Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 31

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Question 1
3 / -1

The equation x² −10x + 2[x] + 16 = x[x] where [x] denotes the greatest integer function. Choose the statement which best reflects the number of solutions x can have.

A
Exactly two solutions in the range (−∞,∞)

B
No solutions

C
Exactly three solutions in the range [0,∞)

D
A unique solution

Solution

x²− 10x + 2[x] + 16 = x[x]

We can rewrite it as x²− 10x + 16 = x[x] − 2[x]

(x−2)(x−8) = [x](x−2)

(x−2)(x−8−[x]) =0

There can be two cases: Case 1-> x - 2 =0

When x - 2 = 0, x becomes +2

Other case, Case 2 is x - 2 is not equal to 0, then,

x − 8 − [x] = 0

x − [x] = 8

We know that x − [x] = {x}

We know that 0 ≤ {x} ≤ 1

Hence, this case becomes invalid.

We get only one value of x, one unique solution.
Question 2
3 / -1

A

B

C

D

Solution
Question 3
3 / -1

Two mutually perpendicular chords AB and CD meet at point P inside the circle such that AP = 12cm, PB = 8 units, and DP = 6 units. Calculate the perimeter of the circle.

A

B

C

D

Solution

As AB and CD are two chords that intersect each other,

AP × PB = CP × PD => 12 × 8 = 6 × CP

CP = 16

From O, we can draw OM, which is perpendicular to AB, and ON, which is perpendicular to CD.

From the centre, a line perpendicular to a chord bisects the chord.

We have, AM = MB = 10cm

MP = 2cm, ON = 2cm, CD = 22cm, CN = 11cm.

ON²+ CN²= OC²

4 + 121 = r² (where r is the radius of the circle)

r²= 125
Question 4
3 / -1

A work is done by hiring some men and some boys. The work requires constructing a wall or breaking a wall. Each of these men or boys can either construct the wall or break the wall. During working, the speed of each man or boy to construct the wall is 10 times the speed of each man or boy respectively to break the wall. It is known that the speed of each man is the same and the speed of each boy is also the same and not equal to that of each man. One man & 7 boys can break the wall in 6 days while 1 man & 3 boys can build the wall in 1 day. In how many days will 1 man and 1 boy together break a wall?

A
10 Days

B
12 Days

C
20 Days

D
15 Days

Solution

Let the time taken by a man to build & to break the wall be x days & 10x days respectively.

Let the time taken by a boy to build & to break the wall y days & 10y days respectively.

Let total bricks in wall be = 10xy

This means that a man can build & break 10x bricks & x bricks respectively in a day and a boy can build & break 10y bricks & y bricks respectively in a day.

Since, 1 man & 7 boys can break the wall in 6 days

i.e. 6(y + 7x) = 10xy ---- (i)

Also, 1 man & 3 boys can build the wall in 1 day

i.e. 1(10y + 30x) = 10xy ---- (ii)

From (i) and (ii), y = 3x --- (iii)

From (i) & (iii), y = 6 & x = 2

Now, time taken by 1 man & 1 boy to break the wall
Question 5
3 / -1

Abhi wrote down several distinct positive integers, not exceeding 60. Their product was not divisible by 36. At most how many numbers could he have written?

A
40

B
30

C
41

D
34

Solution

There are 20 multiples of 3 from 1 to 60.

If we take only one multiple of 3 out of the 20 multiples, then the product will not be divisible by 36.

Hence, he could have written at most 60 - 19 = 41 numbers.
Question 6
3 / -1

The average of 8 numbers is 75. If the smallest number is deleted, then the average of the remaining numbers is A and if the largest number is deleted, then the average of the remaining numbers is B. If A + B = 90, then what is the average of the smallest and largest numbers?

A
275

B
355

C
315

D
285

Solution
Question 7
3 / -1

In a race, cyclist X completes the race 20 meters ahead of cyclist Y, cyclist Y completes 15 meters ahead of cyclist Z, and cyclist X completes 33 meters ahead of cyclist Z. What is the total length of the race in meters?

A
150

B
220

C
70

D
80

Solution

Let the distance be L meters

When cyclist X finishes the race, cyclist Y has covered (L - 20) meters.

When cyclist Y finishes the race, cyclist Z has covered (L - 15) meters.

When cyclist X finishes the race, cyclist Z has covered (L - 15) meters.
Question 8
3 / -1

A was born two years after B, who is four years old. At present, the average age (in years) of A, B, C and D is a perfect square. Also the difference between the ages of any two persons is not more than five years. Which of the following could be the ages of C and D (in years)?

A
8, 2

B
7, 3

C
10, 9

D
6, 3

Solution

A’s and B’s present ages are 2 and 4 years respectively.

The difference between the ages of any two persons is not more than five years.

Thus 1 cannot be considered.

Let it be 4.

A + B + C + D = 16

C + D = 10

Again it is given that the age difference of any two persons is not more than five years. Thus the ages of C and D could be 7 and 3 years, satisfying all the conditions.3

When it be 9

A + B + C + D = 36

C + D = 30

But the age difference of any two persons now becomes more than five years. Hence this case is discarded.
Question 9
3 / -1

The series A0, A1, A2, A3,......A0, A1, A2, A3,...... is a strictly increasing arithmetic progression of positive integers such that

A
32

B
28

C
24

D
19

Solution
Question 10
3 / -1

A cement trader bought cement for Rs.15 lakhs from a cement factory. Each day, he transports 12 lorry loads of cement from the factory to his godown at Rs.2,500 per lorry load. From the 11th day onwards, the factory charged him Rs.2,000 per day for using their godown. It took the trader 20 days to transport all the cement to his site. If the trader sold the cement to a retailer for Rs.26.5 lakhs, then what was his profit percentage?

A
20%

B
25%

C
27.5%

D
12.5%

Solution

It is said that the godowns were cleared in 20 days.

He was charged from the 11th day at Rs.2,000 per day.

Total charge = Rs.20,000

Total cost price of the dealer = 15,00,000 + 12 × 2500 × 20 + 20000

= 1500000 + 600000 + 20000 = Rs.21,20,000

Total selling price = Rs.26,50,000