Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 37

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Question 1
3 / -1

'A', 'B' and 'C' are three friends whose incomes are in ascending arithmetic progression. Incomes of 'A', 'B' and 'C' is Rs. (x² - 1600), Rs. (50x + 19000) and Rs. (x² + 4400), respectively. Out of the given persons whose income is a perfect cube, spends 60% of his income and saves the rest such that his savings is equal to the average of savings of other two persons. The person, whose income is (20/9) times the savings of person whose income is a perfect cube, spends Rs. '90x'.

The sum of the income and expenditure of the person who spends the least is how much percent more/less than savings of the person who saves the most?

A
240%

B
220%

C
160%

D
200%

Solution
Question 2
3 / -1

There are five students in a class - P, Q, R, S, and T. Marks obtained by P is (3M + 2N), by Q is (N + 2M), by R is (2N + M). marks obtained by S is 100 (more than rest 4 students) which is twice as that of T. Marks obtained by T is (N + 2Y). M > N, both the distinct multiple of 10.

Find the average score of five students together if difference between marks of Q and R is at least 10.

I. 2(3M - 2N) - 4²

II. 3Y + 4

III. 3(4N - Y) + 4²

A
II only

B
I and III only

C
I and II only

D
I, II and III

Solution

Difference between marks of Q and R ≥ 10

N + 2M - 2N - M ≥ 10

M - N ≥ 10

M and N are both district multiple of 10. So, minimum possible value of M and N is 20 and 10. If value of M and N is 30 and 10.

Marks of P = 3 x 30 + 2 x 10 = 110 > 100. This is not possible because marks of S are maximum 100.

Marks of T = 1/2 x 100 = 50

10 + 2Y = 50

Value of Y = 20

Required average = (3M + 2N + N + 2M + 2N + M + 100 + 500)/5 = 320/5 = 64

I. 2(3M - 2N) - 4²

Required value = 2 x (3 x 20 - 2 x 10) - 16 = 64

This statement is follows

II. 3Y + 4

Required value = 3 x 20 + 4 = 64

This statement is follows

III. 3(4N - Y) + 4²

3 x (4 x 10 - 20) + 16 = 76

This statement doesn't follow.

So, Only I and II follows

Hence, option 3 is correct.
Question 3
3 / -1

Pradip bought some candles and matches. Sandip spent the same amount as Pradip and bought half as many candles and 30 more matches than Pradip. For the given two persons, the cost of one match is Rs. 10 less than that of one candle and they bought at least one item of both types.

If the cost price of each candle is Rs. 30, then find the number of matches bought by Sandip.

A
25

B
45

C
20

D
Can't be determined

Solution

Let the number of candles bought by Pradip be '2x'

So, number of candles bought by Sandip = 2x ÷ 2 = 'x'

Let number of matches bought by Pradip = 'y'

So, number of matches bought by Sandip = (30 + y)

Let the cost price of each match be Rs. 'z'

So, cost price of each candle = Rs. '(z + 10)'

According to the question,

(y X z) + 2x X (z + 10) = z X (30 + y) + x X (z + 10)

Or, yz + 2xz + 20x = 30z + zy + xz + 10x

Or, xz + 10x = 30z

Or, x(z + 10) = 30z

Or, x = {30z/(z + 10)}

According to the question,

10 + z = 30

So, z = 20

So, x = {(30 X 20)/(10 + 20)} = 20

So, total amount spent by Pradip = y X 20 + (10 + 20) X (2 X 20) = Rs. (20y + 1200)

And total amount spent by Sandip = (30 + y) X 20 + (20 + 10) X 20 = 600 + 20y + 600 = Rs. (1200 + 20y)

So, 20y + 1200 = 1200 + 20y

Hence, option 4 is correct.
Question 4
3 / -1

There are four earners in a family. The highest earner makes Rs. 5 lakh per month, whereas the lowest earner makes Rs. 2.4 lakh per month. Which of the following statement(s) is/are true?

I. The average monthly income of the family can be Rs. 3.3 lakh.

II. The average monthly income of the family can be Rs. 3 lakhs.

III. The difference between maximum and minimum possible average monthly income of the family is Rs. 1.30 lakhs.

A
Only I and II

B
Only II and III

C
Only I and III

D
All I, II and III

Solution

Maximum possible average income of the family will be achieved when three people earn the maximum possible income i.e., Rs. 5 lakh per month.

So, maximum possible average monthly income of the family = (5 + 5 + 5 + 2.4) ÷ 4 = Rs. 4.35 lakh

Minimum possible average income of the family will be achieved when three people earn minimum possible income i.e., Rs. 2.4 lakh per month.

So, minimum possible average monthly income of the family = (2.4 + 2.4 + 2.4 + 5) ÷ 4 = Rs. 3.05 lakh

Statement I:

Since, minimum possible average monthly income of the family = (2.4 + 2.4 + 2.4 + 5) ÷ 4 = Rs. 3.05 lakh

So, statement I is true.

Statement II:

Since, minimum possible average monthly income of the family = (2.4 + 2.4 + 2.4 + 5) ÷ 4 = Rs. 3.05 lakh

So, statement II is false.

