# “20-20” Quantitative Aptitude | Crack NIACL Assistant Prelims 2018 Day-190

Dear Readers, Find the aptitude test questions to crack latest bank exams. We regularly provide 20 aptitude test questions daily for students. Aspirants practice these questions on a regular basis to improve your score in aptitude section. Aspirants preparing for the exams can make use of this “20-20” Quantitative Aptitude Questions.ย Here we have started New Series of Practice Materials specially forย Crack NIACL Assistantย  Prelims 2018. Aspirants those who are preparing for the exams can use this “20-20” Quantitative Aptitude Questions.

โ20-20โ Aptitude Questions | Crack NIACL Assistant prelims 2018 Day-190

maximum of 20 points

Click โStart Quizโ to attend these Questions and view Solutions

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 โฅ y

c) If x < y

d) If x โค y

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

ย 1) I. 8x2 + 28x + 12 = 0

II. 9y2 + 27y + 14 = 0

2) I. 2x2 + 11x + 14 = 0

II. 3y2 + 14y + 15 = 0

3) I. 4x2 + 16x + 7 = 0

II. 5y2 + 32y + 35 = 0

ย 4) I. 7x โ 5y = 55

II. 3x + 7y = – 13

ย 5) I. x2 โ 3x โ 28 = 0

II. y2 โ 17y + 72 = 0

ย Directions (6 โ 10): what value should come in place of question mark (?) in the following questions?

6) 6 ร (?) รท (24 ร 9) – 125 = 23

a) 4756

b) 3274

c) 5328

d) 6120

e) None of these

ย 7) ? โ (212 รท 525) ร 1725 = (57900 รท 15)

a) 4277

b) 3135

c) 4851

d) 5309

e) None of these

8) (5/9) of 3402 โ 16 ร 78 รท 24 =?

a) 2364

b) 1838

c) 3160

d) 2742

e) None of these

9) (256 ร 10-3)2 ร 1024 รท (413 ร 10-4) = 10?

a) -2

b) 3

c) -4

d) 5

e) None of these

ย 10) โ110592 = 113 + 24 % of 1800 – ?

a) 2180

b) 2565

c) 1570

d) 1715

e) None of these

Directions (11 – 15): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

11) Quantity I: 200 m long train crosses a tunnel of length 175 m at the speed of 75 km/hr. Then find the time taken to cross the tunnel?

Quantity II: A train of length 150 m crosses a pole in 15 sec. Find the time taken by another train of length 250 m running in the opposite direction at the speed of 44 km/hr?

a) Quantity I > Quantity II

b) Quantity I โฅ Quantity II

c) Quantity II > Quantity I

d) Quantity II โฅ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

ย 12) Quantity I: The S.I on a certain sum of money for 4 years at 6 % per annum is Rs. 6000. Then find the principle?

Quantity II: The C.I on a certain sum of money for 3 years at 10 % per annum is Rs. 6620. Then find the principle?

a) Quantity I > Quantity II

b) Quantity I โฅ Quantity II

c) Quantity II > Quantity I

d) Quantity II โฅ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

ย 13) (x + y)/(x โ y) = 4 and x โ (y/3) = 4

Quantity I: The value of x is?

Quantity II: The value of y is?

a) Quantity I > Quantity II

b) Quantity I โฅ Quantity II

c) Quantity II > Quantity I

d) Quantity II โฅ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

ย 14) Quantity I: 3 years ago, the ratio of ages of P and Q is 2: 3. Pโs age after 6 years is equal to present age of Q. Find the present age of Q?

Quantity II: 5 years ago, the ratio of ages of A and B is 5: 3. Sum of the ages of A and B, after 6 years is 62 years. Find the present age of C, if C is 5 years younger than A?

a) Quantity I > Quantity II

b) Quantity I โฅ Quantity II

c) Quantity II > Quantity I

d) Quantity II โฅ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

ย 15) A box contains 8 white balls, 6 violet balls and 4 yellow balls.

Quantity I: If 4 balls are drawn randomly, then the probability of getting at least one violet balls?

