Question 1. The Least Perfect Square, Which Is Divisible By Each Of 21, 36 And sixty six Is?

Answer :

L.C.M. Of 21, 36, sixty six = 2772 Now,

2772 = 2 x 2 x three x 3 x 7 x eleven To make it an excellent rectangular,

it must be expanded by using 7 x eleven. So,

required range = 2 x 2 x three x 3 x 7 x 7 x 11 x eleven = 213444.

Question 2. Which Greatest Possible Length Can Be Used To Measure Exactly 15 Meter seventy five Cm, eleven Meter 25 Cm And 7 Meter sixty five Cm?

Answer :

We want to find out the HCF for given period.

15 meter seventy five cm = 1575 cm.

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

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

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

Networking Interview Questions

Question three. Log Xy = 100 And Log X2 = 10, Then The Value Of Y Is?

Answer :

log2x = 10 ⇒ x = 210.

∴ logx y = one hundred

⇒ y = x100

⇒ y = (210)100 [put value of x]

⇒ y = 21000.

Question four. The Simplified Form Of Log(75/16) -2 Log(five/9) +log(32/343) Is ?

Answer :

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

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

= log(25 x 3) – log ( four x four ) – log(25) + log81 + log(sixteen x 2) -log (eighty one x 3)

= log 25 + log three – log 16 – log 25 + log 81 + log 16 + log 2 – log eighty one – log 3

= log 2.

Networking Tutorial

Question five. The Value Of Log2(1/sixty four) Is?

Answer :

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.

Aptitude Interview Questions

Question 6. A Train Of Length a hundred and ten 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?

Answer :

Distance = a hundred and ten m

Relative speed = 60 + 6 = 66 kmph

(Since each the train and the person are in moving in contrary route) = (sixty six*5/18) m/sec = fifty five/three m/sec.

Question 7. 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?

Answer :

Let the full distance included be S km.

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

Average speed =S× 240 5S = forty eight km/hr.

Hardware and Networking Interview Questions

Question 8. 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?

Answer :

Here as in keeping with circumstance let us take into account 1/3 range = x,

then 2nd range = 2x and 3rd quantity = 4x,

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

7x = 3×seventy seven, x = three x 11 = 33,

first variety = 4x = four x 33 = 132.

Question nine. A Merchant Sold An Article At 10% Loss. If He Had Sold It Rs 450 More,eight% Would Have Been Gained On The Cost Price. Find The Cost Price?

Answer :

Let the fee price be a hundred%. It is sold at 10% loss.

So it is sold at ninety% of the price rate.

Ninety % of the value price + 450 = 108% of the Cost fee 18% of the fee rate = Rs 450 Cost charge of the e book = 450/18 x one hundred = Rs 2500.

Network Monitoring Interview Questions

Question 10. 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?

Answer :

1525-1300= 225 for 1.5 yrs (three.5-2)

so for one yr 225/1.Five= 150

then for 2 yrs hobby is one hundred fifty+150=three hundred

Then most important 1300-300=one thousand.

Now one hundred fifty/a thousand*one hundred= 15%.

Question 11. The Mean Daily Profit Made By A Shopkeeper In A Month Of 30 Days Was Rs. 350. If The Mean Profit For The First Fifteen Days Was Rs. 275, Then The Mean Profit For The Last 15 Days Would Be?

Answer :

Find the overall income made in the first 15 days:

Mean day by day profit for the first 15 days = Ra 275

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

Find the full earnings for the ultimate 15 days:

10,500 - 4125 = Rs 6375.

Find the mean profit for the final 15 days:

Mean income = 6375 ÷ 15 = Rs 425.

Network Engineer Interview Questions

Question 12. There Were 35 Students In A Hostel. If The Number Of Students Increases By 7, The Expenses Of The Mess Increase By Rs. Forty two Per Day While The Average Expenditure Per Head Diminishes By Re 1. Find The Original Expenditure Of The Mess?

Answer :

Let d be the common daily expenditure

Original expenditure = 35 × d

New expenditure = 35 × d + 42

New common expenditure may be :

(35 × d + forty two)/forty two = d - 1

On solving, we get d = 12

Therefore authentic expenditure = 35 × 12 = 420.

