## Dell EMC Aptitude Interview Questions & Answers

• 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?

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?

Let C.P.= Rs. 100.

Then, Profit = Rs. 320,

S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295

Required percentage = (295/420) * 100

= 70%(approx).

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• Question 3. 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?

Income = Rs. 100

Expenditure = Rs. 80

Savings = Rs. 20

Present Expenditure 80x(15/100) = Rs. 12 = 80 + 12 = Rs. 92

Present Savings = 100 – 92 = Rs. 8

100 —— 8

? ——— 400 => 5000

His salary = Rs. 5000.

• Question 4. Raghu Earns 25% On An Investment But Loses 10% On Another Investment. If The Ratio Of The Two Investment Is 3:5. What Is The Gain Or Loss On Two Investments Taken Together ?

Taking the 2 investments to be 3x and 5x respectively

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

Therefore, Gain% = 0.25/8 x 100 = 3.125 %.

• 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 ?

Let 1kg of Rs. 100 then 840gm is of Rs. 84.

Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.]

So now S.P = Rs. 96

But actually it contains 840 gm so C.P for shopkeeper = Rs. 84

S.P = Rs. 96

C.P = Rs. 84

Profit% = {(S.P-C.P)/C.P}x100

{(96-84)/84} x 100 = 14.28571429% PROFIT.

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• 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?

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

Loss% = 4

We know that, C.P = 100/(100 – Loss%) x 100

=> 100/100-4 x 10800

=> 1080000/96

C.P = Rs. 11,250.

• 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 ?

Let the cost price of a ball is Rs.x

Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls

The equation is :

17x – 720 = 5x

Solving the equation

we get x = 60

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

• Question 8. Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is?

Let S be the sample space.

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

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

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

Cost Price = Rs. 1950

Selling Price = Rs. 2340

Profit = S.P – C.P

Profit = Rs. 2340 – 1950 = 390

Profit % = (Profit/C.P) x 100

Profit % = (390/1950) x 100

Profit % = 20 %.

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• 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?

P(first letter is not vowel) = 2/4

P(second letter is not vowel) = 1/3

So, probability that none of letters would be vowels is = 2/4×1/3=1/6.

• Question 11. Four Dice Are Thrown Simultaneously. Find The Probability That All Of Them Show The Same Face?

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

= 6*6*6*6=64

n(S) = 64

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

X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}

n(X) = 6

Hence required probability = n(X)n(S)=6/64=1216.

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• Question 12. Three Unbiased Coins Are Tossed. What Is The Probability Of Getting At Most Two Heads?

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

Let E = event of getting at most two heads.

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

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

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• Question 13. Three Unbiased Coins Are Tossed.what Is The Probability Of Getting At Least 2 Heads?

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

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

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

= 4/8= 1/2.

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

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

Let E = event of getting a total more than 7

= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(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?

Amount of Rs. 100 for 1 year

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

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

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• Question 16. Simple Interest On A Certain Sum Of Money For 3 Years At 8% Per Annum Is Half The Compound Interest On Rs. 4000 For 2 Years At 10% Per Annum. The Sum Placed On Simple Interest Is?

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

=Rs.840

sum=Rs.(420 * 100)/3*8=Rs.1750.

• Question 17. What Will Be The Compound Interest On A Sum Of Rs.25,000 After 3 Years At The Rate Of 12 P.c.p.a?

Amount

= Rs.(25000x(1+12/100)³

= Rs.(25000×28/25×28/25×28/25)

= Rs. 35123.20.

C.I = Rs(35123.20 -25000)

= Rs.10123.20.

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

Given,

Compound rate, R = 10% per annum

Time = 2 years

C.I = Rs. 420

Let P be the required principal.

A = (P+C.I)

Amount, A = P(1 + (r/100))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 = 420/0.21 = Rs. 2000.

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

S.I. = Rs.(1200*10*1)/100=rs.120

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

Difference = Rs.(123-120) =Rs.3

• Question 20. Find The Compound Interest On Rs.16,000 At 20% Per Annum For 9 Months, Compounded Quartely?

Principal = Rs.16,000;

Time=9 months = 3 quarters;

Amount

=Rs.[16000x(1+5/100)³] =[16000×21/20×21/20×21/20]

= Rs.18522.

C.I

= Rs.(18522 – 16000)

= Rs.2522.

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• Question 21. In How Many Ways Can The Letters Of The Word ‘leader’ Be Arranged ?

No. of letters in the word = 6

No. of ‘E’ repeated = 2

Total No. of arrangement = 6!/2! = 360.

• Question 22. If The Simple Interest On A Sum Of Money At 5% Per Annum For 3 Years Is Rs. 1200, Find The Compound Interest On The Same Sum For The Same Period At The Same Rate?

Clearly, Rate = 5% p.a .,

Time = 3 years

S.I =Rs.1200.

So,Principal

=Rs.(100 x 1200/3x5)

=Rs.8000.

Amount

=Rs.[8000 x (1+5/100)³]

=Rs(8000×21/20×21/20×21/20)

= Rs.9261

C.I

=Rs.(9261-8000)

=Rs.1261.

• Question 23. Pumps Working 8 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?

Let the required no of working hours per day be x.

More pumps , Less working hours per day  (Indirect Proportion)

Less days, More working hours per day    (Indirect Proportion)

Pumps4 : 3Days1 : 2}?? 8:x

=> (4 * 1 * x) = (3 * 2 * 8)

=> x=12

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• 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?

