Dell Emc Aptitude Placement Papers - Dell Emc Aptitude Interview Questions and Answers
Q1. In What Time A 360 M. Long Train Moving At The Speed Of forty four Km/hr Will Cross A one hundred forty M. Long Bridge ?
Speed = forty four kmph x five/18 = one hundred ten/9 m/s
We realize that, Time = distance/velocity
Time = (360 + 140) / (110/9)
= 500 x nine/a hundred and ten = forty one sec.
Q2. 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 + zero ) = four.
Q3. Two Stations A And B Are two hundred 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
Assume each trains meet after 'p' hours after 7 a.M.
Distance covered by way of train beginning from A in 'p' hours = 20p km
Distance blanketed by means of train starting from B in (p-1) hours = 25(p-1)
Total distance = two hundred
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Me, they meet after 5 hours after 7 am, ie, they meet at 12 p.M.
Q4. The Table Is Bought For Rs. 1950 And Sold At Rs. 234
Cost Price = Rs. 1950
Selling Price = Rs. 2340
Profit = S.P – C.P
Profit = Rs. 2340 – 1950 = 390
Profit % = (Profit/C.P) x one hundred
Profit % = (390/1950) x a hundred
Profit % = 20 %.
Q5. 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 : five 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 cash = (11x / 1) =11x
No. Of fifty paise cash = (9x / zero.Five) = 18x
No. Of 25 paise cash = (5x / zero.25) = 20x
11x + 18x + 9x = 342
38x = 342
x = nine
Therefore, no. Of 1 rupee cash = 11 x nine = ninety nine coins
No. Of 50 paise coins = 18 x nine = 162 cash
No. Of 25 paise coins = 20 x nine = a hundred and eighty coins.
Q6. 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 quantity be 10x + y
10x + y = 7(x + y)
? X = 2y ...(i)
10(x +2 ) + (y + 2) = 6(x + y + 4) + four
or 10x + y + 22 = 6x + 6y + 28
? 4x - 5y = 6 ...(ii)
Solving equations (i) and (ii)
We get x = 4 and y = 2.
Q7. 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
Let 1/2 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 = 8
Solving equations (i) and (ii) , we get
x = 2 and y = 6.
Q8. 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?
= Rs.(25000x(1+12/one hundred)³
= Rs. 35123.20.
C.I = Rs(35123.20 -25000)
Q9. 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 jogging distance in 4 weeks = (24 x 240) + (four x four hundred)
= 5760 + 1600
= 7360 meters
= 7360/a thousand
=> 7.36 kms.
Q10. 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 = occasion of having at maximum two heads.
Then E = TTT, TTH, THT, HTT, THH, HTH, HHT.
Q11. A Shopkeeper Sold A Mobile Phone For Rs. 120
Given that SP = Rs. 12000 - 10% = Rs. 10,800
Loss% = 4
We recognise that, C.P = 100/(100 - Loss%) x a hundred
=> one hundred/one hundred-four x 10800
C.P = Rs. 11,250.
Q12. K And L Starts Walking Towards Each Other At 4 Pm At Speed Of three Km/hr And four Km/hr Respectively. They Were Initially 17.Five Km Apart. At What Time Do They Meet ?
Suppose they meet after 'h' hours
3h + 4h = 17.Five
7h = 17.Five
h = 2.5 hours
So they meet at => 4 + 2.5 = 6:30 pm.
Q13. 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 Reac
Let the space 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 began 6 hours earlier than @p.M. I.E., at 8 a.M.
Required velocity = 60/five = 12 kmph.
Q14. 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 fee of a ball is Rs.X
Given, on promoting 17 balls at Rs. 720, there may be a loss identical to the price price of 5 balls
The equation is :
17x - 720 = 5x
Solving the equation
we get x = 60
Therefore, fee charge of a ball is Rs. 60.
Q15. Running 3/4th Of His Usual Rate, A Man Is 15min Late. Find His Usual Time In Hours ?
Walking at 3/4th of ordinary rate implies that time taken might be four/3th of the standard time. In different words, the time taken is 1/3rd greater than his usual time
so 1/3rd of the standard time = 15min
or common time = 3 x 15 = 45min = 45/60 hrs = three/four hrs.
Q16. 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)
= four/eight= half of.
