Online Quiz in Percentage For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Percentage-Test 4
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Subject :- Quantitative Aptitude
Chapter :- Percentage- Test 4
Questions :- 25
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When the price of rice is increased by 25 percent, a family reduces its consumption
such that the expenditure is only 10 percent more than before. If 40 kg of rice is consumed by family before, then find the new consumption of family.
Correct
Suppose initially price per kg of rice is 100 then
their expenditure is 4000.
Now their expenditure is only increased by only
10% i.e – 4400.
Increased price of rice = 125.
So new consumption = 4400/125 = 35.2
Incorrect
Suppose initially price per kg of rice is 100 then
their expenditure is 4000.
Now their expenditure is only increased by only
10% i.e – 4400.
Increased price of rice = 125.
So new consumption = 4400/125 = 35.2
Question 2 of 25
2. Question
1 points
The price of rice is increased by 20 percent and a person decrease his consumption by 15
percent, so his expenditure on rice is
Correct
Let initial price of rice – 100 and new price of
rice – 120
suppose initial consumption is 100kg and new
consumption is 85kg
Initial expenditure = 10000
New expenditure = 10200
(200/10000)*100 = 2 percent increase
Incorrect
Let initial price of rice – 100 and new price of
rice – 120
suppose initial consumption is 100kg and new
consumption is 85kg
Initial expenditure = 10000
New expenditure = 10200
(200/10000)*100 = 2 percent increase
Question 3 of 25
3. Question
1 points
A salary is 40 percent more than B. B’s salary is 30 percent less than C. If the
difference between the salary of C and A is 1200 rupees, then what is the monthly income
of C
Correct
A = (140/100)*B
B = (70/100)*C
[(100/70) – (140/100)]*B = 1200.
B = 42000.
C = (100/70)*42000 = 60000
Incorrect
A = (140/100)*B
B = (70/100)*C
[(100/70) – (140/100)]*B = 1200.
B = 42000.
C = (100/70)*42000 = 60000
Question 4 of 25
4. Question
1 points
When the price of rice is increased by 30 percent, a family reduces its consumption
such that the expenditure is only 20 percent more than before. If 50 kg of rice is consumed by family before, then find the new consumption of family (approx.)
Correct
Suppose initially price per kg of rice is 100 then
their expenditure is 5000.
Now their expenditure is only increased by only
20% i.e – 6000.
Increased price of rice = 130.
So new consumption = 6000/130 = 46.1
Incorrect
Suppose initially price per kg of rice is 100 then
their expenditure is 5000.
Now their expenditure is only increased by only
20% i.e – 6000.
Increased price of rice = 130.
So new consumption = 6000/130 = 46.1
Question 5 of 25
5. Question
1 points
One type of liquid contains 20 percent of milk and second type of liquid contains 40
percent milk. If 4 part of the first and 6 part of the second are mix, then what is the
percent of water in the mixture.
Correct
Do these type of question by taking 100 litre
water = 80ltr and 60ltr in first and second
mixture respectively
now percent of water = [(80*4 +60*6)/1000]100 = 68%
Incorrect
Do these type of question by taking 100 litre
water = 80ltr and 60ltr in first and second
mixture respectively
now percent of water = [(80*4 +60*6)/1000]100 = 68%
Question 6 of 25
6. Question
1 points
40% of the students like Mathematics, 50% like English and 10% like both Mathematics
and English. What % of the students like neither English nor Mathematics?
Correct
n(M or E) = n(M) + n(E) – n(M and E)
n(M or E) = 40+50-10 = 80
so % of the students who like neither English
nor Mathematics = 100 – 80 = 20%
Incorrect
n(M or E) = n(M) + n(E) – n(M and E)
n(M or E) = 40+50-10 = 80
so % of the students who like neither English
nor Mathematics = 100 – 80 = 20%
Question 7 of 25
7. Question
1 points
A watermelon weighing 20 kg contains 96% of water by weight. It is put in sun for some time and some water evaporates so that now it contains only 95% of water by weight.
