“20-20” Quantitative Aptitude | Crack IBPS RRB PO Prelims 2018 Day-164

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“20-20” Aptitude Questions | Crack IBPS RRB PO Prelims 2018 Day-164

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Directions (Q. 1 – 5) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

a) If x > y

b) If x ≥ y

c) If x < y

d) If x ≤ y

e) If x = y or the relation cannot be established

1) I. X3 =1331

II. 2y2 -21y +55 =0

2) I. 5x = 7y+21

II. 11x +4y+109 =0

3) I. 2x2 -11x +12 =0

II. 2y2 -17y +36 =0

4) I. 6x2 – 32x + 42 = 0

II. y2 + 7y + 12 = 0

5) I. x2 + 6x + 8 = 0

II. 4y2 – 13y + 9 = 0

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

6) (3/7) of (34/13) of 3054 – ? + 56 % of 7549 = 63.21

a) 8150

b) 6830

c) 5670

d) 7690

e) 9560

7) 48 % of 4249 + ? % of 5597 = 52 % of 4648 + 1301.67

a) 30

b) 45

c) 60

d) 80

e) 55

 8) 35 % of ? – 2650 ÷ 17 – 2021 ÷ 7 = 52.21

a) 2150

b) 1870

c) 1420

d) 1630

e) 2340

9) [(45 × 23) ÷ 46 – 749.9 ÷ 149] = 3 × ?

a) 12

b) 6

c) 20

d) 17

e) 25

 10) 27 % of 4495 – 32 % of 247 = ? – √286 % of 1187

a) 2150

b) 2370

c) 1580

d) 1760

e) 1340

 11) A certain sum is divided among A, B and C in such a way that A gets 35 more than the 2/5 of the sum and B gets Rs. 75 less than the 1/5 of the sum and C gets Rs. 350. Find the sum?

a) Rs. 900

b) Rs. 775

c) Rs. 825

d) Rs. 1050

e) None of these

 12) The angles of quadrilateral are in the ratio of 3:2:5:8. The second smallest angle of quadrilateral is equal to second smallest angle of triangle. One of the angles of triangle is equal to twice its smallest angle. Find the largest angle of triangle?

a) 80

b) 60

c) 100

d) 40

e) None of these

 13) The average weight of 8 persons is increased by 2 when one of the person whose weight is 62 kg is replaced by 2 women. Find the average weight of 2 women?

a) 30 kg

b) 38 kg

c) 39 kg

d) 42 kg

e) None of these

 14) A circular road runs around a circular ground. If the radius of the ground is 5.5 m and the difference between the circumference of the outer circle and that of the inner circle is 110m, then the area of the road is?

a) 1354.25 m

b) 1678.75 m

c) 1567.50 m

d) 2119.25 m

e) None of these

15) Ravi spends 30% of his monthly income on food and 20% on education of his children. Of the remaining salary, he spends 25% on entertainment and 15% on conveyance. He is now left with Rs. 12,336. What is the monthly salary of Ravi?

a) Rs. 27600

b) Rs. 28750

c) Rs. 33200

d) Rs. 41120

e) None of these

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

Following pie chart -1 shows the percentage distribution of savings of 5 different friends in January 2018 and pie chart – 2 shows the percentage distribution of savings of 5 friends in February 2018.

16) The savings of Manoj and Pari together in January 2018 is what percentage more/less than the savings of Pari and Subha together in February 2018?

a) 15 % more

b) 25 % less

c) 15 % less

d) 25 % more

e) 20 % more

 17) Find the difference between the average savings of Sree and Moni together in January 2018 to that of the average savings of Moni and Manoj together in February 2018?

a) 725

b) Rs. 900

c) Rs. 850

d) Rs. 675

e) None of these

 18) Find the ratio between the savings of Manoj and Subha together in January 2018 to that of the savings of Pari and Moni together in February 2018?

a) 315 : 223

b) 447 : 335

c) 15 : 7

d) 259 : 312

e) None of these

 19) Total savings of Moni in January 2018 is what percentage of total savings of Pari in February 2018?

a) 150 %

b) 125 %

c) 140 %

d) 105 %

e) None of these

 20) Find the total savings of Manoj and Subha in January and February together?

a) Rs. 26800

b) Rs. 24550

c) 28150

d) Rs. 32500

e) None of these

 Answers:

Direction (1-5)

1) Answer: a)

I. X3 =1331

X = ∛1331 =11

II. 2y2 -21y + 55 =0

2y2 -10y -11y +55 =0

2y(y – 5)-11 (y – 5) =0

(2y – 11) (y – 5) =0

Y= 11/2, 5

X > y

2) Answer: a)

5x – 7y = 21 —> (1)

11x +4y = -109 –> (2)

By solving the equation (1), (2), we get,

X= -7, y = -8

X > y

3) Answer: d)

