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  for detail information about ICSE Board Class-6 .

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

–: Select Topics :–

Exe-14 A,

Exe-14 B,

Exe-14 C,

Exe-14 D,

Exe-14 E,

### 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 = ………

i) 2 out of 7 = .27
(ii) 5 out of 17 = 57
(iii) three-fifths = 35

#### Question -2.

Fill in the blanks :

(i) Proper
(ii) Improper
(iii) Improper
(iv) 1
(v) -1
(vi) Mixed
(vii) Like
(viii) Unlike fraction
(ix) Equal fraction
(x) Like

(xi) +2 = 4113

(xii) – (4×5+3)5

= – 235

Question-3

From the following fractions, separate :
(i) Proper fractions
(ii) Improper fractions :

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:– 29 , 715 ,1120 , 1823  and 2735
(ii) Improper fractions :– 43 ,  2011

#### Question -4.

Change the following mixed fractions to improper fractions :

#### Question- 5.

Change the following improper fractions to mixed fractions :

#### Question -6.

Change the following groups of fractions to like fractions :

32112  9811240112, 63112

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

#### Question -1.

Reduce the given fractions to their lowest terms :

#### Question -2.

State, whether true or false?

#### Question -3.

Which fraction is greater?

#### Question- 4.

Which fraction is smaller ?

#### Question- 5.

Arrange the given fractions in descending order of magnitude :

(iii) …. 57….,….. 88…….. 911

LCM 0f  Numerator 5 , 3, 9 = 45

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

hence

4555….,….. 4563……… 45120….

ie.- 911….,….. 57……….. 38….

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 :

(i) 916….,….. 712……….. 14….
LCM of 16,12,and 4 = 48
2848….,…..  2848…………. 1248..

Arrange the  fractions in ascending order

….1248..<…..2848….<…….2848……

ie ..14..,<..916..<. 712…..

(ii) 56….,….. 27…..89……. 13….

LCM of 6, 7, 9 and 3 = 126
105126….,36126….. 112126……..42126..

Arrange the  fractions in ascending order

….36126..<…..42126……<…….105126…..112126

ie ..27…,<..13….<. 56….89

(iii) 23….,…59.. …..56……. 38….

LCM of 3, 9, 6 and 8 = 72
4872….,4072….. 6072……..2772..

Arrange the  fractions in ascending order

….2772.<…..4072…….<…….4872…..6072

ie ..38…,<..59….<. .23….56

#### Question -7.

I bought one dozen bananas and ate five of them. What fraction of the total number of bananas was left ?

Number of bananas bought = 1
Dozen = 12
Number of bananas eaten by me = 5
Number of bananas left = 12 – 5 = 7
Fraction = 712.

#### Question- 8.

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

(i) ….611………59…….

LCM of 11 and 9 = 99

….5499………5599…..

….5499……<…5599…………..

ie. ….611….<…..59…………..

(ii) ….37………913…………..

LCM of 7 and 13 = 91

….3991………6391…..

3991……<…6391…..

ie ….37…..<….913…………..

(iii) ….5664………78…………..

LCM of 64 and 8 = 64

….5664………5664…..

.5664….=…..5664…..

ie ….5664….=…..78…………..

(iv) ….512………833…………..

LCM of 12 and 33 = 132

….55132………32132…..

….55132…..>….32132…..

.ie…512….>…..833…….

#### 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.

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

= ..5036

= 2518

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

= ..1450

= 725

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

#### Question- 1.

(i) ……1.34…….and  38….

….74…….and  38..

LCM of 4and 8 = 8

…….148…….+  38……..= ……178……

(ii)

(i) …….25……2..315……..710…….

……..25……+..3315….+….710……

LCM of 5 , 15 and 10 = 30

…..1230……+..6630….+….2130……

=  …………..12+66+2130…………………….

= ……9930…….3310

(iii)

(iv)

(vi)

Simplify :

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

Simplify :

Simplify :

Simplify :

#### Question- 4.

Simplify :

= … 112..x.. 196….=…. 11152….….

Simplify :

### Selina Solutions of Exercise – 14 EFractions for ICSE Class-6th Concise Mathematics

#### Question -1.

From a rope of 1012 m long, 4 58 m is cut off. Find the length of the remaining rope.

Length of rope =  1012  m

Length of cut off. rope = 4 58

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

= 1012 m – 4 58

= 212

= ….212….- 378…..

=

#### Question -2.

A piece of cloth is 5 metre long. After washing, it shrinks by 125 of its length. What is the length of the cloth after washing?

Length of piece of cloth is = 5 metre

After washing, it shrinks =  125 of its length

length of the cloth after washing = 5 metre  – 125 of its 5

#### Question -3.

I bought wheat worth Rs. 1212, rice worth Rs. 2534 and vegetables worth Rs. 1014. If I gave a hundred-rupee note to the shopkeeper ; how much did he return to me

wheat worth = Rs. 1212,

rice worth Rs. 2534

vegetables worth Rs. 1014

Total Amount = (wheat+rice+vegetables)

Total Amount = (..1212,..+….Rs. 2534…+…1014….)

= 1944,

#### Question -4.

Out of 500 oranges in a box, 325, are rotten and 15, are kept for some guests. How many oranges are left in the box?

Number of oranges = 500

rotten oranges = 325, of 500 = 60

= 60

kept for some guests = .15, 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 112 of the ornament is copper, find the weight of gold in it.

total weight is = 96 g

copper =  112 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?

Let total Work be = 1

work on Monday = 12

work on Tuesday = 13

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

#### Question- 7.

A man spends 38 of his money and 8 still has Rs. 720 left with him. How much money did he have at first?

Let Money be =1

spends =  38 of his money

money Left = 1 – 38

= 58

According to condition given = Rs. 720 left

…..58…of 1 .= ..720.

= 720 x 85

= 1152

#### Question -8.

In a school, 45 of the students are boys, and the number of girls is 100. Find the number of boys.

Number of boy = Total strength – number of girl

Number of boy = 500 – 100 = 400

#### Question- 9.

After finishing 34 of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

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?

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  310  on food,  14  on house rent,  110  on insurance and  225  on holidays. How much did he save in that month ?

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

he spends on food, =   310  of 7,200 = 2160

on house rent = 14 of 7,200 = 1800

Money spend on insurance= 110  of 7200 = 720

on holidays =  225 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

Evaluate :

#### 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?

#### 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?

= Rs. 84,000 – (37,800) = Rs. 46,200

#### Question- 6.

In one hour Rohit walks 3$\frac { 2 }{ 5 }$ km. How much distance will he cover in 2$\frac { 1 }{ 2 }$ hours?

#### Question -7.

An 84 m long string is cut into pieces each of length 5$\frac { 1 }{ 4 }$ m. How many pieces are obtained?

#### 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

#### 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?

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

Simplify :