Selina Solutions Concise Maths Class 7 Chapter 8 Percent and Percentage Exercise 8C explains the concept of percentage change. The solved examples before the exercise problems help students understand the method of solving problems, with ease. The solutions PDF has answers in a stepwise manner based on the latest syllabus and ICSE exam pattern. It provides a chance to students to analyse their weaknesses and work on them for better academic performance. Students can clarify their doubts using Selina Solutions Concise Maths Class 7 Chapter 8 Percent and Percentage Exercise 8C PDF, from the links which are provided below for free download.
Selina Solutions Concise Maths Class 7 Chapter 8: Percent and Percentage Exercise 8C Download PDF
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Exercise 8C page: 100
1. The salary of a man is increased from ₹ 600 per month to ₹ 850 per month. Express the increase in salary as percent.
Solution:
Salary of a man = ₹ 600
Increased salary of a man = ₹ 850
So the amount of increase = 850 – 600 = ₹ 250
Here the percentage increase = (250 × 100)/ 600
We get
= 125/3
= 41 2/3%
2. Increase:
(i) 60 by 5%
(ii) 20 by 15%
(iii) 48 by 12 ½ %
(iv) 80 by 140%
(v) 1000 by 3.5%
Solution:
(i) 60 by 5%
It is given that
Rate of increase = 5%
So the total increase = 5% of 60
We can write it as
= 5/100 × 60
= 3
Here the increased number = 60 + 3 = 63
(ii) 20 by 15%
Increase on 20 by 15% = 20 × 15/100 = 3
So the increased number = 20 + 3 = 23
(iii) 48 by 12 ½ %
Increase on 48 by 12 ½ % = 48 × 25/2%
We can write it as
= 48 × 25/ (2 × 100)
By further calculation
= 48 × 1/8
= 6
So the increased number = 48 + 6 = 54
(iv) 80 by 140%
Increase on 80 by 140% = 80 × 140/100 = 112
So the increased number = 80 + 112 = 192
(v) 1000 by 3.5%
Increase on 1000 by 3.5% = 1000 × 3.5/100
We can write it as
= 1000 × 35/ (10 × 100)
= 35
So the increased number = 1000 + 35 = 1035
3. Decrease:
(i) 80 by 20%
(ii) 300 by10%
(iii) 50 by 12.5%
Solution:
(i) 80 by 20%
Decrease on 80 by 20% = 80 × 20/100 = 16
So the decreased number = 80 – 16 = 64
(ii) 300 by10%
Decrease on 300 by 10% = 300 × 10/100 = 30
So the decreased number = 300 – 30 = 270
(iii) 50 by 12.5%
Decrease on 50 by 12.5% = 50 × 12.5/100
We can write it as
= (50 × 125)/ (10 × 100)
= 25/4
= 6.25%
So the decreased number = 50 – 6.25 = 43.75
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?
Solution:
