HCL Aptitude Interview Questions & Answers

  1. Question 1. Two Tailors X And Y Are Paid A Total Of Rs. 550 Per Week By Their Employer. If X Is Paid 120 Percent Of The Sum Paid To Y, How Much Is Y Paid Per Week?

    Answer :

    Let, the sum paid to Tailor X per week be Rs. X

    and the sum paid to Tailor Y per week be Rs. Y

    Given, two tailors X and Y are paid a total of Rs. 550 per week

    => X + Y = 550 —> eqn (1)

    Given, X is paid 120 percent of the sum paid to Y

    => X = 120 % of Y

    => X = (120 / 100) * Y

    => X = (6 / 5) * Y

    On substituting this value for X in eqn (1), we get

    => (6 / 5) * Y + Y = 550

    => (6Y + 5Y) / 5 = 550

    => 11Y = 550 * 5

    => 11Y = 2750

    => Y = 2750 / 11

    => Y = 250 Rs.

    Thus, the sum paid to Tailor Y per week be Rs. 250

  2. Question 2. An Article Worth Rs. 1200 Is Given Two Successive Discounts Of 10 % And 10 % Respectively. What Is The Percentage Of Discount Which Is Equivalent To Give As Single Discount?

    Answer :

    Given, two successive discount of 10 %.

    Let, x = First discount = 10%, y = second discount = 10%

    Total Discount % = (x + y – [xy/100]) %

                              = (10 + 10 – [10×10/100]) %

                              = (20 – [100/100]) %

                              = (20 – 1) %

                              = 19%

    Total Discount % which is equivalent for two successive discount of 10 % = 19%

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  4. Question 3. A Family Consists Of Two Grandparents, Two Parents And Three Grandchildren. The Average Age Of The Grandparents Is 67 Years, That Of The Parents Is 35 Years And That Of The Grandchildren Is 6 Years. What Is The Average Age Of The Family?

    Answer :

    Given

    The family consists of two grandparents, two parents and three grandchildren

    The average age of two grandparents = 67 years.

    => Total age of two grandparents = (67 * 2) = 134 years

    The average age of two parents = 35 years

    => Total age of two parents = (35 * 2) = 70 years

    The average age of three grand children = 6 years.

    => Total age of three grand children = (6 * 3) = 18 years.

    Required Average age of the family = Total age of (two grandparents + two parents + three grand children) / Total members in the family

                                                        = (134 + 70 + 18) / (2 + 2 + 3)

                                                        = 222 / 7

                                                        = 31 (5 / 7) years

  5. Question 4. If 13:11 Is The Ratio Of Present Age Of Jothi And Viji Respectively And 15:9 Is The Ratio Between Jothi’s Age 4 Years Hence And Viji’s Age 4 Years Ago. Then What Will Be The Ratio Of Jothi’s Age 4 Years Ago And Viji’s Age 4 Years Hence?

    Answer :

    Let the present age of Jothi and Viji be 13X and 11X respectively.

    Given, Jothi’s age 4 years hence and Viji’s age 4 years ago in the ratio 15:9.

    That is, (13X + 4) / (11X – 4) = 15 / 9

    => 9 (13X + 4) = 15 (11X – 4)

    => 117X + 36 = 165X – 60

    => 165X – 117X = 60 + 36

    => 48X = 96

    => X = 96 / 48

    => X = 2

    Now, Required ratio = (13X – 4) / (11X + 4)

    on substituting value of X = 2 we get,

    = [13(2)-4] / [11(2)+4]

    = 22/26

    = 11/13

    Hence the answer is 11:13

  6. Question 5. The Average Age Of A Group Of 10 Students Was 14. The Average Age Increased By 1 Year When Two New Students Joined The Group. What Is The Average Age Of The Two New Students Who Joined The Group?

    Answer :

    The average age of a group of 10 students is 14.

    Therefore, the sum of the ages of all 10 of them = 10 * 14 = 140

    When two students join the group, the average increase by 1. New Average = 15

    Now there are 12 students Therefore, sum of all the ages of 12 students = 15 x 12 = 180

    Therefore, the sum of the ages of two students who joined = 180 – 140 = 40

    And the average age of these two students = 20

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  8. Question 6. The Average Age Of A Group Of 10 Students Was 10. The Average Age Increased By 1 Year When Two New Students Joined The Group. What Is The Average Age Of The Two New Students Who Joined The Group?

