Fractions ICSE Class-7th Concise Selina Maths Solutions

Fractions ICSE Class-7th Concise Selina Maths Solutions Chapter-3 . We provide step by step Solutions of Exercise / lesson-3 Fractions for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-3 A, Exe-3 B, Exe-3 C  Exe-3 D and  Exe-3 E to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

Fractions ICSE Class-7th Concise Selina Maths Solutions Chapter-3

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–: Select Topics :–

Exe-3 A,

 Exe-3 B,

 Exe-3 C,

 Exe-3 D,

Exe-3 E

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Exercise – 3 A

Fractions ICSE Class-7th Concise

Question 1:-

Classify, each fraction given below, as decimal or vulgar fraction, proper or improper fraction and mixed fraction :

Answer-1

(i) Vulgar and Proper

(ii) Decimal and Improper

(iii) Decimal and proper

 (iv) Vulgar and Improper

(v)Mixed

(vi) Decimal

(vii) Mixed and Decimal

(viii) Vulgar and Proper

Question -2:-

Express the following improper fractions as mixed fractions :

Answer-2

(i) ..¹⁸⁄₅ ..= 3 ³⁄₅

(ii) ..⁷⁄₄ ..= 1 ³⁄₄

(iii) .25⁄₆ ..= 4 ¹⁄₆

(iv) ³⁸⁄₅ ..= 7 ³⁄₅

(v) ²²⁄₅ ..=4 ²⁄₅

Question 3:-

Express the following mixed fractions as improper fractions :

Answer-3

(i) 

(ii)

(iii)

(iv)

(v)

Question 4:-

Reduce the given fractions to lowest terms

Answer 4:-

(i)

(Dividing by 2, the HCF of 8 and 18)

(ii)

(Dividing by 9, the HCF of 27 and 36)

(iii)

(Dividing by 6, the HCF of 18 and 42)

(iv)

(Dividing by 5, the HCF of 35 and 75)

(v)

 

(Dividing by 9, the HCF of 18 and 45)

Question 5:-

State : true or false

Answer:-

(i) True 

(ii) False

(iii) True

Question 6:-

Distinguish each of the following fractions, given below, as a simple fraction or a complex fraction :

Answer:-6

(i)  It is a simple fraction. 

(ii)  It is a simple fraction. 

(iii)  It is a simple fraction. 

(iv)  It is a complex fraction. 

(v)  It is a complex fraction. 

(vi)  It is a complex fraction. 

(vii)  It is a complex fraction. 

(viii) It neither complex nor simple as the denominator is zero.

 

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Exercise- 3 B

Fractions ICSE Class-7th Concise Selina Maths Solutions

Question 1:-

For each pair, given below, state whether it forms like fractions or unlike fractions :

Answer:-1

(i)  These are like fractions.

(ii)  These are unlike fractions.

(iii)  These are unlike fractions.

Question 2:-

Convert given fractions into fractions with equal denominators :

Answer:-

(i)

Hence, ¹⁵⁄₁₈ and ¹⁴⁄₁₈ are the required fractions.

(ii)  

 L.C.M of 3, 6 and 12 = 12

(iii)

L.C.M of 5, 20, 40 and 16 = 80

Hence, the required fractions are  ..⁶⁴⁄₈₀ , ⁶⁸⁄₈₀ , ⁴⁶⁄₈₀ and ⁵⁵⁄₈₀ 

Convert given fractions into fractions with equal numerators :

Answer:-3

(i)  L.C.M. of 8 and 12 = 24 

Hence the required fractions are –⁶⁰⁄₁₃₀ .,. ⁶⁰⁄₉₂.and… ⁶⁰⁄85…

(iii)

_⁄Hence the required fractions are  ²²⁵⁄_285,  ²²⁵⁄_{252,  ²²⁵⁄275 and ²²⁵⁄235}

Hence the required fractions are  ²²⁵⁄_285,  ²²⁵⁄_{252,  ²²⁵⁄275 and ²²⁵⁄235}

Question 4:-

Answer-4:-

(i)

L.C.M. of denominators 3, 5, 4 and 6 = 60

From above we see that

Hence,  ¹⁄₆  ¹⁄_3, ²⁄₅  ³⁄_4,  are in ascending order.

