“20-20” Quantitative Aptitude | Crack NIACL Assistant Prelims 2018 Day-184

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“20-20” Aptitude Questions | Crack NIACL Assistant prelims 2018 Day-184

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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,

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. 3x + 5y = 34

II. 5x – 2y = 5

2) I. 2x2 + 13x + 18 = 0

II. 3y2 + 10y + 8 = 0

3) I. x2 + 9x + 14 = 0

II. 3y2 – 10y + 8 = 0

4) I. √324 x + √11664 = 0

II. (16807)1/5 y + (9261)1/3 = 0

 5) I. 2x – y = 31

II. y = ∛19683

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

6) 56 % of 1200 + 25 % of ? = 3420

a) 15728

b) 14716

c) 10992

d) 18260

e) None of these

 7) (72/167) of 83.5 % of (53/108) of ? = 159

a) 900

b) 750

c) 660

d) 520

e) None of these

 8) (5/29) of 3915 + (7/15) of 6345 = ? + 2765

a) 756

b) 925

c) 657

d) 871

e) None of these

9) (5 × 5)4 ÷ (875 ÷ 7)3 × (125 × 25)2 = (5)? + 5

a) 5

b) 4

c) 7

d) 6

e) None of these

10) [(2464 ÷ 8 ÷ 14) + (28 × 7)1/2] = (?)2

a) 6

b) 10

c) 4

d) 15

e) None of these

11) C is 40% less efficient than A. A and B together can finish a piece of work in 10 days. B and C together can do it in 15 days. In how many days can A alone finish the same piece of work?

a) 15 days

b) 9 days

c) 12 days

d) 18 days

e) None of these

12) The diameter of a smaller circle and larger circle are the side of a square and the diagonal of the square respectively. If the side of the square is 16 cm, then find the ratio of the areas of the smaller circle and the larger circle?

a) 5 : 11

b) 1 : 2

c) 3 : 7

d) 7 : 13

e) None of these

13) The ratio of present ages of son and daughter of a lady is 7 : 8 and the age of lady is sum total ages of her daughter and son. If the ratio of ages of son after 3 years and the age of daughter 6 years ago is 4 : 3 then what will be the ratio of the present age of lady and the age of her son after four years?

a) 4 : 7

b) 5 : 3

c) 6 : 11

d) 9 : 5

e) None of these

14) A shopkeeper offers two successive discounts of 10 % and 15 % on an article and is still able to earn 53 % profit. If he offers only 20 % discount, then what will be his profit percentage?

a) 54 %

b) 60 %

c) 58 %

d) 66 %

e) None of these

15) In a bag, there are 6 red balls and 4 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green colour?

a) 3/7

b) 5/8

c) 1/2

d) 11/19

e) None of these

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

The following table shows the amount invested by six different companies during six different months (in lakhs)

16) Find the difference between the investment made by company A and E together in March and company B and D together in January (in lakhs)?

a) 75 lakhs

b) 65 lakhs

c) 50 lakhs

d) 85 lakhs

e) None of these

17) Find the sum of the amount invested by company B and D in all the given months together (in lakhs)?

a) 670 lakhs

b) 560 lakhs

c) 620 lakhs

d) 590 lakhs

e) None of these

18) In which month, the investment made by these companies were maximum?

a) March

b) April

c) May

d) June

e) None of these

19) The investment made by company A in June is approximately what percentage less than of the investment made by company D in the same month?

a) 23 %

b) 15 %

c) 45 %

d) 5 %

e) 50 %

20) Find the average investment made by company F during all the months (in lakhs)?

a) 35 lakhs

b) 48 lakhs

c) 37 lakhs

d) 44 lakhs

e) 46 lakhs

Answers :

Direction (1-5) :

1). Answer: c)

3x + 5y = 34 –à (1)

5x – 2y = 5—à (2)

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

X = 3, y = 5

X < y

2). Answer: d)

  1. 2x2 + 13x + 18 = 0

2x2 + 4x + 9x + 18 = 0

2x(x + 2) + 9(x + 2) = 0

(2x + 9)(x + 2) = 0

X = -9/2, -2 = -4.5, -2

II. 3y2 + 10y + 8 = 0

3y2 + 6y + 4y + 8 = 0

3y(y + 2) + 4(y + 2) = 0

(3y + 4) (y + 2) = 0

Y = -4/3, -2 = -1.33, -2

x ≤ y

3). Answer: c)

  1. x2 + 9x + 14 = 0

(x+ 7) (x + 2) = 0

X = -7, -2

II. 3y2 – 10y + 8 = 0

3y2 – 6y – 4y + 8 = 0

3y(y -2) – 4 (y – 2) = 0

(3y – 4) (y – 2) = 0

Y = 4/3, 2

X < y

4). Answer: c)

  1. √324 x + √11664 = 0

18x + 108 = 0

18x = -108

X = – 6

II. (16807)1/5 y + (9261)1/3 = 0

(75)1/5 + (213)1/3 = 0

7y + 21 = 0

7y = -21

Y = -3

X < y

5). Answer: a)

