Interview Questions.

Esko Aptitude Placement Papers - Esko Aptitude Interview Questions and Answers

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Esko Aptitude Placement Papers - Esko Aptitude Interview Questions and Answers

Q1. A Boat Can Travel With A Speed Of 13 Km / Hr In Still Water. If The Speed Of The Stream Is four Km / Hr. Find The Time Taken By The Boat To Go sixty eight Km Downstream?

Speed Downstream= (thirteen + 4) km/hr

= 17 km/hr.

Time taken to journey 68 km downstream  =(sixty eight / 17)hrs

= four hrs.

Q2. A Train Left Station At A Hour B Minutes. It Reached Station Y At B Hour C Minutes On The Same Day, After Travelling C Hours A Minutes (clock Shows Time From zero Hours To 24 Hours). Number Of Possible V

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

A, C and B cannot have values extra than or identical to 24

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

Looking at  equation, we get no price of A satisfies both equation

Q3. A Takes 3 Min forty five Seconds To Complete A Kilometre. B Takes four Minutes To Complete The Same 1 Km Track. If A And B Were To Participate In A Race Of 2 Kms, How Much Start Can A Give B In Terms Of Distance

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

B takes four mins to run a km. I.E one thousand/4= 250 m/min = 250/60 m/sec 

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

The begin that A can deliver B in a km race therefore, is 62.Five meters, the gap that B run in 15 seconds. 

Hence in a 2 km race, A can deliver B a start of sixty two.Five * 2 = a hundred twenty five m or 30 seconds.

Q4. Look At This Series: 80, 10, 70, 15, 60, ... What Number Should Come Next ?

This is an alternating addition and subtraction collection. In the primary pattern, 10 is subtracted from each wide variety to arrive at the subsequent. In the second one, five is brought to every variety to arrive at the subsequent

Q5. Alvin, Ben And Clinton Run A Race, With Alvin Finishing 48 Meters Ahead Of Ben And seventy two Meters Ahead Of Clinton, While Runner Ben Finishes 32 Meters Ahead Of Runner Clinton. Each Runner Travels The Enti

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

And the length of race = d

permit in time t alvin finishes the race

A*t=d

B*t=d-forty eight ------- (1)

C*t=d-seventy two --------(2)

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

Now Ben covers forty eight mt extra to complete the race.

In the equal time Clinton covers 40mt (as it's miles for the reason that he's crushed by way of 32mt by means of ben)

So,B:C=forty eight:40....(4) (speed proportional to distance included if time is identical)

By, evaluating equation (three) and (4)

d=192 mt

Q6. In A 200 M Race, A Beats B By 20 M And C By 38m. In A Race Of 300 M B Will Beat C By

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

then c runs 200-38=162m

if b runs 300m then c runs 162/one hundred eighty*three hundred=270

therefore b beats c by 300-270=30

Q7. The Difference Between The Simple Interest And Compound Interest On A Certain Sum Of Money For 2 Years At 15% P. A. Is Rs. 4

Since we know that the interest fee is zero.15, and understanding that the distinction among  years of compound interest is not anything but hobby on hobby, we can find the primary yr’s hobby as –

45/zero.15 = 300.

Now if the interest is three hundred on the give up of three hundred and sixty five days, then the essential is three hundred / zero.15 = 2,000

Q8. Two Students Appeared At An Examination. One Of Them Secured nine Marks More Than The Other And His Marks Was 56% Of The Sum Of Their Marks. The Marks Obtained By Them Are:

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

Then, x + 9 = (fifty six/one hundred)(x + nine + x)

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

3x = ninety nine

x = 33

So, their marks are 42 and 33.

Q9. A Team Of Workers Was Employed By A Contractor Who Undertook To Finish 360 Pieces Of An Article In A Certain Number Of Days. Making Four More Pieces Per Day Than Was Planned, They Could Complete The J

Days taken inside the general scenario = 360/N;

Days taken whilst 4 articles are organized extra according to day = 360/N + 4;

The difference within the day is one, therefore;

360/n - 360/n+4 =1

N2 + 4N – 1440 = 0;

N = 36,

i.E. Range of item organized in wellknown situation is 36, and wherein four articles prepared greater is 40.

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

Q10. A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss ?

Let us count on that the value fee of the article = Rs.One hundred

Therefore,

the merchant might have marked it to Rs.A hundred + 75% of Rs.A hundred = one hundred +75 = one hundred seventy five.

Now, if he sells it at no earnings or loss, he sells it at the value rate.I.E., he gives a

discount of Rs.75 on his promoting price of Rs.One hundred seventy five.

Therefore, his % discount = (75/175) x one hundred= forty two.Eighty five%

Q11. A Can Run three Kms In three Min 18 Sec And B Can Run Same Distance In 3 Min forty Sec, Then By How Much Distance A Can Beat B ?

Given, 3km or 3000m Difference is 22 2d

3 min 40sec = 220 sec

so, 3000*22/220 = 300m

Q12. In A 200 M Race A Beats B By 35 M Or 7 Seconds. A's Time Over The Cause Is ?

