Interview Questions.

Question 1. A Person Sold Two Cows Each For Rs.9900. If He Gained 10% On One And Lost 20% On The Other, Then Which Of The Following Is True?

Answer :

The CP of profitable cow  = 9900/1.1 = 9000

and profit = Rs. 900 

The  CP of loss yielding cow = 9900/0.8 = 12375

and loss = Rs. 2475

so, the net loss = 2475 - 900 = 1575.

Question 2. In A Certain Store, The Profit Is 320% Of The Cost. If The Cost Increases By 25% But The Selling Price Remains Constant, Approximately What Percentage Of The Selling Price Is The Profit?

Answer :

Let C.P.= Rs. 100. 

Then, Profit = Rs. 320, 

S.P. = Rs. 420.

New C.P. = a hundred twenty five% of Rs. One hundred = Rs. One hundred twenty five

New S.P. = Rs. 420.

Profit = Rs. (420 - one hundred twenty five) = Rs. 295

Required percent = (295/420) * one hundred 

= 70%(approx).

Computer Hardware Interview Questions
Question three. A Man Saves 20% Of His Monthly Salary. If An Account Of Dearness Of Things He Is To Increase His Monthly Expenses By 15%, He Is Only Able To Save Rs. 400 Per Month. What Is His Monthly Salary?

Answer :

Income = Rs. 100

Expenditure = Rs. 80

Savings = Rs. 20

Present Expenditure 80x(15/one hundred) = Rs. 12 = eighty + 12 = Rs. 92

Present Savings = 100 – 92 = Rs. 8

a hundred ------ 8

? --------- 400 => 5000

His salary = Rs. 5000.

Question four. Raghu Earns 25% On An Investment But Loses 10% On Another Investment. If The Ratio Of The Two Investment Is three:five. What Is The Gain Or Loss On Two Investments Taken Together ?

Answer :

Taking the two investments to be 3x and 5x respectively

Total income of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25

Therefore, Gain% = 0.25/eight x a hundred = 3.One hundred twenty five %.

Question 5. A Seller Uses 840 Gm In Place Of 1 Kg To Sell His Goods. Find His Actual Profit/loss % When He Sells His Article On 4% Loss On Cost Price ?

Answer :

Let 1kg of Rs. One hundred then 840gm is of Rs. Eighty four.

Now (label on can 1kg but carries 840kg ) so for consumer it's miles of Rs. 100 and similarly gives 4% bargain [he sells his article on 4% loss on cost price.] 

So now S.P = Rs. Ninety six

But simply it carries 840 gm so C.P for shopkeeper = Rs. 84

S.P = Rs. Ninety six

C.P = Rs. 84

Profit% = (S.P-C.P)/C.Px100

(ninety six-eighty four)/eighty four x 100 = 14.28571429% PROFIT.

Aptitude Interview Questions
Question 6. A Shopkeeper Sold A Mobile Phone For Rs. 12000. Had He Offered Discount Of 10% On The Selling Price, There Would Be A Loss Of 4%. What Is The Cost Price Of That Mobile Phone?

Answer :

Given that SP = Rs. 12000 - 10%  = Rs. 10,800

Loss% = four

We understand that, C.P = 100/(one hundred - Loss%) x a hundred

=> a hundred/100-4 x 10800

=> 1080000/96

C.P = Rs. Eleven,250.

Question 7. On Selling 17 Balls At Rs. 720, There Is A Loss Equal To The Cost Price Of 5 Balls. The Cost Price Of A Ball Is ?

Answer :

Let the cost price of a ball is Rs.X

Given, on promoting 17 balls at Rs. 720, there's a loss identical to the price rate of five balls

The equation is :

17x - 720 = 5x

Solving the equation 

we get x = 60

Therefore, cost rate of a ball is Rs. 60.

Hardware design Interview Questions
Question eight. Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is?

Answer :

Let S be the pattern area. 

Then, n(S) = 52C2=(fifty two x 51)/(2 x 1) = 1326. 

Let E = occasion of having 1 spade and 1 heart.

N(E)= quantity of approaches of choosing 1 spade out of thirteen and 1 heart out of thirteen = 13C1*13C1 = 169.

P(E) = n(E)/n(S) = 169/1326 = thirteen/102.

Question 9. The Table Is Bought For Rs. 1950 And Sold At Rs. 2340. Find The Profit Percent?