  1. y = ∛19683

Y = 27

II. 2x – y = 31

2x – 27 = 31

2x = 58

X = 29

X > y

Direction (6-10) :

6). Answer: c)

(56/100)*1200 + (25/100)*x = 3420

672 + (25/100)*x = 3420

(25/100)*x = 3420 – 672

(25/100)*x = 2748

X = 2748*(100/25) = 10992

7). Answer: a)

(72/167)*(83.5/100)*(53/108) *x = 159

X = 900

8). Answer: d)

(5/29)*3915 + (7/15)*6345 = x + 2765

675 + 2961– 2765 = x

X = 871

9). Answer: b)

(52)4 ÷ (875/7)3 × (53 × 52)2 = (5)x + 5

58 ÷ (53)3 × (55)2 = (5)x + 5

58 ÷ 59 × 510 = (5)x + 5

58 – 9 + 10 = (5)x + 5

59 = (5)x + 5

9 = x + 5

X = 4

10). Answer: a)

((2464/8)/14) + (196)1/2 = x2

(308/14) + 14 = x2

X2 = 22 + 14 = 36

X = 6

11). Answer: c)

The efficiency ratio of A and C = 100 : 60 = 5 : 3

The day ratio of A and B = 3 : 5 (3x, 5x)

(A + B)’s one day work = 1/10

(B + C)’s one day work = 1/15

(A – C)’s one day work = (1/10) – (1/15) = 1/30

According to the question,

(1/3x) – (1/5x) = 1/30

2x/(3x*5x) = 1/30

2/15x = 1/30

60 = 15x

X = 4

A alone can finish a work in = 3x = 12 days

12). Answer: b)

The diameter of the smaller circle = Side of the square = 16 cm

The radius of the smaller circle(r) = 8 cm

The diameter of the larger circle = Diagonal of the square = √2*Side = 16√2

The radius of the larger circle = 8√2

The ratio of area of smaller circle and area of larger circle

= > 82 : (8√2)2

= > 64 : 64*2 = 1 : 2

13). Answer: d)

The ratio of present ages of son and daughter of a lady = 7 : 8 (7x, 8x)

The present age of lady = 7x + 8x = 15x

According to the question,

(7x + 3) / (8x – 6) = 4/3

21x + 9 = 32x – 24

11x = 33

x = 3

Son’s present age = 21years

Daughter’s present age = 24 years

Present age of lady = 45 years

Required ratio = 45: (21 + 4) = 45 : 25 = 9 : 5

14). Answer: b)

Let us assume the marked price of the article be Rs. 100,

Sp= 100*(90/100)*(85/100) = Rs. 76.5

(153/100)*Cp = 76.5

Cp = 76.5*(100/153) = Rs. 50

If the shopkeeper offers only 20 % discount, the profit % will be,

Sp = 100*(80/100) = Rs. 80

Profit % = [(80 – 50)/50]*100 = (30/50)*100 = 60 %

15). Answer: c)

Total probability n(S) = 10C3

Required probability n(E) = 6C2 and 4C1

P(E) = n(E)/n(S) = (6C2 and 4C1)/10C3 = 1/2

Direction (16-20) :

16). Answer: a)

Investment made by A and E in March = 52 + 83 = 135 lakhs

Investment made by B and D in January = 28 + 32 = 60 lakhs

Hence, required difference = 135 – 60 = 75 lakhs

17). Answer: c)

The amount invested by company B in all the given months

= > 28 + 40 + 38 + 45 + 20 + 95 = 266 lakhs

The amount invested by company D in all the given months

= > 32 + 55 + 72 + 90 + 25 + 80 = 354 lakhs

Required sum = 266 + 354 = 620 lakhs

18). Answer: b)

The total investment of companies,

In January = 25 + 28 + 50 + 32 + 24 + 35 = 194 lakhs

In February = 45 + 40 + 60 + 55 + 28 + 38 = 266 lakhs

In March = 52 + 38 + 46 + 72 + 83 + 65 = 356 lakhs

In April = 80 + 45 + 70 + 90 + 65 + 76 = 426 lakhs

In May = 15 + 20 + 47 + 25 + 13 + 30 = 150 lakhs

In June = 62 + 95 + 27 + 80 + 50 + 20 = 334 lakhs

April month has the maximum investments.

19). Answer: a)

Required % = [(80 – 62)/ 80] × 100 = (18 / 80) × 100 = 22.5% = 23%

20). Answer: d)

Required average = (35 + 38 + 65 + 76 + 30 + 20)/6 = 264/6 = 44 lakhs

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Topic Daily Publishing Time
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Current Affairs Quiz 9.00 AM
Quantitative Aptitude “20-20” 11.00 AM
Vocabulary (Based on The Hindu) 12.00 PM
General Awareness “20-20” 1.00 PM
English Language “20-20” 2.00 PM
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Daily Current Affairs Updates 5.00 PM
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Reasoning Ability “20-20” 7.00 PM
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