NIACL AO Mains– Quantitative Aptitude Questions Day- 15

Dear Readers, Bank Exam Race for the Year 2019 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for NIACL AO Mains 2019 Exams can practice these questions daily and make your preparation effective.

NIACL AO Mains– Quantitative Aptitude Questions Day- 15

maximum of 10 points
Table is loading

Click here to take NIACL AO Mock Test

 

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) x2 + 19x – 92 = 0

II) 2y2 – 15y – 38 = 0

 

2)

I) 3x2 + 23x + 44 = 0

II) 4y2 – 6y – 28 = 0

 

3)

I) 6x – 7y + 2 = 3x – 2y

II) 2x – 3y = 0

 

4)

I) x = ∛19683 + 52 ÷ 4

II) y = √2401 – 32

 

5)

I) 2x2 – 25x + 77 = 0

II) 3y2 – 23y + 42 = 0

 

Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:

The following table represents number of different coloured balls contained by eight bags

6) Probability of drawing one blue ball from bag A is 3/10. Probability of drawing one green ball from bag B is 1/3 and probability of drawing one blue ball from bag B is 2/9. Find the product of the probability of drawing 2 green balls from bag A and probability of drawing 2 red balls from bag B.

a) 4/855

b) 7/969

c) 5/876

d) 6/931

e) None of these

7) Probability of drawing one blue ball from bag B is 1/7. Probability of drawing one blue ball from bag D 1/9 and probability of drawing one green ball from bag D is 5/18. Find the ratio of the probability of drawing 2 green balls from bag C and probability of drawing 2 white balls from bag D.

a) 53:49

b) 48:25

c) 51:28

d) 37:21

e) None of these

8) Probability of drawing one white ball from bag E is 1/3. Probability drawing one green ball from bag F is 2/9 and probability of drawing one red ball from bag F is 1/6. Probability of drawing one green ball from bag E is what percent of probability of drawing one blue ball from bag F?

a) 72%

b) 68%

c) 64%

d) 56%

e) None of these

9) Probability of drawing one red ball from bag H is 2/5 and probability of drawing one green ball from bag H is 1/5. Find the probability of drawing four balls of different colour from bag H.

a) 5/696

b) 3/712

c) 4/969

d) Cannot be determined

e) None of these

10) Probability of drawing one green ball from bag G is 1/5. Probability of drawing one white ball from bag H is ¼ and probability of drawing one green ball from bag H is 1/5.Find the sum of the probability of drawing two blue balls from bag G and probability of drawing 2 green balls from bag H.

a) 109/515

b) 113/525

c) 111/413

d) 116/665

e) None of these

Answers :

Direction (1-5) : 

1) Answer: e)

I) x2 + 19x – 92 = 0

(x + 23) (x – 4) = 0

X = -23, 4

II) 2y2 – 15y – 38 = 0

2y2 + 4y – 19y – 38 = 0

2y (y + 2) -19 (y + 2) = 0

(2y – 19) (y + 2) = 0

Y = 19/2, -2 = 9.5, -2

Can’t be determined

2) Answer: c)

I) 3x2 + 23x + 44 = 0

3x2 + 12x + 11x + 44 = 0

3x (x + 4) + 11 (x + 4) = 0

(3x + 11) (x + 4) = 0

X = -11/3, -4 = -3.66, -4

II) 4y2 – 6y – 28 = 0

4y2 + 8y – 14y – 28 = 0

4y (y + 2) -14 (y + 2) = 0

(4y – 14) (y + 2) = 0

Y = 14/4, -2 = 3.5, -2

X < y

3) Answer: a)

I) 6x – 7y + 2 = 3x – 2y

3x – 5y = -2 –> (1)

II) 2x – 3y = 0 —> (2)

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

X = 6, y = 4

X > y

4) Answer: e)

I) x = ∛19683 + 52 ÷ 4

X = 27 + (52/4)

X = 27 + 13 = 40

II) y = √2401 – 32

Y = 49 – 9 = 40

X = y

5) Answer: a)

I) 2x2 – 25x + 77 = 0

2x2 – 11x – 14x + 77 = 0

X (2x – 11) – 7 (2x – 11) = 0

(x – 7) (2x – 11) = 0

X = 7, 11/2 = 7, 5.5

II) 3y2 – 23y + 42 = 0

3y2 – 9y – 14y + 42 = 0

3y (y – 3) -14(y – 3) = 0

(3y – 14) (y – 3) = 0

Y = 14/3, 3 = 4.66, 3

X > y

Direction (6-10) :

6) Answer: b)

Let, the number of white balls in bag A = n

Total number of balls in bag A = 5 + 6 + 7 + n = 18 + n

6/(18 + n) = 3/10

=> 60 = 54 + 3n

=> 3n = 60 – 54

=> 3n = 6

=> n = 2

Total number of balls in bag A = 18 + 2 = 20

Probability of drawing 2 green balls from bag A = 7c2/20c2

= (7 x 6)/(20 x 19)

= 21/190

Let, the number of red balls in bag B = x

And number of green balls in bag B = y

Total number of balls in bag B = 4 + 3 + x + y = (7 + x + y)

y/(7 + x + y) = 1/3

=> 3y = 7 + x + y

=>x – 2y = -7 —— (i)

