Interview Questions.

Hcl Aptitude Placement Papers - Hcl Aptitude Interview Questions and Answers

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Hcl Aptitude Placement Papers - Hcl Aptitude Interview Questions and Answers

Q1. A Starts A Business With A Capital Of Rs. Eighty five,zero

Let B joins for x months.

Given, A's capital = Rs. 85,000 for one year

B's capital = Rs. Forty two,500 for x months

Given, Ratio of earnings = 3: 1

=> Ratio of funding =Ratio of earnings

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

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

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

=> 24 / x = three/1

=> x = 24/three

=> x = eight

Therefore, No. Of months "B" inside the commercial enterprise = 8 months

Q2. If thirteen:eleven Is The Ratio Of Present Age Of Jothi And Viji Respectively And 15:nine 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 four Yea

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

Given, Jothi's age four years hence and Viji's age four years ago within the ratio 15:nine.

That is, (13X + 4) / (11X - 4) = 15 / nine

=> 9 (13X + four) = 15 (11X - 4)

=> 117X + 36 = 165X - 60

=> 165X - 117X = 60 + 36

=> 48X = 96

=> X = ninety six / 48

=> X = 2

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

on substituting price of X = 2 we get,

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

= 22/26

= eleven/thirteen

Hence the wer is 11:thirteen

Q3. What Will Be The Amount If Sum Of Rs.10, 00,000 Is Invested At Compound Interest For three Years With Rate Of Interest 11%, 12% And thirteen% Respectively?

Given

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

Each rate of interest is calculated for 365 days.

Hence, N = 1 yr.

Amount after three years,

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

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

= 10, 00,000 * (111/a hundred) * (112/one hundred) * (113/one hundred)

= 111 x 112 x 113

= 14, 04,816

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

Q4. What Would Be The Compound Interest Accrued On An Amount Of ten thousand Rs. At The End Of 2 Years At The Rate Of four % Per Annum?

Given main = 10000

No. Of years = 2

Rate of hobby = 4

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

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

= ten thousand x (104 / one hundred)2

= 10000 x (104 / a hundred) x (104 / one hundred)

= 104 x 104

= 10816

Compound Interest = Amount - Principal

= 10816 - 10000

=816

Q5. A Sum Of Rs. 12,500 Amounts To Rs. 15,500 In four 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 * a hundred / (p x t)

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

Q6. If Two Numbers Are In The Ratio 6: thirteen 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 = four

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

Q7. 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 ==> four: 5 

Total earnings = Rs.9000/- 

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

closing amount = Rs.7200/- 

Total parts = 9 

nine parts -----> Rs.7200/- 

1 component ------> Rs.800/- 

A's proportion ===> Rs.800/- * 4 elements = Rs.3200/

Q8. The Selling Price Of 100 Articles Is The Same As The Cost Price Of one hundred twenty Articles. Find Gain Percent?

Let the fee fee of each article be Rs okay

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

We understand that C.P of 100 articles = one hundred okay

Gain on the acquisition of one hundred articles = S.P - C.P 

=> Gain = 120 okay - a hundred k =20k

Profit percent = (earnings/C.P) x a hundred = (20k/100k) x one hundred = 20%

Q9. A Shopkeeper Gives Two Successive Discounts Of 10% On A Product Worth Rs. 18

Given, Original Price = Rs. 1800

Price after 1st cut price of 10% = 1800 - (10/ a hundred) * 1800

= 1800 - a hundred and eighty

= 1620

Price after second cut price of 10% = 1620 -(10/ a hundred) * 1620

= 1620 - 162

= 1458

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

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

= (342 / 1800) * 100%

= 19 %

Q10. In The First 30 Overs Of A Cricket Game, The Run Rate Was Only 4.

Required Run charge = (320 - (four.Five x 30) )/ (20) = ( 185) /(20) = nine.25

Q11. 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 bargain of 10 %.

Let, x = First cut price = 10%, y = 2d bargain = 10%

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

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

                          = (20 - [100/100]) %

                          = (20 - 1) %

                          = 19%

Total Discount % that's equal for 2 successive cut price of 10 % = 19%

Q12. 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 forty % Water In It?

