# Deloitte Aptitude Placement Papers - Deloitte Aptitude Interview Questions and Answers updated on 19.May.2024

Let the filling capacity of the pump = x m3/ min.

Then the emptying capacity of the pump = (x + 10) m3/ min.

Time required for filling the tank = 2400/x minutes

Time required for emptying the tank = 2400/x+10 minutes

Pump needs 8 minutes lesser to empty the tank than it needs to fill it

⇒2400/x−2400/x+10=8

⇒300/x−300/x+10=1

⇒300/(x+10)−300/x=x(x+10)

⇒3000=x2+10x

⇒x2+10x−3000=0

(x+60)(x−50)=0

x = 50 or -60

Since x can not be negative, x=50

i.e.,filling capacity of the pump = 50 m3/min.

If A covers 400m, B covers 395 m

If B covers 400m, C covers 396 m

If D covers 400m, C covers 384 m

Now if B covers 395 m, then C will cover 396/400×395=391.05m

If C covers 391.05 m, then D will cover 400/384×391.05=407.24

If A and D run over 400 m, then D win by 7.2 m (approx.)

Here as per condition let us consider third number = x,

then 2nd number = 2x and 3rd number = 4x,

Given that (4x+2x+x)/3 = 77,

7x = 3×77, x = 3 x 11 = 33,

first number = 4x = 4 x 33 = 132.

Let the speeds of the two trains be s and 2s m/s respectively.

Also, suppose that the lengths of the two trains are P and Q metres respectively.

Then,

P+Q/2s−s=60------(1)

and

P/2s−s=40------(2)

On dividing these two equation we get:

P+Q/P=60/40

P:Q = 2 : 1.

Average speed of train leaving Delhi = 2004=50 km/hr

Average speed of train leaving Ambala cantt. = 200×27=4007

By the time the other train starts from Ambala cantt, the first train had travelled 100 km

Therefore, the trains meet after:

=200−100/(50+400/7)=14/15hr

=14/15×60=56minutes

Hence they meet at 8:56 am.

Distance = Speed × time

Here time = 2hr 45 min = 114 hr

Distance = 4×114=11 km

New Speed =16.5 kmph

Therefore time = DS=1116.5=40 min.

Let the total distance covered be S km.

Total time taken= S (2×40) + S (2×60) = 5S 240 hr

Average speed =S× 240 5S = 48 km/hr.

Given Exp= log75/16 – 2 log5/9 + log32/343

= log [(25 x 3) / (4 x 4)] – log (25/81) + log [(16 x 2) / (81 x 3)]

= log(25 x 3) – log ( 4 x 4 ) – log(25) + log81 + log(16 x 2) -log (81 x 3)

= log 25 + log 3 – log 16 – log 25 + log 81 + log 16 + log 2 – log 81 – log 3

= log 2.

Find the total profit made in the first 15 days:

Mean daily profit for the first 15 days = Ra 275

Total profit for the first 15 days = 275 x 15 = Rs 4125.

Find the total profit for the last 15 days:

10,500 - 4125 = Rs 6375.

Find the mean profit for the last 15 days:

Mean profit = 6375 ÷ 15 = Rs 425.

log2x = 10 ⇒ x = 210.

∴ logx y = 100

⇒ y = x100

⇒ y = (210)100 [put value of x]

⇒ y = 21000.

Let the speed of goods train be x km/hr.

Then,

(50+x)×( 5 18 )= 187.5 9

⇒ x= 25 km/hr.

A can give B a 100 m start and C a 150m.

Start me when A runs 1000m,

B runs 900m and C runs 850m.

When B runs 1000m,

C will run 1000 x (850/900) m (i.e. 8500/9 m) Thus,

B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.

log2(164)=xlog2(164)=x

Rewrite the equation as x=log2(164)x=log2(164).

x=log2(164)x=log2(164) Logarithm base 22 of 164164 is −6-6.

x=−6.

Time = 50/60=5/6 hr

Speed = 48 mph

Distance  = S×T=48×5/6=40 km

Time = 40/60 hr

New speed = 40×3/2=60 kmph.

Distance = 110 m

Relative speed = 60 + 6 = 66 kmph

(Since both the train and the man are in moving in opposite direction) = (66*5/18) m/sec = 55/3 m/sec.

Let x be the number of passengers and y be the fare taken from passengers.

3xy + 50xy = 1325 => xy = 25

Amount collected from II class passengers = 25 × 50 = Rs. 1250.

et both of them meet after T min

4500 m are covered by Suresh in 60 m.

In T min he will cover 4500T60

Likewise, In T min Suresh's wife will cover 3750T60

Given,

4500T60+3750T60=726

T=5.28 minutes.

Let the normal speed be 's' km/hr

Then new speed = (s+5) km/hr

300/s−2=300/s+5

On solving this equation we get:

s = 25 km/hr.

We need to find out the HCF for given length.

15 meter 75 cm = 1575 cm.

11 meter 25 cm = 1125 cm 7 meter 65 cm = 765 cm.

1575 = 5 *5 *3 *3 *7 1125 = 5 *5 *5 *3 *3 765.

= 5 *3 *3 *17 HCF of 1575, 1125 and 765 is 45 (5 *3*3).

Let Boys in class = B

Girls in class = 20

Now, (20B+15*20)/(B+20) = 18

⇒ B = 30.

Let d be the average daily expenditure

Original expenditure = 35 × d

New expenditure = 35 × d + 42

New average expenditure will be :

(35 × d + 42)/42 = d - 1

On solving, we get d = 12

Therefore original expenditure = 35 × 12 = 420.

Let the cost price be 100%. It is sold at 10% loss.

So it is sold at 90% of the cost price.

90 % of the cost price + 450 = 108% of the Cost price 18% of the cost price = Rs 450 Cost price of the book = 450/18 x 100 = Rs 2500.

L.C.M. of 21, 36, 66 = 2772 Now,

2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square,

it must be multiplied by 7 x @So,

required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 213444.

1525-1300= 225 for 1.5 yrs (3.5-2)

so for one yr 225/1.5= 150

then for 2 yrs interest is 150+150=300

Then principal 1300-300=1000.

Now 150/1000*100= 15%.