Percent and Percentage ICSE Class-7th Concise Selina Maths Solutions

Percent and Percentage ICSE Class-7th Concise Selina Maths Solutions Chapter-8 . We provide step by step Solutions of Exercise / lesson-8 Percent and Percentage for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-8 A , Exe-8 B and Exe-8 C, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

Percent and Percentage ICSE Class-7th Concise Selina  Solutions Chapter-8


–: Select Topics :–

Exe-8 A,

Exe-8 B,

Exe-8 C,


 Exercise – 8 A Solved Questions of Percent and Percentage for ICSE Class-7th

Question 1.

Express each of the following as percent :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-a1
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 1

Question 2.

Express the following percentages as fractions and as decimal numbers :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-a2

Answer

(i)

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 1

(ii) 2.50 %

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 2

= 0.025

(iii)

0.02 %

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 3

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 4

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 3

Question 3.

What percent is :
(i) 16 hours of 2 days ?
(ii) 40 paisa of Rs. 2 ?
(iii) 25 cm of 4 metres
(iv) 600 gm of 5 kg ?
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 4

Question 4.

Find the value of:
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-a4
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 5

Question 5.

In a class of 60 children, 30% are girls. How many boys are there ?
Answer

Total children = 60,

Girls = 30%

∴Total girls = 30% of 60

= 60×(30/100)=18

∴ No. of boys = 60 – 18

= 42

Question 6.

In an election, two candidates A and B contested. A got 60% of the votes. The total votes polled were 8000. How many votes did each get ?
Answer

Total number of votes polled = 8000

A got 60% of the votes

A got total votes = 60% of 8000

= 8000×(60/100)=4800

∴ B got total votes = 8000 – 4800

= 3200

Question 7.

A person saves 12% of his salary every month. If his salary is ₹2,500, find his expenditure.
Answer

Total salary = ₹2500

Saving = 12% of the salary

∴ Total savings = 12% of ₹ 2500

= ₹ 2500×(12/100)

=₹300

∴ Total expenditure = ₹ 2500 – ₹ 300

= ₹ 2200

Question 8.

Seeta got 75% marks out of a total of 800. How many marks did she lose ?
Answer

Total marks = 800

Marks Seeta got = 75% of total marks

∴ Total marks Seeta got = 75% of 800

= 800×(75/100)

=600

∴ Marks Seeta lose = 800 – 600

= 200

Question 9.

A shop worth ₹25,000 was insured for 95% of its value. How much would the owner get in case of any miss happening ?
Answer

Value of shop =₹25,000

Insured amount = 95% of total value

=95% of ₹25,000

=₹25,000×(95/100)

= ₹ 23750

Question 10.

A class has 30 boys and 25 girls. What is the percentage of boys in the class ?
Answer

No. of boys = 30

No. of girls = 25

Total number of children = 30 + 25 = 55

∴Percentage of boys in the class

=(30/55)×100

=(600/11)=54 (6/11)%

Question 11.

Express :
(i) 3 \frac { 2 }{ 5 } as a percent
(ii) 0.0075 as percent
(iii) 3 : 20 as percent
(iv) 60 cm as percent of 1 m 25 cm
(v) 9 hours as a percent of 4 days.
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 12
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 13

Question 12.

(i) Find 2% of 2 hours 30 min.
(ii) What percent of 12 kg is 725 gm?
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 7 Percent and Percentage 14


 Exercise – 8 B Percent and Percentage of  Concise Selina Mathematics

Question 1.

Deepak bought a basket of mangoes containing 250 mangoes 12% of these were found to be rotten. Of the remaining, 10% got crushed. How many mangoes were in good condition ?
Answer

Total mangoes = 250

Rotten mangoes = 12 % of 250

= 250 × (12/100)=30

Remaining mangoes = 250 – 30

= 220

Mangoes which were crushed = 10% of 220

=220×(10/100)

=22

∴ Balance = 220 – 22

= 198

Hence 198 mangoes were in good condition.

Question 2.

In a Maths Quiz of 60 questions, Chandra got 90% correct answers and Ram got 80% correct answers. How many correct answers did each give ?
What percent is Ram’s correct answers to Chandra’s correct answers ?
Answer

No. of . total questions = 60

Chandra got correct answers of the questions

= 90% of 60

=(60×90)/100=54

Ram got correct answer of the questions

= 80% of 60

=(60×80)/100=48

∴ Percentage of Ram’s correct answer of that of Chandra’s

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 5

Question 3.

