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Quiz Description :
Name: Percentage Aptitude Question Answer Test – 2
Subject: Aptitude
Topic: Percentage Word Problem
Questions: 12 Objective type
Time Allowed: 20 minutes
Important for: SSC CGL, CHSL, GD, SBI Clerk, IBPS PO / Clerk, LIC Insurance, Railways, school students of class 8th, 9th, 10th, 11th and 12th, B. Sc. (maths), IITJEE, AIPMT, Teaching entrance, Jobs written test of WIPRO, InfoSys, IBM, UHG, Tech Mahindra, TCS, TCL and other center level and state level competitions.
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 Question 1 of 12
1. Question
1 pointsThe population of a village has increased from 60,000 to 65,000. What is the increased percentage of the village?
CorrectIncrease in population = 65000 – 60000 = 5000
Percentage increase = 5000/60000×100= 25/3%IncorrectIncrease in population = 65000 – 60000 = 5000
Percentage increase = 5000/60000×100= 25/3%UnattemptedIncrease in population = 65000 – 60000 = 5000
Percentage increase = 5000/60000×100= 25/3%  Question 2 of 12
2. Question
1 points. In an election of two candidates, the candidate who gets 41% is rejected by a majority of 2412 votes. Find the total number of votes polled.
CorrectPercentage difference between candidate polled votes = (59% – 41%) = 18 %
As per the question;
candidate rejected by 18% of majority = 2412 votes
∴100% = 2412/18×100 = 13400 votesIncorrectPercentage difference between candidate polled votes = (59% – 41%) = 18 %
As per the question;
candidate rejected by 18% of majority = 2412 votes
∴100% = 2412/18×100 = 13400 votesUnattemptedPercentage difference between candidate polled votes = (59% – 41%) = 18 %
As per the question;
candidate rejected by 18% of majority = 2412 votes
∴100% = 2412/18×100 = 13400 votes  Question 3 of 12
3. Question
1 pointsWyatt’s salary is increased from Rs. 1500 to Rs. 1830. Find the increase percentage.
CorrectIncrease in salary = Rs 1830 – Rs. 1500 = Rs. 330
Percentage in salary = 330/1500×100
= 22%IncorrectIncrease in salary = Rs 1830 – Rs. 1500 = Rs. 330
Percentage in salary = 330/1500×100
= 22%UnattemptedIncrease in salary = Rs 1830 – Rs. 1500 = Rs. 330
Percentage in salary = 330/1500×100
= 22%  Question 4 of 12
4. Question
1 pointsThe population of a town has increased from 5000 to 13000. What is the increased percentage of the town?
CorrectIncrease in population = 13000 – 5000 =8000
Percentage increase = (8000/5000)×100= 800/5=160%IncorrectIncrease in population = 13000 – 5000 =8000
Percentage increase = (8000/5000)×100= 800/5=160%UnattemptedIncrease in population = 13000 – 5000 =8000
Percentage increase = (8000/5000)×100= 800/5=160%  Question 5 of 12
5. Question
1 pointsIf 10 liters of water is evaporated on boiling from 15 liters of sugar solution containing 10% sugar, find the percentage of sugar in the remaining solution.
Correct% of sugar in the original solution = 10% of 15 liters = 10/100×15 = 1.5%
After evaporation of 10 liters of water, the quantity of the remaining solution = 15 – 10 = 5 liters
∴The required percentage of sugar = 1.5/5×100% = 150/5 = 30%Incorrect% of sugar in the original solution = 10% of 15 liters = 10/100×15 = 1.5%
After evaporation of 10 liters of water, the quantity of the remaining solution = 15 – 10 = 5 liters
∴The required percentage of sugar = 1.5/5×100% = 150/5 = 30%Unattempted% of sugar in the original solution = 10% of 15 liters = 10/100×15 = 1.5%
After evaporation of 10 liters of water, the quantity of the remaining solution = 15 – 10 = 5 liters
∴The required percentage of sugar = 1.5/5×100% = 150/5 = 30%  Question 6 of 12
6. Question
1 pointsOne type of liquid contains 25% of milk, the other contains 30% of milk. A cane of filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of milk in the new mixture.
CorrectIncorrectUnattempted  Question 7 of 12
7. Question
1 pointsDue to fall in manpower, the production in a factory decrease by 25%. By what percent should the working hour be increased to restore the original production?
CorrectLet us consider the manpower = 100 units and
Working hours = 100 units
Suppose working hours increase by x%.
Then, (100 25) (100 + x) = 100 × 100
or 100+x= 400/3
∴ x= 400/3IncorrectLet us consider the manpower = 100 units and
Working hours = 100 units
Suppose working hours increase by x%.
Then, (100 25) (100 + x) = 100 × 100
or 100+x= 400/3
∴ x= 400/3UnattemptedLet us consider the manpower = 100 units and
Working hours = 100 units
Suppose working hours increase by x%.
Then, (100 25) (100 + x) = 100 × 100
or 100+x= 400/3
∴ x= 400/3  Question 8 of 12
8. Question
1 pointsA man had some eggs. Out of them 4% were thrown away as they had become rotten, 80% of the remaining eggs were sold and now he was left with only 96 eggs. How many eggs had he in the beginning?
CorrectMethod 1:
Let he had 100 eggs in all.
