1}⁄_{3}….<. ^{5}⁄_{6}….^{8}⁄_{9}…

**(iii) ^{2}⁄_{3}….,…^{5}⁄_{9}.. …..^{5}⁄_{6}……. ^{3}⁄_{8}….**

LCM of 3, 9, 6 and 8 = 72

^{48}⁄_{72}….,^{40}⁄_{72}….. ^{60}⁄_{72}……..^{27}⁄_{72}..

Arrange the fractions in ascending order

….^{27}⁄_{72}.<…..^{40}⁄_{72}…….<…….^{48}⁄_{72}…..^{60}⁄_{72}…

ie ..^{3}⁄_{8}…,<..^{5}⁄_{9}….<. .^{2}⁄_{3}….^{5}⁄_{6}…

**Question -7.**

**Answer-7**

Number of bananas bought = 1

Dozen = 12

Number of bananas eaten by me = 5

Number of bananas left = 12 – 5 = 7

Fraction = ^{7}⁄_{12}.

**Question- 8.**

Insert the symbol ‘=’ or ‘>’ or ‘<’ between each of the pairs of fractions, given below :

**Answer-8**

**(i) …. ^{6}⁄_{11}………^{5}⁄_{9}…….**

LCM of 11 and 9 = 99

….^{54}⁄_{99}………^{55}⁄_{99}…..

….^{54}⁄_{99}……<…^{55}⁄_{99}…………..

ie. ….^{6}⁄_{11}….<…..^{5}⁄_{9}…………..

**(ii) …. ^{3}⁄_{7}………^{9}⁄_{13}…………..**

LCM of 7 and 13 = 91

….^{39}⁄_{91}………^{63}⁄_{91}…..

^{39}⁄_{91}……<…^{63}⁄_{91}…..

ie ….^{3}⁄_{7}…..<….^{9}⁄_{13}…………..

**(iii) …. ^{56}⁄_{64}………^{7}⁄_{8}…………..**

LCM of 64 and 8 = 64

….^{56}⁄_{64}………^{56}⁄_{64}…..

.^{56}⁄_{64}….=…..^{56}⁄_{64}…..

ie ….^{56}⁄_{64}….=…..^{7}⁄_{8}…………..

**(iv) …. ^{5}⁄_{12}………^{8}⁄_{33}…………..**

LCM of 12 and 33 = 132

….^{55}⁄_{132}………^{32}⁄_{132}…..

….^{55}⁄_{132}…..>….^{32}⁄_{132}…..

.ie…^{5}⁄_{12}….>…..^{8}⁄_{33}…….

**Question -9.**

Out of 50 identical articles, 36 are broken. Find the fraction of :

(i) The total number of articles and the articles broken.

(ii) The remaining articles and total number of articles.

**Answer-9**

Total number of articles = 50

Number of articles broken = 36

Remaining articles = 50 – 36 = 14

Now

(i) the fraction of the total number of articles and articles broken

= ..^{50}⁄_{36}

= ^{25}⁄_{18}

(ii) The remaining articles and total number of articles.=

= ..^{14}⁄_{50}

= ^{7}⁄_{25}

**Fractions ICSE Class-6th Concise** Mathematics Selina Solutions Exercise – 14 C

**Question- 1.**

Add the following fractions :

**Answer-1**

(i) ……1.^{3}⁄_{4}…….and ^{3}⁄_{8}….

….^{7}⁄_{4}…….and ^{3}⁄_{8}..

LCM of 4and 8 = 8

…….^{14}⁄_{8}…….+ ^{3}⁄_{8}……..= ……^{17}⁄_{8}……

(ii)

(i) …….^{2}⁄_{5}……2..^{3}⁄_{15}……..^{7}⁄_{10}…….

……..^{2}⁄_{5}……+..^{33}⁄_{15}….+….^{7}⁄_{10}……

LCM of 5 , 15 and 10 = 30

…..^{12}⁄_{30}……+..^{66}⁄_{30}….+….^{21}⁄_{30}……

= …………..^{12+66+21}⁄_{30}…………………….

= ……^{99}⁄_{30}…….^{33}⁄_{10}

(iii)

(iv)

(vi)

**Question -2.**

Simplify :

**Answer-2**

**Fractions Exe-14 D for ICSE Class-6th **Mathematics Selina Solutions

**Question- 1.**

Simplify :

**Answer-1**

**Question -2.**

Simplify :

**Answer-2**

**Question -3.**

Simplify :

**Answer-3**

**Question- 4.**

Simplify :

**Answer-4**

**= … ^{1}⁄_{12}..x.. ^{1}⁄_{96}….=…. ^{1}⁄_{1152}….**….

**Question -5.**

Simplify :

**Answer-5**

### Selina Solutions of Exercise – 14 E **Fractions for ICSE Class-6th Concise** Mathematics

**Question -1.**

From a rope of 10^{1}⁄_{2} m long, 4 ^{5}⁄_{8} m is cut off. Find the length of the remaining rope.

**Answer-1**

Length of rope = 10^{1}⁄_{2} m

Length of cut off. rope = 4 ^{5}⁄_{8}

Length of the remaining rope = Length of rope – Length of cut off. rope

= 10^{1}⁄_{2} m – 4 ^{5}⁄_{8}

= ^{21}⁄_{2}

= ….^{21}⁄_{2}….- ^{37}⁄_{8}…..

=

**Question -2.**

A piece of cloth is 5 metre long. After washing, it shrinks by ^{1}⁄_{25} of its length. What is the length of the cloth after washing?

