**Hello Aspirants**. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in **Inequalities** which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

**The ratio of the present age of Bala to that of Arnav is 3 : 11. Arnav is 12 years younger than Rahim. Rahim’s age after 7 years will be 85 years.**

**Quantity I: The present age of Bala’s father, who is 25 years older than Bala**

**Quantity II: Rahim’s present age**

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**B. Quantity I < Quantity II**

Explanation:

11x = 85 – 7 – 12

x = 6

Present age of Bala = 18

Present age of Bala’s father = 18 + 25 = 43; Rahim’s present age = 78**Mr. Ramesh bought two watches which together cost him Rs.440. He sold one of the watches at a loss of 20% and the other one at a gain of 40%. The selling price of both watches are same.**

**Quantity I: SP and CP one of the watches sold at a loss of 20%**

**Quantity II: SP and CP one of the watches sold at a profit of 40%**

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**D. Quantity I ≤ Quantity II**

Explanation:

80/100 * x = 140/100 * y

x = 7/4y

x + y = 440

7/4 y + y = 440

y = 160 ; x = 280**Ravi, Hari and Sanjay are three typists, who working simultaneously, can type 228 pages in four hours. In one hour, Sanjay can type as many pages more than Hari as Hari can type more than Ravi. During a period of five hours, Sanjay can type as many passages as Ravi can, during seven hours.**

**Quantity I: Number of pages typed by Ravi**

**Quantity II: Number of pages typed by Hari**

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**B. Quantity I < Quantity II**

Explanation:

Let Ravi, Hari and Sanjay can type x, y, and z pages respectively in 1 h.

Therefore, they together can type 4(x + y + z) pages in 4 h

∴ 4(x + y + z) = 228

⇒ x + y + z = 57 …..(i)

Also, z – y = y – x

i.e., 2y = x + z ……(ii)

5z = 7x ……(iii)

From Eqs. (i) and (ii), we get

3y = 57

⇒ y = 19

From Eq. (ii), x + z = 38

x = 16 and z = 22**The length of a rectangle wall is 3/2 times of its height. The area of the wall is 600m².**

Quantity I: Height of the wall

Quantity II: Length of the wall

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**A. Quantity I > Quantity II**

Explanation:

length = 3x

height = 2x

Area of the wall = 3x * 2x = 6x² = 600

Length = 30 & Height = 20**Quantity I: x² – 26x + 168 = 0**

**Quantity II:****y² – 29y + 210 = 0**

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**D. Quantity I ≤ Quantity II**

Explanation:

x² – 26x + 168 = 0

x = 12, 14

y² – 29y + 210 = 0

y = 14, 15**Quantity I: x² – 21x + 110 = 0**

Quantity II: y² – 18x + 80 = 0

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**C. Quantity I ≥ Quantity II**

Explanation:

x² – 21x + 110 = 0

x = 10 11

y² – 18y + 80 = 0

y = 10 8**A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged.**

**Quantity I: X = Inlet Pipe Efficiency**

**Quantity II: Y = Outlet Pipe Efficiency**

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**A. Quantity I > Quantity II**

Explanation:

Inlet pipe Efficiency = 100/(8/6) = 75%

Outlet pipe Efficiency = 100/(6) = 16.66%**Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job.**

**Quantity I: Probability of selecting no woman**

Quantity II: Probability of selecting at least one woman

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**B. Quantity I < Quantity II**

Explanation:

Man only = 8C2 = 14

Probability of selecting no woman = 14/91

Probability of selecting at least one woman = 1 – 14/91 = 77/91**A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random,**

**Quantity I: Probability that at least one is Black.**

Quantity II: Probability that all is Black.

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**A. Quantity I > Quantity II**

Explanation:

Total Balls = 15

Probability = 11c4/15c4 = 22/91

One is black = 1 – 22/91 = 69/91**Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern.**

Quantity I: Due to leakage, time taken to fill the tank

Quantity II: Time taken to empty the full cistern

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or relation cannot be established**B. Quantity I < Quantity II**

Explanation:

Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108).

Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min.

Due to leakage, time taken to fill the tank = 7 hours 12 min + 48 min = 8 hours

Work done by two pipes and leak in 1 hour = 1/8.

Work done by the leak in 1 hour =(15/108)-(1/8)=(1/72).

Leak will empty the full cistern in 72 hours.

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