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 for detail information about ICSE Board Class-7.

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

–: Select Topics :–

Exe-3 A,

Exe-3 B,

Exe-3 C,

Exe-3 D,

Exe-3 E

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 :

(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 :

(i) ..185 ..= 3 35

(ii) ..74 ..= 1 34

(iii) .25⁄6 ..= 4 16

(iv) 385 ..= 7 35

(v) 225 ..=4 25

Question 3:-

Express the following mixed fractions as improper fractions :

#### (i)

(ii)

(iii)

(iv)

(v)

Question 4:-

Reduce the given fractions to lowest terms

#### (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

#### Question 6:-

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

### 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 :

(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 :

(i)

Hence, 1518 and 1418 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  ..6480 , 6880 , 4680 and 5580

#### Question 3:-

Convert given fractions into fractions with equal numerators :

(i)  L.C.M. of 8 and 12 = 24

Hence the required fractions are..2427  and 2434

Hence the required fractions are –60130 .,. 6092.and… 6085

(iii)

Hence the required fractions are  225285,  225252,  225⁄275 and 225⁄235

Hence the required fractions are  225285,  225252,  225⁄275 and 225⁄235

#### Question 4:-

Put the given fractions in ascending order by making denominators equal :

(i)

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

From above we see that

Hence,  16  13, 25  34,  are in ascending order.

Question- 5:-

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

(i)

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

From above we see that

Hence,   89 56,   13,   415, are in descending order..

(ii)

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

It is clear from the above that

Hence,   120165 > 120168, >  120270 >  120⁄ 280

or  811 > 57, >  4⁄ 9 >  3⁄ 7

Hence,81157,  4⁄ 9 ,  3⁄ 7 are in descending order.

(iii)

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

It is clear from the above that

24⁄33,2440, >  24⁄ 44 >  24⁄ 240

or  811 > 35, >  6⁄ 11 >  1⁄ 10

Hence,811 , 35,  6⁄ 11  1⁄ 10 are in descending order.

#### Question -6:-

Find the greater fraction :

(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 :

(i)

(ii)

(iii)

Question-8:-

Insert three fractions between

(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………..

Hence, three fractions between 1/2 and 5/7 will be 3/5,  2/3 and 7/10

and Fraction between  919  and   611,….

…(9+6)(19+11).=..1530….=…,.12

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)..=…...2427 = 89..

and Fraction between   1315  and   23,….

= ..(13+2)(5+3)..=… 1518 = 5⁄ 6.....

Hence, three fractions between 11/12 and 2/3 will be …8⁄9…1315..and…5⁄ 6…….

(v)

Hence, three fractions between 4/7 and 3/4 will be  ..1118.…. .711..and…2⁄ 3…….

Question-9:-

Insert two fractions between

(i)

Hence, two fractions between 1 and 3/11

will be  13  and   27,….

(ii)

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

will be  613 and   717,….

(iii)

Hence, two fractions between 5/6 and 1.15,…... will be  1 and 116,….

### Exercise – 3 C

Fractions ICSE Class-7th Concise Selina Maths Solutions

#### Question-1:-

Reduce to a single fraction :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

#### Question-2:-

Simplify :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(HCF of 1800 and 420 = 60)

= ….307…..=….427…………

#### Question-3:-

Subtract :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Question -4:-

Find the value of

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Question-5:-

Simplify and reduce to a simple fraction :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

#### Question-6:-

A bought 3  3/4 kg of wheat and 2 $\frac { 1 }{ 2 }$ kg of rice. Find the total weight of wheat and rice bought.

#### Question 7.

Which is greater,$\frac { 3 }{ 5 }$ or $\frac { 7 }{ 10 }$ and by how much?

#### Question-8:-

What number should be added to 8 23   to 12 56

For finding the required fraction, we have

to subtract  8 23  from 12 56

Required number =  12 56  .–.8 23 .

#### Question-9:-

What should be subtracted from 8$\frac { 3 }{ 4 }$ to get  2 23

#### Question-10:-

A field is 16 12 m long and  12 25 m wide. Find the perimeter of the field.

#### Question-11:-

Sugar costs ₹37  1/2 per kg. Find the cost of 8 34 kg sugar.

#### 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?

#### Question-13:-

A rectangular park has length = 23  23  m and breadth = 16  23 m. Find the area of the park.

#### Question-14:-

Each of 40 identical boxes weighs 4 45 kg Find the total weight of all the boxes.

