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Name : Aptitude Speed test for bank Clerk exams
Subject : Aptitude
Topic : Quantitative Aptitude
Questions : 33 mcq
Time Limit : 30 minutes
Important for : IBPS Clerk Prelims, SBI Clerk, IBPS RRB Office assistant, RBI Assistant, Scholarship exams etc.
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 Question 1 of 33
1. Question
1 pointsRaju’s age after 15 years will be 5 times his age 5 years back, what is the present age of Raju?
CorrectExplanation – x+15 = 5 (x5)
4x = 40
x = 10.
IncorrectExplanation – x+15 = 5 (x5)
4x = 40
x = 10.
UnattemptedExplanation – x+15 = 5 (x5)
4x = 40
x = 10.
 Question 2 of 33
2. Question
1 pointsThe total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A?
CorrectExplanation – A+B = 12+B+C
A? C = 12+B? B=12
hence C is younger than A by 12 years.
IncorrectExplanation – A+B = 12+B+C
A? C = 12+B? B=12
hence C is younger than A by 12 years.
UnattemptedExplanation – A+B = 12+B+C
A? C = 12+B? B=12
hence C is younger than A by 12 years.
 Question 3 of 33
3. Question
1 pointsRatio between Rahul and Deepak is 4:3, after 6 years Rahul age will be 26 years. What is Deepak present age?
CorrectExplanation – Present age is 4x and 3x
4x+6 = 26
x = 5
So Deepak’s age is 3 (5) which is 15.
IncorrectExplanation – Present age is 4x and 3x
4x+6 = 26
x = 5
So Deepak’s age is 3 (5) which is 15.
UnattemptedExplanation – Present age is 4x and 3x
4x+6 = 26
x = 5
So Deepak’s age is 3 (5) which is 15.
 Question 4 of 33
4. Question
1 pointsThe ages of 2 persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?
CorrectExplanation – x? 6 = 3(x166)
x? 6 = 3x?66
2x = 60 hence x = 30.
IncorrectExplanation – x? 6 = 3(x166)
x? 6 = 3x?66
2x = 60 hence x = 30.
UnattemptedExplanation – x? 6 = 3(x166)
x? 6 = 3x?66
2x = 60 hence x = 30.
 Question 5 of 33
5. Question
1 points(3080+6160) /28
CorrectExplanation – As per BODMAS rule first we solve this equation in brackets the we will go for division
(9240)/28 = 330
IncorrectExplanation – As per BODMAS rule first we solve this equation in brackets the we will go for division
(9240)/28 = 330
UnattemptedExplanation – As per BODMAS rule first we solve this equation in brackets the we will go for division
(9240)/28 = 330
 Question 6 of 33
6. Question
1 points100+50×2200+400÷2 =?
CorrectExplanation – as per BODMAS rule first multiplication wil be done and then addition
So it will be 100+100200+200 = 200
IncorrectExplanation – as per BODMAS rule first multiplication wil be done and then addition
So it will be 100+100200+200 = 200
UnattemptedExplanation – as per BODMAS rule first multiplication wil be done and then addition
So it will be 100+100200+200 = 200
 Question 7 of 33
7. Question
1 points7500+(1250/50) = ?
CorrectExplanation – as per BODMAS rule first we solve the brackets and then other
7500+(25) = 7525
IncorrectExplanation – as per BODMAS rule first we solve the brackets and then other
7500+(25) = 7525
UnattemptedExplanation – as per BODMAS rule first we solve the brackets and then other
7500+(25) = 7525
 Question 8 of 33
8. Question
1 pointsHow many pieces of 0.85 m can be cut from a rod 42.5 m long?
CorrectExplanation – We need to divide
42.5/0.85 = 50.
IncorrectExplanation – We need to divide
42.5/0.85 = 50.
UnattemptedExplanation – We need to divide
42.5/0.85 = 50.
 Question 9 of 33
9. Question
1 pointsSimplify (31/10)×(3/10)+(7/5)/20
CorrectExplanation – (31/10)*(3/10)+(7/5)/20 = (3.1)*(0.3)+(1.4)/20
0.93+0.07 = 1
IncorrectExplanation – (31/10)*(3/10)+(7/5)/20 = (3.1)*(0.3)+(1.4)/20
0.93+0.07 = 1
UnattemptedExplanation – (31/10)*(3/10)+(7/5)/20 = (3.1)*(0.3)+(1.4)/20
0.93+0.07 = 1
 Question 10 of 33
10. Question
1 pointsA man buys an article for Rs. 27.50 and sells it for Rs. 28.60. What will be his gain percent?
