Mathematical Ability Test-4

Result

Mathematical Ability Test-4

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Weekly Quiz Competition

Question 1
1 / -0

To dispose of the old stocks, a person sold a tea-set for Rs 3,540, which was 41% below the cost price. To make a profit of 11%, the seller should have sold the set for ____ more.

A
2664.42

B
2460

C
1812.60

D
3120

Solution
Question 2
1 / -0

In the given figure ∠QPR = 60°. If OQ and OR are the angle bisectors of ∠Q and ∠R, then find the value of ∠QOR.

A
120°

B
150°

C
30°

D
75°

Solution

Given:

∠QPR = 60°

OQ and OR are the bisector of ∠Q and ∠R.

According to the formula used,
Question 3
1 / -0

The polar coordinates of the vertices of a triangle are Then, the triangle is:

A
right angled triangle

B
isosceles triangle

C
equilateral triangle

D
None of these

Solution
Question 4
1 / -0

A
π/4

B
π/2

C
π

D
0

Solution
Question 5
1 / -0

What is the value of tan 75° + cot 75°?

A
2

B
4

C
2√3

D
4√3

Solution
Question 6
1 / -0

In formula Mean deviation = MD = (1/n) ∑|x − M| what does M indicates:

A
Mean

B
Median

C
Mode

D
All of the above

Solution

The mean deviation is the sum of the deviations from the average divided by the number of items in the series.
Question 7
1 / -0

A man purchased an old typewriter at Rs 19200, and spent some amount on its repair. He sold it for Rs 24150, and earned a profit of 15%. Find the amount he spend on the repair of the old typewriter.

A
Rs 1200

B
Rs 1800

C
Rs 2100

D
Rs 1500

Solution

⇒ 19200 + x = 21000

⇒ x = 21000 – 19200

⇒ x = 1800

∴ The amount spend on repairing of old typewriter is Rs 1800.
Question 8
1 / -0

Find the equation of ellipse whose eccentricity is 1/2, length of latus rectum is 4 and the center is (0,0).

A
9x²+ 12y²= 64

B
9x²+ 12y²= 16

C
12x²+ 9y²= 64

D
12x²+ 9y²= 16

Solution
Question 9
1 / -0

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A
20 cm

B
30 cm

C
50 cm

D
60 cm

Solution

Given

length of a rectangle is decreased by 4 cm and the width is increased by 3 cm

Formula used:

Area of rectangle = (l × b) sq. unit and Area of square = (side)² sq. unit

Perimeter of rectangle = 2(l + b) units

Let the length and breadth of the rectangle be x and y

⇒ Area of rectangle = (l × b) sq. unit

⇒ xy cm²

⇒ new length and breadth = (x - 4) and (y + 3)

⇒ (x - 4) = (y + 3)

⇒ x - y = 7 .....(1)

⇒ Area of square = (x - 4) × (y + 3)

According to question

⇒ Area of rectangle = Area of square

⇒ xy = (x - 4) × (y + 3)

⇒ 3x - 4y = 12 .....(2)

⇒ On solving these equation

⇒ x = 16 and y = 9

⇒ Perimeter of rectangle = 2(l + b) units

⇒ 2(16 + 9)

⇒ 2 × 25

⇒ 50 cm

∴ perimeter of the original rectangle is 50 cm
Question 10
1 / -0

The HCF of two numbers a and b is c. What is the LCM of those two numbers?

A
ab/c

B
ac/b

C
b/ca

D
abc

Solution