Exl Service Aptitude Placement Papers - Exl Service Aptitude Interview Questions and Answers updated on 26.May.2022

Lets assume the CP of watch = Rs x 

and sunglasses = Rs y.

2600=96x/100 + 104y/100

2700= 104x/100 + 96y/100

By solving,

y=700

x=1960.

Lets assume quantity of zinc to added = x kg.

zinc quantity in alloy = 5/8*16=10 Kg.

and copper quantity = 3/8*16=6 kg.

Alloy new ratio of zinc and copper = 3:1 

Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.

Lets assume the speed of slower train = x miles/hrs.

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

Given, passed each other at 1:30 PM i.e after 3 1/2 hrs.

Both train travelling towards each other so total relative speed = x+(x+6) = (2x+6)

So, 287/(2x+6) = 7/2 => 574 = 14x + 42 

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

So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs.

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 + 100 = 0

=> (x-2) (x-50) = 0

=> x=2 & x=50

As 10 litres is withdrawn so vessel capacity will be 50 litres.

Lets assume Ritu present age = x years

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

As per 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 = 54 years.

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

so, the radius of 2nd circle = (14-x)cm.

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

By solving above eq. 

x = 11 or 3.

Let the three integers be 2x + 1, 2x + 3 , 2x + 5.

Therefore,  6x + 9 = 117,

x = 18

The three integers are 37, 39, 41.

Lets assume 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 finish race b needs to run another 12 m, 

=> he runs another 12 m. when B finishes race he is 8 m ahead of C. 

so last 12 m B has run, C has run 10 m.as speeds are constant, 

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

Lets assume second grandson = x year.

So, first grandson age = (x+2) year ,

third grandson = (x-2) year

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

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

Amount of work done by 20 men = 24 women = 40 boys

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

According to the 1st condition, 5 men can do a job in 12 days working for 8 hours a day.

Required :- how many more men are required to work with 6 women and 2 boys.

Let the required number of men be M.

According to the given information, 

=>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/@( 6 women= 5men & 2 boys =1 men)

=>32= M + 6.

=>M=26.

Hence, 26 men are working with 6 women and 2 boys.

Total late in three day = 16*3 = 48 min.

Total late in one day =16 min.

so late in 1 min. = 16/(24*60) = 0.01111;

Hence in 5 hour (5:00 AM to 10:00 AM on 4th day) =.0111*5*60 = 3.333;

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

Lets assume speed of boat in still 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

=> 120*[(x-2) + (x+2)] = 11*x^2 -44

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

=> 11* x^2 -240x -44 = 0 ---------(1)

By solving eq. (1) 

x = 22 km/hr.

First time distance= H

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

similarly third time 80% of 4H/5 = H(4^2)/(5^2)

and so on.. 

This will lead to infinite terms of geometric progression

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

So, Sum = H+ 2*4H/(5(1-4/5)) = 9H.

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

So, the difference in tax = (7/2 - 10/3)% = 1/6%.

=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14.

Lets assume sum of money = Rs. x ;

A gets Rs. 33 when sum distributed in the ratio of 3:2:5

so, 33=x*3/10

=> x=110.

Lets assume day required to complete the job = x day

Men Day

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

3(u) 4(d)

4 x

x/4 = 3/4

=> x = 3 days.

[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87

16.100189-X+3.9497=13.87

X=6.179889.

Consider a solution = 1 ltr

X is the certain quantity which has to be replaced

Now,

40%of (1-x)+(25%of x)=35/100 

=> 40/100 * (1-x) + 25x/100 = 35/100 

=> 40/100 - 40x/100 + 25x/100 = 35/100 

=> 15x/100 = 5/100 

=> x = 1/3.

If the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11.

[(Sum of digits at odd places) - (Sum of digits at even places)]/11

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

similarly others number are divisible by 11.

Given Two No, ratio = 3:4 and their HCF = 4

So, No. = 3*4 =12 and 4*4=16 

LCM of 12,16 = 48.

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

So, as per the question:

z+7=x ---(i)

u- 9=z ---(ii)

x+ z=v ---(iii)

v+w=110 ---(iv)

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

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

w=57.

Lets assume additional hours = x hrs.

So total No. hours in journey = (4+x)

[(55*4)+(70*x)]/(4+x)=60 => 

=>x=2

Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs.

Simply appreciates variety 1 by 26% and depreciates variety 2 by 26% as:

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

For n = 2 we get both values equal.

Variety 2 Variety1

3969.00 1369 Initially

2937.06 1724.94 after I year

2173.42 2173.42 after II year

So the price become same after 2 years.

26-22=4

39-35=4

52-48=4

65-61=4

LCM of (26,39,52,65)=780

So smallest number = (780-4) = 776.

Lets assume distance between P and Q = d km. 

Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.

Time taken by B in both side = 2d/12 = d/6 hrs.

B tool 10 min. or 1/6 hrs. less than A. 

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

=> (19d-15d)/90 = 1/6

=> 4d/90 = 1/6

=> d = 15/4 km = 3.75 km.

C.P. of the article = Rs. 100

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

SP after the discount = Rs.(125*88)/100 = Rs. 110

therefore Gain percent = 10.

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then

a+b+c+d=58*4=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=56, e=64.

According to baye's theorem:

p(no of way to drawn a ball from 2nd urns is red and ball trferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/45

Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/45

so probability=(9/45)/(17/45)=9/17.