In an examination, the maximum marks are 900. A student gets 33% of the maximum marks and fails by 45 marks. What is the passing mark ? Also, find the pass percentage.
Answer

Maximum marks = 900

A student got 33% of 900 marks

=900×(33/100)=297

No. of. marks by which he  failed = 45

∴ Pass marks = 297 + 45 = 342

Percentage of pass marks = (342×100)/900

=38%

Question 4.

In a train, 15% people travel in first class, 35% travel in second class. The balance travel in the A.C. class ? Calculate the percentage of A.C. class travellers ?
Answer

Let no. of. people = 100

No. of. people in first class = 15

and no.of people travel in second class = 35

∴ Balance = 100 – (15 + 35) = 100 – 50 = 50

∴ Percent of people travel in AC class = 50 %.

Question 5.

A boy eats 25% of the cake and gives away 35% of it to his friends. What percent of the cake is still left with him ?
Answer

Let total cake = 100

Cake which was eaten by the boy = 25

Cake which was given to his friends = 35

∴ Balance cake = 100 – (25 + 35)

= 100 – 60 = 40

Hence he has 40% of the cake with him.

Question 6.

What is the percentage of vowels in the English alphabet ?
Answer

There are 5 vowels in 26 English alphabets

∴ Percentage of vowels =(5×100)/26

=(250/13)

=19 (3/13)%

Question 7.

Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-b7
Answer

(i) 

Let number be x.

Then 6 (1/4)% of x = 375

⇒ 25/(4×100) of x = 375

⇒ 1/16 x = 375

so x = (375×16)/1

=6000

Hence number = 6000

(ii)

Let number x

then 0.2 % of x = 5

⇒ 2/(10×100) of x = 5

⇒ 1/500 of x = 5

⇒ x = (5×500)1

∴ Number = 2500

(iii) 

Let the number x

then 16 (2/3)% of x = 30

⇒ 50/(3×100) of x = 30

⇒1/6 of x = 30

⇒ x = 30 × 6 = 180

Hence number = 180

Question 8.

The money spent on the repairs of a house was 1% of its value. If the repair, costs Rs. 5,000, find the cost of the house.
Answer

Let cost of house = x

Then cost of repairs = 1 % of x

∴ 1% of x = 5000

⇒(1/100)× x = 5000

⇒ x = 5000 × (100/1)

x = 500000

Hence cost of house = Rs. 500000

Question 9.

In a school out of 300 students, 70% are girls and 30% are boys. If 30 girls leave and no new boy is admitted, what is the new percentage of girls in the school ?
Answer

total number of children in a school = 300

No. of. boys = 30% of 300

=(30/100)×300=90

and no.of.girls = 70% of 300

=(70/100)×300=210

Now no.of girls left = 30

∴ No. of. girls after leaving 30 girls

= 210 – 30 = 180

and No. of. children in the school = 180 +  90 = 270

∴ % of girls now =(180/270)×100=(200/3)%

=66 (2/3)%

Question 10.

Kumar bought a transistor for Rs. 960. He paid 12 \frac { 1 }{ 2 } % cash money. The rest he agreed to pay in 12 equal monthly instalments. How much will he pay each month ?
Answer

Price of transistor = Rs. 960

Amount paid in cash = 12 (1/2)% of Rs. 960

=252×100×960 = Rs. 120

Balance amount = Rs. 960 – Rs. 120

Rs. 840

No. of instalments = 12

∴ Amount of each instalment = Rs. 840 ÷ 12 = Rs. 70

Question 11.

An ore contains 20% zinc. How many kg of ore will be required to get 45 kg of zinc ?
Answer

In an ore, zinc = 20%

Let quantity of ore = x

∴ 20% of x = 45 kg

⇒ (20/100) × x=45

⇒ x/5 = 45

⇒ x=45×5=225

∴ quantity of ore = 225 kg


 Exercise – 8 C Percent and Percentage Concise Class-7th

Question 1.

The salary of a man is increased from Rs. 600 per month to Rs. 850 per month. Express the increase in salary as percent.
Answer

Original salary of a man = Rs. 600

Increased salary = Rs. 850

∴ Amount of increase = Rs. 850 – 600 = Rs. 250

Percentage increase

=(250×100)/600=125/3

=41 (2/3)%

Question 2.

Increase :
(i) 60 by 5%
(ii) 20 by 15%
(iii) 48 by 121 %
(iv) 80 by 140%
(v) 1000 by 3.5%

Answer

(i)

Rate of increase = 5%

∴ Total increase = 5% of 60 =(5/100)×60

∴ Increased number = 60 + 3 = 63

(ii)

Increase on 20 at the rate of 15%

=20×(15/100)=3

∴ Increased number = 20 + 3 = 23

(iii)

48 by 12 (1/2) %

= 48 × (25/2)%

= 48 × 25/(2×100)

=48×18=6

Increased number = 48 + 6 = 54

(iv)

Increase on 80 by 140% = 80×(140/100)=112

increased number = 80 + 112 = 192

(v)

Increase on 1000 by 3.5% = 1000×(3.5/100)

= 1000 × 35(10×100)=35

∴ increased number = 1000 + 35 = 1035

Question 3.