Question- 5:-

Arrange the given fractions in descending order by making numerators equal :

Answer-5:-

(i)

L.C.M. of numerators 5, 4, 8 and 1 = 40

From above we see that

Hence,   ⁸⁄₉ ⁵⁄_6,   ¹⁄_3,   ⁴⁄_15, are in descending order..

(ii)

L.C.M. of numerators 3, 4, 5 and 8 = 120

It is clear from the above that

Hence,   ¹²⁰⁄₁₆₅ > ¹²⁰⁄_168, >  ¹²⁰⁄ 270 >  ¹²⁰⁄ 280 

or  ⁸⁄₁₁ > ⁵⁄_7, >  ⁴⁄ 9 >  ³⁄ 7

(iii)

L.C.M. of numerators 1, 6, 8 and 3 = 24

It is clear from the above that

_²⁴⁄33,> ²⁴⁄_40, >  ²⁴⁄ 44 >  ²⁴⁄ 240

 or  ⁸⁄₁₁ > ³⁄_5, >  ⁶⁄ 11 >  ¹⁄ 10

Question -6:-

Find the greater fraction :


Answer:-6

(i)  
LCM of 5 and 15 = 15

It is clear from above that  ¹¹⁄ 15 >  ⁹⁄ 15

Hence ¹¹⁄ 15 is greater.

(ii)

LCM of 5 and 10 = 10

It is clear from above that  ⁸⁄ 10 >  ³⁄ 10

Hence ⁴⁄ 5 is greater.

(iii)

LCM of 7 and 9 = 63

It is clear from above that  ⁵⁴⁄ 63 >³⁵⁄ 63

Hence  ⁵⁴⁄ 63 or  ⁶⁄ 7 is greater.

(iv)  

 LCM of 8 and 9 = 72

It is clear from above that ³²⁄ 72 >²⁷⁄ 72

Hence ³²⁄ 72 or ⁴⁄9 is greater.

(v)

LCM of 7 and 10 = 70

It is clear from above that  ^-20⁄ 70 > ²¹⁄ 70

Hence  ^-20⁄ 70 or ^-2⁄ 7 is greater.

Question-7:-

Insert one fraction between :

Answer:-7

(i)

(ii)

(iii)

Question-8:-

Insert three fractions between

Answer-8:-

(i)

Fraction between ²⁄ 5 and ³⁄ 7 

….⁽²⁺³⁾⁄ (5+7.)…= .⁵⁄ 12………………

Fraction between  ³⁄ 7 and ⁴⁄ 9

Hence, three fractions between 2/5 and 4/9 will be .⁵⁄ 12… ³⁄ 7…,..⁷⁄ 16..

(ii)

Fraction between ¹⁄ 2 and ²⁄ 3

…⁽¹⁺²⁾ ⁄ (2+3).=…³⁄ 5…

and Fraction between ….²⁄ 3 and ⁵⁄ 7…………………

= …⁽²⁺⁵⁾ ⁄ (3+7).=…⁷⁄10………..

and Fraction between  ⁹⁄₁₉  and   ⁶⁄_11,….

 …⁽⁹⁺⁶⁾⁄_(19+11).=..¹⁵⁄₃₀….=…_,.¹⁄₂…

Hence, three fractions between 3/8 and 6/11 will be  and 4/9  9/19  1/2

(iv)

Fraction between 11/12 and 13/15

= ..^(11+13)⁄₁₅₊₃₎..=…...²⁴⁄27 = …⁸⁄9..

and Fraction between   ¹³⁄₁₅  and   ²⁄_3,….

