Deloitte Aptitude Placement Papers - Deloitte Aptitude Interview Questions and Answers updated on 15.Oct.2019
Deloitte Aptitude Interview Questions and Answers
A Pump Can Be Operated Both For Filling A Tank And For Emptying It. The Capacity Of The Tank Is 2400 M
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
x = 50 or -60
Since x can not be negative, x=50
i.e.,filling capacity of the pump = 50 m3/min.
A Race Course Is 400 M Long. A And B Run A Race And A Wins By 5m. B And C Run Over The Same Course And B Win By 4m. C And D Run Over It And D Wins By 16m. If A And D Run Over It, Then Who Would Win An
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.)
The Average Of Three Numbers Is 77.the First Number Is Twice The Second And The Second Number Is Twice The Third. Find The First Number?
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.
A Good Train And A Passenger Train Are Running On Parallel Tracks In The Same Direction. The Driver Of The Goods Train Observes That The Passenger Train Coming From Behind Overtakes And Crosses His Tr
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.
On dividing these two equation we get:
P:Q = 2 : 1.
A Train Starts From Delhi At 6:00 Am And Reaches Ambala Cantt. At 10am. The Other Train Starts From Ambala Cantt. At 8am And Reached Delhi At 11:30 Am, If The Distance Between Delhi And Ambala Cantt I
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:
Hence they meet at 8:56 am.
Walking At The Rate Of 4 Kmph A Man Cover Certain Distance In 2 Hr 45 Min. Running At A Speed Of 16.5 Kmph The Man Will Cover The Same Distance In?
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.
A Car Travels First Half Distance Between Two Places With A Speed Of40 Km/hr And Rest Of The Half Distance With A Speed Of 60 Km/hr.the Average Speed Of The Car Is?
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.
The Simplified Form Of Log(75/16) -2 Log(5/9) +log(32/343) Is ?
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.
The Mean Daily Profit Made By A Shopkeeper In A Month Of 30 Days Was Rs. 35
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.
Log Xy = 100 And Log X2 = 10, Then The Value Of Y Is?
log2x = 10 ⇒ x = 210.
∴ logx y = 100
⇒ y = x100
⇒ y = (210)100 [put value of x]
⇒ y = 21000.
A Man Sitting On A Train Travelling At The Rate Of 50 Km/hr Observes That It Takes 9 Sec For Goods Train Travelling In The Opposite Direction To Pass Him. If The Goods Train Is 187.5m Long. Find Its S
Let the speed of goods train be x km/hr.
(50+x)×( 5 18 )= 187.5 9
⇒ x= 25 km/hr.
In A Kilometer Race, A Can Give B A 100 M Start And C A 150 M Start. How Many Meters Start Can B Give To C?
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.
The Value Of Log2(1/64) Is?
Rewrite the equation as x=log2(164)x=log2(164).
x=log2(164)x=log2(164) Logarithm base 22 of 164164 is −6-6.
A Train Covers A Distance In 50 Min, If It Runs At A Speed Of 48 Kmph On An Average. The Speed At Which The Train Must Run To Reduce The Time Of Journey To 40 Min Will Be?
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.
A Train Of Length 110 Meters Is Running At A Speed Of 60 Kmph.in What Time, It Will Pass A Man Who Is Running At 6 Kmph In Thedirection Opposite To That In Which The Train Is Going?
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.
The Ratio Between The Number Of Passengers Travelling By I And Ii Class Between The Two Railway Stations Is 1 : 50, Whereas The Ratio Of I And Ii Class Fares Between The Same Stations Is 3 :
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.
The Jogging Track In A Sports Complex Is 726 M In Circumference. Suresh And His Wife Start From The Same Point And Walk In Opposite Direction At 4.5 Km/hr And 3.75 Km/hr Respectively. They Will Meet F
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
A Passenger Train Takes Two Hours Less For A Journey Of 300 Km If Its Speed Is Increased By 5 Km/hr From Its Normal Speed. The Normal Speed Is?
Let the normal speed be 's' km/hr
Then new speed = (s+5) km/hr
On solving this equation we get:
s = 25 km/hr.
Which Greatest Possible Length Can Be Used To Measure Exactly 15 Meter 75 Cm, 11 Meter 25 Cm And 7 Meter 65 Cm?
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).
The Average Age Of All The Student Of A Class Is 18 Years. The Average Age Of Boys Of The Class Is 20 Years And That Of The Girls Is 15 Years. If The Number Of Girls In The Class Is 20, Then Find The
Let Boys in class = B
Girls in class = 20
Now, (20B+15*20)/(B+20) = 18
⇒ B = 30.
There Were 35 Students In A Hostel. If The Number Of Students Increases By 7, The Expenses Of The Mess Increase By Rs. 42 Per Day While The Average Expenditure Per Head Diminishes By Re
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.
A Merchant Sold An Article At 10% Loss. If He Had Sold It Rs 450 More,8% Would Have Been Gained On The Cost Price. Find The Cost Price?
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.
The Least Perfect Square, Which Is Divisible By Each Of 21, 36 And 66 Is?
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.
A Certain Sum Of Money Amounts To Rs.1300 In 2 Years And To Rs. 1525 In 3.5 Years. Find The Sum And The Rate Of Interest?
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%.
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