    Answer :

    The average age of a group of 10 students is 10.

    Therefore, the sum of the ages of all 10 of them = 10 * 10 = 100

    When two students join the group, the average increase by 1.

    => New Average = 11

    Now there are 12 students.

    Therefore, sum of all the ages of 12 students = 11 x 12 = 132

    Therefore, the sum of the ages of two students who joined = 132 – 100 = 32

    And the average age of these two students = (32 / 2) = 16

  9. Question 7. In The First 30 Overs Of A Cricket Game, The Run Rate Was Only 4.5. What Should Be The Run Rate In The Remaining 20 Overs To Reach The Target Of 320 Runs?

    Answer :

    Required Run rate = (320 – (4.5 x 30) )/ (20) = ( 185) /(20) = 9.25

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  11. Question 8. The Ratio Of Number Of Boys And Girls In A School Of 720 Students Is 7:5. How Many More Girls Should Be Admitted To Make The Ratio 1:1?

    Answer :

    Given, boys: girls = 7: 5

    Let, the total number of boys = 7x and total number of girls= 5x

    Given, total students = 720

    => 7x + 5x = 720

    => 12x = 720

    => x = 60

    So, total number of boys = 7x = 7 * 60 = 420 and

    total number of girls= 5x = 5 * 60 = 300

    Let y be the number of girls added to make the ratio 1 : 1

    => 420 / (300 + y) = 1/1

    => 420 = (300 + y)

    => y = 420 – 300

    => y = 120

    So, 120 more girls should be admitted to make the ratio 1:1

  12. Question 9. If Two Numbers Are In The Ratio 6: 13 And Their Least Common Multiple Is 312, The Sum Of The Numbers Is?

    Answer :

    Let the two numbers be 6k and 13k

    LCM of 6k and 13k = 78k

    =>78k = 312

    => k = 4

    Sum of the numbers 6k + 13k = 19k = 19 * 4 = 76

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  14. Question 10. A 20 Liters Mixture Of Milk And Water Comprising 60% Pure Milk Is Mixed With “x” Liters Of Pure Milk. The New Mixture Comprises 80% Milk. What Is The Value Of “x”?

    Answer :

    Original mixture comprises 20 liters of milk and water.

    Out of the 20 liters, 60% is pure milk.

    => (60 / 100) x 20 = pure milk

    => 12 liters = pure milk

    In 20 liters mixture remaining 8 liters = water

    When “x” liters of pure milk is added to 20 liters of mixture

    New mixture = (20 + x) liters

    Milk in new mixture = (12 + x) liters

    Given milk in new mixture = 80% of (20 + x)

    => 12 + x = (80 / 100) * (20 + x)

    => 12 + x = (4 / 5) * (20 + x)

    => 5 (12 + x) = 4 (20 + x)

    => 60 + 5 x = 80 + 4 x

    => 5 x – 4 x = 80 – 60

    => x = 20 liters

  15. Question 11. A Zookeeper Counted The Heads Of The Animals In A Zoo And Found It To Be 80. When He Counted The Legs Of The Animals He Found It To Be 260. If The Zoo Had Either Pigeons Or Horses, How Many Horses Were There In The Zoo?

    Answer :

    Given, Zoo had either pigeons or horses

    Heads of the animals in Zoo = 80

    => pigeons + horses = 80

    Let p = number of pigeons

    h = number of horses

    => p = 80 – h

    Given, Legs of the animals = 260

    Each pigeon has 2 legs and each horse has 4 legs

    => 2p + 4h = 260

    Substitute p = 80 – h

    => 2 (80 – h) + 4h = 260

    => 160 – 2h + 4h = 260

    => 2h = 260 – 160

    => 2h = 100

    h = 50

    So, number of horses in the Zoo = 50

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  17. Question 12. How Many Liters Of Water Should Be Added To A 30 Liter Mixture, Containing Milk And Water In The Ratio 7:3 Such That The Resultant Mixture Has 40 % Water In It?

    Answer :

    Given: 30 Liters of mixture, Milk and water in the ratio 7 : 3 Which means, we have 21 liters of milk and 9 liters of water.

    We add water the resulting solution is 21 liters of milk and 9 + x liters of water.

    Total quantity = 30 +x.