Quantity II: If 3 balls are drawn randomly, then the probability of getting 2 white balls?

a) Quantity I > Quantity II
b) Quantity I โฅ Quantity II
c) Quantity II > Quantity I
d) Quantity II โฅ Quantity I
e) Quantity I = Quantity II or Relation cannot be established

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

The following bar graph shows the price of gold per gram, diamond per gram and platinum per gram in different weeks in a certain month.

ย

16) Total price of gold per gram of 1st and 3rd week is what percentage more/less than the total price of platinum per gram of 2nd and 4th week?

a) 50 %

b) 35 %

c) 25 %

d) 30 %

e) None of these

ย 17) Find the average number of diamond per gram in the month?

a) Rs. 7000

b) Rs. 6000

c) Rs. 6500

d) 7500

e) None of these

ย 18) Find the difference between the total number of gold per gram in the month to that of the total number of platinum per gram in the month?

a) 2500

b) Rs. 4000

c) Rs. 3000

d) Rs. 3500

e) None of these

ย 19) The price of diamond per gram in 4th week is approximately what percentage of the price of gold per gram in 3rd week?

a) 185 %

b) 172 %

c) 243 %

d) 224 %

e) None of these

ย 20) Find the ratio between the total number price per gram of all the given type of jewels in 2nd week to that of 3rd week?

a) 5: 7

b) 7: 8

c) 3: 5

d) 11: 13

e) None of these

Direction (1-5)

I. 8x2 + 28x + 12 = 0

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

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

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

X = -4/8, -3 = -0.5, -3

II. 9y2 + 27y + 14 = 0

9y2 + 6y + 21y + 14 = 0

3y (3y + 2) + 7 (3y + 2) = 0

(3y + 7) (3y + 2) = 0

Y = -7/3, -2/3 = – 2.33, – 0.667

Canโt be determined

I. 2x2 + 11x + 14 = 0

2x2 + 4x + 7x + 14 = 0

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

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

X = -7/2, -2 = -3.5, -2

II. 3y2 + 14y + 15 = 0

3y2 + 9y + 5y + 15 = 0

3y (y + 3) + 5(y + 3) = 0

(3y + 5) (y + 3) = 0

Y = -5/3, -3 = -1.667, -3

Canโt be determined

I. 4x2 + 16x + 7 = 0

4x2 + 2x + 14x + 7 = 0

2x (2x + 1) + 7(2x + 1) = 0

(2x + 7)(2x + 1) = 0

X = -7/2, -1/2 = -3.5, -0.5

II. 5y2 + 32y + 35 = 0

5y2 + 25y + 7y + 35 = 0

5y(y + 5) + 7(y + 5) = 0

(5y + 7) (y + 5) = 0

Y = -7/5, -5 = – 1.4, -5

Canโt be determined

7x โ 5y = 55 –>(1)

3x + 7y = – 13–>(2)

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

X = 5, y = -4

X > y

I. x2 โ 3x โ 28 = 0

(x + 4) (x โ 7) = 0

x = -4, 7

II. y2 โ 17y + 72 = 0

(y โ 8) (y โ 9) = 0

y = 8, 9

x < y

Direction (6-10)

6*(x) รท 216 – 125 = 23

(x) / 36 = 23 + 125

(x) = 148*36

X = 5328

X โ (21*21*1725)/525 = (57900/15)

X โ 1449 = 3860

X = 3860 + 1449

X = 5309

(5/9)*3402 โ (16*78)/24 = x

X = 1890 โ 52

X = 1838

(44 ร 10-3)2 ร 45 รท (413 ร 10-4) = 10x

48 ร 10-6 ร 45 รท (413 ร 10-4) = 10x

(413 ร 10-6) / (413 ร 10-4) = 10x

10-6 + 4 = 10x

10-2 = 10x

X = -2

48 = 1331 + (24/100)*1800 โ x

48 = 1331 + 432 โ x

x = 1331 + 432 – 48

x = 1715

Direction (11-15)

Quantity I:

Time = D/S = (200 + 175)/(75*(5/18))

= > (375*18)/(75*5) = 18 sec

Quantity II:

Speed of first train = 150/15 = 10 m/sec = 10*(18/5) = 36 km/hr

Time = D/S = (150 + 250)/[(36 + 44)*(5/18)]

