Impetus Aptitude Placement Papers - Impetus Aptitude Interview Questions and Answers updated on 03.Feb.2023

Umesh’s 5 day’s work = 5 x 1/15 = 1/3 

Remaining work = (1 – 1/3) = 2/3 (1/10 +1/ 15) work is done by both in 1 day 

Therefore 2/3 work is done by both in (6 x 2/3) = 4days. 

The work was completed in 9 days.

Ratio of speed = 3:4

Ratio of time = 4:3

let A takes 4x hrs, B takes 3x hrs

then 4x-3x = 30/60 hr

x = ½ hr

Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr.

Let the required ratio be x: 1: y. 

Then S.I on Rs x for 6 years at 10% p.a = S.I on re.1 10 years at 12 %p.a 

X × 10/100 × 6 = 1 × 12/100 × 10 

=> X = 120/60 =2 

S.I on Re.1 for 10 years at 12 % p.a = S. I on Rs. Y for 12 years at 15 % p.a 

Therefore, (1 x 12/100 x 10) = (y x 15/100 x 12) 

=> y = 120/180 = 2/3 

Required ratio = 2 : 1 = 2/3 = 6 : 3 : 2.

1 man’s one day’s work = 1/96 12 men’s 3 day’s work = (3 x 1/8) = 3/8 

Remaining work = (1 – 3/8) = 5/8 15 men’s 1 day’s work = 15/96 

Now 15/96 work is done by them in 1day 

Therefore 5/8 work will be done by them in (96/15 x 5/8) i.e., 4 days.

Let the required time be x years. 

Then 800 + 800 x 12/100 × x = 910 + 910 x 10/100 × x 

=> (96x -91 x) = 110 

=> 5x =110 

=> x =22.

Original price = Rs. 100 

Price after first discount = Rs. 90 

Price after second discount = Rs. (80/100 × 90) = Rs. 72 

Price after third discount = Rs. (60/100 × 72) = Rs. 43.20 

Single discount = (100 – 43.20) = 56.8%.

(A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4

A:B = 2:3 and B:C = 1:4 = 3:12

A:B:C = 2:3:12

A;s share = 510 * 2/17 = Rs. 60

B's share = 510 * 3/17 = Rs. 90

C's share = 510 * 12/17 = Rs. 360.

Let the principal be Rs x. 

Then X × (1 + 10/100)3 – x = 331 

=> (x × 11/10 × 11/10 × 11/10 - x) = 331

=> ((1331x-1000x)/1000) = 331 

=> 331x = 331000 

=> X = 1000 

Hence the principal is Rs 1000.

Given x = k/y2, where k is a constant. 

Now, y = 2 and x = 1 gives k = 4. 

x = 4/y2 => x = 4/62, when 

y = 6 => x = 4/36 = 1/9.

CP of 120 calendars = 120 x 3 = Rs. 360

SP of 40 at Rs. 4 each = Rs. 40 x 4 = Rs.160

SP of 60 at Rs 5 each = Rs 60 x 5 = Rs. 300

SP of remaining 20 calendars = Rs 20 x 3 = 60

Total SP = Rs.( 160 + 300 + 60) = 520

Profit % = 520 – 360 = 160

Profit % = 160/360 x 100 = 400/9.

Let x kms be covered in y hours, then first speed = x / y km/hr

Again x/2 km is covered in 2y hrs 

Therefore, new speed = (x/2 x 1/2y)km/hr 

= (x/ 4y) km/hr 

Ratio of speeds = x/y : x/4y = 1 : ¼

= 4 : 1.

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.

1/x + 2/x + 4/x = 1/5

7/x = 1/5 => x = 35 hrs.

Suppose the two trains meet after x hours, then 

=> 30x + 27x = 342 

=> 57x = 342 

=> X = 6 

So the two trains will meet after 6 hours.

Let P, Q and R represent their respective monthly incomes.

Then, we have:

P + Q = (5050 * 2) = 10100 --- (i)

Q + R = (6250 * 2) = 12500 --- (ii)

P + R = (5200 * 2) = 10400 --- (iii)

Adding (i), (ii) and (iii), we get: 

2(P + Q + R) = 33000 = P + Q + R = 16500 --- (iv)

Subtracting (ii) from (iv), we get, P = 4000.