    Water percentage is 40 % = > 40 x (30+x)/100 = 9 + x

    =>4(30+x) = 10(9+x)

    => 120 + 4x = 90 + 10x

    => 10x – 4x = 120-90

    =>6x = 30

    => x = 5

    Thus the quantity of water added = 5 liters

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  19. Question 13. How Many Liters Of A 12 Litre Mixture Containing Milk And Water In The Ratio Of 2: 3 Are Replaced With Pure Milk So That The Resultant Mixture Contains Milk And Water In Equal Proportion?

    Answer :

    The mixture contains 40% milk and 60% water in it.

    That is 4.8 liters of milk and 7.2 liters of water.

    Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.

    That is we will end up with 6 liters of milk and 6 liters of water. Water gets reduced by 1.2 liters.

    To remove 1.2 liters of water from the original mixture containing 60% water, we need to remove 1.2 / 0.6 liters of the mixture = 2 liters.

  20. Question 14. A And B Invested In A Business In The Ratio 2:3 And The Ratio Of Their Period Of Investment Is 4:5. Then Their Profit Ratio Is?

    Answer :

    Share’s ratio = A: B = 2: 3 

    Time ratio = A: B = 4: 5 

    Then, profit ratio = A: B = 2*4: 3*5 = 8: 15

  21. Question 15. A, B Invested Rs.20, 000/- And Rs.25, 000/- Respectively In A Business. The 20% Of Profits Goes To Charities. The Rest Being Divided In Proportion To Their Capitals Out Of A Total Profit Of Rs.9000/-. The A’s Share Is?

    Answer :

    A = Rs.20, 000/- & B = Rs.25, 000/- 

    ==> A: B = 20: 25 ==> 4: 5 

    Total profit = Rs.9000/- 

    20 % of profit goes to charities ===> Rs.9000/- – Rs.1800/- = Rs.7200/- 

    remaining amount = Rs.7200/- 

    Total parts = 9 

    9 parts —–> Rs.7200/- 

    1 part ——> Rs.800/- 

    A’s share ===> Rs.800/- * 4 parts = Rs.3200/

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  23. Question 16. A Starts A Business With A Capital Of Rs. 85,000. B Joins In The Business With Rs.42500 After Some Time. For How Much Period Does B Join, If The Profits At The End Of The Year Are Divided In The Ratio Of 3: 1?

    Answer :

    Let B joins for x months.

    Given, A’s capital = Rs. 85,000 for 12 months

    B’s capital = Rs. 42,500 for x months

    Given, Ratio of profit = 3: 1

    => Ratio of investment =Ratio of profit

    => (85000 * 12): (42500 * x) = 3: 1

    => (850 * 12): (425 * x) = 3: 1

    => (2 * 12): (x) = 3: 1

    => 24 / x = 3/1

    => x = 24/3

    => x = 8

    Therefore, No. of months “B” in the business = 8 months

  24. Question 17. A And B Invest In A Business In The Ratio 3: 2. Assume That 5% Of The Total Profit Goes To Charity. If A’s Share Is Rs. 855, What Is The Total Profit?

    Answer :

    Assume that the total profit is x.

    Since 5% goes for charity, 95% of x will be divided between A and B in the ratio 3: 2

    Given, A’s share is Rs. 855

    => A’s profit = (95x/100) * (3/5) = 855

    => (95x/100) * (3/5) = 855

    => 19x * 3 = 855 *100

    => 57 x = 85500

    => x = 1500

    Hence the total profit = Rs. 1500

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  26. Question 18. A Seller Gains The Cost Of 40 Dozen Apples By Selling 25 Dozen Of Apples. Find Out The Gain Percent?

    Answer :

    Given 

    Cost price (C.P) of 40 dozen of apples is equal to selling (S.P) of 25 dozen of apples.

    Let the C.P of 1 dozen of apple = Rs.1

    Therefore C.P of 25 dozen apples = Rs. 25

    and C.P of 40 dozen apples = Rs.40

    => C.P of 40 dozen of apples = S.P of 25 dozen apples = Rs.40

    Profit % = (S.P of 25 dozen apples – C.P of 25 dozen apples) / C.P of 25 dozen apples * 100%

    = {(40 – 25) / 25} * 100%

    = {15 / 25} * 100%

    = 60%

    Therefore, required Profit % = 60 %

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  28. Question 19. The Selling Price Of 100 Articles Is The Same As The Cost Price Of 120 Articles. Find Gain Percent?