Answer :

Cost Price = Rs. 1950

Selling Price = Rs. 2340

Profit = S.P – C.P

Profit = Rs. 2340 – 1950 = 390

Profit % = (Profit/C.P) x a hundred

Profit % = (390/1950) x 100

Profit % = 20 %.

Dell Oracle Interview Questions
Question 10. If Two Letters Are Taken At Random From The Word Home, What Is The Probability That None Of The Letters Would Be Vowels?

Answer :

P(first letter isn't vowel) = 2/four

P(2nd letter isn't always vowel) = 1/three

So, possibility that none of letters would be vowels is = 2/four×1/3=1/6.

Question eleven. Four Dice Are Thrown Simultaneously. Find The Probability That All Of Them Show The Same Face?

Answer :

The overall variety of basic events associated to the random experiments of throwing four dice concurrently is:

= 6*6*6*6=sixty four

n(S) = sixty four

Let X be the event that all cube show the same face. 

X =  (1,1,1,1,), (2,2,2,2), (three,three,three,three), (4,4,four,4), (5,5,five,5), (6,6,6,6)

n(X) = 6

Hence required probability = n(X)n(S)=6/sixty four=1216.

Dell Boomi Interview Questions
Question 12. Three Unbiased Coins Are Tossed. What Is The Probability Of Getting At Most Two Heads?

Answer :

Here S = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH

Let E = event of having at most two heads.

Then E = TTT, TTH, THT, HTT, THH, HTH, HHT.

P(E) =n(E)/n(S)=7/8.

Computer Hardware Interview Questions
Question thirteen. Three Unbiased Coins Are Tossed.What Is The Probability Of Getting At Least 2 Heads?

Answer :

Here S= TTT, TTH, THT, HTT, THH, HTH, HHT, HHH.

Let E = event of having as a minimum two heads = THH, HTH, HHT, HHH.

P(E) = n(E) / n(S)

      = 4/eight= 1/2.

Question 14. In A Simultaneous Throw Of Pair Of Dice. Find The Probability Of Getting The Total More Than 7?

Answer :

Here n(S) = (6 x 6) = 36

Let E = event of having a total greater than 7

        = (2,6),(3,5),(3,6),(4,four),(four,5),(four,6),(five,3),(5,four),(5,five),(five,6),(6,2),(6,3),(6,4),(6,five),(6,6)

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

Question 15. The Effective Annual Rate Of Interest Corresponding To A Nominal Rate Of 6% Per Annum Payable Half-yearly Is?

Answer :

Amount of Rs. 100 for 1 year

when compounded half of-every year = Rs.[100*(1+3/100)^2]=Rs.106.09

Effective charge=(106.09-100)%=6.09%.

Laptop Repair Interview Questions
Question sixteen. Simple Interest On A Certain Sum Of Money For 3 Years At eight% Per Annum Is Half The Compound Interest On Rs. 4000 For 2 Years At 10% Per Annum. The Sum Placed On Simple Interest Is?

Answer :

C.I.= Rs.[4000*(1+10/100)^2-4000]

 =Rs.840

sum=Rs.(420 * one hundred)/3*8=Rs.1750.

Question 17. What Will Be The Compound Interest On A Sum Of Rs.25,000 After three Years At The Rate Of 12 P.C.P.A?

Answer :

Amount

= Rs.(25000x(1+12/100)³

= Rs.(25000x28/25x28/25x28/25)

= Rs. 35123.20.

C.I = Rs(35123.20 -25000)

= Rs.10123.20.

Cpu Interview Questions
Question 18. Find The Principal If The Interest Compounded At The Rate Of 10% Per Annum For Two Years Is Rs. 420 ?

Answer :

Given, 

Compound rate, R = 10% in line with annum

Time = 2 years

C.I = Rs. 420

Let P be the desired primary.

A = (P+C.I)

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

(P+C.I) = P[1 + (10/100)]2

(P+420) = P[11/10][11/10]

P-1.21P = -420

0.21P = 420

Hence, P = 420/zero.21 = Rs. 2000.

Aptitude Interview Questions
Question 19. The Difference Between Simple Interest And Compound On Rs. 1200 For One Year At 10% Per Annum Reckoned Half-yearly Is?

Answer :

S.I. = Rs.(1200*10*1)/a hundred=rs.A hundred and twenty

C.I. =rs[1200*(1+5/100)2-1200]=rs.123

Difference = Rs.(123-a hundred and twenty) =Rs.Three

Question 20. Find The Compound Interest On Rs.Sixteen,000 At 20% Per Annum For 9 Months, Compounded Quartely?

