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QUESTION: 1

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

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

QUESTION: 2

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%

Solution:

80/100 * x = 140/100 * y

x = 7/4y

x + y = 440

7/4 y + y = 440

y = 160 ; x = 280

QUESTION: 3

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

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

QUESTION: 4

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

Solution:

length = 3x

height = 2x

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

Length = 30 & Height = 20

QUESTION: 5

Quantity I: x² – 26x + 168 = 0

Quantity II: y² – 29y + 210 = 0

Solution:

x² – 26x + 168 = 0

x = 12, 14

y² – 29y + 210 = 0

y = 14, 15

QUESTION: 6

Quantity I: x² – 21x + 110 = 0

Quantity II: y² – 18x + 80 = 0

Solution:

x² – 21x + 110 = 0

x = 10 11

y² – 18y + 80 = 0

y = 10 8

QUESTION: 7

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

Solution:

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

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

QUESTION: 8

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

Solution:

Man only = 8C2 = 14

Probability of selecting no woman = 14/91

Probability of selecting at least one woman = 1 – 14/91 = 77/91

QUESTION: 9

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.

Solution:

Total Balls = 15

Probability = 11c4/15c4 = 22/91

One is black = 1 – 22/91 = 69/91

QUESTION: 10

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

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.

QUESTION: 11

Mr. Ramesh has three daughters namely Rohini, Anita and Keerthi. Rohini is the eldest daughter of Mr. Ramesh while Keerthi is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44.

Quantity I: The age of Rohini and Anita after two years

Quantity II: The age of Rohini and Keerthi after four years

Solution:

Square numbers – a, b, c

(a + 5) + (b + 5)+ (c + 5) = 44

a + b + c = 44 – 15 = 29

Possible values of a, b, c = 4, 9, 16 [Out of 1, 4, 9, 16, 25] The age of Rohini and Anita after two years = 29; The age of Rohini and Keerthi after four years = 28

QUESTION: 12

A sum of Rs. 8800 is to be divided among three brothers Anil, Deepak and Ramesh in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 year respectively remains equal.

Quantity I: Share of Ramesh

Quantity II: Share of Anil

Solution:

x*5*1/100 = y*5*2/100 = z*5*3/100

x:y:z = 6:3:2

Share of Ramesh = 2/11 * 8800 = 1600; Share of Anil = 6/11 * 8800 = 4800

QUESTION: 13

Person A sold his car to Person B at a profit of 20% and B sold it to C at a profit of 10%. Person C sold it to a mechanic at a loss of 9.09%. Mechanic spent 10% of his purchasing price and then sold it at a profit of 8.33% to Person “A” once again

Quantity I: Selling Price of A

Quantity II: Selling Price of C

Solution:

A:

CP = 100

SP = 120

B:

CP = 120

SP = 132

C:

CP = 132

SP = 120

Mechanic:

CP = 120 + 12 = 132

SP = 143

QUESTION: 14

Smallest side of a right-angled triangle is 13 cm less than the side of a square of perimeter 72 cm. The second largest side of the right angled triangle is 2 cm less than the length of the rectangle of area 112 cm² and breadth 8 cm.

Quantity I: Six more than the side of the right-angled triangle(not the smallest)

Quantity II: The Sum of Hypotenuse of the right-angled triangle and the smallest side

Solution:

Side of square = 72/4 = 18 cm

Smallest side of the right angled triangle = 18 – 13 = 5 cm

Length of rectangle = 112/8 = 14 cm

Second side of the right angled triangle = 14 – 2 = 12 cm

Hypotenuse of the right angled triangle = √(25 + 144) = 13cm

QUESTION: 15

Quantity I: (x – 36)² = 0

Quantity II: y² = 1296

Solution:

x² – 72x + 1296 = 0

x = 36, 36

y² = 1296

y = ±36

QUESTION: 16

Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank.

Quantity I: x = Pipe “Q” Efficiency. y = net efficiency

Quantity II: x = Pipe “P” Efficiency. y = net efficiency

Solution:

Pipe P Efficiency = 100/10 = 10%

Pipe Q Efficiency = 100/20 = 5%

Net Efficiency = 15%

QUESTION: 17

In an office there are 40% female employees. 50% of the male employees are UG graduates. The total 52% of employees are UG graduates out of 1800 employees.

Quantity I: Male Employees who are UG Graduates

Quantity II: Female Employees who are UG Graduates

Solution:

Total employees = 1800

female employees = 40%

male employees = 60%

50% of male employess = UG graduates = 30%

Female employees who are UG graduates = 22%

QUESTION: 18

Suresh spends 23% of an amount of money on an insurance policy, 33% on food, 19% on children’s education and 16% on recreation. He deposits the remaining amount of Rs. 504 in bank.

Quantity I: Amount spend on food and insurance policy together

Quantity II: Amount spend on children’s education and recreation together

Solution:

Total amount = x

Savings(%)

[100 – (23 + 33 + 19 + 16 )]% = 9 %

9% of x = 504

⇒ x = 504 * 100/9 = 5600

Amount spend on food and insurance policy together = 56% of 5600 = Rs.3136

QUESTION: 19

Last year there were 610 boys in a school. The number decreased by 20 percent this year. The number of girls is 175 percent of the total number of boys in the school this year.

Quantity I: Total number of boys in the school this year.

Quantity II: Half of total number of Girls in the school this year.

Solution:

No of boys in a school last year = 610

No of boys in a school for this year

610*80/100=122

610-122=488

No of girls =175/100 * 488=854 = 427

QUESTION: 20

Two persons A and B start from the opposite ends of a 450 km straight track and run to and from between the two ends. The speed of the first person is 25 m/s and the speed of other is 35 m/s. They continue their motion for 10 hours.

Quantity I: First person speed

Quantity II: Second person speed

Solution:

First person speed = 25 m/s * 18/5 = 90 kmph

Second person speed = 35 m/s * 18/5 = 126 kmph

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