#### Question-15:-

Out of 24 kg of wheat, 56 th of wheat is consumed. Find, how much wheat is still left?

Total wheat available = 24 kg

Wheat consumed =  5th” of 24 kg

=  5th ×24=20 kg

∴ Remaining wheat = 24 – 20 kg = 4 kg

#### Question-16:-

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

Total length of rod = 2 25 m

Length of rod to be divided into 5 equal parts

∴ Length of each part of rod = 2 25 ÷ 5

#### Question-17:-

If A = 3 3 and B = 6 5  find :
(i) A+B
(ii) B /A

#### Question-18:-

Cost of 3 5  litres of oil is ₹83 12 . Find the
cost of one litre oil.

∴ Cost of 1 liter oil = 83 12../   ..5

#### Question-19:-

The product of two numbers is 20 57. If one of these numbers is 6 23  find the other.

#### Question-20:-

By what number should 5 56 be multiplied 1 to get 3 13 ?

Simplify………..

Simplify

Simplify

= –  4 1150 .

(Using BODMAS)

= 2 419....

.(Using BODMAS)

Using BODMAS

Using BODMAS

### 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 ?

(i) Length of line AB = 6 cm

and length of line CD = 15 cm

Length of line AB to length of CD =  615.=.25.

(ii)

Length of line AB = 6 cm

and length of line CD = 15 cm

1/2 of AB = 12 ×6=3 cm

1/3 of CD = 13 ×15=5 cm

∴1/2 of AB to 1/3 of CD = 35

(iii)

Length of line AB = 6 cm

and length of line CD = 15 cm

1/5 of AB = 15 ×15=3 cm

∴1/5 of CD to that of AB = 36….=……12

#### Question-2:–

Subtract 2/7 – 5/21 from the sum of  34 , 57  and  512

#### Question-3:-

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5 $\frac { 1 }{ 4 }$ kg, 9 $\frac { 1 }{ 2 }$ kg and 9 $\frac { 3 }{ 4 }$ kg respectively.
(i) How many kg did he sell ?
(ii) How many kg are still left in the sack ?

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$\frac { 1 }{ 2 }$ hours, 8$\frac { 1 }{ 4 }$ hours, 6 $\frac { 1 }{ 4 }$4 3hours, 6$\frac { 3 }{ 4 }$ hours and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour ?

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 $\frac { 1 }{ 3 }$ m of yellow ribbon, 6 $\frac { 1 }{ 6 }$ m of red ribbon and 3 $\frac { 2 }{ 9 }$ m of blue ribbon for decorating a room. How many metres of ribbon did he buy ?

Length of yellow ribbon = 13 m=13/3m

Length of red ribbon = 6 16 m=37/6 m

Length of blue ribbon =3 29 m=29/9 m

Total length of ribbon = 133+376+299

=(78+111+58)/18

(LCM of 3, 6, 9 = 18)

=247/ 18

=13 1318  meters

#### Question-6:-

In a business, Ram and Deepak invest $\frac { 3 }{ 5 }$ and $\frac { 2 }{ 5 }$ of the total investment. IfRs. 40,000 is the total investment, calculate the amount invested by each ?

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 $\frac { 2 }{ 5 }$ of them. How many problems were still left to be worked out by her ?

No.of problems of Geeta = 30

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

=93 ×30=20

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

#### Question-8:-

A picture was marked at Rs. 90. It was sold at $\frac { 3 }{ 4 }$ of its marked price. What was the sale price ?

Marked price = Rs. 90

Sale price = 3/4 of Rs. 90 =34×90

= Rs. 270/4

= Rs. 67 12

= Rs. 67.50

#### Question-9:-

Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10$\frac { 1 }{ 2 }$ kg ?

Total no. of parcels = 15

Weight of each parcel = 1012  kg = 21/2 kg

Total weight  = 15 of 212  kg

= 212×15 kg

= 315/15

= 157 115 kg

= 157.5 kg

#### Question-10:-

A rope is 25$\frac { 1 }{ 2 }$ m long. How many pieces , 1 $\frac { 1 }{ 2 }$ each of length can be cut out from it?

Total length of the rope = 25 12 m = 51/2 m

Length of each piece = 12 m = 3/2 m

∴ No. of pieces =  512 ÷  32 = 512..=… ×23 …..=….17..pieces……….….