Answer :

Principal = Rs.16,000;

Time=nine months = 3 quarters;

Amount

=Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20]

= Rs.18522.

C.I

= Rs.(18522 - 16000)

= Rs.2522.

Sony India Aptitude Interview Questions
Question 21. In How Many Ways Can The Letters Of The Word 'leader' Be Arranged ?

Answer :

No. Of letters inside the phrase = 6

No. Of 'E' repeated = 2

Total No. Of association = 6!/2! = 360.

Question 22. If The Simple Interest On A Sum Of Money At 5% Per Annum For three Years Is Rs. 1200, Find The Compound Interest On The Same Sum For The Same Period At The Same Rate?

Answer :

Clearly, Rate = five% p.A .,

Time = 3 years

S.I =Rs.1200.

So,Principal

=Rs.(one hundred x 1200/3x5)

=Rs.8000.

Amount

=Rs.[8000 x (1+5/100)³]

=Rs(8000x21/20x21/20x21/20)

= Rs.9261

C.I

=Rs.(9261-8000)

=Rs.1261.

Question 23. Pumps Working eight Hours A Day, Can Empty A Tank In 2 Days. How Many Hours A Day Must 4 Pumps Work To Empty The Tank In 1 Day?

Answer :

Let the desired no of working hours in step with day be x.

More pumps , Less working hours consistent with day  (Indirect Proportion)

Less days, More running hours in step with day    (Indirect Proportion)

 Pumps4 : 3Days1 : 2?? Eight:x

=> (four * 1 * x) = (three * 2 * eight)

=> x=12

Cognizant Aptitude Interview Questions
Question 24. In A Question Divisor Is 2/3 Of The Dividend And 2 Times The Remainder. If The Remainder Is 5, Find The Dividend?

Answer :

Divisor = 2/3 x dividend

and Divisor = 2 x the rest

or 2/3 x dividend = 2 x five

Dividend = 2 x 5 x 3 / 2 = 15.

Hardware layout Interview Questions
Question 25. How Many Figures (digits) Are Required To Number A Book Containing 200 Pages ?

Answer :

Number of 1 digit pages from 

1 to nine = nine 

Number of  digit pages from

10 to 99 = ninety

Number of three digit pages from

100 to two hundred = one zero one.

Question 26. The Digit In The Units Place Of A Number Is Equal To The Digit In The Tens Place Of Half Of That Number And The Digit In The Tens Place Of That Number Is Less Than The Digit In Units Place Of Half Of The Number By 1, If The Sum Of The Digits Of The Number Is Seven, Then What Is The Number ?

Answer :

Let half of the no. = 10x + y

and the no. = 10v + w

From the given situations, 

w= x and v = y-1

Thus the no. = 10 (y-1) + x

? 2(10x + y ) = 10 (y-1) + x

? 8y - 19x = 10 ...(i)

v + w = 7

? Y-1 + x = 7 

? X + y = eight

Solving equations (i) and (ii) , we get 

x = 2 and y = 6.

Deloitte Aptitude Interview Questions
Question 27. A Two-digit Number Is Seven Times The Sum Of Its Digits, If Each Digit Is Increased By 2, The Number Thus Obtained Is four More Than Six Times The Sum Of Its Digits, Find The Number ?

Answer :

Let the two-digit quantity be 10x + y

10x + y = 7(x + y)

? X = 2y ...(i)

10(x +2 ) + (y + 2) = 6(x + y + 4) + 4

or 10x + y + 22 = 6x + 6y + 28

? 4x - 5y = 6 ...(ii)

Solving equations (i) and (ii)

We get x = 4 and y = 2.

Dell Oracle Interview Questions
Question 28. If A Number Is Decreased By four And Divided By 6 The Result Is 9. What Would Be The Result If 3 Is Subtracted From The Number And Then It Is Divided By five ?

Answer :

(x - 4) / 6 = 9 

Multiply both aspects by using 6: 

x - four = 54 

Add four to both aspects: 

x = fifty eight 

(fifty eight - three) / five = fifty five / 5 = eleven.

Question 29. The Numbers X, Y, Z Are Such That Xy = 96050 And Xz = 95625 And Y Is Greater Than Z By One. Find Out The Number Z ?

Answer :

xy = 96050 ...(i)

and xz = 95625 ...(ii)

and y - z = 1 ... (iii)

Dividing (i) by way of (ii) we get 

y/z = 96050 / 95625 

= 3842 / 3825 

= 226 / 225 ... (iv)

Combining (iii) and (iv) we get z = 225.