= ..⁽¹³⁺²⁾⁄₍₅₊₃₎..=… ¹⁵⁄₁₈ = …⁵⁄ 6.....

Hence, three fractions between 11/12 and 2/3 will be …⁸⁄9…¹³⁄₁₅..and…⁵⁄ 6…….

(v)

Hence, three fractions between 4/7 and 3/4 will be  ..¹¹⁄₁₈.…. .⁷⁄₁₁..and…²⁄ 3…….

Question-9:-

Insert two fractions between

Answer-9:-

(i)

Hence, two fractions between 1 and 3/11

will be  ¹⁄₃  and   ²⁄_7,….

(ii)

Hence, two fractions between 5/9 and 1/4

will be  6⁄₁₃ and   ⁷⁄_17,….

(iii)

Hence, two fractions between 5/6 and 1.¹⁄_5,…... will be  1 and 1¹⁄_6,….

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Exercise – 3 C 

Fractions ICSE Class-7th Concise Selina Maths Solutions

Question-1:-

Reduce to a single fraction :


Answer-1:-

(i)

(ii)

(iii)

(iv)

(v)

(vi) 

(vii)

(viii)

(ix)

Question-2:-

Simplify :

Answer-2:-

(i)  

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

 

(HCF of 1800 and 420 = 60)

= ….³⁰⁄₇…..=….4²⁄₇…………

Question-3:-

Subtract :

Answer-3:-

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Question -4:-

Find the value of

Answer-4:-

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question-5:-

Simplify and reduce to a simple fraction :


Answer-5:-

(i)

(ii)

(iii)

 

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Question-6:-

A bought 3  3/4 kg of wheat and 2  kg of rice. Find the total weight of wheat and rice bought.
Answer-6:-

Question 7.

Which is greater, or  and by how much?
Answer-7:-

Question-8:-

What number should be added to 8 ²⁄₃   to 12 ⁵⁄₆  
Answer-8:-

For finding the required fraction, we have

to subtract  8 ²⁄₃  from 12 ⁵⁄₆  

 Required number =  12 ⁵⁄₆  .–.8 ²⁄₃ .

Question-9:-

What should be subtracted from 8 to get  2 ²⁄₃
Answer-9:-

Question-10:-

A field is 16 ¹⁄₂ m long and  12 ²⁄₅ m wide. Find the perimeter of the field.
Answer-10:-

Question-11:-

Sugar costs ₹37  1/2 per kg. Find the cost of 8 ³⁄4 kg sugar.
Answer-11:-

Question-12:-

A motor cycle runs 31 1/4  km consuming 1 litre of petrol. How much distance will it run consuming  1 3/5  liter of petrol?
Answer-12:-

Question-13:-

A rectangular park has length = 23  ²⁄₃  m and breadth = 16  ²⁄₃ m. Find the area of the park.
Answer-13:-

Question-14:-

Each of 40 identical boxes weighs 4 ⁴⁄₅ kg Find the total weight of all the boxes.
Answer-14:-

Question-15:-

Out of 24 kg of wheat, ⁵⁄₆ ^th of wheat is consumed. Find, how much wheat is still left?

Answer-15:-

Total wheat available = 24 kg

Wheat consumed =  ⁵⁄₆ ^th” of 24 kg

=  ⁵⁄₆ ^th ×24=20 kg

∴ Remaining wheat = 24 – 20 kg = 4 kg

Question-16:-

A rod of length  2 ²⁄₅ m metre is divided into five equal parts. Find the length of each part so obtained.

Answer-16:-

Total length of rod = 2 ²⁄₅ m

Length of rod to be divided into 5 equal parts

∴ Length of each part of rod = 2 ²⁄₅ ÷ 5

Question-17:-

If A = 3 ³⁄₈  and B = 6 ⁵⁄₈   find :
(i) A+B
(ii) B /A
Answer-17:-

Question-18:-

Cost of 3 ⁵⁄₇   litres of oil is ₹83 ¹⁄₂ . Find the
cost of one litre oil.
Answer-18:-

∴ Cost of 1 liter oil = 83 ¹⁄₂../   ..3 ⁵⁄₇ …

Question-19:-

The product of two numbers is 20 ⁵⁄₇. If one of these numbers is 6 ²⁄₃  find the other.