**Answer-2**

Length of piece of cloth is = 5 metre

After washing, it shrinks = ^{1}⁄_{25} of its length

length of the cloth after washing = 5 metre – ^{1}⁄_{25} of its 5

**Question -3.**

I bought wheat worth Rs. 12^{1}⁄_{2}, rice worth Rs. 25^{3}⁄_{4} and vegetables worth Rs. 10^{1}⁄_{4}. If I gave a hundred-rupee note to the shopkeeper ; how much did he return to me

**Answer-3**

wheat worth = Rs. 12^{1}⁄_{2},

rice worth Rs. 25^{3}⁄_{4}

vegetables worth Rs. 10^{1}⁄_{4}

Total Amount = (wheat+rice+vegetables)

Total Amount = (..12^{1}⁄_{2},..+….Rs. 25^{3}⁄_{4}…+…10^{1}⁄_{4}….)

= ^{194}⁄_{4},

**Question -4.**

Out of 500 oranges in a box, ^{3}⁄_{25}, are rotten and ^{1}⁄_{5}, are kept for some guests. How many oranges are left in the box?

**Answer-4**

Number of oranges = 500

rotten oranges = ^{3}⁄_{25}, of 500 = 60

= 60

kept for some guests = .^{1}⁄_{5}, of..500..= 100…

oranges are left in the box = 500-60-100 = 340

**Question -5.**

An ornament piece is made of gold and copper. Its total weight is 96g. If ^{1}⁄_{12} of the ornament is copper, find the weight of gold in it.

**Answer-5**

total weight is = 96 g

copper = ^{1}⁄_{12} of 96 g = 8 g

weight of gold = 96 g – 8 g = 88 g

**Question -6.**

A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work?

**Answer-6**

Let total Work be = 1

work on Monday = ^{1}⁄_{2}

work on Tuesday = ^{1}⁄_{3}

work on Wednesday = 1- (.work on Monday +….work on Tuesday..)

**Question- 7.**

A man spends ^{3}⁄_{8} of his money and 8 still has Rs. 720 left with him. How much money did he have at first?

**Answer-7**

Let Money be =1

spends = ^{3}⁄_{8} of his money

money Left = 1 – ^{3}⁄_{8}

= ^{5}⁄_{8}

According to condition given = Rs. 720 left

…..^{5}⁄_{8}…of 1 .= ..720.

= ^{720 x 8}⁄_{5}

= 1152

**Question -8.**

In a school, ^{4}⁄_{5} of the students are boys, and the number of girls is 100. Find the number of boys.

**Answer-8**

Number of boy = Total strength – number of girl

Number of boy = 500 – 100 = 400

**Question- 9.**

After finishing ^{3}⁄_{4} of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

**Answer-9**

Let the journey = x km

Distance Left = Total dis -distance cover

= 16 – 12= 4 km

**Question -10.**

When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey?

**Answer-10**

Let Length of journey = x

journey travelled =15 km

**Question -11.**

In a particular month, a man earns Rs. 7,200. Out of this income, he spends ^{3}⁄_{10} on food, ^{1}⁄_{4} on house rent, ^{1}⁄_{10} on insurance and ^{2}⁄_{25} on holidays. How much did he save in that month ?

**Answer-11**

particular month, a man earns = Rs. 7,200

he spends on food, = ^{3}⁄_{10} of 7,200 = 2160

on house rent = ^{1}⁄_{4} of 7,200 = 1800

Money spend on insurance= ^{1}⁄_{10 }of 7200 = 720 _{ }

on holidays = ^{2}⁄_{25} of 7200 = 576

Total amount spent = (2160+1800 + 720 + 576)

= 5256 Rs

Amount saved = 7200- 5256

= Rs 1944

### Exercise – 14 F Selina Solutions for Class-6 ICSE Mathematics

**Question- 1.**

**Answer-1**

**Question -2.**

**Answer-2**

**Question -3.**

Evaluate :

**Answer-3**

**Question -4.**

Mr. Mehra gave one-third of his money to his son, one-fifth of his money to his daughter and the remaining amount to his wife. If his wife got Rs. 91,000, how much money did Mr. Mehra have originally?

**Answer-4**

**Question -5.**

A sum of Rs. 84,000 is divided among three persons A, B and C. If A gets one-fourth of it and B gets one-fifth of it; how much did C get?

**Answer-5**

= Rs. 84,000 – (37,800) = Rs. 46,200

**Question- 6.**

In one hour Rohit walks 3 km. How much distance will he cover in 2 hours?

**Answer-6**

**Question -7.**

An 84 m long string is cut into pieces each of length 5 m. How many pieces are obtained?

**Answer-7**

**Question- 8.**

In buying a ready made shirt-two-fifths of my pocket money is spent If Rs. 540 is still left with me, find :

(i) The money I had before I bought the shirt.

(ii) The emit of the shirt

**Answer-8**

**Question -9.**

Mohan leaves Rs. 1,20,000 to his wife and three children such that two-fifths of this money is given to his wife and the remaining is distributed equally among the children. Find, how much each child gets?

**Answer-9**

Total amount = Rs. 12,0,000

Amount given to his wife = 2/5 of Rs. 1,20,000

= Rs. 2 x 24,000 = Rs. 48,000

Remaining amount = Rs. 120000 – Rs. 48000 = Rs. 72000

This amount is distributed among three children equally.

Each’s share = Rs. 72,000 x 1/3 = Rs. 24,000

**Question -10.**

Simplify :

**Answer-10**

— End of **Fractions ICSE Class-6th** Solutions :–

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