4/(7 + x + y) = 2/9

=> 36 = 14 + 2x + 2y

=> 2x + 2y = 22

=> x + y = 11 —— (ii)

Equation (ii) – equation (i)

x + y – x + 2y = 11 + 7

=> 3y = 18

=> y = 6

Putting the value of y in equation (ii)

x + 6 = 11

=> x = 11 – 6

=> x = 5

Total number of balls in bag B = 7 + 5 + 6 = 18

Probability of drawing 2 red balls from bag B = 5c2/18c2

= (5 x 4)/(18 x 17)

= 10/153

Required product = 21/190 x 10/153 = 7/969

7) Answer: c)

Let the number of blue balls in bag C = n

Total number of balls in bag C = (4 + 6 + 8 + n) = 18 + n

n/(18 + n) = 1/7

=> 7n = n + 18

=> 7n – n = 18

=> 6n = 18

=> n = 3

Total number of balls in bag C = 18 + 3 = 21

Probability of drawing 2 green balls from bag C = 6c2/21c2

= (6 x 5)/(21 x 20)

= 1/14

Let, the number of blue balls in bag D = x

And number of white balls in bag D = y

Total number of balls in bag D = 7 + 5 + x + y = 12 + x + y

x/(12 + x + y) = 1/9

=> 9x = 12 + x + y

=> 8x – y = 12 —— (i)

5/(12 + x + y) = 5/18

=> 90 = 60 + 5x + 5y

=> 5x + 5y = 30

=> x + y = 6 —– (ii)

Adding equations (i) and (ii)

8x – y + x + y = 12 + 6

=> 9x = 18

=> x = 2

Putting the value of x in equation (ii)

2 + y = 6

=> y = 6 – 2

=> y = 4

Total number of balls in bag D = 12 + 2 + 4 = 18

Probability of drawing 2 white balls from bag D = 4c2/18c2

= (4 x 3)/(18 x 17)

= 2/51

Required ratio = 1/14: 2/51 = 51:28

8) Answer: a)

Let the number of blue balls in bag E = n

Total number of balls in bag E = 4 + 3 + 5 + n = 12 + n

5/(12 + n) = 1/3

=> 15 = 12 + n

=> n = 3

Total number of balls in bag E = 12 + 3 = 15

Probability of drawing one green ball from bag E = 3/15 = 1/5

Let the number of green balls in bag F = x

And number of white balls in bag F = y

Total number of balls in bag F = 3 + 5 + x + y = 8 + x + y

x/(8 + x + y) = 2/9

=> 9x = 16 + 2x + 2y

=> 7x – 2y = 16 —– (i)

3/(8 + x + y) = 1/6

=> 18 = 8 + x + y

=> x + y = 10 —— (ii)

Equation (i) – 7 x equation (ii)

7x – 2y – 7x – 7y = 16 – 70

=> -9y = -54

=> y = 6

Putting the value of y in equation (ii)

x + 6 = 10

=> x = 4

Total number of balls in bag F = 8 + 4 + 6 = 18

Probability of drawing one blue ball from bag F = 5/18

Required percentage = (1/5)/(5/18) x 100

= 1/5 x 18/5 x 100

= 18/25 x 100

= 72%

9) Answer: d)

Let, the number of blue balls in bag H = x

And number of white balls in bag H = y

Total number of balls in bag H = 8 + 4 + x + y = 12 + x + y

8/(12 + x + y) = 2/5

=> 40 = 24 + 2x + 2y

=> 2x + 2y = 40 – 24

=> 2x + 2y = 16

=> x + y = 8

4/(12 + x + y) = 1/5

=> 20 = 12 + x + y

=> x + y = 8

Hence, required value cannot be find out.

10) Answer: d)

Let, the number of red balls in bag G = n

Total number of balls in bag G = 6 + 3 + 2 + n = 11 + n

3/(11 + n) = 1/5

=> 15 = 11 + n

=> n = 4

Total number of balls in bag G = 11 + 4 = 15

Probability of drawing two blue balls from bag G = 6c2/15c2

= (6 x 5)/(15 x 14)

= 1/7

Let, the number of blue balls in bag H = x

And number of white balls in bag H = y

Total number of balls in bag H = 8 + 4 + x + y = 12 + x + y

y/(12 + x + y) = ¼

=> 4y = 12 + x + y

=>x – 3y = -12 —— (i)

4/(12 + x + y) = 1/5

=> 20 = 12 + x + y

=> x + y = 8 —– (ii)

Equation (i) – equation (ii)

x – 3y – x – y = – 12 – 8

=> -4y = – 20

=> y = 5

Putting the value of y in equation (i)

x – 15 = -12

=> x = 15 – 12

=> x = 3

Total number of balls in the bag = 12 + 3 + 5 = 20

Probability of drawing 2 green balls from bag H = 4c2/20c2

= (4 x 3)/(20 x 19)

= 3/95

Required sum = 1/7 + 3/95

= (95 + 21)/665

= 116/665

 

Online Mock Tests 2019: