Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 39

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Question 1
3 / -1

A tank had n litres of water. A person draws out n% of water from it. A few hours later, he draws out n% of the remaining water in the tank. Later, another person draws out (9n²/1000) litres of water from the tank. Which of the following can be the possible amount of water that was in the tank initially?

A
50 litres

B
15 litres

C
100 litres

D
200 litres

Solution

Amount of water that was in the tank initially = n litres.

After first instance, amount of water left = (100 - n)/100 % of n = n(100 - n)/100

After second instance, amount of water left = (100 - n)/100 % of n(100 - n)/100 = n((100 - n)/100)²

After third instance, amount of water in the tank = n((100 - n)/100)²- (9n²/1000)

= n(10000 - 200n + n² - 90n)/10000

= n(10000 + n² - 290n)/10000

= n((n - 250)(n - 40))/10000

Since this is amount of water left in the tank, so it can either be 0 or greater than 0.

Now, the above expression will be greater than 0 when 0 < n < 40 and n > 250 and it will be equal to 0 when n = 40 or 250. So, n cannot be any value from 40 to 250.

Hence, n can be 15.
Question 2
3 / -1

A vendor sells a shirt at p% discount to Anuj. He sells the same shirt to Ankit at q% discount. He marks up the price of the third shirt by pq% but sells it at 10% discount to Animesh. If he suffers an overall loss of 0.333%, then what cannot be the possible value of p?

A
1

B
8

C
12

D
5

Solution

Let the cost of the shirt be 100.

So, S.P. for Anuj = ((100 - p)/100) * 100 = 100 - p

S.P. for Ankit = ((100 - q)/100) * 100 = 100 - q

S.P. for Animesh = 0.9 * ((100 + pq)/100) * 100 = 90 + 0.9pq

Total C.P. = 300

And, total S.P. = 100 - p + 100 - q + 90 + 0.9pq = 290 + 0.9pq - p - q

Loss percentage = (300 - (290 + 0.9pq - p - q))/300 * 100% = (10 - 0.9pq + p + q)/3 %

(10 - 0.9pq + p + q)/3 = 0.333

(10 - 0.9pq + p + q)/3 = 1/3

10 - 0.9pq + p + q = 1

0.9pq - p - q = 9

0.9pq - (p + q) = 9 ....(i)

Option (a): When p=1

0.9(1)q - (1 + q) = 9

q= - 100 (invalid as q cannot be -ve)

So value of p ≠ 1

Option (b): when p=8

0.9(8)q - (8 + q) = 9

q=2.74

So, value of p can be 8

Option (c): when p=12

0.9(12)q - (12 + q) = 9

q=2.14

so, p can be 12

option (d): When p=5

0.9(5)q - (5 + q) = 9

q=4

so, value of p can be 5.
Question 3
3 / -1

A child consumed an ice cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. If the height of the ice cream-cone was 8 cm, what was the height of the remaining ice cream-cone?

A
2.5 cm

B
3.0 cm

C
3.5 cm

D
4.0 cm

Solution
Question 4
3 / -1

A
10 < S < 11

B
9 < S < 10

C
11 < S < 12

D
12 < S < 13

Solution

Split the given series into two parts:

S = S₁ + S₂ such that,
Question 5
3 / -1

A saint has a magic pot. He puts one gold ball of radius 1 mm daily inside it for 10 days. If the weight of the first ball is 1 gm and if the radius of a ball inside the pot doubles every day, how much gold has the saint made using his magic pot?

A
(2³⁰- 69)/7 gm

B
(2³⁰+ 69)/7 gm

C
(2³⁰- 71)/7 gm

D
(2³⁰+ 71)/7 gm

Solution
Question 6
3 / -1

If x and y are natural numbers, which satisfy the relation, log₂(x-7) =2log₄(5-y), then find the maximum value of log₄ x - log

A
-0.5

B
1.5

C
log₂(√3 -1)

D
- 1 log₂3

Solution

log₄ x = ½ (log₂ x)

log₂ (8/√y) = log₂ 8 - log₂√y

= 3 - ½log₂ y

So, the given equation becomes: ½ (log₂ x + log₂ y) - 3

= ½ (log₂xy) - 3

Now this expression will be maximum when we will have the maximum value of xy.

Now it is given that, log₂(x-7) = 2log₄ (5-y)

⇒ log₂(x-7) = log₂ (5-y)

⇒ x-7=5-y

⇒ x+y=12

In order to define both the terms, (x - 7) > 0 and (5 - y) > 0

⇒ x > 7 and y <5

So xy is maximum when x = 8 and y = 4,

The maximum value of xy = 8 X 4 = 32

So the maximum value of the given expression = ½ [log₂ 32] - 3 = 5/2 - 3 = -0.5
Question 7
3 / -1

If p is the root of equation mx² + nx + q = 0, q is the root of the equation mx² + nx + p = 0 and p ≠ q, then find the equation whose roots are p and q.

A
m²x²+ m(n + 1)x + 1 + n = 0

B
m²x²+ m(n - 1)x + 1 - n = 0

C
m²x²+ m(n + 1)x + 10 - n = 0

D
m²x²+ m(n - 1)x + 12 + n = 0

Solution

Since p and q are the roots of equations so they will satisfy them also.