The new weight of watermelon would be?
Correct
Let new weight be x kg
Since the pulp is not being evaporated, the
quantity of pulp should remain same in both
cases. This gives
(100-96)% of 20 = (100-95)% of x
Solve, x = 16 kg
Incorrect
Let new weight be x kg
Since the pulp is not being evaporated, the
quantity of pulp should remain same in both
cases. This gives
(100-96)% of 20 = (100-95)% of x
Solve, x = 16 kg
Question 8 of 25
8. Question
1 points
If the price of wheat is reduced by 2%. How many kilograms of wheat a person can
buy with the same money which was earlier sufficient to buy 49 kg of wheat?
Correct
Let the original price = 100 Rs per kg
Then money required to buy 49 kg = 49*100 =
Rs 4900
New price per kg is (100-98)% of Rs 100 = 98
So quantity of wheat bought in 4900 Rs is
4900/98 = 50 kg
Incorrect
Let the original price = 100 Rs per kg
Then money required to buy 49 kg = 49*100 =
Rs 4900
New price per kg is (100-98)% of Rs 100 = 98
So quantity of wheat bought in 4900 Rs is
4900/98 = 50 kg
Question 9 of 25
9. Question
1 points
Monthly salary of A is 30% more than B’s monthly salary and B’s monthly salary is
20% less than C’s. If the difference between the monthly salaries of A and C is Rs 800,
then find the annual salary of B.
Correct
Let C’s monthly salary = Rs 100, then B’s =(100-20)% of 100
= 80, and A’s monthly =
(100+30)% * 80 = 104
If difference between A and C’s monthly salary
is Rs 4 then B’s monthly salary is Rs 80
So if difference is Rs 800, B’s monthly salary is
(80/4) * 800 = 16,000
So annual salary = 12*16,000
Incorrect
Let C’s monthly salary = Rs 100, then B’s =(100-20)% of 100
= 80, and A’s monthly =
(100+30)% * 80 = 104
If difference between A and C’s monthly salary
is Rs 4 then B’s monthly salary is Rs 80
So if difference is Rs 800, B’s monthly salary is
(80/4) * 800 = 16,000
So annual salary = 12*16,000
Question 10 of 25
10. Question
1 points
Mixture 1 contains 20% of water and mixture 2 contains 35% of water. 10 parts
from 1st mixture and 4 parts from 2nd mixture is taken and put in a glass. What is
the percentage of water in the new mixture of glass?
Correct
Water in new mixture from 1st mixture =
(20/100) * 10 = 2 parts
Water in new mixture from 2nd mixture =
(35/100) * 4 = 7/5 parts
Required % =[ [2+ (7/5)]/(10+4)] * 100
Incorrect
Water in new mixture from 1st mixture =
(20/100) * 10 = 2 parts
Water in new mixture from 2nd mixture =
(35/100) * 4 = 7/5 parts
Required % =[ [2+ (7/5)]/(10+4)] * 100
Question 11 of 25
11. Question
1 points
3 years ago the population of a town was 1,60,000. In the three respective years the
population increased by 3%, 2.5% and 5% respectively. What is the population of town
after 3 years?
Correct
Incorrect
Question 12 of 25
12. Question
1 points
There are 2500 students who appeared for an examination. Out of these, 35% students
failed in 1 subject and 42% in other subject and 15% of students failed in both the
subjects. How many of the students passed in either of the 2 subjects but not in both?