CorrectExplanation – Gain = (28.6027.50) = 1.10
Gain % = (1.10/27.50)*100% = 4%.
IncorrectExplanation – Gain = (28.6027.50) = 1.10
Gain % = (1.10/27.50)*100% = 4%.
UnattemptedExplanation – Gain = (28.6027.50) = 1.10
Gain % = (1.10/27.50)*100% = 4%.
 Question 11 of 33
11. Question
1 pointsHari purchased 25 kg of Wheat at Rs. 4 per kg and 35 kg of wheat at Rs. 4.50 per kg. He sold the mixture at Rs. 4.25 per kg. Find his gain or loss
CorrectExplanation – Total C.P = (25*4+35*4.50) = 257.50
Total S.P = (60*4.25) = 255
loss = (257.50255) = 2.50 Rs.
IncorrectExplanation – Total C.P = (25*4+35*4.50) = 257.50
Total S.P = (60*4.25) = 255
loss = (257.50255) = 2.50 Rs.
UnattemptedExplanation – Total C.P = (25*4+35*4.50) = 257.50
Total S.P = (60*4.25) = 255
loss = (257.50255) = 2.50 Rs.
 Question 12 of 33
12. Question
1 pointsA coin is tossed once. Find the probability of getting a head
CorrectExplanation – We know that if once coin is tossed than it can come 1 by 2.
IncorrectExplanation – We know that if once coin is tossed than it can come 1 by 2.
UnattemptedExplanation – We know that if once coin is tossed than it can come 1 by 2.
 Question 13 of 33
13. Question
1 pointsA dice is thrown twice. Then the total no. Of cases will be?
CorrectExplanation – Total no. Of cases will be (6*6*6) = 216.
IncorrectExplanation – Total no. Of cases will be (6*6*6) = 216.
UnattemptedExplanation – Total no. Of cases will be (6*6*6) = 216.
 Question 14 of 33
14. Question
1 pointsA vendor sells 10 toffees for a rupee, gaining thereby 20%. How many did he buy for a rupee?
CorrectExplanation – S.P of 10 toffees = Re 1, gain= 20%
Re. 5/6 is C.P of 10
Re. 1 is the C.P of (10*6/5) = 12.
IncorrectExplanation – S.P of 10 toffees = Re 1, gain= 20%
Re. 5/6 is C.P of 10
Re. 1 is the C.P of (10*6/5) = 12.
UnattemptedExplanation – S.P of 10 toffees = Re 1, gain= 20%
Re. 5/6 is C.P of 10
Re. 1 is the C.P of (10*6/5) = 12.
 Question 15 of 33
15. Question
1 points3 , 10 , 32 , 100 , ?
CorrectIncorrect3 Χ 3 + 1 = 10
10Χ 3 + 2 = 32
32Χ3 + 4 = 100
100Χ3 + 8 = 308
Unattempted3 Χ 3 + 1 = 10
10Χ 3 + 2 = 32
32Χ3 + 4 = 100
100Χ3 + 8 = 308
 Question 16 of 33
16. Question
1 points5 , 3 , 4, ? ,38
CorrectIncorrect5Χ1 – 2 = 3
3 Χ 2 – 2 = 4
4 Χ 3 2 =10
10Χ42 = 38
Unattempted5Χ1 – 2 = 3
3 Χ 2 – 2 = 4
4 Χ 3 2 =10
10Χ42 = 38
 Question 17 of 33
17. Question
1 points5 , 6 , ?, 57, 244
Correct5Χ1 + (1)^2 = 6
6 Χ 2+ (2)^2 = 16
16 Χ 3 + (3) ^2 = 57
57 Χ 4 + (4 ) ^2 = 244
IncorrectUnattempted  Question 18 of 33
18. Question
1 points9,27, 31,155,161,1127,?
CorrectExplanation – the pattern is ×3,+4,×5,+6,×7,…..
So. The missing term is 1127+8 = 1135
IncorrectExplanation – the pattern is ×3,+4,×5,+6,×7,…..
So. The missing term is 1127+8 = 1135
UnattemptedExplanation – the pattern is ×3,+4,×5,+6,×7,…..
