## Elico Aptitude Interview Questions & Answers

1. Question 1. If In A Long Division Sum,the Dividend Is 380606 And The Successive Remainders From The First To The Last Are 434, 125 And 413, Then The Divisor Is?

Let d = divisor and q = quotient

First remove the last remainder:

d x q + 413 = 380606

d x q = 380193.

2. Question 2. The Sum Of The Two Position Number P And Q Is 2.5 Time Their Difference .if The Product Of The Number 84 Then The What Is The Sum Of Those Two Number?

Let the number be P and Q

According to the question

(P + Q) = 2.5 (P – Q)

P + Q = 2.5P – 25Q

3.5Q = 1.5P

P / Q = 7/3 …..(i)

Now PQ = 84.

3. Electrical and Electronics Engineering Interview Questions

4. Question 3. Out Of Three Given Number The First Number Is Twice The Second And Thrice Number Is 154 Then What Is The Difference Between The First And Third Number ?

Let the third number = N

Then the first number = 3N

And second number = 3N/2

According to the question [N + 3N + 3N/2] / 3=154.

5. Question 4. The Difference Between Two Number Is 18 If Four Times The Second Number Is Less Then Three What Is The Sum These Two Number?

Let first number = P

And second number = Q

According to the question

P – Q = 10 ….(i)

And 3P – 4Q = 18…(ii)

on multiply eq (i) by 3 and then subtract eq (ii) from it

3P – 3Q = 54

3P – 4Q = 18

Q = 36

On putting the value of Q in the eq (i)

P = 18 + Q = 18 +36

P = 54.

6. Question 5. If The Number 653 Xy Is Divisible By 90, Then (x + Y) = ?

90 = 10 x 9

Clearly, 653xy is divisible by 10, so y = 0

Now, 653×0 is divisible by 9.

So, (6 + 5 + 3 + x + 0) = (14 + x) is divisible by 9. So, x = 4.

Hence, (x + y) = (4 + 0) = 4.

7. Aptitude Interview Questions

8. Question 6. A Boy Multiplied 987 By A Certain Number And Obtained 559981 As His Answer. If In The Answer Both 98 Are Wrong And The Other Digits Are Correct, Then The Correct Answer Would Be?

987 = 3 x 7 x 47

So, the required number must be divisible by each one of 3, 7, 47

==> 553681 (Sum of digits = 28, not divisible by 3)

==> 553181 (Sum of digits = 25, not divisible by 3)

==> 555681 are divisible by 3, 7, 47.

9. Question 7. The Difference Between Two Number S 1365. When The Larger Number Is Divided By The Smaller One The Quotient Is 6 And The Remainder Is 15. The Smaller Number Is?

Let the smaller number be x, then larger number = 1365 + x

Therefore 1365 + x = 6x + 15

5x = 1350 ==> x = 270

Required Number is 270.

10. Electrical Engineering Interview Questions

11. Question 8. In Doing A Division Of A Question With Zero Remainder, A Candidate Took 12 As Divisor Instead Of 21. The Quotient Obtained By Him Was 35. The Correct Quotient Is?

Dividend = 12 * 35 =420

Now dividend = 420 and divisor = 21

Therefore correct quotient = 420/21 = 20.

12. Question 9. Find The Sum Of All Odd Number Up To 100?

The given number is 1, 3, 5………….99

This is an A.P with a = 1, d = 2

Let it contain n term 1 + (n-1)2 = 99

==> n = 50

Then required sum = n/2 (first term + last term)

==> 50/2(1 + 99) = 2500.

13. Instrumentational Engineering Interview Questions

14. Question 10. On Dividing A Number By 357, We Get 39 As Remainder. On Dividing The Same Number 17, What Will Be The Remainder?

Let x be the number and y be the quotient. Then,

x = 357 x y + 39

= (17 x 21 x y) + (17 x 2) + 5

= 17 x (21y + 2) + 5)

Required remainder = 5.