P's monthly income = Rs. 4000.

A’s 10 days work = (10 x 1/80) = 1/8 

Remaining work = (1 -1/8) = 7/8 

Therefore 7/8 work is done by A in 42 days. 

Whole work will be done by A in (42 x 8/7)i.e.., 48 days 

Therefore, (A+ B)’s 1 day work = (1/80 + 1/48) = 8/240 = 1/30. 

A and B together can finish it in 30days.

Work to be done by C = (1 – 7/11) = 4/11 

Therefore, (A +B) = C = 7/11 : 4/11 = 7 : 4 

Therefore C’s share = Rs.(550 x 4/11) = Rs. 200.

Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank.

1/x + 1/(x - 5) = 1/(x - 9)

(2x - 5)(x - 9) = x(x - 5) 

x2 - 18x + 45 = 0 

(x- 15)(x - 3) = 0 => x = 15.

B's 10 day's work = 1/15 * 10 = 2/3

Remaining work = (1 - 2/3) = 1/3

Now, 1/18 work is done by A in 1 day.

1/3 work is done by A in (18 * 1/3) = 6 days.

Let the loan taken be Rs. X , then 

X × 6/100 × 1 + X × 6.5/100 × 1 + X × 7/100 × 1 + X × 7.5/100 × 1 = 3375 

=> (6+ 6.5 +7 +7.5) × X/100 = 3375 

=> X = (3375 × 100/27) = 12500.

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. 

Then, (2x + 4000)/(3x + 4000) = 40/57 

6x = 68000 => 3x = 34000

Sumit's present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000.

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

1/x  + 1/(x + 6) = 1/4 

x2 - 2x - 24 = 0 

(x - 6)(x + 4) = 0 => x = 6.

Let the required period of time be x years. 

Then 500 x 4 x 6.25/100 = 400 x 5/100 × x 

=> 20 x 6.25 = 20 × x 

=> X = 6.25 = 625/100 = 25/4 = 6 ¼ years.

Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively. 

Then, 4x - 3x = 1000 => x = 1000.

B's share = Rs. 2x = 2 * 1000 = Rs. 2000.

(A +B)’s 1 day’s work = (1/30 + 1/40) = 7/120

Time is taken by both to finish the work = 120/7 days 

= 17 1/7 days.

Let average for 10 innings be x. Then,

(10x + 108)/11 = x + 6 

= 11x + 66 = 10x + 108

= x = 42.

New average = (x + 6) = 48 runs.

Let the original earnings of A and B be Rs. 4x and Rs. 7x.

New earnings of A = 150% 0f Rs. 4x = (150/100 * 4x) = Rs. 6x

New earnings of B = 75% of Rs. 7x = (75/100 * 7x) = Rs. 21x/4

6x:21x/4 = 8:7

This does not give x. So, the given data is inadequate.

Original price be Rs. X 

C.P = (X – 25% of X) = 3X/4 

S.P = (3X/4 + 10% of 3X/4) = 33X/40= 660 

=> X = 800.

Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15

Remaining part = 1 - 7/15 = 8/15

Part filled by B in 1 minute = 1/20

1/20 : 8/15 :: 1 ; x

x = 8/15 * 1 * 20 = 10 2/3 min = 10 min 40 sec.

The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.

S.P be Rs. X C.P paid by Subhash = Rs. 9X/10 

 S.P received by Subhash = Rs. (108% of Rs. X) 

 = Rs. 27X/25 

 Gain = Rs. (27X/25 – 9X/10) 

 = Rs. 9X/50 Gain = (9X/50 × 10/9X × 100)% = 20%.

Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24

Suppose the tank is filled in x minutes.

Then, x/2(1/24 + 1/40) = 1

x/2 * 1/15 = 1 => x = 30 min.

Let the train meet at a distance of x km from Delhi.

Then x/60 –x/80 = 2 

=> 4x -3x 

=> x =480 

Therefore, Required distance = 480km.