Decrease :
(i)80 by 20%
(ii) 300 by 10%
(iii) 50 by 12.5%
Answer

(i)

Decrease on 80 by 20% =80×(20/100)=16

∴ Decreased number = 80 – 16 = 64

(ii)

Decrease on 300 by 10% =300×(10/100)=30

∴ Decreased number = 300 – 30 = 270

(iii)

Decrease on 50 by 12.5% = 50×(12.5/100)

=(50×125)/(10×100)=25/4=6.25%

∴ Decreased number = 50 – 6.25 = 43.75

Question 4.

What number :
(i) When increased by 10% becomes 88 ?
(ii) When increased by 15% becomes 230 ?
(iii) When decreased by 15% becomes 170 ?
(iv) When decreased by 40% becomes 480 ?
(v) When increased by 100% becomes 100 ?
(vi) When decreased by 50% becomes 50 ?
Answer

(i)

Let the number be = 100

Increase = 10% = 10

Increased number = 100 + 10 = 110

If increased number is 110, then original number = 100

and if increased number is 88, then original number = (100/110)×88

=80

 

(ii)

Let the number be = 100

Increase = 15% = 15

∴ Increased number = 100 + 15 = 115

If increased number is 115, then original number = 100

and if increased number is 230, then original number =(100/230)×115

=200

(iii)

Let the number be = 100

Increase = 15% = 15

∴ Decreased number = 100 – 15 = 85

If decreased number is 85, then original number = 100

and if decreased number is 170, then original number =(100/85)×170

=200

(iv)

Let the number be = 100

Decrease = 40% = 40

∴ Decreased number = 100 – 40 = 60

If decreased number is 60, then original number = 100

and if decreased number is 480, then original number =(100/480)×60

=800

(v)

Let the number be = 100

Increase = 100% = 100

∴ Increased number = 100 + 100 = 200

If Increased number is 200, then original number = 100

and if Increased number is 100, then original number =(100×100)/200

=50

(vi)

Let the number be = 100

Decrease = 50% = 50

∴ Decreased number = 100 – 50 = 50

If decreased number is 50, then original number = 100

and if decreased number is 50, then original number =(100×50)/50

=100

Question 5.

The price of a car is lowered by 20% to Rs. 40,000. What was the original price ? Also, find the reduction in price.
Answer

Let original price of the car = Rs. 100

Reduction = 20%’ = Rs. 20

∴ Reduced price = Rs. 100 – 20 = Rs. 80

If reduced price is Rs. 80, then original price = Rs. 100

and if reduced price is Rs. 40,000 then original price =(100×40000)/80

= Rs. 50000

and reduction = Rs. 50000 – Rs. 40000

= Rs, 10000

Question 6.

If the price of an article is increased by 25%, The increase is Rs. 10. Find the new price.
Answer

Let the price of an article = Rs. 100

Increase = 25%

∴Increase = Rs. 25

If an increased price = Rs. 100 + 25 = Rs. 125

If increase is Rs. 25 then new price = Rs. 125

and if increase is Rs. 10, then new price = Rs. (125×10)/25

= Rs. 50

Question 7.

If the price of an article is reduced by 10%, the reduction is Rs. 40. What is the old price ?
Answer

Let the original (old) price = Rs. 100

Reduction = 10% = Rs. 10

∴If reduction is Rs. 10, then old price = Rs. 100

and if reduction is Rs. 40, then old price = Rs. (100×40)/10 = Rs. 400

Question 8.

The price of a chair is reduced by 25%. What is the ratio of:
(i) Change in price to the old price.
(ii) Old price to the new price.
Answer

Let old (original) price of a chair = Rs. 100

Reduction = 25% = Rs. 25

∴Reduced price = Rs. 100 – Rs. 25 = Rs. 75

(i) Ratio between change in price and old price = 25 : 100

= 1:4 (Dividing by 25)

(ii) Ratio between old price and new price = 100 : 75

= 4:3 (Dividing by 25)

Question 9.

If x is 20% less than y, find :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-c9
Answer

(i)

Let y = 100

then reduction = 20% = 20

then x = 100 – 20 = 80

x/y = (80/100)=45  …

(Dividing by 20)

(ii)

Let y = 100

then reduction = 20% = 20

then x = 100 – 20 = 80

(y-x)/y=1(00-80)/100

=20/100=1/5  …

(Dividing by 20)

(iii) 

Question 10.