Q17. If The Sum Of Two Numbers Is fifty five And The H.C.F. And L.C.M. Of These Numbers Are five And one hundred twenty Respectively, Then The Sum Of The Reciprocals Of The Numbers Is Equal To ?
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 = five x a hundred and twenty = six hundred.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/six hundred = 11/a hundred and twenty.
Q18. Three Numbers Are In The Ratio Of three: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), (five x 60).
Q19. A Person Sold Two Cows Each For Rs.99
The CP of worthwhile 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 internet loss = 2475 - 900 = 1575.
Q20. Simple Interest On A Certain Sum Of Money For three 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?
sum=Rs.(420 * one hundred)/three*8=Rs.1750.
Q21. A Drink Vendor Has 368 Liters Of Maaza, eighty Liters Of Pepsi And 144 Liters Of Sprite. He Wants To Pack Them In C, So That Each Can Contains The Same Number Of Liters Of A Drink, And Doesn't Want To Mix
The variety of liters in every can = HCF of eighty, 144 and 368 = sixteen liters.
Number of c of Maaza = 368/sixteen = 23
Number of c of Pepsi = 80/16 = 5
Number of c of Sprite = one hundred forty four/16 = 9
The overall number of c required = 23 + five + nine = 37 c.
Q22. 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.One hundred twenty
Difference = Rs.(123-one hundred twenty) =Rs.Three
Q23. The Effective Annual Rate Of Interest Corresponding To A Nominal Rate Of 6% Per Annum Payable Half-yearly Is?
Amount of Rs. A hundred for 1 12 months
while compounded half-every year = Rs.[100*(1+3/100)^2]=Rs.106.09
Effective price=(106.09-a hundred)%=6.09%.
Q24. 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 pace = 80km/hr
Total distance = 80 x 10 = 800km
New pace = 800/four =200km/hr
Increase in speed = 200 - 80 = 120km/hr.
Q25. A Bag Contains Equal Number Of 25 Paise, 50 Paise And One Rupee Coins Respectively. If The Total Value Is Rs one zero five, How Many Types Of Each Type Are Present?
Bag includes 25 paise, 50 paise and 1 rupee (one hundred paise) so the ratio becomes 25 : 50 : a hundred or 1 : 2 : 4
Total fee of 25 paise cash =(1 / 7 ) x 105 = 15
Total value of 50 paise cash = (2 / 7) x 105 = 30
Total price of 100 paise cash = (four / 7) x one hundred and five = 60
No. Of 25 paise coins = 15 x four = 60 coins
No. Of 50 paise cash = 30 x 2 = 60 coins
No. Of one rupee cash = 60 x 1 = 60 coins.
Q26. 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 isn't vowel) = 2/four
P(second letter is not vowel) = 1/3
So, opportunity that none of letters could be vowels is = 2/four×1/three=1/6.
Q27. Two Cards Are Drawn Together From A Pack Of fifty two Cards. The Probability That One Is A Spade And One Is A Heart, Is?
Let S be the sample area.
Then, n(S) = 52C2=(52 x fifty one)/(2 x 1) = 1326.
Let E = event of having 1 spade and 1 heart.
N(E)= range of approaches of selecting 1 spade out of 13 and 1 heart out of thirteen = 13C1*13C1 = 169.
P(E) = n(E)/n(S) = 169/1326 = 13/102.
Q28. If A Number Is Decreased By 4 And Divided By 6 The Result Is
(x - 4) / 6 = 9
Multiply both facets with the aid of 6:
x - four = 54
Add 4 to both aspects:
x = 58
(fifty eight - 3) / 5 = 55 / five = 11.
Q29. 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 four% Loss On Cost Price ?
Let 1kg of Rs. 100 then 840gm is of Rs. Eighty four.
Now (label on can 1kg however contains 840kg ) so for client it's miles of Rs. One hundred and in addition offers 4% discount [he sells his article on 4% loss on cost price.]
So now S.P = Rs. Ninety six
But absolutely 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.Px100
(ninety six-eighty four)/eighty four x a hundred = 14.28571429% PROFIT.
Q30. 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 real distance travelled be x km.
=> 12x = 8x + 160
=> 4x = a hundred and sixty
=> x = 40 km.
Q31. In A Question Divisor Is 2/three Of The Dividend And 2 Times The Remainder. If The Remainder Is five, Find The Dividend?
