SBI Clerk Mains Quantitative Aptitude (Day-50)

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Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Application Sums

1) Anu, Bharathi and Divya started the business and the amount invests in Anu to Bharathi and Anu to Divya in the ratio of 3: 4 and 4: 5 respectively. If the investment period of Anu, Bharathi and Divya in the ratio of 13: 10: 16, the total profit of the business is Rs.8340, then what is the average profit share of Bharathi and Divya?

a) Rs.2000

b) Rs.2400

c) Rs.3000

d) Rs.3600

e) None of these

2) The cost price of milk in vessel A is Rs.66 per liter and the cost price of milk in vessel B is Rs.51. If milk in vessel A and B are mixed, then the shopkeeper sold 37.5 liters of this mixture at the cost price of milk in vessel A while he gets the profit of 10%. If he sold the same mixture at the cost price of milk in vessel B, then what is the percentage of loss or profit earned by shopkeeper?

a) 8% profit

b) 8% loss

c) 15% loss

d) 15% profit

e) None of these

3) If the cost price of the article A and B is same but after getting two successive raises of x% each. If cost price of article A is 50% more than cost price of article B after getting two successive raises of 20% and 25% respectively, then find the value of x?

a) 20%

b) 30%

c) 50%

d) 40%

e) None of these

4) If A gets the loan of Rs.27000 from bank at 10% compound interest for 3 years. If he lent some amount to B at 20% simple interest for 2 years and remaining amount to C at 26(2/3)% simple rate of interest for 3 years. If the interest amount received by B and C is equal, then find profit amount earned by A after 3 years?

a) Rs.5463

b) Rs.5568

c) Rs.5578

d) Rs.5583

e) None of these

5) If the car increases the speed by 20 kmph, then it will take 3 hours less to cover the distance between A and B and the car decreases the speed by 20 kmph, then it take 4.5 hours more to cover the same distance, then find the distance between the A and B?

a) 900 km

b) 1200 km

c) 1500 km

d) 1800 km

e) None of these

Data sufficiency

Directions (6 – 10): In each of the following questions, a question is followed by three statements I, II and III. Read all the statements to find the answer to given question and then answer accordingly that which statement/s can give the answer alone/together.

6) Find the time taken by the boat to cover 90 Km downstream and 80 Km upstream.

I. Speed of the boat in still water is 250% more than the speed of the stream.

II. Sum of the twice the speed of the boat in still water and thrice the speed of the stream is 40 Km/h

III. Upstream speed of the boat is 10 Km/h.

a) All I, II and III

b) Any two of the three

c) Only I and III

d) Any one of the three

e) Even I, II and III together are not sufficient.

7) Find the respective ratio of the ages of Arjun and Rupali after 10 years.

I. Ratio of the ages of Rupali and Shyam before three years was 8: 9 respectively.

II. Ratio of the present ages of Arjun and Deepak is 15: 13 respectively.

III. Ratio of the present ages of Shyam and Deepak is 10: 13 respectively.

a) All I, II and III

b) Any two of the three

c) Only II and III

d) Only I and II

e) Even I, II and III together are not sufficient.

8) If x > 0 and y > 0, find the value of (54872)1/x + (1024)1/y

I. x = y – 2

II. 2x = y + 1

III. x + y = 8

a) All I, II and III

b) Any two of the three

c) Only II and III

d) Only I and II

e) Even I, II and III together are not sufficient.

9) If a12÷ ax x a4÷ a2y x a-z = a2, find the value of (x + y + z).

I. x = 3y

II. x > z > y

III. x – z = 2

a) All I, II and III

b) Any two of the three

c) Only II and III

d) Only I and III

e) Even I, II and III together are not sufficient.

10) If 2a + 3b + 5c = m, find the value of (m – a + b).

I. m – 2b = 27

II. a + c = 6

III. m – a = 31

a) All I, II and III

b) Any two of the three

c) Only II and III

d) Only I and III

e) Even I, II and III together are not sufficient.

Answers:

Directions (1-5):

1) Answer: C

Investment of Anu, Bharathi and Divya = (4 * 3): (4 * 4): (5 * 3)

= 12: 16: 15

Profit share of Anu, Bharathi and Divya = (12 * 13): (16 * 10): (16 * 15)

= 39: 40: 60

Average share of Bharathi and Divya = (100/2)/139 * 8340

= 3000

2) Answer: C

Quantity of vessel B = x

Quantity of vessel A = 37.5 – x

CP of the mixture = x * 51 + (37.5 – x) * 66

= 2475 – 15x

SP of the mixture = 37.5 * 66 = 2475

Profit = (2475 – (2475 + 15x))/(2475 – 15x) * 100 = 10

1500x = 24750 – 150x

x = 15

CP of the mixture = 2475 – 15 * 15 = 2250

New SP of the mixture = 37.5 * 51 = 1912.5

Required loss percentage = (2250 – 1912.5)/2250 * 100

= 15% loss

3) Answer: C

Let us take the cost price of article A and B be a

Article A = a * (100 + x)/100 * (100 + x)/100

Article B = a * (120/100 * 125/100) = 1.5a

a * (1 + x/100)2 = 150/100 * 1.5a

(1 + x/100) = 15/10

100 + x = 150

x = 50%

4) Answer: A

SI = (P * N * R)/100

CI = P * (1 + R/100)n – P

Interest received by B = (x * 20 * 2)/100

Interest received by C = ((27000 – x) * (80/3) * 3)/100

40x/100 = (27000 – x) * 80/100

x = 54000 – 2x

x = 18000

Interest received from B = (18000 * 20 * 2)/100 = 7200

Interest received from C = 7200

CI = 27000 * (1 + 10/100)3 – 27000

CI = 8937

Profit earned by A = (7200 * 2) – 8937 = 5463

5) Answer: D

Distance = x

Initial Speed of car = y

((x/y) – x/(y + 20)) = 3

20x/y2 + 20y = 3

x/(y – 20) – x/y = 4.5

20x/(y2 – 20y) = 4.5

3y2 + 60y = 4.5y2 – 90y

3y + 60 = 4.5y – 90

1.5y = 150

y = 100 kmph

20x/(100 * 100 + 20 * 100) = 3

x = 1800 km

Directions (6-10):

