# Hcl Aptitude Placement Papers - Hcl Aptitude Interview Questions and Answers updated on 21.Aug.2019

## Hcl Aptitude Interview Questions and Answers

#### A Starts A Business With A Capital Of Rs. 85,0

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

#### 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 Yea

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 wer is 11:13

#### 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?

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

#### 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?

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

#### 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?

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 %

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

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

#### 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/-.

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/

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

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%

#### A Shopkeeper Gives Two Successive Discounts Of 10% On A Product Worth Rs. 18

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 %

#### In The First 30 Overs Of A Cricket Game, The Run Rate Was Only 4.

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

#### 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?

Given, two successive discount of 10 %.

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

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

= (10 + 10 - [10x10/100]) %

= (20 - [100/100]) %

= (20 - 1) %

= 19%

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

#### 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?

Given: 30 Liters of mixture, Milk and water in the ratio 7 : 3 Which me, 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

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

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.

#### A Zookeeper Counted The Heads Of The Animals In A Zoo And Found It To Be 8

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

#### 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)?

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

#### A And B Invest In A Business In The Ratio 3:

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

#### 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?

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

#### 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?

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

#### 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"?

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

#### The Average Age Of A Group Of 10 Students Was

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 @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

#### 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?

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

#### The Average Age Of A Group Of 10 Students Was

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

#### 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?

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

#### 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?

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

#### 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?

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.

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

Given, two successive discount of 10 %.

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

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

= (10 + 10 - [10x10/100]) %

= (20 - [100/100]) %

= (20 - 1) %

= 19%

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

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

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 %

#### 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?

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

#### A And B Invested In A Business In The Ratio 2:3 And The Ratio Of Their Period Of Investment Is 4:

Share’s ratio = A: B = 2: 3

Time ratio = A: B = 4: 5

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

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

Let the C.P be k Rs.

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

S.P = C.P loss = k â€“8k/100 = 92k/100

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

Loss = C.P â€“S.P = 30,000 â€“27600 = 2400

#### The Ratio Of Number Of Boys And Girls In A School Of 720 Students Is 7:

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

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

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.

#### 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?

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

#### 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. W

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

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