LIC AAO/SBI PO Prelims Quantitative Aptitude Questions 2019 (Day-04)

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Quantitative Aptitude Questions For LIC AAO/SBI PO Prelims (Day-04)

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Directions (1 – 2): Each question contains a statement followed by Quantity I, II and III. Read the information clearly and answer your questions accordingly.

The options represent the relations between these three quantities

A) >

B) <

C) =

D) ≤

E) ≥

For example:

Quantity I = 200

Quantity II = 300

Quantity III = 100

Hence, Quantity I < Quantity II > Quantity III

a) A, B

b) B, C

c) B, A

d) E, B

e) B, D

Answer is option: c)

1) Present age of P is 6 years more than the age of Q after 7 years. The present age of R is 3 times of present age of Q. The ratio of age of R, 3 years ago to that of the present age of S is 3: 4. Present age of T is one-eleventh of the present age of S.

Quantity I:Find the age of S, 8 years ago, if the present age of Q is 12 years?

Quantity II:Find the present age of R, if the present age of P is square of five?

Quantity III:Find the sum of the present age of P and Q together, if the present age of R is 36 years?

a) D, A

b) B, D

c) C, B

d) A, C

e) D, D

2) A, B and C entered into a partnership by investing in the ratio of 4: 7: 9 respectively. After 4 months, A invested 30 % more than the initial investment. And after another 5 months, B withdraws Rs. 5000 but C invested Rs. 10000 more. The ratio of share of A, B and C is 192: 275: 370.

Quantity I: A and P entered into a partnership by investing in the ratio of 2: 3. Find the share of P, if the total profit at the end of the year is Rs. 200000?

Quantity II: Find the share of B, if the total profit at the end of the year for A, B and C is Rs. 669600?

Quantity III: Find the75 % of the share of C, if the total profit at the end of the year for A, B and C is Rs. 1004400?

a) B, B

b) A, D

c) E, C

d) A, C

e) E, B

Directions (Q. 3 – 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

3)

I) x3 – 133 = [56 % of 1250] ÷ 4 + 113 + 124 × 9 ÷ 2

II) y2 – y – 72 = 0

4)

I) 2x2 – x – 66 = 0

II) 3y2 – 5y – 42 = 0

5)

I) 6x + 4y = 4x + 7y

II) 5x + 3y + 5 = 2x + 5y

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

The following table shows the total number of car spare parts (In lakhs) manufactured by five different companies in a certain year and total number of defective and sold items are given.

6) Total number of defective car spare parts of company P, R and T together is approximately what percentage of total number of defective car spare parts of company Q, S and T together?

a) 150 %

b) 170 %

c) 190 %

d) 210 %

e) 230 %

7) Find the ratio of total number of unsold spare parts in company Q to that of company R?

a) 312: 409

b) 265: 288

c) 124: 133

d) 156: 141

e) None of these

8) Find the difference between the total number of car spare parts sold by Company P and R together to that Company Q and T together?

a) 75420

b) 84564

c) 79128

d) 28956

e) None of these

9) How much percentage of car spare parts sold by Company Q?

a) 73.5 %

b) 81 %

c) 84 %

d) 77.5 %

e) None of these

10) Find the average number of car spare parts manufactured by all the given companies together?

a) 8.6 lakhs

b) 7.7 lakhs

c) 9.1 lakhs

d) 7.3 lakhs

e) None of these

Answers:

Directions (1-2):

1) Answer: c)

Present age of P = The age of Q, after 7 years + 6

The ratio of present age of R and Q = 3: 1

The ratio of age of R, 3 years ago to that of the present age of S = 3: 4

Present age of T = (1/11)*the present age of S

Quantity I:Find the age of S, 8 years ago, if the present age of Q is 12 years?

The present age of Q = 12 years

Present age of R = 12*3 = 36 years

The age of R, 3 years ago = 33 years

Present age of S = (33/3)*4 = 44 years

The age of S, 8 years ago = 44 – 8 = 36 years

Quantity II:Find the present age of R, if the present age of P is square of five?

Present age of P = 52 = 25 years

The age of Q, after 7 years = 25 – 6 = 19 years

Present age of Q = 12 years

The ratio of present age of R and Q = 3: 1

Present age of R = 12*3 = 36 years

Quantity III: Find the sum of the present age of P and Q together, if the present age of R is 36 years?