Answer-19:-

Question-20:-

By what number should 5 ⁵⁄₆ be multiplied 1 to get 3 ¹⁄₃ ?
Answer-20:-

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Exercise – 3 D 

Fractions ICSE Class-7th Concise Selina Maths Solutions

Question-1:-

Simplify………..

Answer-1:-

Question-2:-

Answer-2:-

 

Question-3:-

Answer-3:-

Question-4:-

Simplify

Answer-4:-

Question-5:-

Simplify

Answer-5:-

Question-6:-

Answer-6:-

 

= –  4 ¹¹⁄₅₀ .

Question-7:-

Answer-7:-

(Using BODMAS)

= 2 ⁴⁄₁₉....

Question-8:-

Answer-8:-

.(Using BODMAS)

Question-9:-

Answer-9:-

Using BODMAS

Question-10:-

Answer-10:-

Question-11:-

Answer-11:-

Using BODMAS

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 Exercise – 3 E  

Fractions ICSE Class-7th 

Question-1:-

A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is :
(i) The length of AB to that of CD ?
(ii) 1/2 the length of AB to that of 1/3 of CD ?
(iii) 1/5 of CD to that of AB ?

Answer

(i) Length of line AB = 6 cm

and length of line CD = 15 cm

Length of line AB to length of CD =  ⁶⁄₁₅.=.. ²⁄₅.

(ii)

Length of line AB = 6 cm

and length of line CD = 15 cm

1/2 of AB = ¹⁄₂ ×6=3 cm

1/3 of CD = ¹⁄₃ ×15=5 cm

∴1/2 of AB to 1/3 of CD = ³⁄₅

(iii)

Length of line AB = 6 cm

and length of line CD = 15 cm

1/5 of AB = ¹⁄₅ ×15=3 cm

∴1/5 of CD to that of AB = …³⁄₆….=……¹⁄₂

Question-2:–

Subtract 2/7 – 5/21 from the sum of  ³⁄4 , ⁵⁄₇  and  ⁵⁄₁₂

Answer-2:-

Question-3:-

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5 kg, 9  kg and 9  kg respectively.
(i) How many kg did he sell ?
(ii) How many kg are still left in the sack ?
Answer-3:-

Total quantities of potatoes = 120 kg

(i) Quantity of potatoes he sold

(ii)  Quantity of potatoes left

Question-4:-

If a boy works for six consecutive days for 8 hours, 7 hours, 8 hours, 6 4 3hours, 6 hours and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour ?
Answer-4:-

No. of hours, a boy worked in 6 days

Earning per hour = Rs. 36

∴ Total earnings = Rs (175/4)×36

= Rs. 175 × 9 = Rs, 1575

 

Question-5:- 

A student bought 4  m of yellow ribbon, 6  m of red ribbon and 3  m of blue ribbon for decorating a room. How many metres of ribbon did he buy ?
Answer-5:-

Length of yellow ribbon = 4 ¹⁄₃ m=13/3m

Length of red ribbon = 6 ¹⁄₆ m=37/6 m

Length of blue ribbon =3 ²⁄₉ m=29/9 m

Total length of ribbon = ¹³⁄₃+³⁷⁄₆+²⁹⁄₉

=(78+111+58)/18

(LCM of 3, 6, 9 = 18)

=247/ 18 

=13 ¹³⁄₁₈  meters

Question-6:-

In a business, Ram and Deepak invest  and  of the total investment. IfRs. 40,000 is the total investment, calculate the amount invested by each ?
Answer-6:-

