# Interview Questions.

Exl Service Aptitude Placement Papers - Exl Service Aptitude Interview Questions and Answers

## Exl Service Aptitude Placement Papers - Exl Service Aptitude Interview Questions and Answers

Q1. Nitish Sold His Watch And Sunglasses At A Loss Of 4% And Gain Of 4% Respectively For 2600 To Kamal. Kamal Sold The Same Sun Glasses And Watch At A Loss Of four% And Gain Of 4% Respectively For 27

Lets anticipate the CP of watch = Rs x

and sun shades = Rs y.

2600=96x/a hundred + 104y/a hundred

2700= 104x/100 + 96y/100

By fixing,

y=seven-hundred

x=1960.

Q2. An Alloy Of Zinc And Copper Contains The Metals In The Ratio 5 :

Lets expect amount of zinc to added = x kg.

Zinc quantity in alloy = 5/8*sixteen=10 Kg.

And copper amount = three/eight*16=6 kg.

Alloy new ratio of zinc and copper = three:1

Zinc amount in alloy => (10+x)/6=three/1 => 10 + x = 18 => x = eight kg.

Q3. At 10:00 Am 2 Trains Started Travelling Towards Each Other From Station 287 Miles Apart They Passed Each Other At 1:30 Pm The Same Dayy .If Average Speed Of The Faster Train Exceeded By 6 Miles /hr Wh

Lets count on the speed of slower teach = x miles/hrs.

So, speed of faster educate is = (x+6) miles/hrs.

Given, exceeded each different at 1:30 PM i.E after three half of hrs.

Both teach journeying towards each different so general relative velocity = x+(x+6) = (2x+6)

So, 287/(2x+6) = 7/2 => 574 = 14x + forty two

=> 14x = 542 => x = ~38

So spee of faster educate = (x+6) miles/hrs. = forty four miles/hrs.

Q4. A Vessel Is Filled To Its Capacity With Pure Milk. Ten Litres Are Withdrawn From It And Replaced By Water. This Procedure Is Repeated Again. The Vessel Now Has 32 Litres Of Milk. Find The Capacity Of

Lets assume the capacity of the vessel = x litres.

X (1- 10/x)^2 = 32

x^2 +100-20x = 32x

x^2 +100 - 52x= 0

x^2 - 50x - 2x + a hundred = 0

=> (x-2) (x-50) = zero

=> x=2 & x=50

As 10 litres is withdrawn so vessel capacity might be 50 litres.

Q5. If 6 Is Subtracted From The Present Age Of Ritu And The Remainder Is Divided By 6, Then The Present Age Of Sheema Is Obtained. If Sheema Is four Years Younger To Raju Whose Age Is 12 Years, Then Find The

Lets count on Ritu present age = x years

so sheema age = (x-6)/6 years

As in line with question:

sheema age [(x-6)/6 + 4] = 12

(x-6)/6 = 12 - 4

=> (x-6)/6 = 8

=> x - 6 = 48 => x = 54

So Ritu age = fifty four years.

Q6. Two Circles Touch Each Other Externally. The Distance Between Their Centres Is 14 Cm And The Sum Of Their Areas Is one hundred thirty Cm^

Lets assume the radius of 1st circle = x cm.

So, the radius of second circle = (14-x)cm.

(pi*x*x)+(pi*(14-x)*(14-x))=130

By solving above eq.

X = eleven or 3.

Q7. Find Three Consecutive Odd Integers Whose Sum Is 117?

Let the three integers be 2x + 1, 2x + three , 2x + 5.

Therefore,  6x + 9 = 117,

x = 18

The 3 integers are 37, 39, forty one.

Q8. 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 hr=5/6hr

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New pace = forty* 3/2 kmph= 60kmph.

Q9. Three Runners A, B And C Run A Race, With Runner A Finishing 12 Meters Ahead Of Runner B And 18 Meters Ahead Of Runner C, While Runner B Finishes 8 Meters Ahead Of Runner C. Each Runner Travels The En

Lets expect distance of race = x mtrs.

Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.

=> at this point B is 6 m ahead of C. Now to complete race b desires to run some other 12 m,

=> he runs every other 12 m. Whilst B finishes race he's eight m in advance of C.

So ultimate 12 m B has run, C has run 10 m.As speeds are steady,

=> x-12/ x-18 = 12/10 => x = 48 mtrs.

Q10. The Age Of The Grand Father Is The Sum Of His Three Grandsons.The Second Is 2 Year Younger Than First One And The Third One Is 2 Year Younger Than The Second One. Then What Will Be The Age Of The Gran

Lets count on second grandson = x yr.

