“20-20” Quantitative Aptitude | Crack IBPS RRB PO Prelims 2018 Day-160

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“20-20” Aptitude Questions | Crack IBPS RRB PO Prelims 2018 Day-160

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Directions (1 – 5): What approximate value should come in the place of question mark (?) in the following questions?

1) (5218 + 2154) ÷ 21 + (23.12)2 + (3/7) of 2519 – 28 =?

a) 1930

b) 1850

c) 1760

d) 2180

e) 2370 

2) 52 % of 2149 + (18.13)2 ÷ 12 × 3 + (23.2)2 = 3.9 ×?

a) 565

b) 520

c) 470

d) 430

e) 600 

3) (22/13) ÷ (155/258) × 517 =? – (12.11)3

a) 4850

b) 3210

c) 4670

d) 4150

e) 3690 

4) 15 7/8 + 23 1/3 – 8 1/7 + 1 ¾ =? + 5 5/7

a) 40

b) 65

c) 75

d) 90

e) 30 

5) 34478 ÷ 8 + (3/13) of 12154 + 30 % of 599 =?2 – 274

a) 93

b) 74

c) 87

d) 68

e) 82 

Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,

  1. If x > y
  2. If x ≥ y
  3. If x < y
  4. If x ≤ y
  5. If x = y or the relation cannot be established 

6) I. 1/x + 1/(x-10) = 8/75

II. 132/y + 132/(y + 11) = 1

7) I. x2 +18x +72 =0

II. y2 + 30y +56 =0

8) I. 2x2 – 9x +4 =0

II. y2 – 14y +40 =0

9) I. 4x2 -13x +10 =0

II. 2y2 -15y +22 =0

10) I.  5x2 -28x +39 =0

II. 2y2 -13y +20 =0

11) A and B can finish a work in 18 days while B and C can do it in 12 days. A started the work, worked for 6 days only and then B worked for 8 days and the remaining work was finished by C alone in 15 days. In how many days could C alone have finished the whole work?

a) 22 days

b) 20 days

c) 18 days

d) 26 days

e) None of these

 12) A, B and C started a business by investing in the ratio of 3: 5: 7. After 5 months, A invested Rs. 15000 more. And after 3 months, C withdraws Rs. 30000. Find the initial investment of B, if the share of A, B and C is in the ratio of 45: 60: 76?

a) Rs. 75000

b) Rs. 90000

c) Rs. 65000

d) Rs. 80000

e) None of these

13) Manoj marks an item one-fourth more than the CP and allows successive discounts of x% and (x+5) %. If the CP of an item is Rs. 500 and he losses 15 %, then find the value of ‘x’?

a) 18 %

b) 15 %

c) 12 %

d) 20 %

e) None of these

14) The difference between the ages of Yuvaraj and Ajay, 4 years ago was 8 years. If the ratio of their ages 7 years hence is 5: 4, then find the present age of Yuvaraj’s mother who is 25 years older than Yuvaraj?

a) 64 years

b) 72 years

c) 58 years

d) 54 years

e) None of these

15) A and B together can complete a work in 35 days, while A is 40 % more efficient than B, then find in how many days the work will be complete if they work on alternative days starting with A?

a) 70 days

b) 60 days

c) 55 days

d) 66 days

e) None of these

Directions (16 – 20): Study the following information carefully and answer the question given below.

The following table represents the number of students enrolled in six different colleges, percentage of students left the college and percentage of students who passed out the college among those who continued their study in the college.

