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IBPS > Quantitative Aptitude

Explore popular questions from Quantitative Aptitude for IBPS. This collection covers Quantitative Aptitude previous year IBPS questions hand picked by experienced teachers.

Q 1.

Correct4

Incorrect-1

Direction: In the following series, only one number is wrong. Find out the wrong number.7, 43, 164, 488, 1326, 3093

A

488

1326

C

43

D

3093

E

164

Explanation

Solution :The pattern of given series is:
{tex} → 7{/tex},
{tex}→ 43 = 7 + 6^2,{/tex}
{tex}→ 164 = 43 + 11^2, (6 + 5 = 11) {/tex}
{tex}→ 488 = 164 + 18^2, (11 + 7 = 18) {/tex}
{tex}→ 1329 = 488 + 29^2, (18 + 11 = 29) {/tex}
{tex}→ 3093 = 1329 + 42^2, (29 + 13 = 42){/tex}
Thus, the wrong number is 1326 and the correct number is 1329

Q 2.

Correct4

Incorrect-1

Direction: In the following number series only one number is wrong. Find out the wrong number.13, 16, 21, 30, 39, 52, 69

A

21

B

39

30

D

52

E

16

Explanation

Solution :The pattern of series is as: Hence 30 is the wrong term.

Q 3.

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Incorrect-1

Pipe A and Pipe B can fill a tank alone in 18 minutes and 12 minutes, respectively. Another pipe C alone can empty the full tank in 16 minutes. Pipe A and pipe B are opened together for 7 minutes, then pipe B was closed and pipe C was opened for ‘x’ minutes and then closed. If the total time taken to fill the tank is 39 minutes, then find the value of ‘x’.

A

26 minutes

B

20 minutes

C

32 minutes

28 minutes

E

None of these

Explanation

Solution :Let the capacity of the tank be LCM (18, 12 and 16) = 144 units Number of units of water filled by pipe A in one minute = Number of units of water filled by pipe B in one minute = Number of units of water emptied by pipe C in one minute = So, number of units of water filled by pipe A and pipe B together in one minute = So, number of units of water filled by pipe A and pipe B together in 7 minute = Remaining number of units of water to be filled = 144 – 140 = 4 units Now, pipe A and C are opened together Number of units of water emptied by pipe A and C in one minute = Number of units of water emptied by pipe A and C in ‘x’ minute = Number of units of water filled by pipe A alone = So, =4+x=(32-x)x8 =x=28 minutes So option (d) is the correct answer.

Q 4.

Correct4

Incorrect-1

Out of a jar of 200 jellybeans, 40% of them are red or yellow, 80% are either red or blue while remaining are green in colour. Then, the probability of picking up a blue bean, if the probability of picking a yellow or a blue been is 0.8, is-

A

0.2

0.6

C

0.3

D

0.8

E

0.75

Explanation

Solution :Let no. of red, yellow and blue beans be x, y and z respectively. Given, x + y = 40% of 200 x + y = 80 …(1) and, x + z = 80% of 200 x + z = 160 …(2) Subtracting (1) from (2), z – y = 80 …(3) Also given, Probability of picking yellow or blue bean = 0.8 = (No. of yellow or blue bean) / (total no. of beans) Or, (z + y) / 200 = 0.8 z + y = 160 …(4) Adding (3) & (4), 2z = 240 or, z = 120 blue beans y = 40 yellow beans & x = 40 red beans Prob. of picking a blue bean = 120/(120+40+40) = 120/200 = 0.6

Q 5.

Correct4

Incorrect-1

The average age of people of a club is twice the number of people in the club. A person, P, leaves the club and the average age is still twice the number of people in the club. Now another person, Q, leaves the club and the average age is still twice the number of people in the club. If the ratio of the ages of P and Q is 19 : 17, then find the average age of the club, if one more person, R, of age 16 years, leaves the club.