Networking Interview Questions

Question thirteen. 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 : 1. If On A Particular Day Rs. 1,325 Were Collected From The Passengers Travelling Between These Stations, Then What Was The Amount Collected From The Ii Class Passengers?

Answer :

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

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

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

Question 14. A Pump Can Be Operated Both For Filling A Tank And For Emptying It. The Capacity Of The Tank Is 2400 M3. The Emptying Capacity Of The Pump Is 10m3 Per Minute Higher Than Its Filling Capacity. Consequently, The Pump Needs 8 Minutes Less To Empty The Tank Than To Fill It. Find The Filling Capacity Of The Pump?

Answer :

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

Then the emptying potential 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 mins

Pump desires 8 mins lesser to drain the tank than it needs to fill it

⇒2400/x−2400/x+10=eight

⇒three hundred/x−three hundred/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 potential of the pump = 50 m3/min.

Question 15. In A Kilometer Race, A Can Give B A one hundred M Start And C A one hundred fifty M Start. How Many Meters Start Can B Give To C?

Answer :

A can supply B a one hundred m start and C a 150m.

Start means while A runs 1000m,

B runs 900m and C runs 850m.

When B runs 1000m,

C will run a thousand x (850/900) m (i.E. 8500/nine m) Thus,

B can provide C a start of - one thousand - (8500/nine), i.E. 500/nine m.

Cisco Interview Questions

Question 16. 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 Number Of Boys In The Class?

Answer :

Let Boys in class = B

Girls in elegance = 20

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

⇒ B = 30.

Question 17. 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 Speed?

Answer :

Let the velocity of products train be x km/hr.

Then,

(50+x)×( five 18 )= 187.Five nine

⇒ x= 25 km/hr.

Deloitte JAVA Interview Questions

Question 18. Walking At The Rate Of four Kmph A Man Cover Certain Distance In 2 Hr 45 Min. Running At A Speed Of sixteen.5 Kmph The Man Will Cover The Same Distance In?

Answer :

Distance = Speed × time

Here time = 2hr forty five min = 114 hr

Distance = four×114=11 km

New Speed =sixteen.5 kmph

Therefore time = DS=1116.Five=40 min.

Aptitude Interview Questions

Question 19. 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 forty Min Will Be?

Answer :

Time = 50/60=five/6 hr

Speed = forty eight mph

Distance = S×T=forty eight×five/6=40 km

Time = forty/60 hr

New speed = forty×3/2=60 kmph.

Question 20. 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?

Answer :

Let the normal velocity be 's' km/hr

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

three hundred/s−2=300/s+5

On solving this equation we get:

s = 25 km/hr.

Cisco Network Engineer Interview Questions

Question 21. 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 Train Completely In 60 Sec. Whereas A Passenger On The Passenger Train Marks That He Crosses The Goods Train In 40 Sec. If The Speeds Of The Trains Be In The Ratio 1:2. Find The Ratio Of Their Lengths?

Answer :

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=forty------(2)

On dividing these two equation we get:

P+Q/P=60/40

P:Q = 2 : 1.

Question 22. 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 And By How Much?

Answer :

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/four hundred×395=391.05m

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

If A and D run over four hundred m, then D win via 7.2 m (approx.)

Question 23. 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 four.Five Km/hr And three.Seventy five Km/hr Respectively. They Will Meet For The First Time In?

Answer :

et each of them meet after T min

4500 m are protected by using Suresh in 60 m.

In T min he'll cover 4500T60

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

Given,

4500T60+3750T60=726

T=five.28 mins.

Deloitte Tax Consultant Interview Questions

Question 24. 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 Is two hundred Km, Then At What Time Did The Two Trains Meet Each Other?

Answer :

Average pace of educate leaving Delhi = 2004=50 km/hr

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

By the time the other educate starts offevolved from Ambala cantt, the primary teach had travelled 100 km

Therefore, the trains meet after:

=2 hundred−a hundred/(50+four hundred/7)=14/15hr

=14/15×60=56minutes

Hence they meet at eight:56 am.

Hardware and Networking Interview Questions