Suppose they meet after x hours , then 3x +4x = 17.5 

=>  7x = 17.5 

=> X = 2.5hours 

So they meet at 12: 30 p.m.

S.I on Rs. 600 for 4 Years = Rs (7200 - 600) = Rs 1200 

Therefore Rate = (120000/24000) % p.a = 5 % p.a 

New Rate = (5 x 3/2) % = 15/2 % p.a 

Required Amount = [6000 + (6000 x 5/100 x 15/2)] 

= Rs(6000 + 2250) = Rs 8250.

Let the number of wickets taken till the last match be x. Then,

(12.4x + 26)/(x + 5) = 12

= 12.4x + 26 = 12x + 60

= 0.4x = 34 

= x = 340/4 = 85.

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased sears are (140% of 5x), (150% of 7x) and (175% of 8x)

i.e., (140/100 * 5x), (150/100 * 7x) and (175/100 * 8x)

i.e., 7x, 21x/2 and 14x

Required ratio = 7x:21x/2:14x

= 14x : 21x : 28x = 2:3:4.

Let A, B, C represent their respective weights. 

Then, we have:

A + B + C = (45 * 3) = 135 --- (i)

A + B = (40 * 2) = 80 --- (ii)

B + C = (43 * 2) = 86 --- (iii)

Adding (ii) and (iii),

we get: A + 2B + C = 166 --- (iv)

Subtracting (i) from (iv), we get: B = 31

B's weight = 31 kg.

Let 40% of A = 2/3 B. Then,

40A/100 = 2B/3 => 2A/5 = 2B/3

A/B = (2/3 * 5/2) = 5/3

A:B = 5:3.

Ratio of times taken by A and B = 100:130 = 10:13

Suppose B takes x days to do the work.

x = (23 * 13)/10 = 299/10

A's 1 day work = 1/23; B's 1 day work = 10/299

(A + B)'s 1 day work = (1/23 + 10/299) = 1/13

A and B together can complete the job in 13 days.

Let the prices be 4X, 5X and 7 X rupees 

 7X – 4X = 60000 

=> X = 20000 

Required price = 5X = Rs. 100000.

The time taken by B =(x-20/60)hrs = (x-1/3)hrs 

Ratio of speeds = Inverse ratio of time taken 

Therefore 3: 4 =(x-1/3) : x 

=>3x-1/3x=3/4 

=> 12x -4 =9x 

=> 3x =4 

=> X=4/3 hrs. 

X=1 1/3 hours. 

Required time = 1 1/3 hrs.

Let the average of the whole team be x years.

11x - (26 + 29) = 9(x - 1) 

= 11x - 9x = 46 

= 2x = 46 => x = 23 

So, average age of the team is 23 years.

Amount = Rs[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)] 

= Rs(1000 × 26/25 × 21/20 × 53/50) 

= Rs (57876/5) = Rs 11575.20 

C.I = Rs (11575.20 - 10000) = Rs 1575.20.

1/3 of work is done by A in 5 days

Therefore, whole work will be done by A in 15 days 

2/5 of work is done by B in 10 days 

Whole work will be done by B in (10 x 5/2) i.e.., 25 days 

Therefore (A + B)’s 1 day’s work = (1/15 + 1/25) = 8/75 

So both together can finish 75/8 days i.e.., 9 3/8 days.

Let the speed of the first train be 7x km/hr

Then the speed of the second train is 8x km/hr 

But speed of the second train = 400/5 km/hr = 80km/hr 

=> 8x =80 

=> X=10 

Hence the speed of first train is (7 x 10)km/hr = 70km/hr.

P = Rs 16000, R = (10/2)% 

per quarter, t = 3 quarters 

C.I = Rs [16000 × (1+ 5/100)3-16000) 

= Rs [16000 × 21/20 ×21/20× 21/20- 16000)

=Rs (18522-16000)=Rs 2522.

Originally, let the number of boys and girls in the college be 7x and 8x respectively. 

Their increased number is (120% of 7x) and (110% of 8x).

i.e., (120/100 * 7x) and (110/100 * 8x)

i.e., 42x/5 and 44x/5

Required ratio = 42x/5 : 44x/5 = 21:22.