# 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

–: Select Topics :–

**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.**

** **

**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) Thes****e 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

**are the required fractions.**

^{14}⁄_{18}**(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} **

**Question 3:-**

**Convert given fractions into fractions with equal numerators :**

**Answer:-3**

(**i)**** ****L.C.M. of 8 and 12 = 24 **

Hence the required fractions are**.. ^{24}⁄_{27} ** and

^{24}⁄_{34}**(ii)**

**L.C.M. of 6,15 and 12 = 60**

**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:-**

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

**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**

Hence,**^{8}⁄_{11} , ^{5}⁄_{7, } ^{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,}**> ^{24}⁄_{40, }>_{ } ^{24}⁄ 44 > ^{24}⁄ 240**

** **or ^{8}⁄_{11} > ^{3}⁄_{5, }>_{ } ^{6}⁄ 11 > ^{1}⁄ 10

**Hence, ^{8}⁄_{11} , ^{3}⁄_{5, } ^{6}⁄ 11 ^{1}⁄ 10 are in descending order.**

**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}⁄ 72Hence ^{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}⁄ 12Fraction 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}⁄ 5and 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**

**(iii)**

**.**

^{3}⁄_{8,}and^{9}⁄_{19,}….**.=…**

^{(3+9)}⁄_{(8+19)}**……=…**

^{12}⁄27**.**

_{,}

^{4}⁄_{9}**…**

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 1^{1}⁄_{6,}….**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}…..=….4^{2}⁄_{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:-**

**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

** 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}**

**Questio**n**-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 = **10 ^{1}⁄_{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..=… }×^{2}⁄**

_{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+2)/6**

^{1}⁄_{3}=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 }**×

**= 20 m**

^{6}⁄_{1}**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 }**=

**m**

^{1}⁄_{3 }Land bought by Rajesh = 1 × ** ^{1}⁄_{3 }**=

**m**

^{1}⁄_{3 }Remaining land = 1 – ** ^{1}⁄_{3 }**= 3-

**=**

^{1}⁄_{3 }**m**

^{2}⁄_{3 }Now, land bought by Manoj = ** ^{2}⁄_{3 }**×

**= 2/9 m**

^{1}⁄_{3 }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:-**

Money in hand = **₹ 2400**

Money spent on buying clothes

**₹= ^{3}⁄_{5 }of ₹2400**

**₹= ^{3}⁄_{5 }× 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

**=₹960×**

^{1}⁄_{4 }=₹240— End of **Fractions ICSE Class-7th Concise**** **Solutions :–

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

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