I. 2x2 -11x +12 = 0

2x2 – 8x – 3x +12 = 0

2x (x – 4) -3 (x – 4) = 0

(2x – 3) (x – 4) =0

X = 3/2, 4

II. 2y2 -17y +36 = 0

2y2 – 9y – 8y + 36 = 0

y (2y – 9) – 4 (2y – 9) = 0

(y – 4) (2y – 9) =0

Y = 4, 9/2

X ≤ y

4) Answer: a)

I. 6x2 – 32x + 42 = 0

6x2 – 18x – 14x + 42 = 0

6x(x – 3)-14 (x – 3) =0

(6x – 14)(x – 3) = 0

x = 3, 7/3

II. y2 + 7y + 12 = 0

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

y = -4, -3

x > y

5) Answer: c)

I. x2 + 6x + 8 = 0

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

x = -4, -2

II. 4y2 – 13y + 9 = 0

4y2 – 4y – 9y + 9 = 0

4y(y – 1) – 9 (y – 1) =0

(4y – 9) (y – 1) = 0

y = 1, 9/4

x < y

Direction (6-10)

6) Answer: d)

(3/7)*(35/13)*3055 – x + (56/100)*7550 = 63

3525 – x + 4228 = 63

X = 3525 + 4228 – 63

X = 7690

7) Answer: a)

(48/100)*4250 + (x/100)*5600 = (52/100)*4650 + 1302

2040 + 56x = 2418 + 1302

56x = 2418 + 1302 – 2040

56x = 1680

X = 30

8) Answer: c)

(35/100)*x – (2652/17) – (2023/7) = 52

(7x/20) – 156 – 289 = 52

(7x/20) = 52 + 156 + 289

(7x/20) = 497

X = 497*(20/7) = 1420

9) Answer: b)

(46*23)/46 – (750/150) = 3x

23 – 5 = 3x

18 = 3x

X = 18/3 = 6

10) Answer: e)

(27/100)*4500 – (32/100)*250 = x – (17/100)*1200

1215 – 80 = x – 204

1215 – 80 + 204 = x

X = 1339 = 1340

11) Answer: b)

Let total sum be x,

A= (2/5)*x +35

B = (1/5)*x – 75

C = 350

A + B + C = x

(2/5)*x + 35 + (1/5)*x – 75 + 350 = x

(3/5)*x + 310 = x

310 = x – (3/5)*x

(2/5)*x = 310

X= 310*(5/2) = Rs. 775

12) Answer: b)

The angles of quadrilateral are in the ratio of 3:2:5:8.

Total angles = 360

18’s = 360

1’s = 20

The angles are,

= > 60◦, 40◦, 100◦, 160◦

Second smallest angle of quadrilateral = 60◦ = Second smallest angle of triangle

One of the angles of triangle = 2* 40◦ = 80◦

Total angles of triangle= 180◦

Remaining angle = 180 – (60+80) = 40◦

Largest angle of triangle = 80◦

13) Answer: c)

Increased weight = 8*2 = 16 kg

So 2 women’s weight = 62kg +16 kg = 78 kg

Average weight of 2 women = 78/2 = 39 kg

14) Answer: c)

Radius of the ground(r) = 5.5 m

The difference between the circumference of the outer circle and that of the inner circle is 110m

= > 2πR – 2πr = 110

= > 2*(22/7)*(R – r) = 110

= > R – r = 17.5

= > R- 5.5 = 17.5

= > R = 23 m

The area of the road is,

=> πR2 – πr2

=> (22/7) (232 – 5.52)

= > (22/7) (23 + 5.5) (23 – 5.5)

= > (22/7)*28.5*17.5

= > 1567.5 m

15) Answer: d)

Let Ravi’s monthly salary be x,

According to the question,

X*(50/100)*(60/100) = 12336

X = Rs. 41120

Direction (16-20)

16) Answer: c)

The savings of Manoj and Pari together in January 2018

= > (33/100)*35000 = 11550

The savings of Pari and Subha together in February 2018

= > (34/100)*40000 = 13600

Required % = [(13600 – 11550)/13600]*100 = 15 % less

17) Answer: a)

The average savings of Sree and Moni together in January 2018

= > (45/100)*35000*(1/2) = 7875

The average savings of Moni and Manoj together in February 2018

= > (43/100)*40000*(1/2) = 8600

Required difference = 8600 – 7875 = Rs. 725

18) Answer: d)

The savings of Manoj and Subha together in January 2018

= > (37/100)*35000 = 12950

The savings of Pari and Moni together in February 2018

= > (39/100)*40000 = 15600

Required ratio = 12950 : 15600 = 259 : 312

19) Answer: b)

Total savings of Moni in January 2018

= > (20/100)*35000 = 7000

Total savings of Pari in February 2018

= > (14/100)*40000 = 5600

Required % = (7000/5600)*100 = 125 %

20) Answer: c)

The total savings of Manoj and Subha in January

= > (37/100)*35000 = Rs. 12950

The total savings of Manoj and Subha in February

= > (38/100)*40000 = Rs. 15200

Required sum = 12950 + 15200 = Rs. 28150

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