Total investment = Rs. 40000

Ram’s investment = 3/5 of Rs. 40000

= Rs. (3/5) ×40000

= Rs. 2 × 8000 = Rs. 16000

Question-7:-

Geeta had 30 problems for home work. She worked out  of them. How many problems were still left to be worked out by her ?
Answer-7:-

No.of problems of Geeta = 30

No.of problems worked out = 2/3 of 30

=⁹⁄₃ ×30=20

No.of problems left out = 30 – 20 = 10

Question-8:-

A picture was marked at Rs. 90. It was sold at  of its marked price. What was the sale price ?
Answer-8:-

Marked price = Rs. 90

Sale price = 3/4 of Rs. 90 =³⁄₄×90

= Rs. 270/4 

= Rs. 67 ¹⁄₂ 

= Rs. 67.50

Question-9:-

Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10 kg ?

Answer-9:-

Total no. of parcels = 15

Weight of each parcel = 10¹⁄₂  kg = 21/2 kg

Total weight  = 15 of ²¹⁄₂  kg

= ²¹⁄₂×15 kg

= 315/15

= 157 ¹⁄₁₅ kg

= 157.5 kg

Question-10:-

A rope is 25 m long. How many pieces , 1  each of length can be cut out from it?

Answer-10:-

Total length of the rope = 25 ¹⁄₂ m = 51/2 m

Length of each piece = 1 ¹⁄₂ m = 3/2 m

∴ No. of pieces =  ⁵¹⁄₂ ÷  ³⁄₂ = ⁵¹⁄_2..=… ײ⁄_{3 …..=….17..pieces……….….}

Question-11:-

The heights of two vertical poles, above the earth’s surface, are 14  m and 22  respectively. How much higher is the second pole as compared with the height of the first pole ?
Answer-11:-

Height of one pole above earth’s surface =14 ¹⁄₄  m

and height of d=second pole = 22 ¹⁄₃ 

∴ Second pole is higher than the first pole

=22 ¹⁄₃ -14  ¹⁄₄ =  ⁶⁷⁄₃ –  ⁵⁷⁄₄

Question-12:-

Vijay weighed 65 kg. He gained 1  kg during the first week, 1  kg during the second week, but lost  kg during the 16 third week. What was his weight after the third week ?
Answer-12:-

In the beginning, weight of vijay =65 ¹⁄₂  kg

Gained in first week = 1 ²⁄₅  kg

Gained in second week = 1 ¹⁄₄   kg

Lost in the third week = 5/16 kg

∴ Weight of Vijay after third week

Question-13:-

A man spends  of his salary on food and  on house rent, electricity, etc. What fraction of his salary is still left with him ?
Answer-13:-

Let salary of man = Rs. 1

Amount spent on food = 2/5 of Rs. 1 = Rs. 2/5

and amount spent house rent = 3/10 of Rs. 1

= Rs. 3/10

Question-14:-

A man spends  of his salary on food and  of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him ?
Answer-14:-

Let the total amount of salary = Rs. 1

Amount spent on food = 2/5 of Rs. 1 = Rs. 2/5

Remaining amount = 1- ²⁄₅

=5-²⁄₅=Rs.3/5

Amount spent on house rent etc.

=3/10 of 3/5=Rs 9/50

Remaining amount left = ³⁄₅ – ⁹⁄₅₀

=Rs.³⁄₅ – ⁹⁄₅₀

=Rs.(30-9) 50

=Rs.21/50

Question-15:-

Shyam bought a refrigerator for Rs. 5000. He paid  of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month ?
Answer-15:-

Total amount of the refrigerator = Rs 5000

Amount paid in cash = 1/10 of Rs. 5000

=1/10×5000 = Rs. 500

=¹⁄₁₀ x 5000

= Rs. 500

Balance amount = Rs. 5000 – Rs. 500

= Rs. 4500

No.of equally instalments = 12

∴ Amount of each instalment

=  Rs. 4500 ÷ 12

= Rs. 4500 x ¹⁄₁₂

= Rs. 375

Question-16:-

A lamp post has half of its length in mud, and  of its length in water.
(i) What fraction of its length is above the water ?
(ii) If 3 m of the lamp post is above the water, find the whole length of the lamp post.