16) Find the sum of total number of students from all the colleges, who continued their study.

a) 7432

b) 7223

c) 7199

d) 7351

e) None of these

17) Number of students passed from college B is approximately what percent of the number of students passed from college D?

a) 152

b) 180

c) 187

d) 125

e) None of these

18) Total number of students enrolled in A and B together is what percent more than the number of students enrolled in C and D together?

a) 6.134%

b) 8.163%

c) 9.156%

d) 4.146%

e) 7. 178%

19) Find the ratio of number of failed students of college C and college F respectively.

a) 1137: 450

b) 960: 1457

c) 341: 144

d) 450: 137

e) None of these

20) Passed students of college A is approximately what percent of failed students of college C?

a) 121 %

b) 184 %

c) 211 %

d) 312 %

e) None of these

Answers:

Direction (1-5)

1) Answer: a)

(5218 + 2154) ÷ 21 + 232 + (3/7) of 2520 – 28 =?

(7372/21) + 529 + (3/7)*2520 – 28 = x

351 + 529 + 1080 – 28 = x

X = 1932 = 1930

2) Answer: d)

(52/100)*2150 + (18*18*3)/12 + 232 = 4x

1118 + 81 + 529 = 4x

1728 = 4x

1728/4 = x

X = 432 = 430

3) Answer: b)

(22/13)*(260/154)*518 = x – 1728

1480 + 1728 = x

X = 3208 = 3210

4) Answer: e)

15 7/8 + 23 1/3 – 8 1/7 + 1 ¾ = x + 5 5/7

16 + 23 – 8 + 2 – 6 = x

X = 27 = 30

5) Answer: c)

(34480/8) + (3/13) * 12155 + (30/100)*600 = x2 – 274

4310 + 2805 + 180 + 274 = x2

7569 = x2

X = 87

Direction (6-10)

6) Answer: e)

I. 1/x + 1/(x-10) = 8/75

(2x – 10)/(x2 – 10x) = 8/75

75x – 375 = 4x2 – 40x

4x2 – 115x + 375 = 0

4x2 – 100x – 15x + 375 = 0

4x(x – 25) – 15 (x – 25) = 0

(4x – 15)(x – 25) = 0

X = 15/4, 25 = 3.75, 25

II. 132/y – 132/(y + 11) = 1

[132*(y + 11 – y)]/(y2 + 11y) = 1

132*11 = y2 + 11y

Y2 + 11y – 1452 = 0

(y – 33) (y + 44) = 0

Y = 33, -44

Can’t be determined

7) Answer: e)

I. x2 +18x +72 =0

(x+12)(x+6) =0

X= -12, -6

II. y2 + 30 y +56 =0

(y+28)(y+2) =0

Y= -28, -2

Can’t be determined

8) Answer: d)

I. 2x2 – 9x +4 =0

2x2 – x – 8x +4 =0

x (2x – 1) – 4 (2x – 1)=0

(x – 4) (2x – 1) = 0

x = 4, 1/2

II. y2 – 14y +40 =0

y2 -10y – 4y +40 =0

(y – 10) (y – 4) =0

y = 10, 4

x ≤ y

9) Answer: d)

I. 4x2 -13x +10 =0

4x2 -8x-5x +10 =0

4x(x-2) – 5 (x-2) =0

(4x-5)(x – 2) =0

X = 5/4, 2

II. 2y2 -15y +22 =0

2y2 -4y-11y +22 =0

2y(y-2)-11 (y-2) =0

(2y-11)(y – 2) =0

Y=11/2, 2

x ≤ y

10) Answer: e)

I. 5x2 -28x +39 =0

5x2 -15x-13x +39 =0

5x(x-3)-13 (x-3) =0

(5x – 13)(x – 3) =0

X= 13/5, 3

II. 2y2 -13y +20 =0

2y2 -8y-5y +20 =0

2y(y-4)-5 (y-4) =0

(2y – 5) (y – 4) =0

Y=5/2, 4

Can’t be determined

11) Answer: d)

A and B can finish a work in 18 days = 1/A + 1/B = 1/18

B and C can do it in 12 days = 1/B + 1/C = 1/12

According to the question

6/A + 8/B + 15/C = 1

=> 6(1/A + 1/B) + 2(1/B + 1/C) + 13/C = 1

=> 6/18 + 2/12 + 13/C = 1

=> 13/C = 1 – 1/3 – 1/6

=> 13/C = (6 – 2 – 1)/6

=> (1/C) = (3/6)*(1/13) = 1/26

C alone can finish the whole work in 26 days

12) Answer: a)