Correct
Failed in 1st subject = (35/100) * 2500 = 875
Failed in 1st subject = (42/100) * 2500 = 1050
Failed in both = (15/100) * 2500 = 375
So failed in 1st subject only = 875 – 375 = 500
failed in 2nd subject only = 1050 – 375 = 675
passed in 1st only + passed In 2nd only =
675+500=1175
Incorrect
Failed in 1st subject = (35/100) * 2500 = 875
Failed in 1st subject = (42/100) * 2500 = 1050
Failed in both = (15/100) * 2500 = 375
So failed in 1st subject only = 875 – 375 = 500
failed in 2nd subject only = 1050 – 375 = 675
passed in 1st only + passed In 2nd only =
675+500=1175
Question 13 of 25
13. Question
1 points
A bucket is filled with water such that the weight of bucket alone is 25% its weight when
it is filled with water. Now some of the water is removed from the bucket and now the
weight of bucket along with remaining water is 50% of the original total weight. What part of the water was removed from the bucket?
Correct
Let original weight of bucket when it is filled
with water = x
Then weight of bucket = (25/100) * x = x/4
Original weight of water = x – (x/4) = 3x/4
Now when some water removed, new weight of
bucket with remaining water = (50/100) * x =
x/2
So new weight of water = new weight of bucket
with remaining water – weight of bucket = [(x/2)
– (x/4)] = x/4
So part of water removed = [(3x/4) –
(x/4)]/(3x/4)
Incorrect
Let original weight of bucket when it is filled
with water = x
Then weight of bucket = (25/100) * x = x/4
Original weight of water = x – (x/4) = 3x/4
Now when some water removed, new weight of
bucket with remaining water = (50/100) * x =
x/2
So new weight of water = new weight of bucket
with remaining water – weight of bucket = [(x/2)
– (x/4)] = x/4
So part of water removed = [(3x/4) –
(x/4)]/(3x/4)
Question 14 of 25
14. Question
1 points
In a survey done by a committee, it was found that 4000 people have smoking habit.
After a month this number rose by 5%. However due to continuous advices given by
the committee to the people, the number reduced by 5% in the next month and further
by 10% in the next month. What is the total number of smokers after 3 months?
Correct
Number of smokers after 3 months will be =
4000 * (1 + (5/100)) (1 – (5/100)) (1 – (10/100))
= 3591
Incorrect
Number of smokers after 3 months will be =
4000 * (1 + (5/100)) (1 – (5/100)) (1 – (10/100))
= 3591
Question 15 of 25
15. Question
1 points
There are 5000 students in a school. The next year it was found that the number of
boys and girls increased by 10% and 15% respectively making the total number of
students in school as 5600. Find the number of girls originally in the school?
Correct
Let number of girls = x, then no of boys =
(5000-x). then
10% of (1000-x) + 15% of x = (5600-5000)
Solve, x = 2000
Incorrect
Let number of girls = x, then no of boys =
(5000-x). then
10% of (1000-x) + 15% of x = (5600-5000)
Solve, x = 2000
Question 16 of 25
16. Question
1 points
If x is 20% more than y, then by what percent y is smaller than x.
Correct
x = 120y/100 or x = 6y/5
y = 5x/6. Percentage by which y is smaller
Than x is [(x – 5x/6)/x]*100 = 50/3 %
Incorrect
x = 120y/100 or x = 6y/5
y = 5x/6. Percentage by which y is smaller
Than x is [(x – 5x/6)/x]*100 = 50/3 %
Question 17 of 25
17. Question
1 points
In an alloy, there is 15% of brass, to get 90 kg of brass, how much alloy is needed ?
Correct
Let X kg of alloy is needed. So, 15/100 of X =
90. So X =600 kg
Incorrect
Let X kg of alloy is needed. So, 15/100 of X =
90. So X =600 kg
Question 18 of 25
18. Question
1 points
25 litre of solution contains alcohol and water in the ratio 2:3. How much alcohol
must be added to the solution to make a solution containing 60% of alcohol ?
Correct
Initially alcohol 2/5 * 25 = 10 ltr and water is 15
ltr.
To make a solution of 60% alcohol (10+x)/25+x
= 60/100. X = 12.5
Incorrect
Initially alcohol 2/5 * 25 = 10 ltr and water is 15
ltr.