The number of rotten eggs = 4% of 100 = 4
∴ Remaining eggs = 100 – 4 = 96
The eggs sold = 80% of 96
= 80×96/100= 384/5
∴ Remaining eggs = 96 384/5 = 96/5
If 96/5 eggs are left, then eggsShort Method 2:
Total eggs = (A×100×100)/(100x)(100y)
= (96×100×100)/(1004)(10080) =510.63≈511IncorrectMethod 1:
Let he had 100 eggs in all.
The number of rotten eggs = 4% of 100 = 4
∴ Remaining eggs = 100 – 4 = 96
The eggs sold = 80% of 96
= 80×96/100= 384/5
∴ Remaining eggs = 96 384/5 = 96/5
If 96/5 eggs are left, then eggsShort Method 2:
Total eggs = (A×100×100)/(100x)(100y)
= (96×100×100)/(1004)(10080) =510.63≈511UnattemptedMethod 1:
Let he had 100 eggs in all.
The number of rotten eggs = 4% of 100 = 4
∴ Remaining eggs = 100 – 4 = 96
The eggs sold = 80% of 96
= 80×96/100= 384/5
∴ Remaining eggs = 96 384/5 = 96/5
If 96/5 eggs are left, then eggsShort Method 2:
Total eggs = (A×100×100)/(100x)(100y)
= (96×100×100)/(1004)(10080) =510.63≈511  Question 9 of 12
9. Question
1 pointsThe population of a town has increased from 75000 to 78000. F What is the increased percentage of the town?
CorrectIncrease in population = 78000 – 75000 = 3000
Percentage increase = 3000/75000×100= 300/75=4%IncorrectIncrease in population = 78000 – 75000 = 3000
Percentage increase = 3000/75000×100= 300/75=4%UnattemptedIncrease in population = 78000 – 75000 = 3000
Percentage increase = 3000/75000×100= 300/75=4%  Question 10 of 12
10. Question
1 pointsIn an election 5% of the total voters did take part. The candidate who was elected got 65% of the total votes. He got 3500 votes more that his opponent. Find the total number of voters.
CorrectLet the number of total voters = 100
Number of voters who did not take part in the election = 5
∴ Number of voters who took part in the election = 100 – 5 = 95
Number of votes secured by the elected candidate= 65% of 100 = 65
Number of votes secured by losing candidate = 95 – 65 = 30
∴Difference between the votes of the two candidate = 65 – 30 = 35
If difference in 35, the number of total votes = 100
If difference is 3500, the number of total votes= (100×3500)/35=10000IncorrectLet the number of total voters = 100
Number of voters who did not take part in the election = 5
∴ Number of voters who took part in the election = 100 – 5 = 95
Number of votes secured by the elected candidate= 65% of 100 = 65
Number of votes secured by losing candidate = 95 – 65 = 30
∴Difference between the votes of the two candidate = 65 – 30 = 35
If difference in 35, the number of total votes = 100
If difference is 3500, the number of total votes= (100×3500)/35=10000UnattemptedLet the number of total voters = 100
Number of voters who did not take part in the election = 5
∴ Number of voters who took part in the election = 100 – 5 = 95
Number of votes secured by the elected candidate= 65% of 100 = 65
Number of votes secured by losing candidate = 95 – 65 = 30
∴Difference between the votes of the two candidate = 65 – 30 = 35
If difference in 35, the number of total votes = 100
If difference is 3500, the number of total votes= (100×3500)/35=10000  Question 11 of 12
11. Question
1 pointsIn an local society membership election of two candidates, the candidate who gets 47% is rejected by a majority of 2730 votes. Find the total number of votes polled.
CorrectPercentage difference between candidate polled votes = (53% – 47%) = 6 %
As per the question;
candidate rejected by 6% of majority = 2730 votes
∴100% = 2730/6×100 = 45500 votesIncorrectPercentage difference between candidate polled votes = (53% – 47%) = 6 %
As per the question;
candidate rejected by 6% of majority = 2730 votes
∴100% = 2730/6×100 = 45500 votesUnattemptedPercentage difference between candidate polled votes = (53% – 47%) = 6 %
As per the question;
candidate rejected by 6% of majority = 2730 votes
∴100% = 2730/6×100 = 45500 votes  Question 12 of 12
12. Question
1 pointsIf 2 liters of water are evaporated on boiling from 8 liters of sugar solution containing 5% sugar, find the percentage of sugar in the remaining solution.
Correct% of sugar in the original solution = 5% of 8 liters = 5/100×8 = 0.4%
After evaporation of 2 liters of water, the quantity of the remaining solution = 8 – 2 = 6 liters
∴The required percentage of sugar = 0.4/6×100% = 20/3%Incorrect% of sugar in the original solution = 5% of 8 liters = 5/100×8 = 0.4%
After evaporation of 2 liters of water, the quantity of the remaining solution = 8 – 2 = 6 liters
∴The required percentage of sugar = 0.4/6×100% = 20/3%Unattempted% of sugar in the original solution = 5% of 8 liters = 5/100×8 = 0.4%
After evaporation of 2 liters of water, the quantity of the remaining solution = 8 – 2 = 6 liters
∴The required percentage of sugar = 0.4/6×100% = 20/3%