6) Answer: B

Let, speed of the boat in still water = x km/h

And speed of the stream = y km/h

From I and II:

x = y x (100 + 250)/100

=> x = 350y/100

=> x = 7y/2 —- (i)

2x + 3y = 40 —– (ii)

From (i) and (ii)

=> 14y/2 + 3y = 40

=> (14y + 6y)/2 = 40

=> 20y = 80

=> y = 4 Km/h

From (i)

x = 7/2 x 4 = 14 km/h

Required time = 90/(14 + 4) + 80/(14 – 4)

= 90/18 + 80/10

= 5 + 8

= 13 hours

From I and III:

x = y x (100 + 250)/100

=> x = 350y/100

=> x = 7y/2 —- (i)

x – y = 10 —— (ii)

From (i) and (ii)

=> 7y/2 – y = 10

=> (7y – 2y)/2 = 10

=> 5y = 20

=> y = 4 km/h

From (i)

x = 7/2 x 4 = 14 km/h

Required time = 90/(14 + 4) + 80/(14 – 4)

= 90/18 + 80/10

= 5 + 8

= 13 hours

From II and III:

2x + 3y = 40 —– (i)

x – y = 10 ——- (ii)

Equation (i) – 2 x equation (ii)

2x + 3y – 2x + 2y = 40 – 20

=> 5y = 20

=> y = 4 km/h

From (ii)

x – 4 = 10

=> x = 14 Km/h

Required time = 90/(14 + 4) + 80/(14 – 4)

= 90/18 + 80/10

= 5 + 8

= 13 hours

Hence, any two of the three statements are sufficient.

7) Answer: E

From I:

(Rupali – 3): (Shyam – 3) = 8: 9

From II:

Arjun: Deepak = 15: 13

From III:

Shyam: Deepak = 10: 13

From I, II and III:

(Rupali – 3): (Shyam – 3) = 8: 9

Arjun: Deepak = 15: 13

Shyam: Deepak = 10: 13

Hence, Even I, II and III together are not sufficient.

8) Answer: B

From I and II:

x = y – 2 —- (i)

2x = y + 1 —— (ii)

From (i) and (ii)

2(y – 2) = y + 1

=> 2y – 4 = y + 1

=> 2y – y = 4 + 1

=> y = 5

From (i)

x = 5 – 2 = 3

Now, (54872)1/x + (1024)1/y =∛54872 + (1024)(1/5) = 38 + 4 = 42

From II and III:

2x = y + 1 ——- (i)

x + y = 8

=> x = 8 – y —— (ii)

From (i) and (ii)

2(8 – y) = y + 1

=> 16 – 2y = y + 1

=> 3y = 15

=> y = 5

From (i)

2x = 5 + 1

=> 2x = 6

=> x = 3

Now, (54872)1/x + (1024)1/y =∛54872 + (1024)(1/5) = 38 + 4 = 42

From I and III:

x = y – 2 ——– (i)

x + y = 8 ——- (ii)

From (i) and (ii)

y – 2 + y = 8

=> 2y = 8 + 2

=> 2y = 10

=> y = 5

From (i)

x = 5 – 2 = 3

Now, (54872)1/x + (1024)1/y =∛54872 + (1024)(1/5) = 38 + 4 = 42

Hence, any two of the three statements are sufficient.

9) Answer: D

a12÷ ax x a4÷ a2y x a-z = a2

=> a(12 – x + 4 – 2y – z) = a2

=> a(16 – x – 2y – z) = a2

=> 16 – x – 2y – z = 2

=> x + 2y + z = 14 —— (i)

From I:

x = 3y

From II:

x > z > y

From III:

x – z = 2

From I and III:

x = 3y

=> y = x/3 —– (ii)

x – z = 2

=> z = x – 2 —– (iii)

From (i), (ii) and (iii)

x + 2x/3 + x – 2 = 14

=> (3x + 2x + 3x – 6)/3 = 14

=> 8x – 6 = 42

=> 8x = 48

=> x = 6

From (ii)

y = 6/3 = 2

From (iii)

z = 6 – 2 = 4

Now, (x + y + z) = 6 + 2 + 4 = 12

Hence, only I and III are required.

  1. 10) Answer: A

2a + 3b + 5c = m —– (i)

From I:

m – 2b = 27

From II:

a + c = 6

From III:

m – a = 31

From I, II and III:

m – 2b = 27

=> m = 2b + 27—– (ii)

a + c = 6

=> c = 6 – a ——– (iii)

m – a = 31

=> m = a + 31——- (iv)

From (ii) and (iv)

2b + 27 = a + 31

=> a = 2b – 4

=> b = (a + 4)/2 —–(v)

From (i), (iii), (iv) and (v), we have

2a + 3 x (a + 4)/2 + 5(6 – a) = a + 31

=> (4a + 3a + 12 + 60 – 10a)/2 = a + 31

=> -3a + 72 = 2a + 62

=>5a = 10

=> a = 2

From (v)

b = (2 + 4)/2 = 3

From (iv)

m = 2 + 31 = 33

Now,

m – a + b = 33 – 2 + 3 = 34

Hence, all I, II and III together are sufficient.

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