15. Question 11. 2056 X 987 =?

2056 x 987 = 2056 x (1000 – 13)

= 2056 x 1000 – 2056 x 13

= 2056000 – 26728

= 2029272.

16. Electronics Interview Questions

17. Question 12. In A Division Sum, The Divisor Is 10 Times The Quotient And 5 Times The Remainder. If The Remainder Is 46, What Is The Dividend?

Divisor = (5 x 46) = 230

10 x Quotient = 230=230= 23/10

Dividend = (Divisor x Quotient) + Remainder

= (230 x 23) + 46

= 5290 + 46

= 5336.

18. Electrical and Electronics Engineering Interview Questions

19. Question 13. What Was The Day Of The Week On 12th January, 1979?

Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.

78 years = (19 leap years + 59 ordinary years)

==> (38 + 59) odd days = 6 odd days

12 days of January have 5 odd days.

Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.

20. Question 14. A Number When Divide By 6 Leaves A Remainder 3. When The Square Of The Number Is Divided By 6, The Remainder Is?

Let x = 6q + 3.

Then, x2 = (6q + 3)2

= 36q2 + 36q + 9

= 6(6q2 + 6q + 1) + 3

Thus, when x2 is divided by 6, then remainder = 3.

21. Question 15. 310 X 999 =?

310 x 999 = 310 x (1000-1)

= 310 x 1000 -310 x 1

= 310000 – 310

= 309690.

22. BHEL Interview Questions

23. Question 16. Persons Can Repair A Road In 12 Days, Working 5 Hours A Day. In How Many Days Will 30 Persons, Working 6 Hours A Day, Complete The Work?

Let the required number of days be x.

Less persons, More days (Indirect Proportion)

More working hours per day, Less days (Indirect Proportion)

Persons30:39Working hours/day6:5}?? 12:x

=> 30 * 6 * x = 39 * 5 * 12

=> x= 13.

24. Question 17. Men Can Complete A Piece Of Work In 18 Days. In How Many Days Will 27 Men Complete The Same Work ?

Less Men, means more Days {Indirect Proportion}

Let the number of days be x then,

27 : 36 :: 18 : x

[Please pay attention, we have written 27 : 36 rather than 36 : 27, in indirect proportion, if you get it then chain rule is clear to you :)]

=>27x = 36 * 18

=> x = 24

So 24 days will be required to get work done by 27 men.

25. Manufacturing Industries Interview Questions

26. Question 18. Some Persons Can Do A Piece Of Work In 12 Days. Two Times The Number Of Such Persons Will Do Half Of That Work In?

Let x men can do the in 12 days and the required number of days be z

More men, Less days     [Indirect Proportion]

Less work, Less days     [Direct Proportion  ]

men2x:xwork1:12} :: 12 : z

(2x×1×z)=(x×12×12)

z=3.

27. Aptitude Interview Questions

28. Question 19. If The Cost Price Of A Certain Object Doubles, Then The Loss Gets Tripled Of What It Was Initially. The Initial Loss % Was ?

Let the cost price be x and selling price be y.

Loss = x – y

When the cost price doubles, the loss gets tripled.

So it becomes like this, 2x – y = 3(x-y)

=> x = 2y

Loss % = (loss/ C.P) x 100 = [(2y-y)/2y] x 100= 50 %.

29. Question 20. A Quantity Of Tea Is Sold At Rs. 5.75 Per Kilogram. The Total Gain By Selling The Tea At This Rate Is Rs. 60. Find The Quantity Of Tea Being Sold If A Profit Of 15% Is Made On The Deal ?

Say total cost price of tea is x.

Then total profit at a rate of 15% is = (15x/100)

According to question,

15x/100 = 60

so x = 400

C.p of the tea is Rs. 400.

so total selling price will be = (400+60) = Rs.460

so the quantity of the tea will be = (460/5.75) = 80kg.