    Answer :

    Let the cost price of each article be Rs k

    We have, S.P of 100 articles = C.P of 120 articles = 120k

    We know that C.P of 100 articles = 100 k

    Gain on the purchase of 100 articles = S.P – C.P 

    => Gain = 120 k – 100 k =20k

    Profit percentage = (profit/C.P) x 100 = (20k/100k) x 100 = 20%

  29. Question 20. A Man Sold A Horse At A Loss Of 7%. Had He Been Able To Sell It At A Gain Of 9%, It Would Have Fetched Rs. 64 More Than It Did. What Was His Cost Price?

    Answer :

    In the given problem, let C.P denote the cost price, then

    (100+9)% of CP – (100-7) % of C.P = Rs. 64

    => (109) % of CP – (93) % of C.P = Rs. 64

    =>16 % of CP = 64

    => CP = 64 x 100 / 16 = 400

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  31. Question 21. A Farmer Sells His Product At A Loss Of 8 %. If His S.p Was Rs 27600, What Was His Actual Loss?

    Answer :

    Let the C.P be k Rs.

    Loss = 8 %. => loss = 8k/100

    S.P = C.P loss = k –8k/100 = 92k/100

    92k/100 = 27600 => k = 27600 x (100/92) => k = 30000,

    Loss = C.P –S.P = 30,000 –27600 = 2400

  32. Question 22. The Marked Price Of A Ceiling Fan Is Rs. 1250 And The Shopkeeper Allows A Discount Of 6% On It. Find The Selling Price Of The Fan (in Rs)?

    Answer :

    Given, Marked Price (M.P) = Rs. 1250

    Discount = 6 % of M.P

    = (6/100) * 1250

    = 75 Rs.

    Selling price (S.P) = Marked Price – Discount

    = 1250 – 75

    = 1175

    Therefore, Selling price = Rs. 1175

  33. Question 23. Find The Discount Percentage Which Is Equivalent For Two Successive Discounts Of 10 % On A Product Worth Rs. 10,800?

    Answer :

    Given, two successive discount of 10 %.

    Let, x = First discount = 10%, y = second discount = 10%

    Total Discount % = (x + y – [xy/100])%

    = (10 + 10 – [10×10/100]) %

    = (20 – [100/100]) %

    = (20 – 1) %

    = 19%

    Total Discount % which is equivalent for two successive discount of 10 % = 19%

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  35. Question 24. A Shopkeeper Gives Two Successive Discounts Of 10% On A Product Worth Rs. 1800. Without Successive Discount Find The Equivalent Discount In Percentage To Avail The Same Amount As Discount?

    Answer :

    Given, Original Price = Rs. 1800

    Price after 1st discount of 10% = 1800 – {(10/ 100) * 1800}

    = 1800 – 180

    = 1620

    Price after 2nd discount of 10% = 1620 -{(10/ 100) * 1620}

    = 1620 – 162

    = 1458

    Now, Single Discount Amount = 1800 – 1458 = 342 Rs.

    Required Percentage =Single Discount Amount /Original Price * 100%

    = (342 / 1800) * 100%

    = 19 %

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  37. Question 25. A Sum Of Rs. 12,500 Amounts To Rs. 15,500 In 4 Years At The Rate Of Simple Interest. What Is The Rate Of Interest?

    Answer :

    S.I = Amount to be paid – Principal

    => S.I. = Rs. (15500 – 12500) = Rs. 3000.

    Simple Interest, S.I = ( p x t x r) / 100

    => Rate = S.I * 100 / (p x t)

    =>Rate = (100 x 3000 / 12500 x 4) % = 6 %

  38. Question 26. Nalini Borrowed Rs. 1075 From Her Friend At 7% Per Annum. She Returned The Amount After 7 Years. How Much Amount Did She Pay?

    Answer :

    Given principal, p = 1075 Rs,

    rate of interest r= 7%,

    time, t=7 years

    Simple interest, SI = (p x r x t)/100

    => SI = (1075 x 7 x 7)/100

    => SI = 526.75

    Amount = Principal + S.I 

    = 1075 + 526.75

    = 1601.75

    Amount paid by Nalini to her friend is 1601.75 Rs.

