# Esko Aptitude Placement Papers - Esko Aptitude Interview Questions and Answers updated on 26.May.2022

Speed Downstream= (13 + 4) km/hr

= 17 km/hr.

Time taken to travel 68 km downstream  =(68 / 17)hrs

= 4 hrs.

A hours + C ours = B hours .....(i)

A, C and B cannot have values greater than or equal to 24

B minutes + A minutes = C minutes .....(ii)

Looking at two equation, we get no value of A satisfies both equation

A can give B a start of 15 seconds in a km race.

B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec

Therefore, B will cover a distance of = 62.5 meters in 15 seconds.

The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds.

Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds.

This is an alternating addition and subtraction series. In the first pattern, 10 is subtracted from each number to arrive at the next. In the second, 5 is added to each number to arrive at the next

Let us assume the speeds of alvin,ben,clinton = A,B,C resp.

and the length of race = d

let in time t alvin finishes the race

A*t=d

B*t=d-48 ------- (1)

C*t=d-72 --------(2)

So,B:C=(d-48):(d-72) ------- (3)

Now Ben covers 48 mt more to complete the race.

In the same time Clinton covers 40mt (as it is given that he is beaten by 32mt by ben)

So,B:C=48:40....(4) (speed proportional to distance covered if time is same)

By, comparing equation (3) and (4)

d=192 mt

If a runs 200m then b runs 200-20=180m

then c runs 200-38=162m

if b runs 300m then c runs 162/180*300=270

therefore b beats c by 300-270=30

Since we know that the interest rate is 0.15, and knowing that the difference between two years of compound interest is nothing but interest on interest, we can find the first year’s interest as –

45/0.15 = 300.

Now if the interest is 300 at the end of one year, then the principal is 300 / 0.15 = 2,000

Let their marks be (x + 9) and x.

Then, x + 9 = (56/100)(x + 9 + x)

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Days taken in the general scenario = 360/N;

Days taken when 4 articles are prepared extra per day = 360/N + 4;

The difference in the day is one, therefore;

360/n - 360/n+4 =1

N2 + 4N – 1440 = 0;

N = 36,

i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40.

Therefore no of days taken to complete the job = 360/40 = 9.

Let us assume that the cost price of the article = Rs.100

Therefore,

the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e., he offers a

discount of Rs.75 on his selling price of Rs.175.

Therefore, his % discount = (75/175) x 100= 42.85%

Given, 3km or 3000m Difference is 22 second

3 min 40sec = 220 sec

so, 3000*22/220 = 300m

B covers 35m in 7 seconds B take time = (200*7)/35=40

A takes time = (40-7)= 33 Sec.

By that time Anusha covered 100m, Bhanu covered 90m.

So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m.

So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81

By that time Anusha covered 100m, Esha Covers only 81.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will

be 6 significant digits to the right of the decimal point

Let us assume that the cost price of the article = Rs.100

Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 + 75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175

Therefore, his % discount = (75/175) x 100= 42.85%

Let the boat speed in still water be b.

Let the stream speed be x.

2(b+ x) = 3(b-x)

5x=b

b/x=5/1

Work done by all the tanks working together in 1 hour.

=> 1/10 + 1/12 - 1/20 = 2/15

Hence, tank will be filled in 15/2 = 7.5 hour

B covers 100m in 25 seconds B take time =(4000*25)/100=1000 sec=16 min 40 sec.

A takes time =1000 sec-25sec=975 sec= 16 min 25 sec.

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

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.

1992 being a leap year, it has 2 odd days.

So, the day on May, 1993 is 2 days beyond the day on May 6, 1992.

But, on May 6, 1993 it was Thursday.

So, on May 6, 1992 it was Tuesday.

we  can do  a by percentage  method

A can beat b by

100*75/100=75

25 km

75*96/100=72

100-72=28

A can give B 5 points A / B = 80/75

A can give C 15 points A / C = 80/ 65

B can give C a points of ( B / A ) * ( A / C)

(75 / 80) * ( 80 / 65)

(75 / 65) = 15/13

Need to be find in a game of 60 , so multiply numerator and denominator with 4 such that the ratio becomes 60/52

Therefore B can give C of (60-52) = 8 points

A takes time 2.20 min. = 140 Sec. B takes time 2.30 min. = 150 Sec.

Diffrence = (150-140) = 10 Sec.

Now we are to find distance covered in 10 sec by B

150 Sec. = 1500 m.

1 Sec=10 m.

10 Sec = 10*10=100 m

Let the 6 consecutive odd no.’s are:

X, X+2, X+4, X+6, X+8, X+10

Avg. of 1st three no’s is X+2.

Avg. of Last three no’s is X+8.

Given that (X+8)2-(X+2)2=288

X=19

Last Odd no. is X+10= 29.

Ratio of times taken by Sakshi and Tanya = 125:100 = 5:4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x ? x =(4 x 20)/5

? x = 16 days

Hence, Tanya takes 16 days to complete the work.