30. ELECTRONICS & INSTRUMENTATION Engineering Interview Questions

31. Question 21. A Trader Marked The Price Of The T.v. 30% Above The Cost Price Of The T.v. And Gave The Purchaser 10% Discount On The Marked Price, Thereby Gaining Rs.340. Find The Cost Price Of The T.v ?

Let ‘x’ be the cost price.

Now Marked price = x + 30x/100 = 13x/10

10% discount = 10/100 x 13x/10 = 13x/100

Selling price = 13x/10 – 13x/100 = 117x/100

Given gain = 340

Here gain = 117x/100 – x = 17x/100 = 340 => x = Rs. 2000.

32. Question 22. A Driver Of Auto Rickshaw Makes A Profit Of 20% On Every Trip When He Carries 3 Passengers And The Price Of Petrol Is Rs. 30 A Litre. Find The % Profit For The Same Journey If He Goes For 4 Passengers Per Trip And The Price Of Petrol Reduces To Rs. 24 Litres ?

When 3 passengers income was 3x

expense= Rs.30

profit =20% of 30 = Rs.6

That means his earning is Rs.36. so that per passenger fare must be Rs.12.

When 4 passengers

earning = 12×4=Rs.48.

expense =Rs.24.

profit = Rs.24 = 100%.

33. Question 23. The Cost Price Of An Article Is 54% Of The Marked Price. Calculate The Gain Percent After Allowing A Discount Of 15%?

Let marked price = Rs. 100.

Then, C.P. = RS. 54,

S.P. = Rs. 85

Gain % = 31/64 x 100 = 48.4%.

34. Analogue electronics Interview Questions

35. Question 24. A Shopkeeper Who Deals In Books Sold A Book At 16% Loss. Had She Charged An Additional Rs.60 While Selling It , Her Profit Would Have Been 14%. Find The Cost Price, In Rupees, Of The Book ?

Let the C.P be ‘x’

Then, the selling price S.P = x – 16x/100

= 84x/100 = 21x/25

Now, if the S.P is 60 more, then the profit is 14%

=> 21x/25 + 60 = x + 14x/100

=> 114x/100 – 21x/25 = 60

=> (57 – 42)x/50 = 60

=> 15x/50 = 60

x = 3000/15 = 200

Therefore, the Cost price C.P = x = Rs. 200.

36. Electrical Engineering Interview Questions

37. Question 25. Sambhu Buys Rice At Rs. 10/kg And Puts A Price Tag On It So As To Earn A Profit Of 20%. However, His Faulty Balance Shows 1000 Gm When It Is Actually 800 Gm. What Is His Actual Gain Percentage ?

CP of 1000gm = Rs. 10

SP of 800gm = Rs. 12

SP of 1000gm =12×1000/800 = Rs. 15

Now take 1000gm as reference to calculate profit.

Profit=SP-CP=15-10=Rs. 5

Profit % = 5×100/10 = 50%.

38. Question 26. In A Certain Business, The Profit Is 220% Of The Cost. If The Cost Increases By 25% But The Selling Price Remains Constant, Approximately What Percentage Of The Selling Price Is The Profit ?

Let C.P.= Rs. 100.

Then, Profit = Rs.220,

S.P. = Rs.320.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs.320.

Profit = Rs. (320 – 125) = Rs. 195

Required percentage = 195/320×100== 60.9 =~ 61%.

39. BHEL Aptitude Interview Questions

40. Question 27. Arun Purchased 30 Kg Of Wheat At The Rate Of Rs. 11.50 Per Kg And 20 Kg Of Wheat At The Rate Of 14.25 Per Kg. He Mixed The Two And Sold The Mixture. Approximately What Price Per Kg Should Be Sell The Mixture To Make 30% Profit ?

C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.

S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.

S.P. per kg = 819/50 = Rs. 16.38 = 16.30.

41. Instrumentational Engineering Interview Questions

42. Question 28. A Merchant Buys Two Items For Rs. 7500. One Item He Sells At A Profit Of 16% And The Other Item At 14% Loss. In The Deal, The Merchant Makes Neither Any Profit Nor Any Loss. What Is The Difference Between The Selling Price Of Both The Items?

