Percentage
Q1. 50% of ? = 250
(a) 400
(b) 500
(c) 800
(d) 1000
Answer: (b) 500
Let, 50% of x = 250
⇒ (50/100) X x = 250
⇒ x = (250 X 100)/50 = 500
Q2. Rs. 395 are divided among A, B and C in such a manner that B gets 25% more than A and 20% more than C. The share of A is
(a) Rs. 80
(b) Rs. 95
(c) Rs. 120
(d) Rs. 135
Answer: (c) Rs. 120
Let A gets Rs. x
Then B gets 125% of x = 125/100 X x = Rs. 5x/4
If B gets Rs. 5x/4, then C gets = Rs. ( | 100 | X | 5x | ) = Rs. | 25x |
120 | 4 | 24 |
∴ x + | 5x | + | 25x | = 395 |
4 | 24 |
⇒ 24x + 30x + 25x = 395 X 24
⇒ 79x = 395 X 24
⇒ x = 120 = Rs. 120
Q3. The price of a certain item is increased by 10%. If a consumer wants to keep his expenditure on the item same as before, how much percent must he reduce his consumption of that item
(a) 3(1/11)%
(b) 5(1/11)%
(c) 9(1/11)%
(d) 12(1/11)%
Answer: (c) 9(1/11)%
Reduction % in consumption = { | r | X 100% } |
100 + r |
= ( | 10 | X 100 )% |
100 + 10 |
= | 10 | X 100 % |
110 |
= | 100 | % |
11 |
= 9 | 1 | % |
11 |
Q4. Three candidates in all election received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the wining candidate get?
(a) 45%
(b) 51%
(c) 53%
(d) 57%
Answer: (d) 57%
Total votes = 1136 + 7636 + 11628 = 20400
Required Percentage = (11628/20400) X 100 % = 57%
Q5. If the numerator of a fraction is increased by 150% and the denominator is increased by 300% the resultant fraction is 5/16. What is the original fraction
(a) ½
(b) ¼
(c) ¾
(d) 1/5
Answer: (a) ½
Let the original fraction be x/y, then
x + x X 150% | = | 5 |
y + y X 300% | 16 |
⇒ | x + (150x/100) | = | 5 |
y + (300y/100) | 16 |
⇒ | 250x | = | 5 |
400y | 16 |
⇒ | x | = | 5 | X | 400 | = ½ |
y | 16 | 250 |
Q6. The value of a plant depreciates by 10% annually. If the present value of the plant is Rs. 100000, then what will be its value after 3 years
(a) Rs. 43800
(b) Rs. 55900
(c) Rs. 68100
(d) Rs. 72900
Answer: (d) Rs. 72900
Value after 3 years = Rs. [100000 X ( 1 - | 10 | )3] |
100 |
= Rs. [100000 X | 90 | X | 90 | X | 90 | ] |
100 | 100 | 100 |
= Rs. 72900
Q7. In an election between two candidates, 75% of the voters cast their votes, out of which 4% of the voters were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. The total number of voters enrolled in that election was
(a) 16832
(b) 17150
(c) 18460
(d) 18865
Answer: (b) 17150
Let the enrolled vote be x
Votes cast = (75/100) X x = 3x/4
Valid votes = (3x/4) X (96/100) = 18x/25
A/Q | 18x | X 75% = 9261 |
25 |
⇒ | 18x | X | 75 | = 9261 |
25 | 100 |
⇒ x = (9261 X 100)/(18X3) = 17150
Q8. Fresh grapes contains 80% water while dry grapes contain 10% water. If the weight of dry grapes is 250 kg, what is the total weight when it was fresh ?
(a) 980 Kg
(b) 1020 Kg
(c) 1125 Kg
(d) 1235 Kg
Answer: (c) 1125 Kg
Let the weight of the fresh grapes be x kg.
Quantity of water in it = (80/100) X x = 4x/5 Kg
Quantity of pulp in it = (x - 4x/5) Kg = x/5 Kg
Quantity of water in 250 kg dry grapes = (10/100) X 250 Kg = 25 Kg
Quantity of pulp in it = 250 - 25 = 225 kg
∴ x/5 = 225
⇒ x = 1125
Weight of fresh grapes = 1125 Kg
Q9. If the numerator of a fraction is increased by 120% and the denominator is increased by 130%, the resultant fraction is 4/15. What is the original fraction ?
(a) 33/165
(b) 41/165
(c) 46/165
(d) 51/165
Answer: (c) 46/165
Let, the original function be x/y
A/Q | 220x | = | 4 |
230y | 15 |
⇒ | x | = | 4 | X | 230 |
y | 15 | 220 |
= 46/165
Q10. Fresh fruit contains 60% water and dry fruit contains 20% water. How much dry fruit can be obtained from 200Kg of fresh fruit ?
(a) 90 Kg
(b) 100 Kg
(c) 110 Kg
(d) 120 Kg
Answer: (b) 100 Kg
Quantity of water in 200 kg of fresh fruits = 60/100 X 200 Kg = 120 Kg
Quantity of pulp in it = (200 - 120) Kg = 80 Kg
Let the dry fruit be x Kg
Water in it = (20/100) X x = x/5 Kg
Quantity of pulp in it = (x - x/5) = 4x/5 Kg
Now 4x/5 = 80
⇒ x = (80X5)/4 = 100 Kg
Practice Test Exam