# SBI Clerk Mains Quantitative Aptitude (Day-30)

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Quantity I and II

Directions (1-5): Each question below contains a statement followed by Quantity I and Quantity II. Find the relationship among them and mark your answer accordingly. Give answer as,

a)Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c)Quantity II > Quantity I

d)Quantity II ≥ Quantity I

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

1) Quantity I: The ratio of present age of Nive and Renuka is 4: 7. Nive is 8 years younger than Bhairavi. After 7 years, Bhairavi’s age is 35. Then find the present age of Renuka?

Quantity II: The sum of the present age Rahul and his son is 40 years. 15 years after, the ratio of Rahul and his son is 5: 2. Then find the present age of Rahul?

2) Quantity I: A boat can cover 80 km downstream in 48 minutes. If the speed of the current is 3/7 of the boat in still water, then how much distance (in km) can the boat cover upstream in 300 minutes?

Quantity II: Speed of a man in still water is 12 km/hr and the river is running at 6 km/hr. The total time taken to go to a place and come back is 40 hours. Find the distance travelled by the man?

3) Quantity I: A necklace was sold for Rs. 5500 with a profit of 10 %. If it were sold for Rs. 5900, then what would have been the percentage of profit?

Quantity II: A person bought two showcases for Rs. 5000 each. He sold one at a profit of 10 % and the other at a loss of 5 %. What would be his overall profit or loss percentage in the whole transaction?

4) Quantity I: A bucket contains some quantity of milk and water in the ratio of 3: 2. If 15 litres of mixture is drawn out and replaced with water and the ratio of milk and water becomes 21: 29, then find the initial quantity of mixture?

Quantity II: How many litres of water should be added to a 150 litres mixture containing milk and water in the ratio of 2: 1 such that the resultant mixture has 40 % milk in it?

5) Quantity I: The difference between the simple interest and compound interest on a certain sum of money for 2 years at 10 % per annum is Rs. 600. Find the sum?

Quantity II: A certain sum of money invested for a period of 5 years at 8 % per annum, simple interest earned is Rs. 30000. Find the principle?

Application Sums

6) A box contains 8 Pink balls, 5 black balls, 4 orange balls and 3 white balls. If four balls are drawn out randomly, then the probability of getting a different colour balls?

A) 11/35

B) 32/323

C) 18/189

D) 34/77

E) None of these

7) The ratio of the age of P after 3 years and 6 years ago, the age of Q is 3: 4. The present age of R is 2 times the present age of P. The difference between the present age of Q and R is 6 years. Find the present age of P?

A) 22 years

B) 20 years

C) 24 years

D) 16 years

E) None of these

8) The volume of cylinder 1 and 2 is in the ratio of 3: 1 and the height of the cylinder 1 and 2 is in the ratio of 3: 4. Find the ratio between the radius of cylinder 1 and cylinder 2?

A) 7: 5

B) 5: 2

C) 4: 3

D) 2: 1

E) None of these

9) The present age of Janvi and her husband is 5: 6. The present age Janvi’s daughter is (1/5) of Janvi’s present age. Janvi’s daughter is 3 years younger than her son. Find the sum of the present age of Janvi and her husband, if the age of Janvi’s son, after 7 years is 15 years?

A) 58 years

B) 66 years

C) 55 years

D) 62 years

E) None of these

10) The length of train A is 200 m and that of train B is 300 m. These two trains are running in the opposite direction with the velocities of 60 km/hr and 40 km/hr respectively. Find the time taken to cross each other?

A) 14 sec

B) 21 sec

C) 18 sec

D) 25 sec

E) None of these

Directions (1-5) :

Quantity I:

The ratio of present age of Nive and Renuka = 4: 7

Nive = Bhairavi – 8

After 7 years, Bhairavi’s age = 35

Present age of Bhairavi = 28

Nive = 28 – 8 = 20

4’s = 20

1’s = 5

Present age of Renuka = 7’s = 35 years

Quantity II:

The sum of the present age Rahul and his son = 40 years

Rahul + Rahul’s son = 40

15 years after, the ratio of the age of Rahul and his son = 5: 2

According to the question,

5x – 15 + 2x – 15 = 40

7x = 40 + 30

7x = 70

x = 10

Present age of Rahul= 5x – 15 = 35 years

Quantity I = Quantity II

Quantity I:

