“20-20” Quantitative Aptitude | Crack IBPS RRB PO Prelims 2018 Day-158

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“20-20” Aptitude Questions | Crack IBPS RRB PO Prelims 2018 Day-158

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Directions (1 – 5): 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 

1) I. 4x2 – (4 + 2√3)x + 2√3 = 0

II. 8y2 – (24 + 3√3)y + 9√3 = 0

2) I. x2 +9x – 52 = 0

II. y2 – 15y + 54 = 0

3) I. 4x2 + 20x + 24 = 0

II. 3y2 + 18x + 24 = 0

4) I. 7x2 – 6x – 81 = 0

II. 9y2 – 10y – 56 = 0

 5) I. 2x – 5y = 12

II. 5x – 2y = -12

Directions (6 – 10): What approximate value should come in the place of question mark (?) in the following questions?

6) ? = (4128 + 574) ÷ 25 + (18.23)2 + (2/15) of 12524 – 24

a) 2540

b) 3120

c) 2160

d) 3480

e) 3950 

7) 1399.99 ÷ 20.21 = ? + 425.5 – (310 ÷ 6 × 20.3)

a) 750

b) 520

c) 840

d) 430

e) 680 

8) 21.85 × 5.18 + (64.13)2 ÷ 15.91 =? – 28 % of 899

a) 560

b) 620

c) 750

d) 440

e) 800 

9) (15/43) ÷ (119/300) × 20 – 619.67 + 62 % of 1999 =?

a) 750

b) 820

c) 900

d) 640

e) 980 

10) 42 % of 2149 + (3/11) of 18149 – 30 % of 1699 =?

a) 5340

b) 5850

c) 5060

d) 5520

e) 4825

11) The area of a square is 1764 sq cm whose side is 2 times of radius of a circle. The circumference of the circle is equal to the perimeter of rectangle. What is the breadth of rectangle, if the length of the rectangle is 42 cm?

a) 18 cm

b) 25 cm

c) 32 cm

d) 24 cm

e) None of these

12) P, Q and R started a business by investing Rs. 15000, Rs. 18000 and Rs. 24000 respectively. After 3 months, P invested 40 % of the initial investment more. And after 5 months, R withdraws Rs. 4000. Find the share of Q, if the total profit at the end of the year is Rs. 61370?

a) Rs. 18360

b) Rs. 19520

c) Rs. 20480

d) Rs. 17500

e) None of these 

13) 4 years ago, the ratio of ages of P and Q is 4: 3. The present age of R is one-fifth of the present age of Q. Sum of the ages of P, Q and R, after 4 years will be 74 years. Find the present age of Q?

a) 22 years

b) 25 years

c) 28 years

d) 20 years

e) None of these

14) Two pipes can fill a cistern in 15 and 21 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 15 minutes extra are taken for the cistern to be filled up. If the cistern is full, in what time would the leak empty it?

a) 285 hours

b) 260 hours

c) 240 hours

d) 315 hours

e) None of these

15) A customer sold an article at 20% discount on the marked price, yet he got 12% profit on the cost price. If the cost price of the article is Rs. 250, then find the marked price

a) 280

b) 360

c) 350

d) 420

e) 400

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

The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.

16) Marks obtained by Madan in Hindi is what percent of marks obtained by Krishna in General Knowledge?

a) 35.62%

b) 31.25%

c) 21.22%

d) 16.25%

e) None of these

17) Find the total marks obtained by all the students in Science.

a) 465

b) 645

c) 654

d) 456

e) None of these

18) Find the overall percentage of marks obtained by Kunal in all the subjects.

a) 45%

b) 76%

c) 74%

d) 67%

e) None of these

19) Find the respective ratio of marks obtained by Nitin in Science, Maths and English together and marks obtained by Mohan in Hindi, Social Science and General Knowledge together.

a) 455 : 344

b) 344 : 455

c) 341 : 435

d) 134 : 344

e) None of these

20) Find the difference between total marks obtained by Krishna and Rohan in all the subjects together.

a) 135

b) 154.25

c) 145.5

d) 125.75

e) None of these

Answers:

1) Answer: e)

I. 4x2 – (4 + 2√3)x + 2√3 = 0

4x2 – 4x – 2√3x + 2√3 = 0

4x (x -1) – 2√3 (x -1) = 0

(4x – 2√3) (x – 1) = 0

X = √3/2, 1 = 0.866, 1

II. 8y2 – (24 + 3√3)y + 9√3 = 0

8y2 – 24y – 3√3y + 9√3 = 0

8y(y – 3) – 3√3 (y – 3) = 0

(8y – 3√3) (y – 3) = 0

Y = 3√3/8, 3 = 0.64, 3

Can’t be determined

2) Answer: c)

I. x2 + 9x – 52 = 0

(x + 13) (x – 4) = 0

X = -13, 4

II. y2 – 15y + 54 = 0

(y – 9) (y – 6) = 0

Y = 9, 6

X < y

3) Answer: e)

I. 4x2 + 20x + 24 = 0

4x2 + 12x + 8x + 24 = 0

4x(x + 3) + 8(x + 3) = 0

(4x + 8) (x + 3) = 0

X = -2, -3

II. 3y2 + 18y + 24 = 0

3y2 + 12y + 6y + 24 = 0

3y(y + 4) + 6(y + 4) = 0

(3y + 6) (y + 4) = 0

Y = -2, -4

Can’t be determined

4) Answer: e)

