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Number of Questions: 20
Time: 20 Minutes
Category: Aptitude
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Question 1 of 20
1. Question
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Correct
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = 50 x 100 =
Incorrect
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = 50 x 100 =

Question 2 of 20
2. Question
A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
Correct
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
Incorrect
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.

Question 3 of 20
3. Question
A number increased by 20% gives 480. The number is
Correct
Answer: Option ‘C’
Formula = TOTAL=100% ,INCRESE = “+” DECREASE= “”
A number means = 100 %
That same number increased by 20 % = 120 %
120 % ——> 480 (120 × 4 = 480)
100 % ——> 400 (100 × 4 = 400)
Incorrect
Answer: Option ‘C’
Formula = TOTAL=100% ,INCRESE = “+” DECREASE= “”
A number means = 100 %
That same number increased by 20 % = 120 %
120 % ——> 480 (120 × 4 = 480)
100 % ——> 400 (100 × 4 = 400)

Question 4 of 20
4. Question
What percent is 5 gm of 1 kg?
Correct
1 kg = 1000 gm
5/1000 × 100 = 500/1000
=1/2 = 0.5 %
Incorrect
1 kg = 1000 gm
5/1000 × 100 = 500/1000
=1/2 = 0.5 %

Question 5 of 20
5. Question
If 20% of a = b, then b% of 20 is the same as:
Correct
20% of a = b
=> b = 20/100a
b% of 20 = (b/100)20 = [(20/100a)/100]20
= (20 × 20 × a)/(100 × 100)
= 4a/100 = 4% of a
Incorrect
20% of a = b
=> b = 20/100a
b% of 20 = (b/100)20 = [(20/100a)/100]20
= (20 × 20 × a)/(100 × 100)
= 4a/100 = 4% of a

Question 6 of 20
6. Question
In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.
Correct
Answer: Option C
Explanation:
Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42
n(A∪B)=n(A)+n(B)−n(A∩B)=34+42−20=56Failed in either or both subjects are 56Percentage passed = (100−56)%=44%Incorrect
Answer: Option C
Explanation:
Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42
n(A∪B)=n(A)+n(B)−n(A∩B)=34+42−20=56Failed in either or both subjects are 56Percentage passed = (100−56)%=44% 
Question 7 of 20
7. Question
Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calcualte average marks of the remaining students of the class.
Correct
Answer: Option C
Explanation:
Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)=> 70x = 8000 – (950 + 1800) = 5250
=> x = 75.
Incorrect
Answer: Option C
Explanation:
Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)=> 70x = 8000 – (950 + 1800) = 5250
=> x = 75.

Question 8 of 20
8. Question
What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
Correct
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = *100)%
=20%
Incorrect
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = *100)%
=20%

Question 9 of 20
9. Question
Income of A is 25% more than the income of B. What is the income of B in terms of income of A?
Correct
ANSWER: A
Explanation:
One of the most basic questions.Let income of B be 100. Income of A is 25% more than income of B which means Income of A becomes 125
Now income of B in terms of A = 100/125 *100 = 80%
Incorrect
ANSWER: A
Explanation:
One of the most basic questions.Let income of B be 100. Income of A is 25% more than income of B which means Income of A becomes 125
Now income of B in terms of A = 100/125 *100 = 80%

Question 10 of 20
10. Question
Evaluate :
28% of 400 + 45 % of 250
Correct
ANSWER: B
Explanation:
28% of 400 + 45 % of 250
= (28/100 *400 + 45/100 * 250)
= (112 + 112.5)
= 224.5Incorrect
ANSWER: B
Explanation:
28% of 400 + 45 % of 250
= (28/100 *400 + 45/100 * 250)
= (112 + 112.5)
= 224.5 
Question 11 of 20
11. Question
In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.
Correct
Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively .
Then , number of students passed in one or both the subjects
= n(AÈB)=n(A)+n(B) n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x(9x/10)]=x/10.
So, x/10=40 of x=400 .
Hence ,total number of students = 400.
Incorrect
Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively .
Then , number of students passed in one or both the subjects
= n(AÈB)=n(A)+n(B) n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x(9x/10)]=x/10.
So, x/10=40 of x=400 .
Hence ,total number of students = 400.

