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

Answer :

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

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

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

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

Given, X is paid a hundred and twenty percent of the sum paid to Y

=> X = one hundred twenty % of Y

=> X = (120 / a hundred) * Y

=> X = (6 / five) * Y

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

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

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

=> 11Y = 550 * 5

=> 11Y = 2750

=> Y = 2750 / eleven

=> Y = 250 Rs.

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

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, successive cut price 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 equal for 2 successive cut price of 10 % = 19%

Aptitude Interview Questions

Question 3. A Family Consists Of Two Grandparents, Two Parents And Three Grandchildren. The Average Age Of The Grandparents Is sixty seven 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 own family consists of grandparents, mother and father and three grandchildren

The common age of grandparents = 67 years.

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

The common age of dad and mom = 35 years

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

The average age of 3 grand children = 6 years.

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

Required Average age of the circle of relatives = Total age of ( grandparents + dad and mom + three grand youngsters) / Total participants inside the own family

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

= 222 / 7

= 31 (5 / 7) years

Question four. If thirteen: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 prevailing age of Jothi and Viji be 13X and 11X respectively.

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

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

=> nine (13X + four) = 15 (11X - four)

=> 117X + 36 = 165X - 60

=> 165X - 117X = 60 + 36

=> 48X = 96

=> X = ninety six / 48

=> X = 2

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

on substituting price of X = 2 we get,

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

= 22/26

= 11/13

Hence the solution is 11:13

Question five. 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 a long time of all 10 of them = 10 * 14 = a hundred and forty

When college students join the institution, the common growth by way of 1. New Average = 15

Now there are 12 college students Therefore, sum of all of the a long time of 12 students = 15 x 12 = 180

Therefore, the sum of the ages of college students who joined = 180 - 140 = forty

And the common age of these students = 20

Value Labs Aptitude Interview Questions

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 collection of 10 students is 10.

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

When two students be a part of the group, the average boom through 1.

=> New Average = eleven

Now there are 12 students.

Therefore, sum of all the a long time of 12 college students = eleven x 12 = 132

Therefore, the sum of the a while of students who joined = 132 - one hundred = 32

And the average age of those two college students = (32 / 2) = sixteen

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

Answer :

Required Run rate = (320 - (four.Five x 30) )/ (20) = ( 185) /(20) = 9.25

SAP Aptitude Interview Questions

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

Answer :

Given, boys: ladies = 7: 5

Let, the full range of boys = 7x and overall number of women= 5x

Given, general college students = 720

=> 7x + 5x = 720

=> 12x = 720

=> x = 60

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

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

Let y be the quantity of girls brought to make the ratio 1 : 1

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

=> 420 = (300 + y)

=> y = 420 – 300

=> y = one hundred twenty

So, a hundred and twenty greater girls must be admitted to make the ratio 1:1

Question nine. 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

=> okay = four

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

Oracle Aptitude Interview Questions

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 eighty% 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 / a hundred) x 20 = pure milk

=> 12 liters = natural milk

In 20 liters aggregate closing 8 liters = water

When "x" liters of natural milk is brought to twenty liters of combination

New combination = (20 + x) liters

Milk in new combination = (12 + x) liters

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

=> 12 + x = (eighty / a hundred) * (20 + x)

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

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

=> 60 + 5 x = 80 + four x

=> 5 x - 4 x = 80 - 60

=> x = 20 liters

Question eleven. 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 both pigeons or horses

Heads of the animals in Zoo = eighty

=> pigeons + horses = eighty

Let p = number of pigeons

h = wide variety of horses

=> p = 80 - h

Given, Legs of the animals = 260

Each pigeon has 2 legs and each horse has four legs

=> 2p + 4h = 260

Substitute p = eighty - h

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

=> one hundred sixty - 2h + 4h = 260

=> 2h = 260 - a hundred and sixty

=> 2h = one hundred

h = 50

So, wide variety of horses inside the Zoo = 50

BHEL Aptitude Interview Questions

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 forty % Water In It?

Answer :

Given: 30 Liters of mixture, Milk and water inside the ratio 7 : three Which way, we've 21 liters of milk and 9 liters of water.

We upload water the resulting answer is 21 liters of milk and 9 + x liters of water.

Total quantity = 30 +x.

Water percent is 40 % = > forty x (30+x)/one hundred = 9 + x

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

=> 120 + 4x = ninety + 10x

=> 10x - 4x = 120-ninety

=>6x = 30

=> x = five

Thus the amount of water brought = 5 liters

Aptitude Interview Questions

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

Answer :

The combination contains forty% milk and 60% water in it.

That is four.Eight liters of milk and seven.2 liters of water.

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

That is we will grow to be with 6 liters of milk and 6 liters of water. Water receives reduced by means of 1.2 liters.