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  40. Question 27. What Would Be The Amount To Be Paid On The Principal Of 6500 Rs. At The End Of 2 Years At Compound Interest At The Rate Of 15 % Per Annum?

    Answer :

    Given principal = Rs. 6500

    No. of years = 2

    Rate of interest = 15

    Amount = P x (1+r/100)n, 

    = 6500 x (1+15/100)2

    = 6500 x (115/100)2

    = 8596.25

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  42. Question 28. What Would Be The Amount To Be Paid On The Principle Of 12500 Rs. At The End Of 3 Years At Compound Interest At The Rate Of 10 % Per Annum?

    Answer :

    Given principal = 12500 No. of years = 3 Rate of interest = 10

    Amount = P x (1 + r/100) n,

    We get Amount = 12500 x (1+10/100)3 = 12500 * (11/10)3

  43. Question 29. Find The Compound Interest Accrued On The Principal Of Rs. 4000 At The End Of 2 Years At 10 % Per Annum?

    Answer :

    Principal = Rs. 4000, t= 2 years, rate of percent, r = 10 % 

    Amount = P (1 + r/100) ^t = 4000 x (1 + 10/100) ^2 = 4000 x (11/10) x (11/10) = 40 x 121 = 4840 Rs. 

    Amount = Principal + CI => 4000 + CI = 4840 => CI = 840 Rs.

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  45. Question 30. What Will Be The Amount If Sum Of Rs.10, 00,000 Is Invested At Compound Interest For 3 Years With Rate Of Interest 11%, 12% And 13% Respectively?

    Answer :

    Given

    Here, P = Rs.10, 00,000, R1 = 11, R2 = 12, R3 = 13.

    Each rate of interest is calculated for one year.

    Hence, N = 1 year.

    Amount after 3 years,

    = P(1 + R1/100) (1 + R2/100) (1 + R3/100)

    = 10, 00,000 * (1 + 11/100) * (1 + 12/100) * (1 + 13/100)

    = 10, 00,000 * (111/100) * (112/100) * (113/100)

    = 111 x 112 x 113

    = 14, 04,816

    Hence the total amount after 3 years is Rs.14, 04,816

  46. Question 31. What Would Be The Compound Interest Accrued On An Amount Of 12500 Rs. At The End Of 3 Years At The Rate Of 10 % Per Annum?

    Answer :

    Given principal = 12500

    No. of years = 3

    Rate of interest = 10

    Amount = P x (1+r/100) ^n, 

    = 12500 x (1+10/100) ^3

    = 12500 x (11/10)^3

    = 12500 x (11/10)x (11/10)x (11/10)

    = 16637.5

    Compound Interest, C. I = Amount – Principal = 16637.5 – 12500 = 4137.5

  47. Question 32. What Would Be The Compound Interest Accrued On An Amount Of 10000 Rs. At The End Of 2 Years At The Rate Of 4 % Per Annum?

    Answer :

    Given principal = 10000

    No. of years = 2

    Rate of interest = 4

    Amount = P [ 1 + ( r / 100 )n]

    = 10000 x [1 + (4 / 100)2]

    = 10000 x (104 / 100)2

    = 10000 x (104 / 100) x (104 / 100)

    = 104 x 104

    = 10816

    Compound Interest = Amount – Principal

    = 10816 – 10000

    =816

  48. Question 33. Alan Can Complete A Work In 10 Days B Is 25% More Efficient Than A. In How Many Days B And A Together Can Complete The Work?

    Answer :

    A completes in 10 days

    B completes in 6 days.

    A + B 1 day work = (1/10 + 1/6) = (3+5) /30 = 8 / 30 = 4 / 15

    A + B can complete in 15 / 4 days

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  50. Question 34. Two Tapes Can Fill An Empty Tank In 12 And 15 Minutes Respectively. If Both The Taps Are Opened Simultaneously In How Many Minutes The Tank Would Be Full?

    Answer :

    Let the two taps be A and B.

    Given, tap a fill the tank in 12 mins

    Tap B fill the tank in 15 mins

    To find the Time taken by both taps opened together to fill the tank:

    1/(A + B) = (1/A) + (1/B)

    => 1/ (A + B) = (1/ 12) + (1/ 15)

    => 1/ (A + B) = 27 / 180

    Taking reciprocal on both sides

    => A + B = 180 / 27

    =>A + B = 20 / 3 mins