(i) Consider 100 as the number
So the increase = 10% = 10
Increased number = 100 + 10 = 110
If the increased number is 110 then the original number = 100
If the increased number is 88 then the original number = 100/110 × 88 = 80
(ii) Consider 100 as the number
So the increase = 15% = 15
Increased number = 100 + 15 = 115
If the increased number is 115 then the original number = 100
If the increased number is 230 then the original number = (100 × 230)/ 115 = 200
(iii) Consider 100 as the number
So the decrease = 15% = 15
Decreased number = 100 – 15 = 85
If the decreased number is 85 then the original number = 100
If the decreased number is 170 then the original number = 100/85 × 170 = 200
(iv) Consider 100 as the number
So the decrease = 40% = 40
Decreased number = 100 – 40 = 60
If the decreased number is 60 then the original number = 100
If the decreased number is 480 then the original number = (100 × 480)/ 60 = 800
(v) Consider 100 as the number
So the increase = 100% = 100
Increased number = 100 + 100 = 200
If the increased number is 200 then the original number = 100
If the increased number is 100 then the original number = (100 × 100)/ 200 = 50
(vi) Consider 100 as the number
So the decrease = 50% = 50
Decreased number = 100 – 50 = 50
If the decreased number is 50 then the original number = 100
If the decreased number is 50 then the original number = (100 × 50)/ 50 = 100
5. The price of a car is lowered by 20% to ₹ 40,000. What was the original price? Also, find the reduction in price.
Solution:
Consider ₹ 100 as the original price of the car
The price reduction = 20% = ₹ 20
So the reduced price = 100 – 20 = ₹ 80
If the reduced price of the car is ₹ 80 then the original price = ₹ 100
If the reduced price of the car is ₹ 40,000 then the original price = (100 × 40000)/ 80 = ₹ 50, 000
Reduction = 50000 – 40000 = ₹ 10, 000
6. If the price of an article is increased by 25%, the increase is ₹ 10. Find the new price.
Solution:
Consider ₹ 100 as the price of an article
The price of the article is increased = 25% = ₹ 25
So the increased price = 100 + 25 = ₹ 125
If the increase in the price is ₹ 25 then the new price = ₹ 125
If the increase in the price is ₹ 10 then the new price = (125 × 10)/ 25 = ₹ 50
7. If the price of an article is reduced by 10%, the reduction is ₹ 40. What is the old price?
Solution:
Consider ₹ 100 as the original price of an article
The price is reduced = 10% = ₹ 10
If the reduced price is ₹ 10 then the old price = ₹ 100
If the reduced price is ₹ 40 then the old price = (100 × 40)/ 10 = ₹ 400
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.
Solution:
Consider ₹ 100 as the original price of the chair
The price of the chair is reduced = 25% = ₹ 25
So the reduced price = 100 – 25 = ₹ 75
(i) Ratio of change in price to the old price = 25: 100
Dividing by 25
= 1: 4
(ii) Ratio of old price to the new price = 100: 75
Dividing by 25
= 4: 3
9. If x is 20% less than y, find:
(i) x/y
(ii) y – x/ y
(iii) x/ y – x
Solution:
Consider y = 100
Reduction = 20% = 20
x = 100 – 20 = 80
(i) x/ y = 80/ 100
Dividing by 20
= 4/5
(ii) (y – x)/ y = (100 – 80)/ 100
So we get
= 20/100
Dividing by 20
= 1/5
(iii) x/ (y – x) = 80/ (100 – 80)
So we get
= 80/20
Dividing by 20
= 4 /1
= 4
10. If x is 30% more than y; find:
(i) x/y
(ii) y + x/ x
(iii) y/ y – x
Solution:
Consider y = a
We know that
x = a × (100 + 30)/ 30
By further calculation
= a × 130/100
= 13/10 a
(i) x/y = 10/ 13/10 a
We can write it as
= (a × 10)/ 13a
= 10/13
(ii) y + x/ x = (a + 13/10 a)/ 13/10 a
We can write it as
= (10 + 13)a/ (10 × 13/10 a)
By further calculation
= 23a/10 × 10/13a
So we get
= 23/13
(iii) y/ y – x = a/ (a – 13/10 a)
We can write it as
= a/ -3/10 a
So we get
= (a × 10)/ – 3a
= – 10/3
11. The weight of a machine is 40 kg. By mistake, it was weighed as 40.8 kg. Find the error percent.
Solution:
Weight of the machine = 40 kg
Error weight of the machine = 40.8 kg
Error in weight = 40.8 – 40 = 0.8 kg
So the error percent = (0.8 × 100)/ 40
We can write it as
= (8 × 100)/ (10 × 40)
= 2%
12. From a cask, containing 450 litres of petrol, 8% of the petrol was lost by leakage and evaporation. How many litres of petrol were left in the cask?
Solution:
Petrol in the cask = 450 litres
Petrol lost by leakage and evaporation = 8%
So the petrol lost = 8% of 450 litres
We can write it as
= (8 × 450)/ 100
= 36 litres
Petrol left in the cask = 450 – 36 = 414 litres
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?
Solution:
In an alloy
Copper = 13 parts
Zinc = 7 parts
Nickel = 5 parts
So the total alloy = 13 + 7 + 5 = 25 parts
Percentage of copper = 13/25 × 100 = 52%
Percentage of zinc = 7/25 × 100 = 28%
Percentage of nickel = 5/25 × 100 = 20%
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.
Solution:
Marks for first division = 60%
A student gets 530 marks and misses the first division by 2 marks
Marks for first division = 538 + 2 = 540
60% of total marks = 540
We can write it as
60/100 × total marks = 540
So we get
Total marks = (540 × 100)/ 60 = 900
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.
Solution:
Number of pupils = 1200
Number of boys = 900
Number of girls = 1200 – 900 = 300
Number of boys who wear spectacles = 20% of 900
We can write it as
= 20/100 × 900
= 180
Number of girls who wear spectacles = 30% of 300
We can write it as
= 30/100 × 300
= 90
(i) Number of pupils in all wear spectacles = 180 + 90 = 270
(ii) Percent of the total number of pupils wear spectacles = (270 × 100)/ 1200
So we get
= 270/12
= 22.5%
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