“20-20” Quantitative Aptitude | Crack IBPS RRB PO Prelims 2018 Day-148

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“20-20” Aptitude Questions | Crack IBPS RRB PO Prelims 2018 Day-148

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

1) 47% of 1898 + 31% of 4502 = ? × 27 + 1452 + 913.89 – 293.89

  1. 14
  2. 20
  3. 8
  4. 12
  5. 26 

2) 6 1/5 of 19475 + 7 2/7 of 308.29 =? % of 604 + 858.85

  1. 20355
  2. 18750
  3. 16425
  4. 8920
  5. 23575

3) 516 × 3719 ÷ 31 % of 2399 = 64 × 125.89 ÷ 7.21 + ? – 1899

  1. 4170
  2. 4950
  3. 3890
  4. 3330
  5. 5160

4) 514 × 6298. 81 ÷ 24 % of 749 = 18 × 439.71 ÷ 11.12 + ? + 2599

  1. 18920
  2. 14670
  3. 21560
  4. 23780
  5. 26790 

5) (4/11) of 176.18 × 45.12 ÷ 14.78 – ? = 855.31 ÷ 9 – 434.98

  1. 780
  2. 600
  3. 700
  4. 660
  5. 530

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. 3x2 + 4√3x + 4 = 0

II. 5y2 = 2√5y + 8

7) I. 8x2 + 42x + 45 = 0

II. 2y2 + 34y + 104 = 0

8) I. 3√x = (23/√x) + (13/√x)

II. (√y/4) + (4√y/8) = (9/√y)

9) I. 5x2 × x1/2 = 15625

II. y = ∛9261

10) I. x2+ 18x + 45 = 0

II. y2 + 19y + 60 = 0

11) A box contains 6 Pink balls, 3 black balls, 4 orange balls and 2 white balls. If four balls are drawn out randomly, then the probability of getting balls of different colour?

  1. 11/135
  2. 48/455
  3. 23/479
  4. 34/477
  5. None of these

12) The ratio of 3 years after, the age of P and 5 years ago, the age of Q is 6: 5. The present age of R is 2 times the present age of P. The difference between the present age of Q and R is 10 years. Find the present age of P?

  1. 12 years
  2. 20 years
  3. 25 years
  4. 15 years
  5. 18 years

13) A, B and C can do a work in 18, 24 and 30 days respectively. They undertook to finish the work together for Rs. 11750. Find the difference between the shares of A and C?

  1. Rs. 2000
  2. Rs. 1800
  3. Rs. 2400
  4. Rs. 1200
  5. None of these 

14) A, B and C started a business by investing Rs. 20000, Rs. 35000 and Rs. 25000 respectively. After 4 months, A invested 15 % more than the initial investment and after 5 months, C invested Rs. 10000 more but B withdraw 20 % of the initial investment. Find the share of C, if the total profit at the end of the year is Rs. 82750?

  1. Rs. 24600
  2. Rs. 35400
  3. Rs. 27500
  4. Rs. 32800
  5. None of these

15) The difference between the two numbers is 360. 28 % of one number is equal to 42 % of another number. Find the two numbers?

  1. 1160 and 800
  2. 1080 and 720
  3. 1230 and 870
  4. 950 and 590
  5. None of these 

Directions (16 – 20) Study the following table carefully and answer the given questions:

Following table shows the total no of students appeared for the certain examination from different schools.

School Total no of students appeared Appeared boys : girls Pass % No of girls passed
A 2500 2 : 3 72 % 1100
B 3200 5 : 3 76 % 840
C 2800 3 : 4 85 % 1330
D 1600 5 : 3 68 % 513
E 1500 3 : 2 90 % 520

 

16) Total number of boys appeared for an examination from School A and B together is approximately what percentage of total number of girls appeared for an examination from School C and D together?

  1. 115 %
  2. 136 %
  3. 100 %
  4. 85 %
  5. 150 %

17) Find the ratio between the total number of boys passed from School A to that of School C?

  1. 4: 5
  2. 5: 6
  3. 3: 4
  4. 2: 3
  5. None of these 

18) Find the difference between the total number of students passed the examination from School C and E together to that of School B and D together?

  1. 210
  2. 350
  3. 420
  4. 280
  5. None of these

19) Find the average number of boys passed the examination from School A, B, C and E together?

  1. 2348
  2. 1579
  3. 1043
  4. 2156
  5. None of these 

20) Total number of girls passed from School A and B together is approximately what percentage more/less than the total number of girls passed from School C and E together?

  1. 10 % less
  2. 5 % less
  3. 15 % more
  4. 15 % less
  5. 5 % more

Answers:

Direction (1-5)

1). Answer: c)

47% of 1900 + 31% of 4500 = 27x + 1452 + 914 – 294

(47/100)*1900 + (31/100)*4500 = 27x + 1452 + 914 – 294

893 + 1395 – 1452 – 914 + 294 = 27x

27x = 216

X = 8

2). Answer: a)

(31/5)*19475 + (51/7)*308 = (x/100)*600 + 859

120745 + 2244 = 6x + 859

120745 + 2244 – 859 = 6x

6x = 122130

X = 20355

3). Answer: d)

516 × 3720 ÷ 31 % of 2400 = 64 × 126 ÷ 7 + x – 1900

516 × 3720 ÷ [(31/100)*2400] = 64*(126/7) + x – 1900

516 × 3720 ÷ 744 = 1152 + x – 1900

516*(3720/744) = 1152 + x – 1900

2580 – 1152 + 1900 = x

X = 3328 = 3330

4). Answer: b)