Question 30. A Bus Covers Its Journey At The Speed Of 80km/hr In 10hours. If The Same Distance Is To Be Covered In four Hours, By How Much The Speed Of Bus Will Have To Increase ?

Answer :

Initial speed = 80km/hr

Total distance = 80 x 10 = 800km

New speed = 800/four =200km/hr

Increase in pace = 200 - eighty = 120km/hr.

Question 31. Robert Is Traveling On His Cycle And Has Calculated To Reach Point A At 2 P.M. If He Travels At 10 Km/hr; He Will Reach There At 12 Noon If He Travels At 15 Km/hr. At What Speed Must He Travel To Reach A At 1 P.M. ?

Answer :

Let the gap traveled be x km.

Then, x/10 - x/15 = 2

3x - 2x = 60 => x = 60 km.

Time taken to journey 60 km at 10 km/hr = 60/10 = 6 hrs.

So, Robert commenced 6 hours earlier than 2. P.M. I.E., at 8 a.M.

Required velocity = 60/5 = 12 kmph.

Question 32. In What Time A 360 M. Long Train Moving At The Speed Of forty four Km/hr Will Cross A 140 M. Long Bridge ?

Answer :

Speed = forty four kmph x five/18 = one hundred ten/9 m/s

We realize that, Time = distance/velocity

Time = (360 + 140) / (110/nine)

= 500 x nine/110 = forty one sec.

Question 33. K And L Starts Walking Towards Each Other At four Pm At Speed Of 3 Km/hr And four Km/hr Respectively. They Were Initially 17.5 Km Apart. At What Time Do They Meet ?

Answer :

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.Five hours

So they meet at => 4 + 2.5 = 6:30 pm.

Dell Boomi Interview Questions
Question 34. If Hema Walks At 12 Km/hr Instead Of 8 Km/hr, She Would Have Walked 20 Km More. The Actual Distance Travelled By Hema Is ?

Answer :

Let the actual distance travelled be x km. 

Then x/8=(x+20)/12 

=> 12x = 8x + one hundred sixty

=> 4x = a hundred and sixty

=> x = forty km.

Question 35. Kamal Consistently Runs 240 Meters A Day And On Saturday He Runs For four hundred Meters. How Many Kilometers Will He Have To Run In Four Weeks ?

Answer :

Total walking distance in 4 weeks = (24 x 240) + (four x 400)

= 5760 + 1600

= 7360 meters

= 7360/a thousand

=> 7.36 kms.

Question 36. Two Trains Start From Same Place At Same Time At Right Angles To Each Other. Their Speeds Are 36km/hr And 48km/hr Respectively. After 30 Seconds The Distance Between Them Will Be ?

Answer :

Using pythagarous theorem,

distance travelled by using first educate = 36x5/18x30 = 300m

distance travelled by way of 2d teach = 48x5/18x30 = 400m

so distance among them =v( 90000 + 160000) = v250000 = 500mts.

Laptop Repair Interview Questions
Question 37. In A Daily Morning Walk Three Persons Step Off Together. Their Steps Measure seventy five Cm, eighty Cm And 85 Cm Respectively. What Is The Minimum Distance Each Should Walk So That Thay Can Cover The Distance In Complete Steps ?

Answer :

To locate the minimum distance, we need to get the LCM of seventy five, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x sixteen x 17 = 20400

Hence, the minimum distance every have to walk in order that thay can cover the space in entire steps = 20400 cms = 20400/a hundred = 204 mts.

Question 38. Running 3/4th Of His Usual Rate, A Man Is 15min Late. Find His Usual Time In Hours ?

Answer :

Walking at three/4th of regular fee implies that point taken would be 4/3th of the usual time. In different phrases, the time taken is 1/3rd greater than his normal time

so 1/third of the same old time = 15min

or usual time = 3 x 15 = 45min = forty five/60 hrs = three/four hrs.

Question 39. Two Stations A And B Are 200 Km Apart On A Straight Track. One Train Starts From A At 7 A.M. And Travels Towards B At 20 Kmph. Another Train Starts From B At eight A.M. And Travels Towards A At A Speed Of 25 Kmph. At What Time Will They Meet?

Answer :

Assume each trains meet after 'p' hours after 7 a.M.

Distance blanketed by means of teach starting from A in 'p' hours = 20p km

Distance blanketed by educate starting from B in (p-1) hours = 25(p-1)

Total distance = 200

=> 20x + 25(x-1) = 2 hundred

=> 45x = 225

=> p= 5

Means, they meet after five hours after 7 am, ie, they meet at 12 p.M.

