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
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Exercise – 8 A Solved Questions of Percent and Percentage for ICSE Class-7th
Question 1.
Express each of the following as percent :
Answer
Question 2.
Express the following percentages as fractions and as decimal numbers :
Answer
(i)
(ii) 2.50 %
= 0.025
(iii)
0.02 %
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
Question 4.
Find the value of:
Answer
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 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
Question 12.
(i) Find 2% of 2 hours 30 min.
(ii) What percent of 12 kg is 725 gm?
Answer
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
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.
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 % 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 :
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 :
Answer
(i)
Let y = a
(ii)
Let y = a
(iii)
Let y = a
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
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 :–
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