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 101⁄2 m long, 4 5⁄8 m is cut off. Find the length of the remaining rope.
Answer-1
Length of rope = 101⁄2 m
Length of cut off. rope = 4 5⁄8
Length of the remaining rope = Length of rope – Length of cut off. rope
= 101⁄2 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. 121⁄2, rice worth Rs. 253⁄4 and vegetables worth Rs. 101⁄4. If I gave a hundred-rupee note to the shopkeeper ; how much did he return to me
Answer-3
wheat worth = Rs. 121⁄2,
rice worth Rs. 253⁄4
vegetables worth Rs. 101⁄4
Total Amount = (wheat+rice+vegetables)
Total Amount = (..121⁄2,..+….Rs. 253⁄4…+…101⁄4….)
= 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
Answer-6
Question -7.
An 84 m long string is cut into pieces each of length 5
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
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