Question 40. A Drink Vendor Has 368 Liters Of Maaza, 80 Liters Of Pepsi And 144 Liters Of Sprite. He Wants To Pack Them In Cans, So That Each Can Contains The Same Number Of Liters Of A Drink, And Doesn't Want To Mix Any Two Drinks In A Can. What Is The Least Number Of Cans Required ?

Answer :

The number of liters in every can = HCF of 80, one hundred forty four and 368 = 16 liters.

Number of cans of Maaza = 368/16 = 23

Number of cans of Pepsi = 80/16 = five

Number of cans of Sprite = 144/sixteen = nine

The total wide variety of cans required = 23 + 5 + 9 = 37 cans.

Cpu Interview Questions
Question forty one. Which Of The Following Has Most Number Of Divisors?

Answer :

ninety nine  = 1 x 3 x 3 x 11

a hundred and one= 1 x one zero one

176= 1 x 2x 2 x 2 x 2 x 11

182= 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, eleven, 33, ninety nine

divisors of a hundred and one are 1,one zero one

divisors of 176 are 1, 2, 4, 8, 16, 22, forty four, 88, 176

divisors of 182 are 1, 2, 7, 13, 14, 26, ninety one, 182

Hence , 176 hasthe maximum variety of divisors. 

Question 42. If The Sum Of Two Numbers Is fifty five And The H.C.F. And L.C.M. Of These Numbers Are 5 And 120 Respectively, Then The Sum Of The Reciprocals Of The Numbers Is Equal To ?

Answer :

Let the numbers be a and b.

We recognize that fabricated from  numbers = Product of their HCF and LCM

Then, a + b = fifty five and ab = 5 x 120 = 600.

=> The required sum = (1/a) + (1/b) = (a+b)/ab

=fifty five/600 = 11/120.

Sony India Aptitude Interview Questions
Question forty three. If N Is The Greatest Number That Will Divide 1305, 4665 And 6905, Leaving The Same Remainder In Each Case. What Is The Sum Of The Digits Of N ?

Answer :

N = H.C.F. Of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. Of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + zero ) = 4.

Question 44. Three Numbers Are In The Ratio Of 3:four:5 And Their L.C.M Is 3600.Their Hcf Is?

Answer :

Let the numbers be 3x, 4x, 5x.

Then, their L.C.M = 60x.

So, 60x=3600 or x=60.

Therefore,  The numbers are (three x 60), (four x 60), (5 x 60).

Hence,required H.C.F=60.

Question forty five. Find The Lowest Common Multiple Of 24, 36 And 40?

Answer :

To find the LCM of 24, 36 and 40

24 = 2 x 2 x 2 x three

36 = 2 x 2 x 3 x three

forty = 2 x 2 x 2 x 5 

Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x three x three x five

= 8 x nine x five

= 72 x five

= 360.

Question 46. A Bag Contains Equal Number Of 25 Paise, 50 Paise And One Rupee Coins Respectively. If The Total Value Is Rs 105, How Many Types Of Each Type Are Present?

Answer :

Bag includes 25 paise, 50 paise and 1 rupee (100 paise) so the ratio will become 25 : 50 : 100 or 1 : 2 : 4

Total value of 25 paise coins =(1 / 7 ) x 105 = 15 

Total cost of 50 paise cash = (2 / 7) x one hundred and five = 30 

Total price of 100 paise cash = (four / 7) x one zero five = 60 

No. Of 25 paise coins = 15 x four = 60 coins

No. Of fifty paise cash = 30 x 2 = 60 cash

No. Of 1 rupee cash = 60 x 1 = 60 coins.

Question forty seven. A Purse Contains 342 Coins Consisting Of One Rupees, 50 Paise And 25 Paise Coins. If Their Values Are In The Ratio Of eleven : 9 : 5 Then Find The Number Of 50 Paise Coins?

Answer :

Let the value of 1 rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.

No. Of 1 rupee coins = (11x / 1) =11x

No. Of fifty paise coins = (9x / 0.5) = 18x

No. Of 25 paise cash = (5x / 0.25) = 20x

11x + 18x + 9x = 342 

38x = 342 

x = nine

Therefore, no. Of 1 rupee cash = eleven x nine = ninety nine coins

No. Of 50 paise coins = 18 x 9 = 162 cash

No. Of 25 paise cash = 20 x 9 = a hundred and eighty coins.