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

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

Q13. Anusha, Banu And Esha Run A Running Race Of one hundred Meters. Anusha Is The Fastest Followed By Banu And Then Esha. Anusha, Banu And Esha Maintain Constant Speeds During The Entire Race. When Anusha Reached

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

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

So ratio in their speeds = 10 : nine Ratio of the velocity of all the 3 = 100 : 90 : eighty one 

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

Q14. How Many Digits Will Be There To The Right Of The Decimal Point In The Product Of ninety five.Seventy five And .02554?

Sum of decimal places = 7.

Since the final digit to the intense right could be zero (since 5 x four = 20), so there'll

be 6 huge digits to the right of the decimal factor

Q15. A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss?

Let us expect that the price rate of the thing = Rs.A hundred

Therefore, the service provider might have marked it to Rs.One hundred + 75% of Rs.A hundred = 100 + 75 = a hundred seventy five.

Now, if he sells it at no profit or loss, he sells it at the price rate.I.E. He gives a reduction of Rs.Seventy five on his promoting price of Rs.175

Therefore, his % bargain = (75/175) x one hundred= 42.85%

Q16. Twice The Speed Of A Boat Downstream Is Equal To Thrice The Speed Upstream. The Ratio Of Its Speed In Still Water To The Speed Of Current Is

Let the boat pace in nevertheless water be b.

Let the movement speed be x.

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

5x=b

b/x=five/1

Q17. Two Pipes Can Fill The Cistern In 10hr And 12 Hr Respectively, While The Third Empty It In 20hr. If All Pipes Are Opened Simultaneously, Then The Cistern Will Be Filled In

Work finished by means of all the tanks operating together in 1 hour.

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

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

Q18. In A Race Of four Kms A Beats B By one hundred M Or 25 Seconds, Then Time Taken By A Is

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

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

Q19. A Sum Of Money At Simple Interest Amounts To Rs. 815 In 3 Years And To Rs. 854 In 4 Years. The Sum Is:

S.I. For 1 12 months = Rs. (854 - 815) = Rs. 39.

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

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

Q20. 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 eight Meters Ahead Of Runner C. Each Runner Travels The E

Lets anticipate distance of race = x mtrs. 

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

=> at this point B is 6 m beforehand of C. Now to finish race b needs to run another 12 m, 

=> he runs another 12 m. Whilst B finishes race he is eight m ahead of C. 

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

=> x-12/ x-18 = 12/10 => x = forty eight mtrs.

Q21. May 6, 1993 Was Thursday. What Day Of The Week Was On May 6, 1992 ?

1992 being a soar year, it has 2 extraordinary days.

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

But, on May 6, 1993 it was Thursday.

So, on May 6, 1992 it become Tuesday.

Q22. In A a hundred M Race, A Can Beat B By 25 M And B Can Beat C By 4 M. In The Same Race, A Can Beat C By ?

We  can do  a through percent  method 

A can beat b via

a hundred*75/one hundred=seventy five

25 km

75*96/a hundred=seventy two

one hundred-72=28

Q23. In A Game Of eighty Points. A Can Give B 5 Points And C 15 Points. Then How Many Points B Can Give C In A Game Of 60 ?

A can deliver B five points A / B = 80/seventy five

A can supply C 15 factors A / C = eighty/ 65

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

(75 / eighty) * ( 80 / sixty five)

(75 / sixty five) = 15/13

Need to be find in a sport of 60 , so multiply numerator and denominator with four such that the ratio will become 60/fifty two

Therefore B can give C of (60-fifty two) = eight points

Q24. A Can Run 1.5 Km Distance In 2 Min 20 Seconds, While B Can Run This Distance In 2 Min 30 Sec. By How Much Distance Can A Beat B ?

A takes time 2.20 min. = 140 Sec. B takes time 2.30 min. = one hundred fifty Sec.

Diffrence = (a hundred and fifty-a hundred and forty) = 10 Sec.

Now we're to discover distance included in 10 sec through B

a hundred and fifty Sec. = 1500 m.

1 Sec=10 m.

10 Sec = 10*10=a hundred m

Q25. There Are 6 Consecutive Odd Numbers. The Difference Between The Square Of The Average Of The First Three Numbers And The Square Of The Average Of The Last Three Numbers Is 28

Let the 6 consecutive unusual no.’s are:

X, X+2, X+four, X+6, X+eight, X+10

Avg. Of 1st 3 no’s is X+2.

Avg. Of Last three no’s is X+8.

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

X=19

Last Odd no. Is X+10= 29.

Q26. Sakshi Can Do A Piece Of Work In 20 Days. Tanya Is 25% More Efficient Than Sakshi. The Number Of Days Taken By Tanya To Do The Same Piece Of Work Is:

Ratio of times taken with the aid of Sakshi and Tanya = a hundred twenty five:a hundred = five:4.

Suppose Tanya takes x days to do the paintings.

Five : 4 :: 20 : x ? X =(4 x 20)/5 

? X = sixteen days

Hence, Tanya takes sixteen days to finish the paintings.




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