So. The missing term is 1127+8 = 1135
 Question 19 of 33
19. Question
1 pointsTwo pipes P and Q can fill cistern in 12 and 15 minutes respectively.Both are opened together but at the end of 3 minutes P is turned off. Then in how many more minutes will Q fill the cistern
CorrectExplanation – P can fill in 1 minute = 100/12 = 8.33
Q can fill in 1minute = 100/15 = 6.66
P+Q in 1 minute = 15%
In 3 minutes cistern is fill 45%
then, 55/6.66 = 8 (1/4)
IncorrectExplanation – P can fill in 1 minute = 100/12 = 8.33
Q can fill in 1minute = 100/15 = 6.66
P+Q in 1 minute = 15%
In 3 minutes cistern is fill 45%
then, 55/6.66 = 8 (1/4)
UnattemptedExplanation – P can fill in 1 minute = 100/12 = 8.33
Q can fill in 1minute = 100/15 = 6.66
P+Q in 1 minute = 15%
In 3 minutes cistern is fill 45%
then, 55/6.66 = 8 (1/4)
 Question 20 of 33
20. Question
1 pointsWhat will be the difference between simple and compound interest at 10% per annum on a sum of Rs. 1000 after 4 years?
CorrectExplanation – S.I = 1000*100*4/100 => 400
C.I = 1000 (1+10/100) 4(power) 1000
=> 464.10
hence the difference is 464.10400 = 64.10
IncorrectExplanation – S.I = 1000*100*4/100 => 400
C.I = 1000 (1+10/100) 4(power) 1000
=> 464.10
hence the difference is 464.10400 = 64.10
UnattemptedExplanation – S.I = 1000*100*4/100 => 400
C.I = 1000 (1+10/100) 4(power) 1000
=> 464.10
hence the difference is 464.10400 = 64.10
 Question 21 of 33
21. Question
1 pointsTwo dice are tossed. The probability that the total score is a prime no. Is
CorrectExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
IncorrectExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
UnattemptedExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
 Question 22 of 33
22. Question
1 pointsSix bells commence tolling together and toll at intervals of 2,4,6,8,10 & 12 sec respectively. In 30 minutes how many times do they roll together?
CorrectExplanation – L.C.M of 2,4,6,8,10,&12 is 120
So, the bell toll together after every 120 sec, 2 minutes So, in 30 minutes they will toll together = 30/2+1 = 16 times.
IncorrectExplanation – L.C.M of 2,4,6,8,10,&12 is 120
So, the bell toll together after every 120 sec, 2 minutes So, in 30 minutes they will toll together = 30/2+1 = 16 times.
UnattemptedExplanation – L.C.M of 2,4,6,8,10,&12 is 120
So, the bell toll together after every 120 sec, 2 minutes So, in 30 minutes they will toll together = 30/2+1 = 16 times.
 Question 23 of 33
23. Question
1 pointsA shopkeeper expects a gain of 45/2% on his CP. If his sale was Rs. 392, then find his profit
CorrectExplanation – We know, SP = (100+gain%/100*CP)
CP = 100/122.50* 392 = 320
Profit = 392320= Rs. 72.
IncorrectExplanation – We know, SP = (100+gain%/100*CP)
CP = 100/122.50* 392 = 320
Profit = 392320= Rs. 72.
UnattemptedExplanation – We know, SP = (100+gain%/100*CP)
CP = 100/122.50* 392 = 320
Profit = 392320= Rs. 72.
 Question 24 of 33
24. Question
1 pointsFind compound interest on Rs. 7500 at 4% per annum for 2 years, compounded anually
CorrectExplanation – Please apply the formula –
A = P (1+r/100)n
C.I = AP
IncorrectExplanation – Please apply the formula –
A = P (1+r/100)n
C.I = AP
UnattemptedExplanation – Please apply the formula –
A = P (1+r/100)n
C.I = AP
 Question 25 of 33
25. Question
1 pointsThe banker’s discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time is Rs. 60. The sum due is:
CorrectExplanation: Sum = (B.D. × T.D.)/ (B.D. – T.D.)
=Rs. ( 72 ×60)/ (7260) = Rs 360IncorrectExplanation: Sum = (B.D. × T.D.)/ (B.D. – T.D.)
=Rs. ( 72 ×60)/ (7260) = Rs 360UnattemptedExplanation: Sum = (B.D. × T.D.)/ (B.D. – T.D.)
