# Hcl Aptitude Placement Papers - Hcl Aptitude Interview Questions and Answers updated on 02.Mar.2024

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

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

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

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

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 %

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 = 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/

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%

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 %

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

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%

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

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.

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

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

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

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

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

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 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

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 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

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

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

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.

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%

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 %

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

Share’s ratio = A: B = 2: 3

Time ratio = A: B = 4: 5

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

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

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

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.

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

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