Given: 30 Liters of mixture, Milk and water inside the ratio 7 : three Which me, we've got 21 liters of milk and nine liters of water.

We add water the ensuing answer is 21 liters of milk and 9 + x liters of water.

Total quantity = 30 +x.

Water percentage is 40 % = > forty x (30+x)/a hundred = 9 + x

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

=> a hundred and twenty + 4x = 90 + 10x

=> 10x - 4x = 120-90

=>6x = 30

=> x = five

Thus the quantity of water added = five liters

Q13. 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, fee of percent, r = 10 % 

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

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

Q14. 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 = eighty

=> pigeons + horses = eighty

Let p = wide variety of pigeons

h = variety of horses

=> p = eighty - h

Given, Legs of the animals = 260

Each pigeon has 2 legs and every horse has four legs

=> 2p + 4h = 260

Substitute p = 80 - h

=> 2 (eighty - h) + 4h = 260

=> 160 - 2h + 4h = 260

=> 2h = 260 - a hundred and sixty

=> 2h = one hundred

h = 50

So, range of horses within the Zoo = 50

Q15. 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 rate (S.P) = Marked Price - Discount

= 1250 - seventy five

= 1175

Therefore, Selling fee = Rs. 1175

Q16. A And B Invest In A Business In The Ratio 3:

Assume that the entire profit is x.

Since 5% goes for charity, ninety five% of x may be divided between A and B within the ratio three: 2

Given, A's percentage is Rs. 855

=> A's earnings = (95x/100) * (3/5) = 855

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

=> 19x * three = 855 *one hundred

=> fifty seven x = 85500

=> x = 1500

Hence the entire income = Rs. 1500

Q17. 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+five) /30 = 8 / 30 = four / 15

A + B can whole in 15 / 4 days

Q18. 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 fundamental = Rs. 6500

No. Of years = 2

Rate of interest = 15

Amount = P x (1+r/a hundred)n, 

= 6500 x (1+15/one hundred)2

= 6500 x (a hundred and fifteen/a hundred)2

= 8596.25

Q19. 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 aggregate accommodates 20 liters of milk and water.

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

=> (60 / a hundred) x 20 = pure milk

=> 12 liters = pure milk

In 20 liters mixture closing eight liters = water

When "x" liters of pure milk is added to twenty liters of combination

New aggregate = (20 + x) liters

Milk in new combination = (12 + x) liters

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

=> 12 + x = (80 / one hundred) * (20 + x)

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

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

=> 60 + 5 x = 80 + 4 x

=> 5 x - four x = 80 - 60

=> x = 20 liters

Q20. The Average Age Of A Group Of 10 Students Was

The average age of a set of 10 college students is 14.

Therefore, the sum of the a long time of all 10 of them = 10 * 14 = 140

When two students be a part of the institution, the average boom by using @New Average = 15

Now there are 12 students Therefore, sum of all the ages of 12 college students = 15 x 12 = one hundred eighty

Therefore, the sum of the a while of two college students who joined = one hundred eighty - one hundred forty = forty

And the common age of these  students = 20

Q21. What Would Be The Compound Interest Accrued On An Amount Of 12500 Rs. At The End Of three Years At The Rate Of 10 % Per Annum?

Given primary = 12500

No. Of years = 3

Rate of interest = 10

Amount = P x (1+r/a hundred) ^n, 

= 12500 x (1+10/one hundred) ^three

= 12500 x (eleven/10)^three

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

= 16637.5

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

Q22. The Average Age Of A Group Of 10 Students Was

The common age of a group of 10 college students is 10.

Therefore, the sum of the a while of all 10 of them = 10 * 10 = a hundred

When  college students be a part of the institution, the average growth by 1.

=> New Average = eleven

Now there are 12 college students.

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

Therefore, the sum of the a long time of  college students who joined = 132 - 100 = 32

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

Q23. What Would Be The Amount To Be Paid On The Principle Of 12500 Rs. At The End Of three Years At Compound Interest At The Rate Of 10 % Per Annum?