I. 7x2 – 6x – 81 = 0

7x2 + 21x – 27x – 81 = 0

7x(x + 3) – 27(x + 3) = 0

(7x – 27) (x + 3) = 0

X = 27/7, -3 = 3.85, -3

II. 9y2 – 10y – 56 = 0

9y2 + 18y – 28y – 56 = 0

9y (y + 2) -28 (y + 2) = 0

(9y – 28) (y + 2) = 0

Y = 28/9, -2 = 3.11, -2

Can’t be determined

5) Answer: e)

2x – 5y = 12 –> (1)

5x – 2y = -12 —> (2)

By substituting (1) and (2), we get,

X = -4, y = -4

X = y

Direction (6-10)

6) Answer: c)

X = (4128 + 574)/25 + 182 + (2/15)*12525 – 24

X = (4702/25) + 324 + 1670 – 24

X = 188 + 324 + 1670 – 24

X = 2158 = 2160

7) Answer: e)

(1400/20) = x + 426 – (312/6)*20

70 – 426 + 1040 = x

X = 684 = 680

8) Answer: b)

(22*5) + 642 ÷ 16 = x – (28/100)*900

110 + (64*64)/16 = x – 252

110 + 256 + 252 = x

X = 618 = 620

9) Answer: d)

(15/43)*(301/120)*20 – 620 + (62/100)*2000 = x

X = 18 – 620 + 1240

X = 638 = 640

10) Answer: a)

42 % of 2150 + (3/11) of 18150 – 30 % of 1700 = x

X = (42/100)*2150 + (3/11) * 18150 – (30/100)*1700

X = 903 + 4950 – 510

X = 5343 = 5340

11) Answer: d)

Area of square = 1764 Sq cm = a2

Side (a) = 42 cm

Side of square = 2 * Radius of circle

Radius = 42/2 = 21 cm

Circumference of the circle = 2πr = 2*(22/7)*21 = 132 cm

Length = 42 cm

Perimeter of rectangle = 2(42 + b) = 132

42 + b = 66

Breadth = 66 – 42 = 24 cm

12) Answer: a)

The share of P, Q and R

= > [15000*3 + 15000*(140/100)*9]: [18000*12]: [24000*8 + 20000*4]

= > 234000: 216000: 272000

= > 117: 108: 136

361’s = 61370

1’s = 170

The share of Q = 108’s = Rs. 18360

13) Answer: b)

4 years ago, the ratio of ages of P and Q = 4: 3 (4x, 3x)

Present age of P and Q = 4x + 4, 3x + 4

The present age of R = (1/5)*(3x + 4) = (3x + 4)/5

Sum of the ages of P, Q and R, after 4 years = 74 years

4x + 4 + 3x + 4 + (3x + 4)/5 = 74 – 12

35x + 40 + 3x + 4 = 62*5

38x = 310 – 44

X = 266/38 = 7

The present age of Q = 3x + 4 = 25 years

14) Answer: d)

If both filling pipes are working together, portion of the cistern filled in an hour

= 1/15 + 1/21 = (21 + 15)/(15*21) = 4/35

Hence time taken to fill the cistern = 35/4 hours = 8 3/4 hours = 8 hours 45 minutes.

But due to leak it takes 15 min more.

Hence total time taken = 8 hours 45 minutes + 15 min = 9 hours.

Let the leak take n hours to empty the cistern.

So, 1/15 + 1/21 – 1/n = 1/9

=> 4/35 – 1/n = 1/9

=› 1/n = 4/35 – 1/9 = 1/315

Hence the cistern will be emptied in 315 hours.

15) Answer: c)

MP*(80/100) = 250*(112/100)

MP = 250*(112/100)*(100/80)

MP = Rs. 350

Direction (16-20)

16) Answer: b)

Marks obtained by Madan in Hindi = 55/100 x 125 = 68.75

Marks obtained by Krishna in General Knowledge = 88/100 x 250 = 220

Required percentage = 68.75/220 x 100 = 31.25%

17) Answer: b)

Total marks obtained by all the students in Science

= (60 + 45 + 86 + 65 + 84 + 90) * 150/100

= 430 x 150/100 = 645

18) Answer: b)

Marks obtained by Kunal in,

Science = (86/100) x 150 = 129

Maths = (95/100) x 200 = 190

English = (48/100) x 100 = 48

Hindi = (86/100) x 125 = 107.5

Social Science = (68/100) x 125 = 85

General Knowledge = (65/100) x 250 = 162.5

Overall percentage

= [(129 + 190 + 48 + 107.5 + 85 + 162.5)/(150 + 200 + 100 + 125 + 125 + 250)] x 100

= 722/950 x 100 = 76%

19) Answer: b)

Marks obtained by Nitin in:

Science = (60/100) x 150 = 90

Maths = (85/100) x 200 = 170

English = (84/100) x 100 = 84

Marks obtained by Mohan in:

Hindi = (92/100) x 125 = 115

Social Science = (92/100) x 125 = 115

General Knowledge = (90/100) x 250 = 225

Required ratio = (90 + 170 + 84) : (115 + 115 + 225) = 344 : 455

20) Answer: b)

Marks obtained by Krishna in all the subjects

= (65/100) x 150 + (45/100) x 200 + (65/100) x 100 + (48/100) x 125 + (78/100) x 125 + (88/100) x 250

= 97.5 + 90 + 65 + 60 + 97.5 + 220

= 630

Marks obtained by Rohan in all the subjects

= (90/100) x 150 + (80/100) x 200 + (58/100) x 100 + (88/100) x 125 + (87/100) x 125 + (85/100) x 250

= 135 + 160 + 58 + 110 + 108.75 + 212.5

= 784.25

Required difference = 784.25 – 630 = 154.25

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