Present age of R = 36 years

Present age of Q = (36/3) = 12 years

Present age of P = 19 + 6 = 25 years

Required sum = 12 + 25 = 37 years

36 = 36 < 37

Quantity I = Quantity II < Quantity III

2) Answer: a)

The share of A, B and C

= > [4x*4 + 4x*(130/100)*8]: [7x*9 + (7x – 5000)*3]: [9x*9 + (9x + 10000)*3] = 192:275: 370

= > [16x + 208x/5]: [63x + 21x – 15000]: [81x + 27x + 30000] = 192:275: 370

= > [288x/5]: [84x – 15000]: [108x + 30000] = 192:275: 370

According to the question,

= > (84x – 15000)/(108x + 30000) = (275/370)

= > 6216x – 1110000 = 5940x + 1650000

= > 276x = 2760000

= > x = 10000

Initial investment of A = 4x = Rs. 40000

Initial investment of B = 7x = Rs. 70000

Initial investment of C = 9x = Rs. 90000

Quantity I: A and P entered into a partnership by investing in the ratio of 2: 3. Find the share of P, if the total profit at the end of the year is Rs. 200000?

The share of A and P

= > [40000*12] : [(40000/2)*3*12]

= > 2 : 3

5’s = 200000

1’s = 40000

The share of P = Rs. 120000

(Or)

Period of investment is same for both of them. So,

The ratio of profit = 2 : 3

5’s = 200000

1’s = 40000

The share of P = Rs. 120000

Quantity II: Find the share of B, if the total profit at the end of the year for A, B and C is Rs. 669600?

The share of A, B and C

= >192:275: 370

Total profit = 837’s = 669600

1’s = 800

The share of B = 275’s = Rs. 220000

Quantity III: Find the75 % of the share of C, if the total profit at the end of the year for A, B and C is Rs. 1004400?

The share of A, B and C

= > 192 : 275 : 370

Total profit = 837’s = 1004400

1’s = 1200

The share of C = 370’s = Rs. 444000

Required share = 444000*(75/100) = Rs. 333000

120000 < 220000 < 333000

Directions (3-5):

3) Answer: a)

I) x3 – 133 = [56 % of 1250] ÷ 4 + 113 + 124 × 9 ÷ 2

X3 = (56/100)*(1250/4) + 1331 + (124*9)/2 + 133

X3 = 175 + 1331 + 558 + 133 = 2197

X = 13

II) y2 – y – 72 = 0

(y – 9) (y + 8) = 0

Y = 9, -8

X > y

4) Answer: e)

I) 2x2 – x – 66 = 0

2x2 – 12x + 11x – 66 = 0

2x (x – 6) + 11(x – 6) = 0

(2x + 11) (x – 6) = 0

X = -11/2, 6 = -5.5, 6

II) 3y2 – 5y – 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

Can’t be determined

5) Answer: c)

I) 6x + 4y = 4x + 7y

2x – 3y = 0 —> (1)

II) 5x + 3y + 5 = 2x + 5y

3x – 2y = -5 —> (2)

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

X = -3, y = -2

X < y

Directions (6-10):

6) Answer: c)

Total number of defective car spare parts of company P, R and T together

= > 32500 + 33600 + 14400 = 80500

Total number of defective car spare parts of company Q, S and T together

= > 21200 + 6800 + 14400 = 42400

Required % = (80500/42400)*100 = 190 %

7) Answer: b)

Total number of unsold car spare parts in company Q

= > 1060000 – (21200 + 779100) = 259700

Total number of unsold car spare parts in company R

= > 840000 – (33600 + 524160) = 282240

Required ratio = 259700 : 282240 = 265 : 288

8) Answer: d)

The total number of car spare parts sold by Company P and R together

= > 876600 + 524160 = 1400760

The total number of car spare parts sold by Company Q and T together

= > 779100 + 592704 = 1371804

Required difference = 1400760 – 1371804 = 28956

9) Answer: a)

Required % = (779100/1060000)*100 = 73.5 %

10) Answer: c)

The total number of car spare parts manufactured by all the given companies together

= > 12.5 + 10.6 + 8.4 + 6.8 + 7.2 = 45.5 lakhs

Required average = (45.5/5) = 9.1 lakhs

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