For statement III:

Required difference = 4.35 - 3.05 = Rs. 1.30 lakh

Statement III:

Since, difference between maximum and minimum monthly income of the family = 1.35 - 3.05 = Rs. 1.30 lakh

So, statement III is true.

Hence, option 3 is correct.
Question 5
3 / -1

Two cyclists start from the same place to ride in the same direction. Ravi starts at noon with a speed of 20 km/hr and Hari starts at 1:30 pm with a speed of 25 km/hr. At what times, will Ravi and Hari be exactly 7 kms apart?

A
5:23 pm and 8:22 pm

B
6:06 pm and 8:54 pm

C
7:12 pm and 10:28 pm

D
None of these

Solution

The distance between Ravi and Hari when Hari starts = (20)*1.5 = 30 km

Relative speed of Ravi and Hari = 25 - 20 = 5 km/hr

Since Ravi is already ahead then the distance between Ravi and Hari will be exactly 7 kms once when Hari is behind Ravi and once when Hari is leading Ravi by 7 kms.

At 1:30 pm, distance between them = 30 km

Case 1: When Hari is behind Ravi by 7 kms

Time taken by Hari to cover 23 km at a relative speed of 5 km/hr = 23/5 = 4.6 hours or 4 hours and 36 minutes

Hence the time when Hari is behind Ravi by 7 kms = 06:06 pm

Case 2: When Hari is leading Ravi by 7 kms

Both Hari and Ravi will meet after = 30/5 = 6 hours. Hence we can say that they will meet each other at 7:30 pm

After 07:30 pm, Hari will lead Ravi with a relative speed of 5 km/ hr

Hence time taken by Hari to cover 7 km at a relative speed of 5 km/hr = 7/5 = 1.4 hours or 1 hours and 24 minutes

Hence the time when Hari is leading Ravi by 7 kms = 08:54 pm

Hence, option 2 is correct.
Question 6
3 / -1

Harshit mixed 'P' litres water in 120 litres alcohol and sold the entire mixture to Punit. Punit further diluted the mixture by adding 20 litres water. If the percentage of quantity of alcohol in the resultant mixture is not less than that of water, then which of the following statement(s) can/could be true? [Minimum value of 'P' is 65]

I. Maximum value of 'P' is 120.

II. If Punit did not dilute the mixture further, then quantity of water in the mixture would be half that of alcohol.

III. If Punit further dilutes the mixture by adding 40 more litres of water, then the quantity of water and alcohol in the mixture would become equal.

A
Only II

B
Only I

C
Only I and III

D
Only I and II

Solution

For statement I:

Maximum value of 'P' will be achieved when quantity of both water and alcohol will be equal.

So, {(P + 20)/120} = (1/1)

Or, P + 20 = 120

So, P = 100

So, maximum value of 'P' = 100

So, statement I is false.

For statement II:

(P/120) = (1/1)

So, P = 120 > 65

So, statement II might be true, depending on the value of 'P'.

So, statement II could be true.

For statement III:

{(P + 20 + 40)/120} = (1/1)

Or, P + 60 = 120

So, P = 60 < 65

Since, minimum value of 'P' is 65, statement III is false.

Hence, option 1 is correct.
Question 7
3 / -1

Two real numbers are randomly selected between 0 and 100. Find the probability that the difference between the two numbers is less than or equal to 30.

A
43/100

B
51/100

C
17/100

D
None of these

Solution

The probability is given by the following graph. The required probability is the area of the shaded region divided by the area of the square.

Area of unshaded region = 2*(1/2)*70*70 = 4900

Area of square = 10000

Required probability = (10000 - 4900)/10000 = 5100/10000 = 51/100

Hence, option 2 is correct.
Question 8
3 / -1

In the figure given below, a rectangle EFGH is inscribed in a circle. ADBC and HIJK are rectangles such that points A, I, J and B lie on the circumference of the circle. If AD = EF = HI = 2 cm and ED = 1 cm, then find the diameter (in cm) of the circle.

A
√85

B
√35

C
4√5

D
2√45

Solution

Draw the lines DH, BI and FH. It is obvious that BI and FH are the diameters of the circle. Assume that the length of DC and the diameter of the circle are b cm and d cm respectively.
Question 9
3 / -1

A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of side of the equilateral triangle to the radius of the circle.

A
2/√3 - 2

B
4/√3 + 2

C
1/√3 + 4

D
None of these

Solution

Let the length of the side of the triangle ABC be 2x and the side of the square DEFG inscribed be 2a

Triangle ADG is an equilateral triangle as DG is parallel to BC and angle A = 60°.
Question 10
3 / -1

In a quiz competition, six teams with two members, one man and one woman each, participated. During dinner, all the 12 members were seated around a round table. Find the probability that no man sat to the left of the woman from his team, provided that men and women were alternately seated.

A
47/121

B
53/144

C
51/104

D
35/178

Solution

Let us first place the women around the circular table

The number of ways in which the women can sit is is 5!

Number of ways in which men can sit under no constraint = 6!

Total number of ways possible = 5!*6!

Number of ways in which the men can sit such that no man sat to the left of the woman from his team = dearrangement of 6 = 265

Total number of ways such that no man sat to the left of the woman from his team = 5!*265