July 2, 2013

Percentages

Percentages Concept test…

1. A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7: 8: 9: 10. In all papers together, the candidate obtained 60% of the total marks. Then, the number of papers in which he got more than 50% marks is :
(a) 1
(b) 3
(c) 4
(d) 5

Solution: let the marks scored in 5 subjects be 6x, 7x, 8x, 9x, and 10x (on a scale of 1)
Average score = 60%

6x+7x+8x+9x+10x = 60/100
5
8x = 0.6 x = 0.075

So, the marks are 0.45, 0.525, 0.6, 0.675 and 0.75
number of times the marks exceed 50% is 4

2. In a municipal election, there were two candidates. On who got 30% of the votes polled was defeated by 16000 votes. Calculate the total number of votes polled.
(a) 24000
(b) 28000
(c) 30000
(d) 40000

Solution :
Let the total number of votes polled = x
The winning candidate got 70% of the votes polled.
Hence 40% of x = 16000 x = 40000.
30% of increase in Y
15% of increase in Z i.e. 32.25% increase in Z2

Expenditure by Ajay on batteries
= 80% of 150 + 150/2

= 120 + 75 = Rs. 195.
Suppose X = k% of 4/5 X
X = k/100 x 4/5 x k = 125
100 – 20% = 80
80 – 30% = 56
Single discount = 44%

3. Ajay and Vikas are sharing a flat in Delhi, with an arrangement of equally dividing the household expenses. Ajay went to Pune, where a sale was going on and bought batteries for the house, worth Rs. 150 all 20% discount. But he lost them on his WG back and had to buy new ones, after he reached Delhi. How much did he end up
spending on the batteries?
(a) Rs. 280
(b) Rs. 195
(c) Rs. 270
(d) Rs. 75

Solution :
Expenditure by Ajay on batteries

= 80% of 150 + 150/x

= 120 + 75 = Rs. 195.

4. A manufacturer offers a 20% rebate on the marked price of a product. The retailer offers another 30% rebate on the reduced price. The two reductions are equal to a single reduction of
(a) 50%
(b) 44%
(c) 46%
(d) 40%

Solution :

100 – 20% = 80
80 – 30% = 56
Single discount = 44%

5. The population of a variety of tiny bush in an experimental field increased by 10% in the first year, increased by 8% in the second year but decreased by 10% in the third year. If the present number of bushes in the experimental field is 26730, then the number of variety of bushes in the beginning was
(a) 35000
(b) 27000
(c) 25000
(d) 36000

Solution :
Suppose the number of bushes in the experimental field = x
No. of bushes after one year

= x + 10% of x = 11x/10
No. of bushes after 2nd year

= 11x/10+ 8% of x = 11x/10 =
No. of bushes after 3rd year

= – 10% of

= x

x = 26730 x = 25000

6. In a certain shop, which stocks four types of caps, there are 1/3 as many red caps as blue and 1/2 as many green caps as red caps.
There are equal numbers of green caps and yellow caps. If there are 42 blue caps, then what percent of the total caps in the shop are blue?
(a) 70%
(b) 28%
(c) 60%
(d) 14%

Solution:
R = 1/3 B, G = 1/2 R = G = Y

Since B = 42, R = 14, G = 7 and Y = 7

Percentage of blue caps = 42/70 X 100 = 60

7. A bag contains 600 pens of brand A and 1200 pens of brand B. If 12% of brand A pens and 25% of brand B pens are removed, then what is the approximate percentage of total pens removed from the bag ?
(a) 37%
(b) 36%
(c) 22%’
(d) 18%

Solution :
No. of pens removed
= 12% of 600 + 25% of 1200
= 72 + 300 = 372
Percentage of total pens removed

= 372/1800 X 100 = 20.67 = 22

8. A shopkeeper offers to sell 331/3% more quantity of a product for a price that is 20% higher. What is the effective discount that he is offering ?
(a) 10.8%
(b) 13%
(c) 16.67%
(d) 10%