16

B

18

C

20

D

22

E

24

Explanation

Solution :Let, initial no. of people in the club be ‘n’. so, average age of people = 2nTotal sum of people = 2n^2Let 19x and 17x be ages of P and Q respectively, A.T.Q., 2n2 - 19X =2(n-1)2……..(i) And, 2n2 - 19x - 17x = 2(n-2)2 2n2 - 36x = 2(n-2)2………….(ii) Solving (i) and (ii), X=2,n=10Average age of group after R leaves the club= 
(2 × 102 – 19 × 2 – 17 × 2 – 16)/7 =112/7=16 years.

Q 6.

Correct4

Incorrect-1

Direction:  In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

A

C

D

E

or relation cannot be determined

Explanation

Solution :

Q 7.

Correct4

Incorrect-1

Two vessels, A and B contain 360 litres and 320 litres mixture (milk and water) with milk to water in the ratio 5: 4 and 5: 3, respectively. Mixture from both the vessels mixed together and the ratio of milk to water in the mixture in its simplest form is a: b, what is the value of (a+b)?

A

12

17

C

19

D

21

E

27

Explanation

Solution :Quantity of milk in vessel A = Quantity of water in vessel A = Quantity of milk in vessel B = Quantity of water in vessel B = According to the question, (a: b) = (200 + 200): (160 + 120) = 400: 280 = 10: 7 Therefore, (a+b) = 10 + 7 = 17 So option (b) is the correct answer.

Q 8.

Correct4

Incorrect-1

Direction: In the following number series only one number is wrong. Find out the wrong number.35, 118, 280, 600, 1240, 2504, 5036

A

118

B

280

C

600

1240

E

2504

Explanation

Solution :The pattern is as follows

Q 9.

Correct4

Incorrect-1

Santosh bought an article for Rs. (x + 1500), and marked it up by 20%. If he sold it at a discount of (x – 35)% for Rs. 1581, then at what price it should be sold to earn a profit of 40%? [Note: ‘x’ is a multiple of 10]

A

Rs. 1860

Rs. 2170

C

Rs. 2450

D

Rs. 2740

E

None of these

Explanation

Solution :According to question, (x + 1500) x 1.2 x (100 – x + 35) % = 1581 (x + 1500) (135 – x) = 131750 So, cost price of an article = 50 + 1500 = Rs. 1550 Therefore, required selling price = 1550 x 1.4 = Rs. 2170 So option (b) is the correct answer.

Q 10.

Correct4

Incorrect-1

Direction:  In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.
 

A

B

C

E

or relation cannot be determined.

Explanation

{tex}x^2-11x-2\sqrt{26}x+22\sqrt{26}=0{/tex} {tex}\implies x((x-11)-2\sqrt{26}(x-11)=0{/tex} {tex}\implies (x-2\sqrt{26})(x-11)=0{/tex} {tex}\implies x=11, 2\sqrt{26}{/tex}. {tex}y^2-23y+132=0{/tex} {tex}\implies (y-11)(y-12)=0{/tex} {tex}\implies y=11, 12{/tex}. Hence, {tex}x\leq y.{/tex}

Q 11.

Correct4

Incorrect-1

Direction: Given below are two quantities named A and B. Based on the given information, you have to determine relation between the two quantities. You should use the given data and knowledge of Mathematics to choose between the possible answers. 
Quantity A: 262
Quantity B: Number of ways in which 4 socks taken out from a drawer having 4 pair of socks which are unbundled, does not make even a single correct pair.

