**Quadratic Equation Questions & Answers PDF Download.**A quadratic equation is a part of the Quantitative Aptitude section. Candidates are you seeking to practice the multiple-choice questions based on this Quadratic equation often make an appearance in the SSC, Railways Exams, and other Competitive exams. It is one of the essential and highest-scoring topics which you can easily face in the banking exam

__Quadratic Equation- Part I__

__Quadratic Equation- Part I__

__Download Reasoning Questions with Answers Pdf__

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**Directions (1 – 5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.**

**I. x² – 34x + 288 = 0**

II.y² – 28y + 192 = 0

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation cannot be established

**Correct option is : C**

**Solution:**

x² – 34x + 288 = 0

x = 18, 16

II. y² – 28y + 192 = 0

y = 12, 16

**I. x² – 26x + 168 = 0**

**II.y² – 32y + 252 = 0**

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation cannot be established

**Correct option is : D**

**Solution:**

x² – 26x + 168 = 0

x = 12, 14

II. y² – 32y + 252 = 0

y = 14, 18

**I. x² + 26x + 168 = 0**

**II.y² + 23y + 132 = 0**

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation cannot be established

**Correct option is : D**

**Solution:**

x² + 26x + 168 = 0

x = -12, -14

II. y² + 23y + 132 = 0

y = -12, -11

**I. x² – 28x + 195 = 0**

**II. y² – 30y + 216 = 0**

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation cannot be established

**Correct option is : E**

**Solution:**

x² – 28x + 195 = 0

x = 15, 13

II. y² – 30y + 216 = 0

y = 18, 12

5.** I. (x – 19)² = 0**

II. y² = 361

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation cannot be established

**Correct option is : C**

**Solution:**

x² – 38x + 361 = 0

x = 19, 19

II. y² = 361

y = ±19

__Quadratic Equation Part – II__

__Quadratic Equation Part – II__

**Directions(1- 5): In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. (only quantity is to be considered)**

**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

**Correct option is : B**

**Solution:**

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

**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

**Correct option is : B**

**Solution:**

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

**Correct option is : B**

**Solution:**

length = 3x

height = 2x

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

Length = 30 & Height = 20

**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

**Correct option is : A**

**Solution :**

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

Outlet pipe Efficiency = 100/(6) = 16.66%

**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

**Correct option is : B**

**Solution:**

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.

**Quadratic Equation Questions & Answers**

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