514 × 6300 ÷ (24/100)*750 = 18*(440/11) + x + 2600

514*6300/180 = 720 + x + 2600

17990 -720 – 2600 = x

X = 14670

5). Answer: e)

(4/11)*176*(45/15) – x = (855/9) – 435

192 – x = 95 – 435

192 – 95 + 435 = x

X = 532 = 530

Direction (6-10)

6). Answer: c)

I. 3x2 + 4√3x + 4 = 0

3x2 + 2√3x + 2√3x + 4 = 0

√3x (√3x + 2) + 2(√3x + 2) = 0

(√3x + 2) (√3x + 2) = 0

x = -2/√3, -2/√3

II. 5y2 – 2√5y – 8 = 0

5y2 – 4√5y + 2√5y – 8 = 0

√5y (√5y – 4) + 2(√5y – 4) = 0

(√5y + 2) (√5y – 4) = 0

y = -2/√5, 4/√5

x < y

7). Answer: a)

I. 8x2 + 42x + 45 = 0

8x2 + 12x + 30x + 45 = 0

4x (2x + 3) + 15(2x + 3) = 0

(4x + 15) (2x + 3) = 0

x = -15/4, -3/2 = -3.75, -1.5

II. 2y2 + 34y + 104 = 0

2y2 + 8y + 26y + 104 = 0

2y(y + 4) + 26(y + 4) = 0

(2y + 26) (y + 4) = 0

y = -26/2, -4 = -13, -4

x > y

8). Answer: e)

I. 3√x = (23/√x) + (13/√x)

3√x = (23 + 13)/ √x

x = 36/3 = 12

II. (√y/4) + (4√y/8) = (9/√y)

(2√y + 4√y)/8 = 9/√y

6√y /8 = 9/√y

y = (9/6)*8 = 12

x = y

9). Answer: a)

I. 5x2 × x1/2 = 15625

x2 × x1/2 = 15625/5

x2 + (½) = 3125

x5/2 = 3125

x = (3125)2/5 = (55)2/5

x = 52 = 25

II. y = ∛9261

Y = 21

X > y

10). Answer: e)

I. x2+ 18x + 45 = 0

(x + 15) (x + 3) = 0

X = -15, -3

II. y2 + 19y + 60 = 0

(x + 15) (x + 4) = 0

X = -15, -4

Can’t be determined

11). Answer: b)

Total probability n(S) = 15C4 = (15*14*13*12) / (1*2*3*4)

Required probability n(E) = 6C1 and 3C1 and 4C1 and 2C1

P(E) = n(E)/n(S)

P(E) = (6C1 and 3C1 and 4C1 and 2C1) / 15C4

P(E) = 48/455

12). Answer: d)

The ratio of 3 years after, the age of P and 5 years ago, the age of Q = 6: 5 (6x, 5x)

Present age of P and Q = 6x – 3, 5x + 5

The age of R = 2*P = 2*(6x – 3) = 12x – 6

12x – 6 – (5x + 5) = 10

12x – 6 – 5x – 5 = 10

7x = 10 + 11

7x = 21

X = 3

Present age of P = 6x – 3 = 18 – 3 = 15 years

13). Answer: a)

Ratio of Shares of A: B: C = (1/18): (1/24): (1/30) = 20: 15: 12

Difference between the shares of A and C = (8/47)*11750 = Rs. 2000

14). Answer: c)

The share of A, B and C

= > [20000*4 + 20000*(115/100)*8]: [35000*9 + 35000*(80/100)*3]: [25000*9 + 35000*3]

= > 264000: 399000: 330000

= > 88: 133: 110

331’s = 82750

1’s = 250

The share of C = 110’s = Rs. 27500

15). Answer: b)

Let the two numbers be x and y,

X – Y = 360

(28/100)*X = (42/100)*Y

(X/Y) = 3/2

1’s = 360

The two numbers be,

= > 3’s = (360*3) = 1080

= > 2’s = (360*2) = 720

Direction (16-20)

16). Answer: b)

Total number of boys appeared for an examination from School A and B together

= > 2500*(2/5) + 3200*(5/8)

= > 1000 + 2000 = 3000

Total number of girls appeared for an examination from School C and D together

= > 2800*(4/7) + 1600*(3/8)

= > 1600 + 600 = 2200

Required % = (3000/2200)*100 = 136 %

17). Answer: d)

The total number of boys passed from School A

= > 2500*(72/100) – 1100

= > 1800 – 1100 = 700

The total number of boys passed from School C

= > 2800*(85/100) – 1330

= > 2380 – 1330 = 1050

Required ratio = 700: 1050 = 2: 3

18). Answer: a)

The total number of students passed the examination from School C and E together

= > 2800*(85/100) + 1500*(90/100)

= > 2380 + 1350 = 3730

The total number of students passed the examination from School B and D together

= > 3200*(76/100) + 1600*(68/100)

= > 2432 + 1088 = 3520

Required difference = 3730 – 3520 = 210

19). Answer: c)

The total number of boys passed the examination from School A, B, C and E together

= > [2500*(72/100) – 1100] + [3200*(76/100) – 840] + [2800*(85/100) – 1330] + [1500*(90/100) – 520]

= > [1800 – 1100] + [2432 – 840] + [2380 – 1330] + [1350 – 520]

= > 700 + 1592 + 1050 + 830 = 4172

Required average = 4172/4 = 1043

20). Answer: e)

Total number of girls passed from School A and B together

= > 1100 + 840 = 1940

Total number of girls passed from School C and E together

= > 1330 + 520 = 1850

Required % = [(1940 – 1850)/1850]*100 = 5 % more

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