Question 12 of 20
12. Question
In an examination , 35% of total students failed in Hindi , 45% failed in English and 20% in both . Find the percentage of those who passed in both subjects .
Correct
Let A and B be the sets of students who failed in Hindi and English respectively .
Then , n(A) = 35 , n(B)=45 , n(AÇB)=20.
So , n(AÈB)=n(A)+n(B) n(AÇB)=35+4520=60.
Percentage failed in Hindi and English or both=60%
Hence , percentage passed = (10060)%=40%
Incorrect
Let A and B be the sets of students who failed in Hindi and English respectively .
Then , n(A) = 35 , n(B)=45 , n(AÇB)=20.
So , n(AÈB)=n(A)+n(B) n(AÇB)=35+4520=60.
Percentage failed in Hindi and English or both=60%
Hence , percentage passed = (10060)%=40%

Question 13 of 20
13. Question
Due to reduction of 25/4% in the price of sugar , a man is able to buy 1kg more for Rs.120. Find the original and reduced rate of sugar.
Correct
Let the original rate be Rs.x per kg.
Reduced rate = Rs.[(100(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)(120/x)=1 ó (128/x)(120/x)=1
ó x=8.
So, the original rate = Rs.8 per kg
Reduce rate = Rs.[(15/16)*8]per kg = Rs.7.50 per kg
Incorrect
Let the original rate be Rs.x per kg.
Reduced rate = Rs.[(100(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)(120/x)=1 ó (128/x)(120/x)=1
ó x=8.
So, the original rate = Rs.8 per kg
Reduce rate = Rs.[(15/16)*8]per kg = Rs.7.50 per kg

Question 14 of 20
14. Question
How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to 10% solution ?
Correct
Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then , (0.6+x)/(30+x)=10/100ó60+100x=300+10x
ó90x=240 ó x=8/3.
Incorrect
Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then , (0.6+x)/(30+x)=10/100ó60+100x=300+10x
ó90x=240 ó x=8/3.

Question 15 of 20
15. Question
If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?
Correct
Required percentage = [(20*100)/(10020)]%=25%.
Incorrect
Required percentage = [(20*100)/(10020)]%=25%.

Question 16 of 20
16. Question
During one year, the population of town increased by 5% . If the total population is 9975 at the end of the second year , then what was the population size in the beginning of the first year ?
Correct
Population in the beginning of the first year
= 9975/[1+(5/100)]*[1(5/100)] = [9975*(20/21)*(20/19)]=10000.
Incorrect
Population in the beginning of the first year
= 9975/[1+(5/100)]*[1(5/100)] = [9975*(20/21)*(20/19)]=10000.

Question 17 of 20
17. Question
When the price fo a product was decreased by 10% , the number sold increased by 30%. What was the effect on the total revenue ?
Correct
Let the price of the product be Rs.100 and let original sale be 100 pieces.
Then , Total Revenue = Rs.(100*100)=Rs.10000.
New revenue = Rs.(90*130)=Rs.11700.
Increase in revenue = ((1700/10000)*100)%=17%.
Incorrect
Let the price of the product be Rs.100 and let original sale be 100 pieces.
Then , Total Revenue = Rs.(100*100)=Rs.10000.
New revenue = Rs.(90*130)=Rs.11700.
Increase in revenue = ((1700/10000)*100)%=17%.

Question 18 of 20
18. Question
The salary of a person was reduced by 10% .By what percent should his reduced salary be raised so as to bring it at par with his original salary ?
Correct
Let the original salary be Rs.100 . New salary = Rs.90.
Increase on 90=10 , Increase on 100=((10/90)*100)%
= (100/9)%
Incorrect
Let the original salary be Rs.100 . New salary = Rs.90.
Increase on 90=10 , Increase on 100=((10/90)*100)%
= (100/9)%

Question 19 of 20
19. Question
Raman`s salary was decreased by 50% and subsequently increased by 50%.How much percent does he lose?
Correct
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%
Incorrect
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%

Question 20 of 20
20. Question
Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?
Correct
Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.
Incorrect
Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.