Solution :
Suppose the list price of 6 kg is Rs. 600
Price of 1 kg = Rs. 100
The shopkeeper offers to sell 8 kg for Rs. 720, i.e. Rs. 90 per kg.
Hence, Discount = 10%

9. At a school, 20% of the students are seniors. If all of the seniors attended the school play, and 60% of all the students attended the play, then what percent of the non-seniors attended the play?
(a) 20%
(b) 40%
(c) 50%
(d) 100%

Solution : Suppose total number of students = 100
No. of seniors who attended the play = 20
Total number of students who attended the play = 60
No. of non-seniors who attended the play = 20
= 60 – 20 = 40 i.e. 40%

10. The population of a city increases at the rate of 4% per annum. There is an additional annual increase of 1 % in the population due to the influx of job seekers. Therefore, the % increase in the population after 2 years will be:
(a) 10
(b) 10.25
(c) 10.50
(d) 10.75

Solution : 100 105 110.25, I.e. 10.25% .
11. In a society, there are 100 members. Each of them has been allotted membership number from 1 to 100. They started a business in which the nth member contributed Rs. (10 x 2n-S). After one year, 4th member gets Rs. 62 as his share. Find the total profit in the business after one year.
(a)Rs.8 [2100-26]
(b) Rs.4 [299-26]
(c) Rs. 2 [2100 – 26]
(d) None of these

Solution :
nth member contributed Rs. (10 X 2n – 5)
1st member contributed Rs. 15
2nd member contributed Rs. 35
3rd member contributed Rs. 75
4nd member contributed Rs. 155
And so on.
Since 4th member gets Rs. 62 as his share in the profit, therefore we conclude that 40% profit is earned by each member.
Total profit earned
= 40% of [15 + 35 + 75 + ………. + up to 100 terms]

= 2 [3 + 7 + 15 + 31 + ………. + up to 100 terms]

= 2 [(4 + 8 + 16 + 32 + ………. + up to 100 terms]

= 8 [(1 + 2 + 4 + 8 + ………. + up to 100 terms) – 25]
= 8 – 25

= 8 (2100 – 1 – 25)
= 8 (2100 – 26)

12. Avinash spends 30% of his income on scooter petrol, ¼ o the remaining on house rent and the balance on food. If he spends Rs.300 on petrol, then what is the expenditure on house rent?

a) 175
b) 200
c) 125
d) 180
Ans : 175

Solution :

From the given information, Method: 1

For scooter petrol 30x / 100 = 3x / 10
remaining  x – 3x / 10 = 7x / 10 for house  ¼ x 7x / 10 = 7x / 40

remaining for food — > 7x / 10 – 7 x / 40

=> 28 x – 7x / 40 = 21x /40
Amount for petrol = 30/100 x = 30
= 1000
House rent,
7x / 40 = 7 x 1000 / 40 = 175
Method :2
Total income x x – 300
X – 30/100 x =
(in %) = (in Rs.)
30 / 100 x = 300
x = 1000
rem
¼ x 100 = 175Rs.
13. A house owner was having his house painted. He was advised that he would require 25kg of paint. Allowing for 15% wastage and assuming that the paint is available in 2kg cans, what would be the cost of pain purchased, if one can cost Rs.16?

a) 240
b) 480
c) 120
d) 260
Ans: 240Rs.

Solution :

Required no of kgs 25kgs

10% 2.5

5% / 15%  1.25 / 28.75kg

1 can contains 2kg paint
= 28.75

14 can 1 can

= 15 cans

= 15 x 16 = 240

1 can = 16 Rs.
14. A number on subtracting 15 from it, reduces to its 80% what is 40% of that number?

a) 15
b) 30
c) 25
d) 12
Solution :
Let the number be x
x-15 = 80%x

x – 15 – 80/100 x x – 15 = 4/5 x

x – 4/5 x = 15

x / 5 = 15

75

Now,

40% of 75

= 40 /100 x 75

=2/5 x 15
=30
Hint:
30% (75) = 80/100 x 75 = 60 —(1)
= 75- 15 = 60 —(2)
(1)=(2)