To make a solution of 60% alcohol (10+x)/25+x
= 60/100. X = 12.5
Question 19 of 25
19. Question
1 points
In an examination if a person get 20% of the marks then it is fail by 30 marks. Another
person who gets 30% marks gets 30 marks more than the passing marks. Find out the
total marks and the passing marks.
Correct
20% of X = P – 30 (X = Maximum marks and P
= passing marks)
30% of X = P + 30. Solve for X and P.
Incorrect
20% of X = P – 30 (X = Maximum marks and P
= passing marks)
30% of X = P + 30. Solve for X and P.
Question 20 of 25
20. Question
1 points
A company has produced 900 pieces of transistor out of which 15% are defective and
out of remaining 20 % were not sold. Find out the number of sold transistor.
Correct
No of transistor sold = 900*(85/100)*(80/100) = 612
Incorrect
No of transistor sold = 900*(85/100)*(80/100) = 612
Question 21 of 25
21. Question
1 points
In an election the votes between the winner and loser candidate are in the ratio
5:1. If total number of eligible voters are 1000, out of which 12% did not cast their vote
and among the remaining vote 10% declared invalid. What is the number of votes the
winner candidate get ?
Correct
Ratio b/w winner and loser 5:1
Total no of votes casted actually =
1000*(88/100)*(90/100) = 792
5x + x = 792, X =132
Votes of winner candidate = 5*132 = 660
Incorrect
Ratio b/w winner and loser 5:1
Total no of votes casted actually =
1000*(88/100)*(90/100) = 792
5x + x = 792, X =132
Votes of winner candidate = 5*132 = 660
Question 22 of 25
22. Question
1 points
If the price of a commodity is increased by 30%, by how much % a consumer must
reduce his consumption so to keep the expenditure same ?
Correct
If commodity price is increased then reduction
in consumption will be
[(increase in price)/100 + increase in price]*100.
(30/130)*100 = 300/13%
Incorrect
If commodity price is increased then reduction
in consumption will be
[(increase in price)/100 + increase in price]*100.
(30/130)*100 = 300/13%
Question 23 of 25
23. Question
1 points
1000 sweets need to be distributed equally among the school students in such a way that
each student gets sweet equal to 10% of total students. Then the number of sweets, each
student gets.
Correct
No of students = T. Each student gets 10% of T.
So , T students get T^2/10 sweets.
T^2/10 = 1000. So T = 100. So each student gets
10 sweets
Incorrect
No of students = T. Each student gets 10% of T.
So , T students get T^2/10 sweets.
T^2/10 = 1000. So T = 100. So each student gets
10 sweets
Question 24 of 25
24. Question
1 points
Rishi salary is first increased by 20% and then decreased by 25%. How much percent
the salary increased/decreased ?
Correct
Take 100 as rishi salary.
Increased by 20% percent = 120.
Then decreased by 25%, i.e = (75/100)*120 = 90.
So percentage decrease is 10%.
Incorrect
Take 100 as rishi salary.
Increased by 20% percent = 120.
Then decreased by 25%, i.e = (75/100)*120 = 90.
So percentage decrease is 10%.
Question 25 of 25
25. Question
1 points
The income of a person is 10000 and its expenditure is 6000 and thus saves 4000 rs. In
the next year his income is increased by 10% and its expenditure increased by 20%. Now
his saving is what percent lower than the previous saving.
Correct
Initially I-E = S (I = Income, E = expenditure, S
= saving)
10000-6000 = 4000(saving)
Now, I = 11000 and E = 7200. So saving = I – E
= 3800.
[(4000-3800)/4000]*100 = 5%
Incorrect
Initially I-E = S (I = Income, E = expenditure, S
= saving)
10000-6000 = 4000(saving)
Now, I = 11000 and E = 7200. So saving = I – E
= 3800.
[(4000-3800)/4000]*100 = 5%