Quadratic Equation - MCQ 3


20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Quadratic Equation - MCQ 3


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QUESTION: 1

The respective ratio between the present age of Mohan and David is 5:x. Mohan is 9 years younger than Preethi. Preethi’s age after 9 years will be 33 years. The difference between David’s and Mohan’s age is same as the present age of Preethi.

Quantity I: Mohan’s present age
Quantity II: The value of x

Solution:

Preethi’s age after 9 years = 33 years
Preethi’s present age = 33 – 9 = 24 years
Mohan’s present age = 24 – 9 = 15 years
David’s present age = 15 + 24 = 39 years
Ratio between Mohan and David = 15 : 39 = 5 : 13 X = 13

QUESTION: 2

Sri invested some amount(x) at the rate of 12% simple interest and a certain amount(y) at the rate of 10% simple interest. He received yearly interest of Rs.140. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest.

Quantity I: The value of x
Quantity II: The value of y

Solution:

Amount invested at 12% = Rs. x Amount invested at 10% = Rs. y
140 = x*12*1/100 + y*10*1/100
12x + 10y = 14000 -(i)
144 = x*10*1/100 + y*12*1/100
10x + 12y = 14400 -(ii)
x = 545.45; y = 745.45

QUESTION: 3

Ajith can do a piece of work in 10 days, Bala in 15 days. They work together for 5 days, the rest of the work is finished by Chand in two more days. They get Rs. 6000 as wages for the whole work.

Quantity I: What is the sum of Rs.100 and the daily wage of Bala?
Quantity II: What is the daily wage of Chand?

Solution:

Ajith’s 5 days work = 50%
Bala’s 5 days work = 33.33%
Chand’s 2 days work = 16.66%[100- (50+33.33)] Ratio of contribution of work of Ajith, Bala and Chand = 3 : 2 : 1
Ajith’s total share = Rs. 3000
Bala’s total share = Rs. 2000
Chand’s total share = Rs. 1000
Ajith’s one day’s earning = Rs.600
Bala’s one day’s earning = Rs.400
Chand’s one day’s earning = Rs.500

QUESTION: 4

A Bike is available at 40% discount at showroom “A” and the same is available at only 25% discount at showroom “B”. Mr. Arun has just sufficient amount of Rs. 60,000 to purchase it at showroom “A”.

Quantity I: Difference between Marked Price and SP at Show Room “A”
Quantity II: Difference between Marked Price and SP at Show Room “B”

Solution:

Let the marked price be x.
Cost price (CP) = 40 % discount on MP = 0.6y = 60000
⇒ y= Rs.100000 MP
SP at Show Room “A” = Rs. 60000
SP at Show Room “B” = 100000 X 0.75 = 75000
Difference between Marked Price and SP at Show Room “A” = 40000
Difference between Marked Price and SP at Show Room “B” = 25000

QUESTION: 5

Quantity I: (x – 18)² = 0
Quantity II: y² = 324

Solution:

x² – 36x + 324 = 0
x = 18, 18
y² = 324
y = ±18

QUESTION: 6

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 4 hours when the inlet pipe is plugged.

Quantity I: Inlet pipe Efficiency
Quantity II: 3 times of Outlet pipe Efficiency

Solution:

Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(4) = 25%
3 times of Outlet pipe Efficiency = 75%

QUESTION: 7

Harish took an educational loan from a nationalized bank for his 2 years course of MBA. He took the loan of Rs.5 lakh such that he would be charged at 7% p.a. at CI during his course and at 9% CI after the completion of the course. He returned half of the amount which he had to be paid on the completion of his studies and remaining after 2 years.

Quantity I: He returned half of the amount which he had to be paid on the completion of his studies
Quantity II: He returned remaining amount after 2 years

Solution:

5,00,000 * (1.07)² = 572450
Returned amount = 286225
After two years = 286225 * (1.09)² = 340063

QUESTION: 8

The average salary of the entire staff in an office is Rs 250 per month. The average salary of officers is Rs 520 and that of non-officers is Rs. 200.

Quantity I: Number of Officers = 15
Quantity II: Number of Non-Officers

Solution:

Let the required number of non–officers = x
200x + 520 x 15 = 250 (15 + x)
250x – 200x = 520 * 15 – 250 x 15
50x = 4050
x = 81

QUESTION: 9

The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 600 cm.

Quantity I: Area of Square
Quantity II:Area of Rectangle

Solution:

Perimeter of rectangle = Perimeter of Square = 160
4a = 160 ⇒ a = 40
Area of square = 1600
1600 – lb = 600
lb = 1000 cm²

QUESTION: 10

Shree started traveling from a place A to B and Priya started traveling from a place B to A which are 576 km apart. They meet after 12 hours. After their meeting, Shree increased her speed by 2 km/hr and Priya reduced her speed by 2 km/hr, they arrived at B and A respectively at the same time.

Quantity I: Initial Speed of Shree
Quantity II: Initial Speed of Priya

Solution:

Sum of their speeds = Distance/time = 576/12 = 48 kmph
Respective Speed of Shree and Priya = (23 + 25) = 48 kmph

QUESTION: 11

x² – 11x + 28 = 0

y² – 14y + 48 = 0

Solution:

x² – 11x + 28 = 0
x = 7, 4
y² – 14y + 48 = 0
y = 6, 8

QUESTION: 12

x² – 31x + 228 = 0

y² – 21y + 108 = 0

Solution:

x² – 31x + 228 = 0
x = 12, 19
y² – 21y + 108 = 0
y = 12, 9

QUESTION: 13

x² + 31x + 234 = 0

y² + 21y + 104 = 0

Solution:

x² + 31x + 234 = 0
x = -13, -18
y² + 21y + 104 = 0
y = -13, -8

QUESTION: 14

x² – 20x + 84 = 0

y² – 24y + 135 = 0

Solution:

x² – 20x + 84 = 0
x = 14, 6
y² – 24y + 135 = 0
y = 15, 9

QUESTION: 15

(x – 14)² = 0

y² = 196

Solution:

x² – 28x + 196 = 0
x = 14, 14
y² = 196
y = ±14

QUESTION: 16

x² – 43x + 450 = 0

y² – 32y + 255 = 0

Solution:

x² – 43x + 450 = 0
x = 25, 18
y² – 32y + 255 = 0
y = 17, 15

QUESTION: 17

x² – 27x + 182 = 0

y² – 36y + 323 = 0

Solution:

x² – 27x + 182 = 0
x = 14, 13
y² – 36y + 323 = 0
y = 17, 19

QUESTION: 18

x² – 37x + 322 = 0

y² – 22y + 120 = 0

Solution:

x² – 37x + 322 = 0
x = 23, 14
y² – 22y + 120 = 0
y = 10, 12

QUESTION: 19

x² = 81

y² – 30y + 225 = 0

Solution:

x² = 81
x = 9, – 9
y² – 30y + 225 = 0
y = 15, 15

QUESTION: 20

x² – 30x + 221 = 0

y² – 33y + 270 = 0

Solution:

x² – 30x + 221 = 0
x = 13, 17
y² – 33y + 270 = 0
y = 15, 18

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