Let the C.P of one item is Rs. P

and that of other is Rs. (7500 – P)

According to the data given

C.P = S.P

=> Px(116/100) + (7500-P)x(86/100) = 7500

=> 30P = 105000

=> P = 3500

Required difference between selling prices

= Rs. [(3500/100) x 116] – [(4000/100) x 86]

= 4060-3440

= Rs. 620.

43. Question 29. If Books Bought At Prices Ranging From Rs 200 To Rs 350 Are Sold At Prices Ranging From Rs 300 To Rs 425, What Is The Greatest Possibleprofit That Might Be Made In Selling Eight Books?

least cost price      = 200*8 = 1600

greatest sold price = 425 * 8 = 3400

profit required = 3400- 1600 = 1800.

44. Sony India Aptitude Interview Questions

45. Question 30. A Supplier Supplies Cartridges To A News Paper Publishing House. He Earns A Profit Of 25% By Selling Cartridges For Rs. 1540. Find The Cost Price Of The Cartridges ?

Let Cost Price(C.P) = P

gain% = {(S.P-C.P)/C.P} x 100

25 = {(1540-P)/P} x 100

25/100 = (1540-P)/P

=> P = 4(1540)-4P

=> 5P = 4(1540)

=> P = 1232

So, Cost Price = Rs. 1232.

46. Question 31. Selling An Article At A Profit Of 20%, Aman Gets Rs. 400 More Than Selling At A Loss Of 20%. The Cost Price Of The Article Is ?

Cost price of the article is given by

= 400×100/(20+20)

= Rs.1000.

47. Question 32. Uma Sold An Article For 3400 And Got A Profit Of 25%. If He Had Sold The Article For Rs. 3265. How Much Profit% Would Uma Have Got ?

125% —- 3400

=> 100% —- ?

=> ? = 3400×100/125 = 2720

=> Cost price of the article = Rs. 2720

Profit when article sold at Rs. 3265 = 3265 – 2720 = 545

Hence, Profit% = Gain x 100/cost price

=> P% = 545 x 100/2720

=> P% = 20%.

48. ELGI Aptitude Interview Questions

49. Question 33. If The Simple Interest On A Sum Of Money For 2 Years At 5% Per Annum Is Rs. 50, What Is The Compound Interest On The Same At The Same Rate And For The Same Time?

Sum = Rs.(50*100)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

C.I = Rs.(551.25-500)= Rs.51.25.

50. Electronics Interview Questions

51. Question 34. Albert Invested An Amount Of Rs.8000 In A Fixed Deposit Scheme For 2 Years At Compound Interest Rate 5 P.c.p.a. How Much Amount Will Albert Get On Maturity Of The Fixed Deposit ?

Amount

=Rs.[8000x(1+5/100)²]

= Rs.[8000 x 21/20×21/20]

= Rs.8820.

52. Question 35. Simon Deposits \$400 In An Account That Pays 3% Interest Compounded Annually. What Is The Balance Of Simon’s Account At The End Of 2 Years?

I=Prt

I=12

Balance = P +Prt

412

Find the balance at the end of the second year.

I = Prt=12.36

Balance =P + Prt

424.36.

53. Question 36. The Compound Interest On Rs. 30,000 At 7% Per Annum Is Rs. 4347. The Period (in Years) Is?

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000*(1+7/100)^n=34347

n= 2 years.

54. BHEL Interview Questions

55. Question 37. An Amount Of 5,000 Is Invested At A Fixed Rate Of 8 Per Cent Per Annum. What Amount Will Be The Value Of The Investment In Five Years Time, If The Interest Is Compounded Annually?

The only part of this type of calculation that needs particular

care is that concerning the interest rate. The formula assumes that

r is a proportion, and so, in this case:

r = 0.08

In addition, we have P = 5,000 and n = 5, so:

V = P(1 + r)5 = 5,000 x (1 + 0.08)5 = 5,000 x 1.469328 = 7,346.64

Thus the value of the investment will be 7,346.64.