A

Quantity A ≤ Quantity B

B

Quantity A = Quantity B or no relation

Quantity A < Quantity B

D

Quantity A > Quantity B

E

Quantity A {tex} \geq{/tex} Quantity B

Explanation

Solution :Quantity A:  262 Quantity B: ⇒Total number of socks = 4 pairs = 8 socks Now, first socks can be chosen in 8 ways. Now, 7 socks are remaining and since no pair of socks can be completed, hence second socks can be selected from 6 socks only, excluding the one which is similar to first drawn socks, ⇒Number of ways of selecting second socks = 6 Now, 6 socks are remaining, but to fulfil the condition, the third socks can be selected from only 4 socks, ⇒Number of ways in which third socks can be drawn = 4 Now, 5 socks are remaining, and 3 socks are similar to the socks that has been taken out, so to fulfil the condition, the fourth socks can be taken out from remaining 2 socks only ⇒Number of ways of selecting fourth socks = 2 ⇒Total number of ways of selecting 4 socks = 8 × 6 × 4 × 2 = 384 ⇒Quantity B = 384 ∴ Quantity A < Quantity B

Q 12.

Correct4

Incorrect-1

Direction: In the following number series only one number is wrong. Find out the wrong number.26, 28, 34, 46, 66, 95

A

26

B

28

C

46

D

66

95

Explanation

Solution :The pattern of given series is:
{tex}26 + (1^2 + 1) = 28{/tex}
{tex}28 + (2^2 + 2) = 34{/tex}
{tex}34 + (3^2 + 3) = 46{/tex}
{tex}46 + (4^2 + 4) = 66{/tex}
 {tex}66 + (5^2 + 5) = 96{/tex}
 Thus, the wrong number is 95 and the correct number is 96

Q 13.

Correct4

Incorrect-1

Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Quantity I: Milk and water is filled into two vessels of same capacity in which water and milk are in the ratio of 3: 7 and 2: 5. If the mixtures from both the vessels are poured into another container then what will be the ratio of milk and water in the new container?
Quantity II: 119/41

A

Quantity I > Quantity II

Quantity I < Quantity II

C

Quantity I Quantity II

D

Quantity I Quantity II

E

Quantity I = Quantity II or No relation

Explanation

Solution :Quantity I: Let the capacity of each vessel = n litres Amount of milk in 1st vessel = Amount of milk in 2nd vessel = Total amount of milk in both the vessel = Amount of water in the container after mixing = Therefore, ratio of milk and water in the resulting mixture = Quantity II: 119/41Therefore, Quantity I Quantity II So option (b) is the correct answer.

Q 14.

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Incorrect-1

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers. A, B and C enter into a partnership. A contributes Rs 320 for 4 months, B contributes Rs 510 for 3 months and C contributes Rs 270 for 5 months. If the total profit is Rs 208.
Quantity I: profit of C
Quantity II: average profit of A, B and C

A

Quantity I Quantity II

Quantity I Quantity II

C

Quantity I Quantity II

D

Quantity I Quantity II

E

Quantity I Quantity II or No relation

Explanation

Solution :Ratio of the profits of A, B and C = ratio of their partnership = 320 × 4 : 510 × 3 : 270 ×5 = 128: 153: 135 Let the profit of A, B and C be 128 x, 153x and 135x respectively. Then, 128x + 153x + 135 = 208 416x = 208 x= Hence, share of A = 128 × = Rs 64 Share of B = 153 Share of C = 135 × =Rs 67.5 Quantity I = 67.5 Rs. Quantity II = 69 Rs (Quantity I Quantity II)

Q 15.

Correct4

Incorrect-1

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Quantity I: Piyush finds that only 1/3 of the work has been done after working regularly for 8 days. He employs Rakesh who is 60% as efficient as Piyush. How many more days required by Piyush to complete the job
Quantity II:  45 days

A

Quantity I Quantity II

Quantity I Quantity II

C

Quantity I Quantity II

D

Quantity I Quantity II

E

Quantity I Quantity II or No relation

Explanation

Solution :In 8 days, Piyush does = rd work So Piyush complete the work in 24 days alone Efficiency of Piyush and Rakesh together: 1 + 0.6 = 1.6 As 1/3rd work is completed so remaining work i.e. 16 is completed by both in =16/1.6=10 days Quantity I 10 days Quantity II 45 days (Quantity I Quantity II)

Q 16.