The initial investment of A, B and C = 3x, 5x and 7x

The share of A, B and C

= > [3x*5 + (3x + 15000)*7]: [5x*12]: [7x*8 + (7x – 30000)*4] = 45: 60: 76

= > [15x + 21x + 105000]: [60x]: [56x + 28x – 120000] = 45: 60: 76

= > [36x + 105000]: [60x]: [84x – 120000] = 43: 60: 76

According to the question,

(60x)/(84x – 120000) = (60/76)

76x = 84x – 120000

8x = 120000

X = 15000

The initial investment of B = 5x = Rs. 75000

13) Answer: b)

SP = 500*(85/100) = 425

MP = 500*(5/4) = Rs. 625

According to the question,

625*[(100 – x)/100]*[(95 – x)/100] = 425

9500 – 100x – 95x + x2 = 6800

X2 – 195x + 2700 = 0

(x – 180) (x – 15) = 0

X = 180, 15 (180 should be eliminated)

X = 15 %

14) Answer: c)

4 years ago, Yuvaraj – Ajay = 8

7 years hence, the ratio of Yuvaraj and Ajay = 5: 4 (5x, 4x)

Present age of Yuvaraj and Ajay = 5x – 7, 4x – 7

According to the question,

5x – 4x = 8

x = 8

Present age of Yuvaraj = 5x – 7 = 33

Present age of Yuvaraj’s mother = 33 + 25

= > 58 years

15) Answer: a)

Efficiency of A and B = 140: 100 = 7: 5

The ratio of time taken by A and B = 5: 7 (5x, 7x)

(1/5x) + (1/7x) = 1/35

12x/(5x*7x) = 1/35

X = 12

The time taken by A and B = 5x, 7x = 60 and 84 days

Total units = 420 (LCM of 60, 84)

A’s one day work = 7 units

B’s one day work = 5 units

Work done in 2 days = 7 + 5 = 12 units

Work done in 70 days = 12*35 = 420 units

Work will be completed in 70 days

Direction (16-20)

16) Answer: c)

A = (100 – 5)/100 x 1400 = 1330

B = (100 – 8)/100 x 1250 = 1150

C = (100 – 12)/100 x 1550 = 1364

D = (100 – 10)/100 x 900 = 810

E = (100 – 15)/100 x 1300 = 1105

F = (100 – 4)/100 x 1500 = 1440

Required No of students = 1330 + 1150 + 1364 + 810 + 1105 + 1440 = 7199

17) Answer: b)

Number of students who continued study in B = (100 – 8)/100 x 1250 = 1150

Passed Students = 76/100 x 1150 = 874

Number of students who continued study in D = (100 – 10)/100 x 900 = 810

Passed Students = 60/100 x 810 = 486

Required percentage = (874/486) x 100 = 179.83% = 180 %

18) Answer: b)

Total number of students enrolled in A and B together = 1400 + 1250 = 2650

Total number of students enrolled in C and D together = 1550 + 900 = 2450

Required percentage = (2650 – 2450)/2450 x 100 = 8.163%

19) Answer: c)

Number of students who continued study in C = (100 – 12)/100 x 1550 = 1364

Failed = (100-75)/100 x 1364 = 341

Number of students who continued study in F = (100 – 4)/100 x 1500 = 1440

Failed = (100 – 90)/100 x 1440 = 144

Required ratio = 341: 144

20) Answer: d)

Number of students who continued their study from college A = (100 – 5)/100 x 1400 = 1330

Passed = 80/100 x 1330 = 1064

Number of students who continued their study from college C

= (100 – 12)/100 x 1550 = 1364

Failed = (100 – 75)/100 x 1364 = 341

Required percentage = (1064/341) x 100 = 312.023% = 312 %

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