Given main = 12500 No. Of years = 3 Rate of hobby = 10

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

We get Amount = 12500 x (1+10/a hundred)three = 12500 * (11/10)three

Q24. A Man Sold A Horse At A Loss Of 7%. Had He Been Able To Sell It At A Gain Of nine%, It Would Have Fetched Rs. Sixty four More Than It Did. What Was His Cost Price?

In the given hassle, allow C.P denote the value charge, then

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

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

=>sixteen % of CP = sixty four

=> CP = sixty four x a hundred / 16 = four hundred

Q25. 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 aggregate incorporates forty% milk and 60% water in it.

That is 4.8 liters of milk and seven.2 liters of water.

Now we're changing the aggregate with pure milk so that the quantity of milk and water within the mixture is 50% and 50%.

That is we are able to end up with 6 liters of milk and 6 liters of water. Water receives reduced by using 1.2 liters.

To put off 1.2 liters of water from the authentic aggregate containing 60% water, we need to do away with 1.2 / zero.6 liters of the aggregate = 2 liters.

Q26. Find The Discount Percentage Which Is Equivalent For Two Successive Discounts Of 10 % On A Product Worth Rs. 10,800?

Given, two successive bargain of 10 %.

Let, x = First bargain = 10%, y = 2nd discount = 10%

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

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

= (20 - [100/100]) %

= (20 - 1) %

= 19%

Total Discount % that's equivalent for two successive cut price of 10 % = 19%

Q27. 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 forty dozen of apples is same 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.Forty

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

Profit % = (S.P of 25 dozen apples - C.P of 25 dozen apples) / C.P of 25 dozen apples * one hundred%

= (40 - 25) / 25 * 100%

= 15 / 25 * one hundred%

= 60%

Therefore, required Profit % = 60 %

Q28. 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, faucet a fill the tank in 12 mins

Tap B fill the tank in 15 minutes

To discover the Time taken by way of each faucets 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 = a hundred and eighty / 27

=>A + B = 20 / 3 mins

Q29. A And B Invested In A Business In The Ratio 2:3 And The Ratio Of Their Period Of Investment Is four:

Share’s ratio = A: B = 2: three 

Time ratio = A: B = four: five 

Then, income ratio = A: B = 2*four: 3*5 = 8: 15

Q30. 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/a hundred

S.P = C.P loss = okay –8k/one hundred = 92k/one hundred

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

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

Q31. The Ratio Of Number Of Boys And Girls In A School Of 720 Students Is 7:

Given, boys: girls = 7: five

Let, the overall wide variety of boys = 7x and total range of ladies= 5x

Given, total college students = 720

=> 7x + 5x = 720

=> 12x = 720

=> x = 60

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

total range of women= 5x = five * 60 = 300

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

=> 420 / (three hundred + y) = 1/1

=> 420 = (300 + y)

=> y = 420 – 300

=> y = 120

So, one hundred twenty extra women must be admitted to make the ratio 1:1

Q32. 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 main, p = 1075 Rs,

charge of hobby r= 7%,

time, t=7 years

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

=> SI = (1075 x 7 x 7)/one hundred

=> SI = 526.Seventy five

Amount = Principal + S.I 

= 1075 + 526.Seventy five

= 1601.75

Amount paid through Nalini to her pal is 1601.Seventy five Rs.

Q33. 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 in step with week be Rs. X

and the sum paid to Tailor Y in step with week be Rs. Y

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

=> X + Y = 550 ---> eqn (1)

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

=> X = 120 % of Y

=> X = (a hundred and twenty / 100) * Y

=> X = (6 / five) * Y

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

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

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

=> 11Y = 550 * five

=> 11Y = 2750

=> Y = 2750 / eleven

=> Y = 250 Rs.

Thus, the sum paid to Tailor Y in step with week be Rs. 250

Q34. 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 includes two grandparents,  parents and three grandchildren

The common age of  grandparents = 67 years.

=> Total age of  grandparents = (sixty seven * 2) = 134 years

The common age of two dad and mom = 35 years

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

The common age of three grand kids = 6 years.

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

Required Average age of the family = Total age of (two grandparents +  dad and mom + three grand children) / Total participants in the family

 = (134 + 70 + 18) / (2 + 2 + 3)

 = 222 / 7

= 31 (five / 7) years




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