So, first grandson age = (x+2) 12 months ,

0.33 grandson = (x-2) year

So, x+2+x+x-2=3x

i.E grandfather age is 3 times as older as his second grandson.

Q11. If 20 Men Or 24 Women Or forty Boys Can Do A Job In 12 Days Working For eight Hours A Day, How Many Men Working With 6 Women And 2 Boys Take To Do A Job Four Times As Big Working For five Hours A Day For 12 Day

Amount of work achieved by way of 20 men = 24 women = forty boys

i.E 5man = 6 woman = 10 boys. -----------------(i)

According to the 1st condition, five men can do a activity in 12 days operating for 8 hours an afternoon.

Required :- how many more guys are required to work with 6 women and a pair of boys.

Let the required number of men be M.

According to the given information,

=>(5 x 12 x eight )/1=[(M + 5 + 1) x 5 x 12 ]/@( 6 girls= 5men & 2 boys =1 men)

=>32= M + 6.

=>M=26.

Hence, 26 men are operating with 6 women and a couple of boys.

Q12. Exact Time Of A Clock Was Set Right At 5:00 Am Then It Loses sixteen Min. Every Day, On The 4th Day When It Shows 10:00 Am, What Is The Exact Time?

Total past due in three day = 16*3 = 48 min.

Total overdue in one day =16 min.

So overdue in 1 min. = sixteen/(24*60) = 0.01111;

Hence in five hour (5:00 AM to ten:00 AM on 4th day) =.0111*5*60 = 3.333;

So, Exact time may be = 10:52 AM.

Q13. In A Stream Running At 2 Kmph,a Motar Boat Goes 6 Km Upstream And Back Again To The Starting Point In 33 Minutes. Find The Speed Of The Motarboat In Still Water?

Lets expect speed of boat in nonetheless water = x km/hr

So boat downstream speed = (x+2) km/hr.

Boat upstream speed = (x-2) km/hr.

6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20

=> a hundred and twenty*[(x-2) + (x+2)] = eleven*x^2 -forty four

=> 240 x = 11*x^2-44

=> 11* x^2 -240x -forty four = 0 ---------(1)

By solving eq. (1)

x = 22 km/hr.

Q14. A Ball Dropped From H Height And Moves 80% Of Height Each Time. Total Distance Covered Is?

First time distance= H

Second time = 80H/100 = 4H/5

in addition 1/3 time 80% of 4H/five = H(four^2)/(5^2)

and so on..

This will result in limitless terms of geometric development

i.E H+2*4H/5+2*16H/25..

So, Sum = H+ 2*4H/(five(1-four/five)) = 9H.

Q15. If The Sales Tax Reduced From three half% To three 1/three%, Then What Difference Does It Make To A Person Who Purchases An Article With Market Price Of Rs. 8400 ?

Tax reduced from three 1/2 or 7/2% to 3 1/3% or 10/three%.

So, the difference in tax = (7/2 - 10/three)% = 1/6%.

=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/one hundred) = Rs. 14.

Q16. A Gets Rs.33 When A Sum Of Money Was Distributed Among A, B And C In The Ratio three:2:five, What Will Be The Sum Of Money?

Lets anticipate amount of money = Rs. X ;

A receives Rs. 33 while sum disbursed in the ratio of three:2:five

so, 33=x*3/10

=> x=one hundred ten.

Q17. A Constructor Estimates That three People Can Paint Mr. Kh House In 4 Days. If He Uses four People Instead Of 3,how Long Will They Take To Complete The Job?

Lets assume day required to complete the job = x day

Men Day

---------------------

3(u) 4(d)

four x

x/4 = three/four

=> x = three days.

Q18. Find The Approximate Value Of The Following Equation 6.23% Of 258.Forty three - ? + 3.Eleven% Of 127 = 13.87?

[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=thirteen.87

16.100189-X+3.9497=13.87

X=6.179889.

Q19. A Certain Quantity Of 40% Solution Is Replaced With 25% Solution Such That The New Concentration Is 35%. What Is The Fraction Of The Solution That Was Replaced?

Consider an answer = 1 ltr

X is the sure quantity which has to get replaced

Now,

40percentof (1-x)+(25%of x)=35/one hundred

=> forty/a hundred * (1-x) + 25x/a hundred = 35/one hundred

=> 40/100 - 40x/one hundred + 25x/100 = 35/a hundred

=> 15x/a hundred = 5/100

=> x = 1/3.

Q20. Which Of The Following Numbers Are Completely Divisible By 11?