#### Question-11:-

The heights of two vertical poles, above the earth’s surface, are 14 $\frac { 1 }{ 4 }$ m and 22 $\frac { 1 }{ 3 }$ respectively. How much higher is the second pole as compared with the height of the first pole ?

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

and height of d=second pole = 22 1

∴ Second pole is higher than the first pole

=22 13 -14  14 =  673 –  574

#### Question-12:-

Vijay weighed 65 $\frac { 1 }{ 2 }$ kg. He gained 1 $\frac { 2 }{ 5 }$ kg during the first week, 1 $\frac { 1 }{ 4 }$ kg during the second week, but lost $\frac { 5 }{ 16 }$ kg during the 16 third week. What was his weight after the third week ?

In the beginning, weight of vijay =65 1 kg

Gained in first week = 1 2 kg

Gained in second week = 1 14   kg

Lost in the third week = 5/16 kg

∴ Weight of Vijay after third week

#### Question-13:-

A man spends $\frac { 2 }{ 5 }$ of his salary on food and $\frac { 3 }{ 10 }$ on house rent, electricity, etc. What fraction of his salary is still left with him ?

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 $\frac { 2 }{ 5 }$ of his salary on food and $\frac { 3 }{ 10 }$ of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him ?

Let the total amount of salary = Rs. 1

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

Remaining amount = 1- 25

=5-25=Rs.3/5

Amount spent on house rent etc.

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

Remaining amount left = 35 950

=Rs.3– 950

=Rs.(30-9) 50

=Rs.21/50

Question-15:-

Shyam bought a refrigerator for Rs. 5000. He paid $\frac { 1 }{ 10 }$ of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month ?

Total amount of the refrigerator = Rs 5000

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

=1/10×5000 = Rs. 500

=110 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 112

= Rs. 375

#### Question-16:-

A lamp post has half of its length in mud, and $\frac { 1 }{ 3 }$ of its length in water.
(i) What fraction of its length is above the water ?
(ii) If 3$\frac { 1 }{ 3 }$ m of the lamp post is above the water, find the whole length of the lamp post.

(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 – ( 1213) = 1 – (3+2)/6

=1- 56 = (6-5)/6 = 1/6 m

(ii)

But length of post above water

= 3 13 m=10/3 m

∴ 1/6th of total length = 10/3 m

∴ Total length = 103 × 61 = 20 m

#### Question-17:-

I spent $\frac { 3 }{ 5 }$ of my savings and still have Rs. 2,000 left. What were my savings ?

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

∴ Part left = 1-35 = (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, $\frac { 4 }{ 5 }$ of the children are boys. If the number of girls is 200, find the number of boys.

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

∴ No. of. girls = (1 – 45) 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= 1000

∴ No. of boys = 4/5 of 1000

=  4/5 of 1000 = (4/5) x 1000

= 800

#### Question-19:-

If $\frac { 4 }{ 5 }$ of an estate is worth Rs. 42,000, find the worth of whole estate. Also, find the value of $\frac { 3 }{ 7 }$ of it.

4/5 of an estate = Rs. 42000

∴ Total value of estate = Rs. 42000 × 54

= Rs. 105000 × 5 = Rs. 52500

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

=3/7 of  52500 = 37 × 52500

= 3 × 7500 = 22500

#### Question-20:-

After going $\frac { 3 }{ 4 }$ of my journey, I find that I have covered 16 km. How much Journey is still left ?

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

= 13

#### Question-21:-

When Krishna travelled 25 km, he found that $\frac { 3 }{ 5 }$ of his journey was still left. What was the length of the whole journey.

3/5 of the total journey was left

∴ Journey travelled by him = 1-45

= (5-3) 5 = 25

2/5 of total journey = 235 km

∴ Total journey = 25 km ×5/2=125/2 km

= 62 1 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.

Let the piece by Rajesh = 1 × 13 = 1 m

Land bought by Rajesh = 1 × 13 = 1 m

Remaining land = 1 – 13 = 3- 13 =23 m

Now, land bought by Manoj = 23 × 13 = 2/9 m

Land unsold =23 29

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

Land of 4/9 m remain unsold = 600 m²

∴ Total area of the land = 600 × 94

= 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?

Money in hand = ₹ 2400

Money spent on buying clothes

₹= 3of ₹2400

₹= 35 × 2400=3×480=₹1440

Remaining money = ₹ 2400 – ₹ 1440 = ₹ 960

Now, boy spent 1/4 of the remaining money on buying shoes.

∴ Money spent on buying shoes

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

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