If x is 30% more than y; find :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage-c10

Answer

(i)

Let y = a

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 6

(ii)

Let y = a

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 7

(iii)

Let y = a

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 8

Question 11.

The weight of a machine is 40 kg. By mistake it was weighed as 40.8 kg. Find the error percent.
Answer

Actual weight of machine = 40 kg

Errored weight = 40.8 kg

∴ Error in weight = 40.8 – 40 = 0.8 kg

Error % = (0.8×100)/40=

(8×100)/(10×40)=2%

Question 12.

From a cask, containing 450 litres of petrol, 8% of the petrol was lost by leakage and evaporation. How many litres of petrol was left in the cask ?
Answer

Original petrol in the cask = 450 litres

Leakage and evaporation = 8 %

∴ Lost petrol = 8 % of 450 litres =(8×450)/100 = 36 litres

Question 13.

An alloy consists of 13 parts of copper, 7 parts of zinc and 5 parts of nickel. What is the percentage of each metal in the alloy?
Answer

copper = 13 parts, zinc – 7 parts Nickel = 5 Parts

Total alloy = 13 + 7 + 5 = 25 parts

Now, percentage of copper = (13/25)×100 = 52%

Percentage of zinc = (7/25)×100 = 28%

and percentage of nickel = (5/25)×100 = 20%

Question 14.

In an examination, first division marks are 60%. A student secures 538 marks and misses the first division by 2 marks. Find the total marks of the examination.
Answer

Percentage for first division = 60%

A student secures 538 marks but misses the first division by 2 marks.

∴ Marks for first division = 538 + 2 = 540

∴ 60% of total marks = 540

⇒ (60/100)×total marks

=540

⇒ total marks=(540×100)/60

=900

Question 15.

Out of 1200 pupils in a school, 900 are boys and the rest are girls. If 20% of the boys and 30% of the girls wear spectacles, find :
(i) how many pupils in all, wear spectacles ?
(ii) what percent of the total number of pupils wear spectacles ?
Answer

Total number of pupils = 1200

No.of. boys = 900

and no. of girls = 1200 – 900 = 300

No. of. boys who wear spectacles

= 20% of 900 = (20/100)×900=180

No.of. girls who wear spectacles

= 30% of 300 = (30/100)×300=90

(i) ∴ Total number of pupils who  were spectacles = 180 + 90 = 270

(ii) Percentage of pupils who were spectacles

= (270×100)/1200

=27012

=22.5%

Question 16.

Out of 25 identical bulbs, 17 are red, 3 are black and the remaining are yellow. Find the difference between the numbers of red and yellow bulbs and express this difference as percent.
Answer

Total number of bulbs = 25

Number of red bulbs = 17

Number of black bulbs = 3

= 17 + 3 = 20

∴ Number of yellow bulbs = 25 – 20 = 5 bulbs

Difference between the number of red and yellow bulbus

= No.of red bulbs – No. of yellow bulbs

= 17 – 5 = 12

∴ Percentage difference

Exercise - 8 A Solved Questions of Percent and Percentage for ICSE Class-7th img 9

Question 17.

A number first increases by 20% and then decreases by 20%. Find the percentage increase or decrease on the whole.
Answer

Let the number be 100

In 1st condition,

increase = 20% of 100

= (20/100)×100=20

∴ Number after this increase = 100 + 20 = 120

In 2nd condition,

Decrease = 20 % of 120

= (20/100)×120=24

∴ Number after this decrease = 120 – 24 = 96

⇒ Total decrease on the whole = 24 – 20 = 4

and the percentage decrease on the whole

=4/100×100=4%

Question 18.

A number is first decreased by 40% and then again decreased by 60%. Find the percentage increase or decrease on the whole.
Answer

Let the number be 100

In first condition:

Decrease = 40% of 100

= (40/100)×100=40

∴ Number after this decrease = 100 – 40 = 60

In second condition:

Decrease = 60% of 40

= (60/100)×40=24

∴ Number after the 2nd decrease

= 40 – 24 = 16

⇒ Total decrease on the whole = 40 + 24 = 64

∴ Percentage decrease on the whole

= (64/100)×100=64%

Question 19.

If 150% of a number is 750, find 60% of this number.
Answer

Let the required number be x

Now, 150% of x = 750

⇒(150/100) × x = 750

 x = (750×100)/150

=500

Hence,

the required number = 500

Now, 60% of 500 = 500×(60/100)=300

— End of Percent and Percentage Solutions :–

Return to  Concise Selina Maths Solutions for ICSE Class -7 


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