If the distinction of the sum of digits at extraordinary places and the sum of its digits at even locations, is either zero or divisible with the aid of eleven, then truely the number is divisible through eleven.

[(Sum of digits at odd places) - (Sum of digits at even places)]/eleven

=> [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/eleven = 0/eleven = 0

further others quantity are divisible by eleven.

Q21. The Ratio Of Two Numbers Is 3:four And Their Hcf Is four.Their Lcm Is?

Given Two No, ratio = three:four and their HCF = 4

So, No. = three*four =12 and 4*4=sixteen

LCM of 12,16 = 48.

Q22. In An Exam, Ajith, Sachu, Karna, Saheep And Ramesh Scored An Average Of 39 Marks. Saheep Scored 7 Marks More Than Ramesh. Ramesh Scored 9 Fewer Than Ajith. Sachu Scored As Many As Saheep And Ramesh Sc

Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y

So, as in keeping with the question:

z+7=x ---(i)

u- 9=z ---(ii)

x+ z=v ---(iii)

v+w=a hundred and ten ---(iv)

Given, u=32 ---- (v)

By fixing eq. (i), (ii), (iii), (iv) and (v)

w=57.

Q23. A Car Averages fifty five Mph For The First 4 Hours Of A Trip And Averages 70 Mph For Each Additional Hour. The Average Speed For The Entire Trip Was 60 Mph. How Many Hours Long Is The Trip?

Lets anticipate additional hours = x hrs.

So general No. Hours in adventure = (four+x)

[(55*4)+(70*x)]/(four+x)=60 =>

=>x=2

Therefore, Total No. Of hrs. In Journey = (x+4) = 6 hrs.

Q24. The Cost Of Two Varieties Of Paint Is Rs.3969 Per 2 Kg And Rs.1369 Per 2 Kg Respectively. After How Many Years, The Value Of Both Paint Will Be The Same, If Variety 1 Appreciates At 26% Per Annum And

Simply appreciates range 1 by 26% and depreciates range 2 via 26% as:

3969(1-(1/26))^n=1369(1+)1/26))^n

For n = 2 we get each values identical.

Variety 2 Variety1

3969.00 1369 Initially

2937.06 1724.Ninety four after I yr

2173.42 2173.42 after II 12 months

So the fee turn out to be identical after 2 years.

Q25. Find The Smallest Number Which Leaves 22,35, forty eight And sixty one As Remainders When Divided By 26, 39, fifty two And sixty five Respectively?

26-22=4

39-35=four

52-forty eight=4

65-61=four

LCM of (26,39,52,sixty five)=780

So smallest number = (780-four) = 776.

Q26. A Cycled From P To Q At 10 Kmph And Returned At The Rate Of 9 Kmph. B Cycled Both Ways At 12 Kmph. In The Whole Journey B Took 10 Minutes Less Than A. Find The Distance Between P And Q?

Lets count on distance between P and Q = d km.

Time taken by A in each facet = d/10 + d/nine = 19d/90 hrs.

Time taken by way of B in both side = second/12 = d/6 hrs.

B device 10 min. Or 1/6 hrs. Less than A.

So, 19d/90 - d/6 = 1/6

=> (19d-15d)/ninety = 1/6

=> 4d/90 = 1/6

=> d = 15/four km = 3.Seventy five km.

Q27. The Cost Price Of An Article Ls 80% Of Its Marked Price For Sale. How Much Percent Does The Trades-guy Gain After Allowing A Discount Of 12%?

C.P. Of the thing = Rs. A hundred

So Marked price = (one hundred*100)/80 = Rs. 125

SP after the cut price = Rs.(125*88)/100 = Rs. One hundred ten

consequently Gain percent = 10.

Q28. The Average Number Of Visitors Of A Library In The First 4 Days Of A Week Was five

If range of visitors on 1st, 2nd, third, 4th & 5th day are a, b, c, d & e respectively, then

a+b+c+d=fifty eight*four=232 ----(i)

b+c+d+e=60*4=240 ----(ii)

Subtracting eq. (i) from (ii)

e-a=8 ---(iii)

Given, a/e=7/8 ---(iv)

So from eq. (iii) & (iv)

a=fifty six, e=sixty four.

Q29. Two Urns Contain Respectively 2 Red 3 White And three Red,5 White Balls.One Ball Is Drawn At Random From The First Urn And Trferred Into The Second.A Ball Is Now Drawn From The Second Urn And It Turns Out

According to baye's theorem:

p(no of way to drawn a ball from 2d urns is pink and ball trferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/forty five

Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/forty five

so opportunity=(nine/forty five)/(17/forty five)=9/17.