Speed of downstream = (80 * 60)/48 = 100 km/hr

Speed of current: Speed of still water = 3: 7

Speed of upstream = Speed of boat in still water – speed of Current

100 = 10x

= > x = 10

Speed of current = 30 km/hr

Speed of boat in still water = 70 km/hr

Speed of upstream= Speed of boat in still water – speed of Current

= > 70 – 30 = 40 km/hr

Speed = 40 km/hr, Time = 300 minutes

Distance = 40 * (300/60) = 200 km

Quantity II:

Speed of downstream = 12 + 6 = 18 km/hr

Speed of upstream = 12 – 6 = 6 km/hr

Let the distance travelled be x,

(x/18) + (x/6) = 40

x = 180 km

(Or)

Distance = Time *[(Speed of Still water2 – speed of Stream2)/(2*Speed of Still water)]

= > Distance = 40 * [(122 – 62)/(2 * 12)]

= > Distance = 40 * [(144 – 36)/24]

= > Distance = 180 km

Quantity I > Quantity II

Quantity I:

According to the question,

(110/100) * CP = 5500

CP = 5500 * (10/11) = Rs. 5000

SP = 5900

Profit % = (Profit/CP) * 100

= > (900/5000) * 100

= > 18 %

Quantity II:

According to the question,

CP1= 5000, Profit = 10 %

SP1 = 5000 * (110/100) = 5500

CP2 = 5000, Loss = 5 %

SP2 = 5000 × (95/100) = 4750

Total selling price = S.P1 + S.P2 = 5500 + 4750 = 10250

Total cost price = 5000 + 5000 = 10000

Profit % = (250/10000) * 100 = 2.5 %

Quantity I > Quantity II

Quantity I:

Total quantity of mixture = 15 litres

Water = 15 * (2/5) = 6 lit, Milk = 15 * (3/5) = 9 lit

Given,

= > (3x – 9)/(2x – 6 + 15) = 21/29

= > 87x – 261 = 42x + 189

= > 45x = 450

= > x = 10

Initial quantity of milk = 5x = 50 litres

Quantity II:

Total mixture = 150 litres

Milk = 100 lit, water = 50 lit

According to the question,

100/(50 + x) = 40/60

150 = 50 + x

x = 100 litres

Quantity I < Quantity II

Quantity I:

The difference between the simple interest and compound interest for 2 years,

Diff = Sum * (r/100)2

600 = Sum * (10/100)2

(600 * 100 * 100)/100 = Sum

Sum = Rs. 60000

Quantity II:

I = Pnr/100

30000 = (P * 5 * 8)/100

(30000 * 100)/40 = P

Principle = Rs. 75000

Quantity II > Quantity I

Directions (6-10) :

Total probability n(S) = 20C4

Required probability = n(E) = 8C1 and 5C1 and 4C1 and 3C1

P(E) = n(E)/n(S)

P(E) = (8C1 and 5C1 and 4Cand 3C1)/20C4

P(E) = 32/323

The ratio of 3 years after, the age of P and 6 years ago, the age of Q = 3: 4 (3x, 4x)

Present age of P and Q = 3x – 3, 4x + 6

The age of R = 2 * P = 2 * (3x – 3) = 6x – 6

(6x – 6) – (4x + 6) = 6

6x – 6 – 4x – 6 = 6

2x = 6 + 12

2x = 18

x = 9

Present age of P = 3x – 3 = 27 – 3 = 24 years

Volume of the cylinder = πr2h

The ratio of volume of cylinder 1 and 2 = 3: 1

The ratio of height of cylinder 1 and 2 = 3: 4

The ratio between the heights of cylinder 1 to that of cylinder 2,

= > [π * (3x) * r12]/[π * (4x) * r22] = (3/1)

= > r12/r22 = (4/1)

= > r1: r2 = 2: 1

The present age of Janvi and her husband = 5: 6 (5x, 6x)

The present age Janvi’s daughter = (1/5) * Janvi’s present age

Janvi’s daughter = Janvi’s son – 3

Janvi’s son after 7 years = 15 years

Janvi’s son’s present age = 8 years

Janvi’s daughter = 8 – 3 = 5 years

Janvi’s present age = 5 * 5 = 25 years

5’s = 25

1’s = 5

Present age of Janvi’s husband = 6x = 30 years

Required sum = 25 + 30 = 55 years

T = D/S

T = (200 + 300)/[(60 + 40) * (5/18)]

T = (500 * 18)/(100 * 5)

T = 18 sec

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