Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 22

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Question 1
3 / -1

In the above figure, PT and RS are 2 parallel lines which are tangents to the circle with centre O. PQ is also a tangent to the circle ∠ PSO = 30^º and PT = 6 units. What is the value of the area of PTOQ?

A
18√2 sq units

B
18√3 sq units

C
36 sq units

D
6√2 sq units

Solution
Question 2
3 / -1

How many positive integral pairs (x, y) exists such that 4 < ∣x − 5∣ × ∣y − 7∣ < 7

A
18

B
20

C
22

D
24

Solution

Since it is given that xx and yy are integers then |x - 5|×|y - 7| also needs to be an integer.

Thus |x − 5| × |y − 7| = 5 or 6

Case I: |x − 5| × |y − 7| = 5

a) |x - 5| = 5 and |y - 7| = 1 It gives x = 0 or 10 and y = 6 or 8. (10, 6) and (10, 8) are only solution as we are given that x, yx, y are positive

b) |x - 5| = 1 and |y - 7| = 5. It gives x = 6 or 4 and y = 12 or 2. Total of 4 such pairs are possible

Total 6 such pairs possible for case I

Case II: |x − 5| × |y − 7| = 6

a) |x - 5| = 1 and |y - 7| = 6. It gives x = 4 or 6 and y = 1 or 13 Thus 4 pairs are possible

b) |x - 5| = 2 and |y - 7| = 3. It gives x = 3 or 7 and y = 4 or 10. 4 pairs are possible

c) |x - 5| = 3 and |y - 7| = 2. It gives x = 2 or 8 and y = 5 or 9. 4 pairs are possible

d) |x - 5| = 6 and |y - 7| = 1. It gives x = -1 or 11 and y = 6 or 8. Here 2 pairs are possible

Total of 4 + 4 + 4 + 2 = 14 such pairs are possible

Overall 6 + 14 = 20 cases are possible.
Question 3
3 / -1

How many non-zero integer values of x satisfy the equation given below such that x <= 10.

A
8

B
9

C
10

D
11

Solution
Question 4
3 / -1

Cost of 1 pen, 3 pencils and 5 markers is Rs 100. Cost of 18 pencils, 2markers and 6 pens is Rs 320. What is the minimum possible cost of 1 marker, 1 pen and 1 pencils if it is given that the cost of each of the items is a natural number?

A
40

B
32

C
58

D
28

Solution

Let the cost of pen be xx, a pencil be y and a markers be z

x + 3y + 5z = 100.......(I)

6x + 18y + 2z = 320....(II)

6 × (I) − (II) gives us,

28z = 280 or z = 10

putting z = 10 in (I) we get x + 3y = 50

We have to get the minimum value of x + y + z. z = 10 thus we need to minimize x + y

when y = 16 then x = 2 and x + y + z = 2 + 16 + 10 = 28

When y = 15 then x = 5 and x + y + z = 5 + 15 + 10 = 30

We see that as we keep on decreasing y the sum will increase.

Minimum possible value = 28
Question 5
3 / -1

If t satisfies the following equation:

log₃ 78 + t − 6 = log₃ (1 − 3^t−5)

What is the value of |t|

A
2

B
3

C
4

D
5

Solution
Question 6
3 / -1

An entrepreneur borrows an amount of Rs 20 lakhs from a local bank for a duration of 3 years at a 20% interest rate, compounded annually. At the end of each of these 3 years, he returns a certain amount of money to the bank. The amount returned at the end of each of these 3 years is the same. At the end of 3 years, the bank additionally demanded a cheque of Rs. 1.8 lakhs to settle the loan. What was the amount he used to pay at the end of each year?

A
900000

B
800000

C
700000

D
600000

Solution

According to the question

2000000(1.2)³ = x(1.2)² + x(1.2) + x + 180000

3456000 − 180000 = x(1.44 + 1.2 + 1)

3276000 = x(1.44 + 1.2 + 1)

3276000 = x × 3.6

3276000/3.64 = x

x = 900000.
Question 7
3 / -1

A milk merchant buys milk at Rs 50 per litre, but while selling, he marks up the price by 20% and also mixes water to the milk such that the milk-water mixture he sells has 80% milk and 20% water. Assuming that the water has zero cost price, what discount percentage should he give to the mixture such that by selling it, he earns a 10% profit?

A
20

B
80/3

C
40/3

D
30

Solution

CP of 1 litre = 50

MP of 1 litre = 1.2 x 50 = 60

Also, he mixes water to milk in the ratio of 1:4, so in 1 litre of milk, he mixes 250 mL of water.

Hence, the volume of liquid he is selling = 1.25

MP of 1.25 litres = 60 x 1.25 = 75 Rs.

So, to have a 10% profit, SP = Rs 50 x 1.1 = 55 Rs.

Hence, discount percent = 20/75 x 100 = 80/3%
Question 8
3 / -1

Ram was asked to calculate the sum of an arithmetic progression whose first term was 2 and common difference was 7. The total number of terms was 30. He wrote down all the 30 terms in ascending order and then started adding from smallest number. After calculating the sum of first x terms correctly, Ram started taking common difference as 14 instead of 7 and ended up with sum of 3210. How many factors does x − 4 has?

A
4

B
5

C
6

D
7

Solution

Let us calculate the actual sum which Ram would have got if he did not have had made the mistake.

We can use the formula to calculate the sum = 30/2 (2(2) + (30 − 1)7)

Thus the actual sum = 15(4 + 29 × 7) = 3105

Excess sum calculated = 3210 - 3105 = 105

Let last n terms be the terms where the common difference was taken as 14 instead of 7

Thus 7 + 14 + ...n terms = 105

n/2(2(7) + (n − 1)7) = 105

n(n + 1)7 = 210

Or 2 + n − 30 = 0 or (n + 6)(n − 5) = 0.

Since n can not be negative, n = 5, Thus last 5 terms had incorrect common differece. x = 30 - n = 30 - 5 = 25

x − 4 = 21 = 3 × 7

Total 4 factors are there.
Question 9
3 / -1

What is the number of common terms between the following series

A₁: 4,10,16,22,28,..........,610

A₂ : 2,9,16,23,............... 632

A
13

B
14

C
15

D
16

Solution

Both of the series are an Arithmetic Progression

Common difference of the Series are 6 and 7 respectively.

Upon looking, we can see that the first common term is 16.

The series which will have common terms will have first term as 16 and common difference = lcm(6, 7) = 42

Let a_n be the n^thand final term of the series

a_n= 16 + (n − 1)42 ≤ 610

42(n − 1) = 594

n − 1 = 14.14 or n = 15.14. Thus their are total 15 number of common terms.
Question 10
3 / -1

X = Highest power of 144 in 25!

Y = Highest power of 96 in 50!

How many of the following sentences are correct about the value Y - X?

1. It is a perfect square

2. It is a perfect cube

3. It is prime

4. It is even

A
2

B
4

C
6

D
8

Solution