Correct4

Incorrect-1

Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Quantity I: A man sold an article at a loss of 12%, while another man sold the same article having the same cost price at a profit of 8%. The total selling price of both the articles is Rs. 4312. If the man who incurs 12% loss sold the article after giving a discount of Rs. 424, then find the marked price of the article.
Quantity II: 2360 Rs.

A

Quantity I > Quantity II

B

Quantity I < Quantity II

C

Quantity I Quantity II

D

Quantity I Quantity II

Quantity I = Quantity II or No relation

Explanation

Solution :Quantity I: Let, the cost price of both article be Rs. x According to the question, Selling price of the article sold at 12% loss = Rs. 0.88x And, selling price of the article sold at 8% profit = Rs. 1.08x According to the question, 0.88x + 1.08x = 4312 1.96x = 4312 x = Rs. 2200 So, the selling price of the article sold at 12% loss = 0.88 x 2200 = Rs. 1936 Therefore, marked price of the article = 1936 + 424 = Rs. 2360 Quantity II:  Rs. 2360 Therefore, Quantity I = Quantity II So option (e) is the correct answer.

Q 17.

Correct4

Incorrect-1

Direction: In the following series, only one number is wrong. Find out the wrong number.40, 20, 20, 40, 30, 1280

A

20

B

40

C

160

D

1280

30

Explanation

Solution :

Q 18.

Correct4

Incorrect-1

Direction:  In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

A

B

C

E

or relation cannot be determined

Explanation

Solution :

Q 19.

Correct4

Incorrect-1

Direction:  In the following question two equations are given in variables p and q. You have to solve these equations and determine relation between p and q.

A

B

C

D

or relation cannot be determined.

Explanation

Solution : ---------eq.1 ----------eq.2 From eq.1 and eq.2 , we get Thus, the correct option is E

Q 20.

Correct4

Incorrect-1

Direction:  In the following question two equations are given in variables X and Y. You have to solve these equations and determine relation between X and Y.

A

B

D

E

or relation cannot be determined.

Explanation

Solution : ---------------eq.1 y= ,-1----------------eq.2 From eq.1 and eq.2 , we get Thus, the correct option is option (C)

Q 21.

Correct1

Incorrect-0.25

In each of these questions, two equations numbered I and II are given. You have to solve both the equations and give answer
I. {tex} x ^ { 2 } - 22 x + 120 = 0 {/tex}
II. {tex} y ^ { 2 } - 26 y + 168 = 0 {/tex}

A

if {tex} x < y {/tex}

if {tex} x \leq y {/tex}

C

if {tex} x > y {/tex}

D

if {tex} x \geq y {/tex}

E

if {tex} x = y {/tex} or the relationship cannot be established

Explanation

Q 22.

Correct1

Incorrect-0.25

In each of these questions, two equations numbered I and II are given. You have to solve both the equations and give answer
I. {tex} \frac { 3 } { \sqrt { x } } + \frac { 4 } { \sqrt { x } } = \sqrt { x } {/tex}
II. {tex} y ^ { 2 } = \frac { ( 7 ) ^ { \frac { 5 } { 2 } } } { \sqrt { y } } = 0 {/tex}

A

If {tex} x > y {/tex}

B

If {tex} x \geq y {/tex}

C

If {tex} x < y {/tex}

D

If {tex} x \leq y {/tex}

If {tex} x = y {/tex} or the relationship cannot be established.

Explanation


Q 23.

Correct1

Incorrect-0.25

In each of the following questions equation I and equation II have been given. You have to solve both of these equations and give answer
I. {tex} 4 x ^ { 2 } - 32 x + 63 = 0 {/tex}
II. {tex} 2 y ^ { 2 } - 11 y + 15 = 0 {/tex}

A

{tex} x < y {/tex}

{tex} x > y {/tex}

C

{tex} \mathrm { x } \leq \mathrm { y } {/tex}

D

{tex} x \geq y {/tex}

E

{tex} { x } = { y } {/tex} or no relation between two can be established.

Explanation