= > (400*18)/(80*5) = 18 sec

Quantity I = Quantity II

Quantity I:

S.I = (P*n*r)/100

6000 = (P*4*6)/100

Principle = (6000*100)/24 = Rs. 25000

Quantity II:

C.I = P*[(1 + (r/100))n โ 1]

6620 = P*(1 + (10/100))3 โ 1]

6620 = P*[(110/100)3 โ 1]

6620 = P*[(1331/1000) โ 1]

6620 = P*(331/1000)

P = 6620*(1000/331) = Rs. 20000

Quantity I > Quantity II

(x + y)/(x โ y) = 4

X + y = 4x โ 4y

3x โ 5y = 0 –ร  (1)

x โ (y/3) = 4

3x โ y = 12 –ร  (2)

Solving (1) and (2), we get,

X = 5, y = 3

Quantity I > Quantity II

Quantity II:

3 years ago, the ratio of ages of P and Q = 2: 3 (2x, 3x)

Present age of P and Q = 2x + 3, 3x + 3

Pโs age after 6 years = Present age of Q

2x + 3 + 6 = 3x + 3

X = 6

Present age of Q = 3x + 3 = 21 years

Quantity II:

5 years ago, the ratio of ages of A and B = 5: 3 (5x, 3x)

Present age of A and B = 5x + 5, 3x + 5

Sum of the ages of A and B, after 6 years = 62 years

According to the question,

5x + 11 + 3x + 11 = 62

8x = 62 โ 22

8x = 40 = > x = 5

Present age of A = 5x + 5 = 30 years

Present age of C = Present age of A – 5 = 30 – 5 = 25 years

Quantity II > Quantity I

Quantity I:

Total probability n(S) = 18C4

Required probability n(E) = 1 โ P(none is violet)

Probability of getting none is violet balls,

P(E) = n(E)/n(S) = 12C4/18C4

= > 11/68

Required probability = 1 โ (11/68) = 57/68

Quantity II:

Total probability n(S) = 18C3

Required probability n(E) = 8C2 and 10C1ยญ

P(E) = n(E)/n(S) = (8C2 and 10C1)ยญ / 18C3

= > 35/102

Quantity I > Quantity II

Direction (16-20)

Total price of gold per gram of 1st and 3rd week

= > 3250 + 3500 = 6750

Total price of platinum per gram of 2nd and 4th week

= > 2750 + 2250 = 5000

Required % = [(6750 โ 5000)/5000]*100 = 35 %

The average number of diamond per gram in the month

= > (8000 + 6500 + 7000 + 8500)/4

= > 30000/4 = Rs. 7500

The total number of gold per gram in the month

= > 3250 + 3000 + 3500 + 3750 = Rs. 13500

The total number of platinum per gram in the month

= > 2500 + 2750 + 3500 + 2250 = Rs. 11000

Required difference = 13500 โ 11000 = Rs. 2500

The price of diamond per gram in 4th week = 8500

The price of gold per gram in 3rd week = 3500

Required % = (8500/3500)*100 = 242.85 % = 243 %

The total number price per gram of all the given type of jewels in 2nd week

= > 3000 + 6500 + 2750 = Rs. 12250

The total number price per gram of all the given type of jewels in 3rd week

= > 3500 + 7000 + 3500 = Rs. 14000

Required ratio = 12250: 14000 = 7: 8

Daily Practice Test Schedule | Good Luck

 Topic Daily Publishing Time Daily News Papers & Editorials 8.00 AM Current Affairs Quiz 9.00 AM Quantitative Aptitude โ20-20โ 11.00 AM Vocabulary (Based on The Hindu) 12.00 PM General Awareness โ20-20โ 1.00 PM English Language โ20-20โ 2.00 PM Reasoning Puzzles & Seating 4.00 PM Daily Current Affairs Updates 5.00 PM Data Interpretation / Application Sums (Topic Wise) 6.00 PM Reasoning Ability โ20-20โ 7.00 PM English Language (New Pattern Questions) 8.00 PM
Online Mock Tests 2019:
SHARE