56. Question 38. A Sum Is Equally Invested In Two Different Schemes On Ci At The Rate Of 15% And 20% For Two Years. If Interest Gained From The Sum Invested At 20% Is Rs. 528.75 More Than The Sum Invested At 15%, Find The Total Sum?

Let Rs. K invested in each scheme

Two years C.I on 20% = 20 + 20 + 20×20/100 = 44%

Two years C.I on 15% = 15 + 15 + 15×15/100 = 32.25%

Now,

(P x 44/100) – (P x 32.25/100) = 528.75

=> 11.75 P = 52875

=> P = Rs. 4500

Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.

57. Question 39. Find The Compound Interest On Rs. 2680 At 8% Per Annum For 2 Years ?

We know Compound Interest = C.I. = P1+r100t – 1

Here P = 2680, r = 8 and t = 2

C.I. = 26801 + 81002-1= 268027252-12= 26802725+12725-1= 2680 5225×225

= (2680 x 52 x 2)/625

= 445.95

Compound Interest = Rs. 445.95.

58. Question 40. Find The Simple Interest Of Rs.14000 At 10% Per Annum For 3 Months ?

The formula to calculate simple interest is :

SI=(P x T x R)/100

In the given question Principal(P)=Rs. 14000, Time(T)=(3/12)yr. and Rate of interest(R)=10%

So, SI=(14000 x 1/4 x 10)/100

SI= Rs. 350.

59. Manufacturing Industries Interview Questions

60. Question 41. A Train Travels 325 Km In 3.5 Hours And 470 Km In 4 Hours. Find The Average Speed Of Train ?

As we know that Speed = Distance / Time

for average speed = Total Distance / Total Time Taken

Thus, Total Distance = 325 + 470 = 795 km

Thus, Total Speed = 7.5 hrs

Average Speed = 795/7.5 => 106 kmph.

61. Question 42. If A Student Walks At The Rate Of 4 Mts/min From His Home, He Is 4 Minutes Late For School, If He Walks At The Rate Of 6 Mts/min He Reaches Half An Hour Earlier. How Far Is His School From His Home ?

Let the distance between home and school is ‘x’.

Let actual time to reach be ‘t’.

Thus, x/4 = t + 4 —- (1)

and x/6 = t – 30 —–(2)

Solving equation (1) and (2)

=> x = 98 mts.

62. ELECTRONICS & INSTRUMENTATION Engineering Interview Questions

63. Question 43. An Express Traveled At An Average Speed Of 100 Km/hr, Stopping For 4 Min After Every 75 Km. How Long Did It Take To Reach Its Destination 600 Km From The Starting Point ?

Time taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 – 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, total time taken = 6 hrs 28 min.

64. Question 44. ‘k’ Cycled From A To B At 10 Kmph And Returned At The Rate Of 9 Kmph. ‘l’ Cycled Both Ways At 12 Kmph. In The Whole Journey ‘l’ Took 10 Minutes Less Than ‘k’. Find The Distance Between A And B ?

Let ‘d’ be the distance between A and B

K -time = d/10 + d/9 = 19d/90 hours

L -time = 2d/12 = d/6 hours

We know that, 10 min = 1/6 hours

Thus, time difference between K and L is given as 10 minutes.

=> 19d/90 – d/6 = 1/6

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

=> 4d/90 = 1/6

Thus,

d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

65. Question 45. You Drive To The Store At 20 Kmph And Return By The Same Route At 30 Kmph. Discounting The Time Spent At The Store, What Was Your Average Speed ?

Average speed=total distance/total time

Let distance to store be  K

then, total time =(K/20)+(K/30)=K/12

and, total time =(2K)

so average speed= 2K / (K/12) = 24kmph.