Answer-16:-

(i) 

Let length of the post = 1 m

then length of post in mus =1/2 m

and length of post in water = 1/3 m

∴ Length of post above the water

=1 – ( ¹⁄₂+ ¹⁄₃) = 1 – (3+2)/6

=1- ⁵⁄₆ = (6-5)/6 = 1/6 m

(ii)

But length of post above water

= 3 ¹⁄₃ m=10/3 m

∴ 1/6th of total length = 10/3 m

∴ Total length = ¹⁰⁄₃ × ⁶⁄₁ = 20 m

Question-17:-

I spent  of my savings and still have Rs. 2,000 left. What were my savings ?
Answer-17:-

Lett my saving = 1, Part spent = 3/5 of savings

∴ Part left = 1-³⁄₅ = (5-3)/5 = 2/5 of savings

But he left = Rs. 2000

∴ 2/5 of savings = Rs. 2000

∴ Total savings = Rs. 2000×5/2

= Rs. 5000

Question-18:-

In a school,  of the children are boys. If the number of girls is 200, find the number of boys.
Answer-18:-

No. of boys = 4/5 of the total children

∴ No. of. girls = (1 – ⁴⁄₅) of total children

=(5-4)/5 = 1/5 of total children

But no. of girls = 200

∴ 1/5 of total children = 200

Hence total number of children = 200 × ⁵⁄₁ = 1000

∴ No. of boys = 4/5 of 1000

=  4/5 of 1000 = (4/5) x 1000

= 800

Question-19:-

If  of an estate is worth Rs. 42,000, find the worth of whole estate. Also, find the value of  of it.
Answer-18:-

4/5 of an estate = Rs. 42000

∴ Total value of estate = Rs. 42000 × ⁵⁄₄

= Rs. 105000 × 5 = Rs. 52500

and value of 3/7 of it = 3/7 of its value

=3/7 of  52500 = ³⁄₇ × 52500

= 3 × 7500 = 22500

Question-20:-

After going  of my journey, I find that I have covered 16 km. How much Journey is still left ?
Answer-20:-

3/4 of journey = 16 km.

∴ Total journey = 16 km× C = 64/3 km

∴ Journey left = 64/3 – 16/1

= (64-48)/3

=16/3 Km

= 5 ¹⁄₃

Question-21:-

When Krishna travelled 25 km, he found that  of his journey was still left. What was the length of the whole journey.
Answer-21:-

3/5 of the total journey was left

∴ Journey travelled by him = 1-⁴⁄₅

= (5-3) 5 = 25

∴ 2/5 of total journey = 235 km

∴ Total journey = 25 km ×5/2=125/2 km

= 62 ¹⁄₂  km

Question-22:-

From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.

Answer-22:-

Let the piece by Rajesh = 1 × ¹⁄₃ = ¹⁄₃  m

Land bought by Rajesh = 1 × ¹⁄₃ = ¹⁄₃  m

Remaining land = 1 – ¹⁄₃ = 3- ¹⁄₃ =²⁄₃ m

Now, land bought by Manoj = ²⁄₃ × ¹⁄₃ = 2/9 m

Land unsold =²⁄₃ – ²⁄₉

= (6-29)/9 = 4/9 m

Land of 4/9 m remain unsold = 600 m²

∴ Total area of the land = 600 × ⁹⁄₄

= 150 × 9 m² = 1350 m²

Question-23:-

A boy spent 3/5 of his money on buying 1 cloth and 1/4  of the remaining on buying shoes. If initially he has ?2,400; how much did he spend on shoes?
Answer-23:-

— End of Fractions ICSE Class-7th Concise Solutions :–

Return to – Concise Selina Maths Solutions for ICSE Class -7 

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