=Rs. ( 72 ×60)/ (7260) = Rs 360  Question 26 of 33
26. Question
1 pointsIn a 100m race , A can beat B can beat C by 4 m. In the same race, A can beat C by:
CorrectA : B = 100 : 75 and B : C = 100 : 96
A: C=(A/B)×(B/C)=(100/75)×(100/96)=100/72=100:72
A beats C by (10072)m=28m.IncorrectA : B = 100 : 75 and B : C = 100 : 96
A: C=(A/B)×(B/C)=(100/75)×(100/96)=100/72=100:72
A beats C by (10072)m=28m.UnattemptedA : B = 100 : 75 and B : C = 100 : 96
A: C=(A/B)×(B/C)=(100/75)×(100/96)=100/72=100:72
A beats C by (10072)m=28m.  Question 27 of 33
27. Question
1 pointsA man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of boat (in still water) and the stream is :
CorrectLet man’s rate upstream be x kmph. Then, his rate downstream = 2x kmph.
(Speed in still water)
(Speed of stream) = [(2X+X)/2] : [(2XX)/2] = 3X/2 : X/2
= 3:1IncorrectLet man’s rate upstream be x kmph. Then, his rate downstream = 2x kmph.
(Speed in still water)
(Speed of stream) = [(2X+X)/2] : [(2XX)/2] = 3X/2 : X/2
= 3:1UnattemptedLet man’s rate upstream be x kmph. Then, his rate downstream = 2x kmph.
(Speed in still water)
(Speed of stream) = [(2X+X)/2] : [(2XX)/2] = 3X/2 : X/2
= 3:1  Question 28 of 33
28. Question
1 pointsThe difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
CorrectWe have: (l – b) = 23 and 2 (l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = ( l × b) = (63 × 40) m2 = 2520 m2.IncorrectWe have: (l – b) = 23 and 2 (l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = ( l × b) = (63 × 40) m2 = 2520 m2.UnattemptedWe have: (l – b) = 23 and 2 (l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = ( l × b) = (63 × 40) m2 = 2520 m2.  Question 29 of 33
29. Question
1 pointsPipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will take be filled?
CorrectExplanations: Part filled by (A+B+C) in 1 hour = (1/5+1/10+1/30) = 1/3.
All the three pipes together will fill the tank in 3 hours.IncorrectExplanations: Part filled by (A+B+C) in 1 hour = (1/5+1/10+1/30) = 1/3.
All the three pipes together will fill the tank in 3 hours.UnattemptedExplanations: Part filled by (A+B+C) in 1 hour = (1/5+1/10+1/30) = 1/3.
All the three pipes together will fill the tank in 3 hours.  Question 30 of 33
30. Question
1 pointsIn an examination, 35% of the students passed and 455 failed. How many students appeared for the examination?
CorrectExplanations: Let the number of students appeared be x.
Then, 65% of x = 455
<==> 65x/100 = 455
<==> x = (45500/65) = 700.IncorrectExplanations: Let the number of students appeared be x.
Then, 65% of x = 455
<==> 65x/100 = 455
<==> x = (45500/65) = 700.UnattemptedExplanations: Let the number of students appeared be x.
Then, 65% of x = 455
<==> 65x/100 = 455
<==> x = (45500/65) = 700.  Question 31 of 33
31. Question
1 pointsAt an election involving two candidates, 68 votes were declared invalid. The winning candidate secures 52 % and wins by 98 votes. The total number of votes polled is:
CorrectExplanations: Let the number of valid votes be x.
Then, 52% of x – 48% of x = 98
4% of x = 98
4x/100 = 98
x = 98 × 25 = 2450.Total number of votes polled = (2450 + 68) = 2518.
IncorrectExplanations: Let the number of valid votes be x.
Then, 52% of x – 48% of x = 98
4% of x = 98
4x/100 = 98
x = 98 × 25 = 2450.Total number of votes polled = (2450 + 68) = 2518.
UnattemptedExplanations: Let the number of valid votes be x.
Then, 52% of x – 48% of x = 98
4% of x = 98
4x/100 = 98
x = 98 × 25 = 2450.Total number of votes polled = (2450 + 68) = 2518.
 Question 32 of 33
32. Question
1 pointsA library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
CorrectExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
IncorrectExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
UnattemptedExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
 Question 33 of 33
33. Question
1 pointsThe average of ten numbers is 7. If each number is multiplied by 12, then the average of the new set of numbers is:
CorrectExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
IncorrectExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
UnattemptedExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
It will be helpfull for us if you develop and android app where we can practice in offline also and also update the database by connecting it to the internet.
Thanks for your suggestion Sayan. We will try to provide an offline app of our database asap.