Hence, p²m + np + q = 0 ............(1)

And, q²m + nq + p = 0 .........(2)

Equation (2) - Equation (1)

⇒ m (p² - q²) + n (p - q) + (q - p) = 0

⇒ (p - q) {m (p + q) + n -1} = 0

Since in the question it is given that p ≠ q

So, {m (p + q) + n -1} = 0

or p+q = (1-n)/m

Now dividing Eq (1) by p and Eq (2) by q we will get,

pm + n + (q/p) = 0 ............(3)

qm + n + (p/q) = 0 ............(4)

Equating n from both the equations, we will get

-pm - q/p = -qm -p/q

pm - qm = p/q - q/p

m(p-q) = (p+q)(p-q) / pq

pq = (p+q)/m = (1-n)/m²

Now the equation with roots p and q can be written as, x² - (p + q) x + pq = 0

x² - [(1-n)/m]x + (1-n)/m² = 0

⇒m²x² + m(n -1)x + 1 - n = 0
Question 8
3 / -1

Two inlet pipes A and B together can fill a tank in 7.5 min while with the help of an outlet pipe, tank can be filled in 15 min. If efficiency of pipe A is at least twice as that of pipe B, then which of the following cannot be the time taken by pipe A and outlet pipe together to fill the tank? [Time taken (in min) by A is an integer.]

A
120/7 min

B
240/11 min

C
45/2 min

D
30 min

Solution

Let time taken (min) by pipes A and B alone to fill the tank is 'a' and 'b' respectively.

Since efficiency of pipe A is at least twice as that of pipe B which means time taken by pipe B alone to fill the tanks is at least twice as that of pipe A.

b ≥ 2a

Time taken by outlet pipe alone to empty the tank = 1/[(1/7.5) - (1/15)] = 15/(2 - 1) = 15 min

Now,

(1/a) + (1/b) = 1/7.5 = 2/15

(1/b) = (2/15) - (1/a)

(1/b) = (2a - 15)/15a

b = 15a/(2a - 15)

b ≥ 2a

15a/(2a - 15) ≥ 2a

a = (15/2, 45/4]

Integer values of a = 8, 9, 10, and 11

Time taken by pipe A and outlet pipe together to fill the tank when a is 8 = 1/[(1/8) - (1/15)] = 120/7 min

Time taken by pipe A and outlet pipe together to fill the tank when a is 9 = 1/[(1/9) - (1/15)] = 45/2 min

Time taken by pipe A and outlet pipe together to fill the tank when a is 10 = 1/[(1/10) - (1/15)] = 30 min

Time taken by pipe A and outlet pipe together to fill the tank when a is 11 = 1/[(1/11) - (1/15)] = 165/4 min
Question 9
3 / -1

A bag contains 4 red balls, 6 blue balls, and 5 green balls. Ashish picks 2 balls at random without replacement such that the probability that he claims both the balls are of the same color is (97/280). If it is known that Ashish does not always speak the truth, then find the probability of Ashish telling the truth.

A
(5/7)

B
(3/4)

C
(7/8)

D
(1/15)

Solution

Let the probability that Ashish speaks the truth = 'x'

Then, the probability that Ashish lies = (1 - x)

Probability of picking two balls of the same colour = (⁴C₂+ ⁶C₂ + ⁵C₂) ÷ ¹⁵C₂ = (6 + 15 + 10) ÷ 105

= (31/105)

Probability of picking two balls not of the same colour = {¹⁵C₂- (⁴C₂+ ⁶C₂ + ⁵C₂)} ÷ ¹⁵C₂

= (74/105)

The probability that Ashish claims he picked two balls of the same colour = Probability that Ashish picks two balls of the same colour X probability that Ashish speaks the truth + Probability that Ashish picks two balls of different colour X Probability that Ashish lies.

= (31/105) X x + (74/105) X (1 - x)

Or, (31x/105) + {74/105 - 74x/105} = (97/280)

Or, {(74 - 43x)/105} = (97/280)

Or, 10185 = 20720 - 12040x

Or, x = 10535 ÷ 12040 = 0.875

Or, x = (7/8)

Hence, option 3 is correct.
Question 10
3 / -1

A = (x + 2)% of C and B = (x - 2)% of C. If A² - B² < (A + B)² < A² + B², then how many integer values 'x' can take?

A
4

B
1

C
3

D
2

Solution

(x + 2)² - (x - 2)² < (x + 2 + x - 2)² < (x + 2)² + (x - 2)²

(x² + 4 + 4x) - (x² + 4 - 4x) < (2x)² < (x² + 4 + 4x) + (x² + 4 - 4x)

x² + 4 + 4x - x² - 4 + 4x < 4x² < x² + 4 + 4x + x² + 4 - 4x

8x < 4x² < 2x² + 8

2x < x² < 0.5x² + 2

2x < x²

x² - 2x > 0

x (x - 2) > 0

x < 0 and x > 2

x² < 0.5x² + 2

0.5x² < 2

x² < 4

- 2 < x < 2

Range of values of 'x' = ( - 1, 0, 1)

Total values 'x' can take is 3.