Divisor = 2/3 x dividend

and Divisor = 2 x remainder

or 2/3 x dividend = 2 x 5

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

• Question 25. How Many Figures (digits) Are Required To Number A Book Containing 200 Pages ?

Number of one digit pages from

1 to 9 = 9

Number of two digit pages from

10 to 99 = 90

Number of three digit pages from

100 to 200 = 101.

• 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 ?

Let 1/2 of the no. = 10x + y

and the no. = 10v + w

From the given conditions,

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 = 8

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

x = 2 and y = 6.

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• 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 4 More Than Six Times The Sum Of Its Digits, Find The Number ?

Let the two-digit number 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.

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• Question 28. If A Number Is Decreased By 4 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 5 ?

(x – 4) / 6 = 9

Multiply both sides by 6:

x – 4 = 54

x = 58

(58 – 3) / 5 = 55 / 5 = 11.

• 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 ?

xy = 96050 …(i)

and xz = 95625 …(ii)

and y – z = 1 … (iii)

Dividing (i) by (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 4 Hours, By How Much The Speed Of Bus Will Have To Increase ?

Initial speed = 80km/hr

Total distance = 80 x 10 = 800km

New speed = 800/4 =200km/hr

Increase in speed = 200 – 80 = 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. ?

Let the distance traveled be x km.

Then, x/10 – x/15 = 2

3x – 2x = 60 => x = 60 km.

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

So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph.

• Question 32. In What Time A 360 M. Long Train Moving At The Speed Of 44 Km/hr Will Cross A 140 M. Long Bridge ?

Speed = 44 kmph x 5/18 = 110/9 m/s

We know that, Time = distance/speed

Time = (360 + 140) / (110/9)

= 500 x 9/110 = 41 sec.

• Question 33. K And L Starts Walking Towards Each Other At 4 Pm At Speed Of 3 Km/hr And 4 Km/hr Respectively. They Were Initially 17.5 Km Apart. At What Time Do They Meet ?

Suppose they meet after ‘h’ hours

Then< /p>

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

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

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• 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 ?

Let the actual distance travelled be x km.

Then x/8=(x+20)/12

=> 12x = 8x + 160

=> 4x = 160

=> x = 40 km.

• Question 35. Kamal Consistently Runs 240 Meters A Day And On Saturday He Runs For 400 Meters. How Many Kilometers Will He Have To Run In Four Weeks ?

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

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 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 ?

Using pythagarous theorem,

distance travelled by first train = 36×5/18×30 = 300m

distance travelled by second train = 48×5/18×30 = 400m

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

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• Question 37. In A Daily Morning Walk Three Persons Step Off Together. Their Steps Measure 75 Cm, 80 Cm And 85 Cm Respectively. What Is The Minimum Distance Each Should Walk So That Thay Can Cover The Distance In Complete Steps ?

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.

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

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min

or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 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 8 A.m. And Travels Towards A At A Speed Of 25 Kmph. At What Time Will They Meet?

Assume both trains meet after ‘p’ hours after 7 a.m.

Distance covered by train starting from A in ‘p’ hours = 20p km

Distance covered by train starting from B in (p-1) hours = 25(p-1)

Total distance = 200

=> 20x + 25(x-1) = 200

=> 45x = 225

=> p= 5

Means, they meet after 5 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 ?

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.

Number of cans of Maaza = 368/16 = 23

Number of cans of Pepsi = 80/16 = 5

Number of cans of Sprite = 144/16 = 9

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

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• Question 41. Which Of The Following Has Most Number Of Divisors?

99  = 1 x 3 x 3 x 11

101= 1 x 101

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, 11, 33, 99

divisors of 101 are 1,101

divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176

divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182

Hence , 176 hasthe most number of divisors.

• Question 42. If The Sum Of Two Numbers Is 55 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 ?

Let the numbers be a and b.

We know that product of two numbers = Product of their HCF and LCM

Then, a + b = 55 and ab = 5 x 120 = 600.

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

=55/600 = 11/120.

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• Question 43. 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 ?

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 + 0 ) = 4.

• Question 44. Three Numbers Are In The Ratio Of 3:4:5 And Their L.c.m Is 3600.their Hcf Is?

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

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

So, 60x=3600 or x=60.

Therefore,  The numbers are (3 x 60), (4 x 60), (5 x 60).

Hence,required H.C.F=60.

• Question 45. Find The Lowest Common Multiple Of 24, 36 And 40?

To find the LCM of 24, 36 and 40

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

40 = 2 x 2 x 2 x 5

Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5

= 8 x 9 x 5

= 72 x 5

= 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?

Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4

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

Total value of 50 paise coins = (2 / 7) x 105 = 30

Total value of 100 paise coins = (4 / 7) x 105 = 60

No. of 25 paise coins = 15 x 4 = 60 coins

No. of 50 paise coins = 30 x 2 = 60 coins

No. of 1 rupee coins = 60 x 1 = 60 coins.

• Question 47. A Purse Contains 342 Coins Consisting Of One Rupees, 50 Paise And 25 Paise Coins. If Their Values Are In The Ratio Of 11 : 9 : 5 Then Find The Number Of 50 Paise Coins?

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

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

No. of 50 paise coins = (9x / 0.5) = 18x

No. of 25 paise coins = (5x / 0.25) = 20x

11x + 18x + 9x = 342

38x = 342

x = 9

Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins

No. of 50 paise coins = 18 x 9 = 162 coins

No. of 25 paise coins = 20 x 9 = 180 coins.