66. Question 46. K Is 50% Faster Than L. If L Starts At 9 A.m. And K Starts At 10 A.m. L Travels At A Speed Of 50 Km/hr. If L And K Are 300 Kms Apart, The Time When They Meet When They Travel In Opposite Direction Is ?

let ‘t’ be the time after which they met since L starts.

Given K is 50% faster than L

50 t + 1.5*50(t-1) = 300

50 t +75 t = 300 + 75

t = 375 / 125 = 3 hrs past the time that L starts

So they meet at (9 + 3)hrs = 12:00 noon.

67. Question 47. Laxmi And Prasanna Set On A Journey. Laxmi Moves Northwards At A Speed Of 20kmph And Prasanna Moves Southward At A Speed Of 30 Kmph. How Far Will Be Prasanna From Laxmi After 60 Minutes ?

We know 60 min = 1 hr

Total northward Laxmi’s distance = 20kmph x 1hr = 20 km

Total southward Prasanna’s distance = 30kmph x 1hr = 30 km

Total distance between Prasanna and Laxmi is = 20 + 30 = 50 km.

68. Question 48. Joel Travels The First 3 Hours Of His Journey At 60 Mph Speed And The Remaining 5 Hours At 24 Mph Speed. What Is The Average Speed Of Joel’s Travel In Kmph ?

Average speed = Total distance / Total time.

Total distance traveled by Joel = Distance covered in the first 3 hours + Distance covered in the next 5 hours.

Distance covered in the first 3 hours = 3 x 60 = 180 miles

Distance covered in the next 5 hours = 5 x 24 = 120 miles

Therefore, total distance traveled = 180 + 120 = 300 miles.

Total time taken = 3 + 5 = 8 hours.

Average speed = 300/8 = 37.5 mph.

we know that 1 mile = 1.6 kms

=> 37.5 miles = 37.5 x 1.6 = 60 kms

Average speed in Kmph = 60 kmph.

69. Question 49. A Train-a Passes A Stationary Train B And A Pole In 24 Sec And 9 Sec Respectively. If The Speed Of Train A Is 48 Kmph, What Is The Length Of Train B?

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 – 120

=> 320 – 120 = 200 mts.

70. Question 50. Tilak Rides On A Cycle To A Place At Speed Of 22 Kmph And Comes Back At A Speed Of 20 Kmph. If The Time Taken By Him In The Second Case Is 36 Min. More Than That Of The First Case, What Is The Total Distance Travelled By Him (in Km)?

Let the distance travelled by Tilak in first case or second case = d kms

Now, from the given data,

d/20 = d/22 + 36 min

=> d/20 = d/22 + 3/5 hrs

=> d = 132 km.

Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.

71. Question 51. Two Trains Are Running With Speeds 30 Kmph And 58 Kmph Respectively In The Same Direction. A Man In The Slower Train Passes The Faster Train In 18 Seconds. Find The Length Of The Faster Train?

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 – 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts.

72. Question 52. A Car Covers Its Journey At The Speed Of 80km/hr In 10hours. If The Same Distance Is To Be Covered In 4 Hours, By How Much The Speed Of Car Will Have To Increase?

Initial speed = 80km/hr

Total distance = 80 * 10 = 800km

new speed = 800/4 =200km/hr

Increase in speed = 200 – 80 = 120km/hr.

73. Question 53. A Monkey Climbs Up A Greased Pole, Ascends 20m And Slip 4m In Alternate Minutes. If The Pole Is 96m High, How Many Minutes Will It Take To Reach The Top?

net height climbed in 2 min = 20m – 4m = 16m

In 10 min the net height climbed = 16 * 10/2 = 80m

Remaining height = 96m – 80m = 16m

In the 11th min , the monkey will be ascending up.

Time taken to ascend the last 16min

16/20 = 4/5min

Total time taken = 10 + 4/5 = 54/5 min.

74. Question 54. A Car Is Running At 7/10 Of Its Own Speed Reached A Place In 22 Hours. How Much Time Could Be Saved If The Train Would Run At Its Own Speed ?