Quantitative Aptitude (QA) Test

Result

Quantitative Aptitude (QA) Test - 20

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Question 1
3 / -1

A wire which was used to form a square of side 30 cm is divided into 3 parts such that the length of the longest part is 25% more than length of the second longest part. If the shortest part is 20 cm shorter than the longest part, then the area (in cm²) of the triangle formed by these three parts is

A
540

B
480

C
800

D
600

Solution

The length of the wire = 4*30 = 120 cm.

Let 'x' be the length of the smallest part of the wire.

Length of the longest part = x + 20

Length of the second longest part*1.25 = Length of the longest part

⇒ Length of the second longest part = 0.8*Length of the longest part

⇒ x + 20 + 0.8*(x+20) + x = 120

⇒ x + 20 + 0.8x + 16 + x = 120

2.8x + 36 = 120

2.8x = 84

⇒ x = 30 cm

The sides of the triangle are 30 cm, 40 cm, and 50 cm.

We can see that the sides of the triangle form a Pythagorean triplet.

Therefore, the area of the triangle is 0.5*30*40 = 600 cm²

Therefore, option D is the right answer.
Question 2
3 / -1

When the number of sides of a polygon increases from 'n' to 'n+1', A is the change in the sum of exterior angles and B is the change in the sum of interior angles. Which of the following statements is true?

A
A = 180, B = 180

B
A = 0, B = 360

C
A = 180n, B = 360

D
A = 0, B = 180

Solution

The sum of the exterior angles of any polygon is 360 degrees. So, there will be no change in the sum of the exterior angles. The sum of the interior angles of an 'n' sided polygon is (n-2)*180. Hence, if the number of sides is increased by 1, the sum of the interior angles increases by 180.

Therefore, A = 0 and B = 180.
Question 3
3 / -1

For any point, in the first quadrant, on the line 4x + 9y = 36, what is the maximum value of x⁶ * y⁶ ?

A
3⁹

B
3¹²

C
3¹⁰

D
3¹¹

Solution
Question 4
3 / -1

Find the area of the region enclosed between the lines y=2|x|+3 and y=-2|x|+8.

A
23/4

B
30/7

C
25/4

D
32/7

Solution
Question 5
3 / -1

If it is known that xx and yy are two positive numbers such that what is the minimum value of x + y ?

A
6

B
24

C
18

D
54

Solution
Question 6
3 / -1

If log m=300\times×log1.0033333..

What is the approximate value of m?

A
2.5

B
2.6

C
2.7

D
3

Solution
Question 7
3 / -1

A
95

B
96

C
97

D
100

Solution
Question 8
3 / -1

N is a multiple of 72. Also, it is known that all the digits of N are different. What is the largest possible value of N?

A
9876543120

B
9876543125

C
9876543150

D
9876543200

Solution

We know that the number is a multiple of 72. Therefore, the number must be a multiple of both 8 and 9.

If the number contains all the digits from 0 to 9, the sum of the digits will be 9*10/2 = 45.

Therefore, a number containing all the ten digits will be divisible by 9.

For a number to be divisible by 8, the last 3 digits should be divisible by 8.

Now, we have to find the largest possible number. Therefore, the left-most digit must be 9, the second digit from the left should be 8 and so on.

We'll get 9876543210 as the number. However, the last 3 digits are not divisible by 8. We must try to make the last 3 digits divisible by 8 without altering the position of other numbers. '120' is divisible by 8.

Therefore, the largest number with different digits divisible by 72 is 9876543120.
Question 9
3 / -1

The number of ways in which 33 identical pens can be distributed among three boys such that each of them receives an odd number of pens is

A
136

B
156

C
150

D
105

Solution

Let 2a+1, 2b+1, 2c+1 be the number of pens received by each of the three boys.

2a+1+2b+1+2c+1=33 [0 ≤ a, b, c ≤ 15]

2a+2b+2c=30

a+b+c=15

Required number of ways =^{15 + 3 - 1}C_{3 - 1}

= ¹⁷C₂

= 136

A is the correct answer.
Question 10
3 / -1

A shopkeeper has 5 dozens of apples. On selling 12 apples for Rs.600, he incurs a loss equal to the cost price of 2 apples. At what total price should he mark the remaining 4 dozens so that even after giving 20% discount, he would have an overall gain of 10%?

A
4200

B
4300

C
4500

D
5000

Solution

CP = SP + loss

So, CP of 12 apples = SP of 12 apples+ CP of 2 apples

⇒ CP of 10 apples = SP of 12 apples

SP of 12 apples = Rs.600

⇒ CP of 10 apples = Rs.600

CP of 1 apple = Rs. 60

Total CP = Rs. (5 * 12 * 60) = Rs. 3600

Required overall gain = 10%

Required total SP = Rs. 3600 + 10% of Rs. 3600 = Rs. 3960

Remaining apples = 4 dozens

Required SP for 4 dozens = Rs. (3960 - 600) = Rs.3360

According to the question,

MP - 20% of MP = Rs. 3360

Or, MP = Rs. 4200

Hence, 4200 is the correct answer.