15. From a news article: “In the first six months of 1996, the number of deaths [from AIDS] fell 12% to 22,000 [compared to the same period a year earlier].” How many AIDS deaths were there during the first six months of 1995?

a) 20000
b) 24000
c) 29000
d) 25000

Solution:
1995 → 1996
old new
old 22,000

Because there was a 12% decrease, we have to use –12%

–12% = (22,000 – old) / old

–0.12 * old = 22,000 – old

–0.12 * old + old = 22,000 0.88 * old = 22,000

old = 22,000 / 0.88 old = 25,000

16. Suppose you are earning a salary of $1,000 per week. Your company experiences a slowdown in earnings, and asks all workers to take a 20% pay cut. What is your salary after the cut? Six months later, the company recovers and it offers you a 20% increase on your current salary. How much will you be earning after the increase?
a) $800
b) $920
c) $850
d) $960

Solution:

New salary = (1 – 20%)(1 + 20%) – 1 New salary = (1 – 0.20)(1 + 0.20) – 1 New salary = (0.80)(1.20) – 1

New salary = 0.96 – 1

New salary = –0.04 * 100%

New salary will have a decrease of 4%, or ($1000 – $1000 * 0.04) = $960

17. A chemist has a 20% and a 40% acid solutions. What amount of each solution should be used in order to make 300 ml of a 28% acid solution?
a) x =120,y=180
b) x =100,y=200
c) x =80,y=120
d) x=130,y=170

Solution:
Let x be the solution at 20% and y be the solution at 40%, hence x + y = 300 ml

We now write an equation that expresses that the total acid in the final 300 ml is equal to the sum of the amounts of acid in x and y

28% * 300 = 20% * x + 40% * y

Solve the above system of equations to find x = 180 and y = 120

18. Cassandra invested one part of her $10,000 at 7.5% per year and the other part at 8.5% per year. Her income from the two investment was $820. How much did she invest at each rate?

a) X= $3000 ,Y =$820
b) X= $2500 ,Y =$920
c) X= $1400 ,Y =$1020
d) X= $2200 ,Y =$800
Solution :

Let x and y be the amount invested at 7.5% and 8.5% respectively Income = $820 = 7.5% * x + 8.5% * y

The total amount invested is also known 10,000 = x + y

Solve the system of the equations to find x and y. x = $3000 and y = $7000

As a practice check that 7.5% of $3000 and 8.5% of $7000 gives $820
19. Out of a world population of approximately 6.6 billion, 1.2 billion people live in the richer countries of Europe, North America, Japan and Oceania and is growing at the rate of 0.25% per year, while the other 5.4 billion people live in the lees developed countries and is growing at the rate of 1.5%. What will be the world population in 5 years if we assume that these rates of increase will stay constant for the next 5 years. (round answer to 3 significant digits)

a) 7.03billion
b) 8.00billion
c) 9.03billion
d) 12.05billion

Solution

Let us first calculate the population PR in 5 years in the richer countries PR = (1.2 + 0.25% * 1.2) = 1.2(1 + 0.25%) after one year

PR = 1.2(1 + 0.25%) + 0.25% * 1.2(1 + 0.25%) = 1.2(1 + 0.25%) 2after two years

Continue with the above and after 5 years, PR will be PR = 1.2(1 + 0.25%) 5 after 5 years

Similar calculations can be used to find the population PL in less developed countries after 5 years.
PL = 5.4(1 + 1.5%) 5 after 5 years
The world population P after 5 years will be
P = PR + PL = 1.2(1 + 0.25%) 5 + 5.4(1 + 1.5%) 5 = 7.03 billion
20. In a test of 80 questions, Jyothsna answered 75% of the first 60 questions correctly. What % of the remaining questions she has to answer correctly so that she can secure an overall

percentage of 80 in the test?
a. 80% b. 90% c. 85% D. 95%
Solution :
[(75% X 60) + (x% X 20)] / 80 = 80% => x = 95. (since required is 80%) (OR) 60 out of 80 is 3/4. So, (3/4 X 75) + (1/4 X x) = 80 => x =95.

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