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

Question 14. A And B Invested In A Business In The Ratio 2:three And The Ratio Of Their Period Of Investment Is four:five. Then Their Profit Ratio Is?

Answer :

Share’s ratio = A: B = 2: three

Time ratio = A: B = 4: five

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

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 income = Rs.9000/-

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

remaining quantity = Rs.7200/-

Total parts = nine

9 components -----> Rs.7200/-

1 element ------> Rs.800/-

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

Sony India Aptitude Interview Questions

Question sixteen. 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 three: 1?

Answer :

Let B joins for x months.

Given, A's capital = Rs. 85,000 for 365 days

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

Given, Ratio of profit = 3: 1

=> Ratio of funding =Ratio of profit

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

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

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

=> 24 / x = 3/1

=> x = 24/3

=> x = eight

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

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

Answer :

Assume that the total income is x.

Since five% goes for charity, ninety five% of x may be divided among A and B within the ratio 3: 2

Given, A's percentage is Rs. 855

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

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

=> 19x * 3 = 855 *one hundred

=> 57 x = 85500

=> x = 1500

Hence the overall profit = Rs. 1500

Mu Sigma Aptitude Interview Questions

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 forty dozen of apples is equal to selling (S.P) of 25 dozen of apples.

Let the C.P of one 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.Forty

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

= (forty - 25) / 25 * 100%

= 15 / 25 * one hundred%

= 60%

Therefore, required Profit % = 60 %

Value Labs Aptitude Interview Questions

Question 19. The Selling Price Of 100 Articles Is The Same As The Cost Price Of one hundred twenty Articles. Find Gain Percent?

Answer :

Let the price charge of every article be Rs ok

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

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

Gain on the acquisition of 100 articles = S.P - C.P

=> Gain = 120 k - one hundred k =20k

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

Question 20. 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?

Answer :

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

(a hundred+9)% of CP – (100-7) % of C.P = Rs. Sixty four

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

=>16 % of CP = sixty four

=> CP = 64 x a hundred / sixteen = four hundred

Infosys Aptitude Interview Questions

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 okay Rs.

Loss = eight %. => loss = 8k/one hundred

S.P = C.P loss = ok â€“8k/a hundred = 92k/a hundred

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

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

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 charge (S.P) = Marked Price - Discount

= 1250 - seventy five

= 1175

Therefore, Selling fee = Rs. 1175

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 cut price = 10%

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

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

= (20 - [100/100]) %

= (20 - 1) %

= 19%

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

IBM Aptitude Interview Questions

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 cut price of 10% = 1800 - (10/ one hundred) * 1800

= 1800 - a hundred and eighty

= 1620

Price after 2nd cut price of 10% = 1620 -(10/ 100) * 1620

= 1620 - 162

= 1458

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

Required Percentage =Single Discount Amount /Original Price * one hundred%

= (342 / 1800) * 100%

= 19 %

SAP Aptitude Interview Questions

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) / one hundred

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

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

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 important, p = 1075 Rs,

rate of hobby r= 7%,

time, t=7 years

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

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

=> SI = 526.75

Amount = Principal + S.I

= 1075 + 526.Seventy five

= 1601.Seventy five

Amount paid by way of Nalini to her friend is 1601.75 Rs.

Capgemini Aptitude Interview Questions

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 hobby = 15

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

= 6500 x (1+15/100)2

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

= 8596.25

Oracle Aptitude Interview Questions

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 important = 12500 No. Of years = 3 Rate of interest = 10

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

We get Amount = 12500 x (1+10/one hundred)three = 12500 * (eleven/10)three

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, charge of percentage, r = 10 %

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

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

Cognizant Aptitude Interview Questions

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 eleven%, 12% And 13% Respectively?

Answer :

Given

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

Each fee of hobby is calculated for 12 months.

Hence, N = 1 year.

Amount after three years,

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

= 10, 00,000 * (1 + eleven/one hundred) * (1 + 12/100) * (1 + thirteen/100)

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

= 111 x 112 x 113

= 14, 04,816

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

Question 31. 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?

Answer :

Given main = 12500

No. Of years = 3

Rate of hobby = 10

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

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

= 12500 x (11/10)^3

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

= 16637.5

Compound Interest, C. I = Amount - Principal = 16637.Five - 12500 = 4137.5

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 = four

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

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

= ten thousand x (104 / a hundred)2

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

= 104 x 104

= 10816

Compound Interest = Amount - Principal

= 10816 - 10000

=816

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 paintings = (1/10 + 1/6) = (3+five) /30 = 8 / 30 = four / 15

A + B can whole in 15 / 4 days

BHEL Aptitude Interview Questions

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 faucets be A and B.

Given, tap a fill the tank in 12 minutes

Tap B fill the tank in 15 mins

To locate the Time taken by means of each 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 each facets

=> A + B = one hundred eighty / 27

=>A + B = 20 / three minutes