Divisor = 2/three x dividend
and Divisor = 2 x the rest
or 2/3 x dividend = 2 x five
Dividend = 2 x five x three / 2 = 15.
Q32. In A Daily Morning Walk Three Persons Step Off Together. Their Steps Measure 75 Cm, 80 Cm And eighty five Cm Respectively. What Is The Minimum Distance Each Should Walk So That Thay Can Cover The Distance In C
To discover the minimum distance, we need to get the LCM of 75, 80, 85
Now, LCM of seventy five, eighty, eighty five = 5 x 15 x 16 x 17 = 20400
Hence, the minimal distance every have to walk so that thay can cowl the space in complete steps = 20400 cms = 20400/100 = 204 mts.
Q33. If The Simple Interest On A Sum Of Money At five% Per Annum For three Years Is Rs. 1200, Find The Compound Interest On The Same Sum For The Same Period At The Same Rate?
Clearly, Rate = five% p.A .,
Time = three years
=Rs.(100 x 1200/3x5)
=Rs.[8000 x (1+5/100)³]
Q34. 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 specified no of working hours in line with day be x.
More pumps , Less running hours per day (Indirect Proportion)
Less days, More running hours per day (Indirect Proportion)
Pumps4 : 3Days1 : 2?? 8:x
=> (4 * 1 * x) = (3 * 2 * 8)
Q35. Four Dice Are Thrown Simultaneously. Find The Probability That All Of Them Show The Same Face?
The general quantity of simple events related to the random experiments of throwing four cube simultaneously is:
= 6*6*6*6=sixty four
n(S) = sixty four
Let X be the occasion that all dice show the equal face.
X = (1,1,1,1,), (2,2,2,2), (3,three,3,three), (four,4,4,four), (five,five,five,5), (6,6,6,6)
n(X) = 6
Hence required opportunity = n(X)n(S)=6/sixty four=1216.
Q36. Find The Lowest Common Multiple Of 24, 36 And 40?
To locate 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 forty = 2 x 2 x 2 x three x three x five
= eight x 9 x 5
= seventy two x five
Q37. 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. = a hundred twenty five% of Rs. One hundred = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - one hundred twenty five) = Rs. 295
Required percentage = (295/420) * 100
Q38. In How Many Ways Can The Letters Of The Word 'chief' Be Arranged ?
No. Of letters in the word = 6
No. Of 'E' repeated = 2
Total No. Of association = 6!/2! = 360.
Q39. How Many Figures (digits) Are Required To Number A Book Containing 200 Pages ?
Number of 1 digit pages from
1 to 9 = nine
Number of two digit pages from
10 to ninety nine = ninety
Number of three digit pages from
100 to 2 hundred = one zero one.
Q40. 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. Four hundred Per Month. What Is His Monthly Salary?
Income = Rs. One hundred
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80x(15/one hundred) = Rs. 12 = eighty + 12 = Rs. Ninety two
Present Savings = a hundred – 92 = Rs. 8
one hundred ------ eight
? --------- four hundred => 5000
His profits = Rs. 5000.
Q41. 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 = occasion of having a complete more than 7
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Q42. 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;
= Rs.(18522 - 16000)
Q43. Find The Principal If The Interest Compounded At The Rate Of 10% Per Annum For Two Years Is Rs. 420 ?
Compound charge, R = 10% according to annum
Time = 2 years
C.I = Rs. 420
Let P be the required primary.
A = (P+C.I)
Amount, A = P(1 + (r/one hundred))n
(P+C.I) = P[1 + (10/100)]2
(P+420) = P[11/10][11/10]
P-1.21P = -420
zero.21P = 420
Hence, P = 420/0.21 = Rs. 2000.
Q44. 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) through (ii) we get
y/z = 96050 / 95625
= 3842 / 3825
= 226 / 225 ... (iv)
Combining (iii) and (iv) we get z = 225.
Q45. Which Of The Following Has Most Number Of Divisors?
99 = 1 x three x 3 x eleven
one 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 thirteen
So, divisors of ninety nine are 1, three, 9, eleven, 33, ninety nine
divisors of 101 are 1,101
divisors of 176 are 1, 2, 4, 8, sixteen, 22, 44, 88, 176
divisors of 182 are 1, 2, 7, 13, 14, 26, ninety one, 182
Hence , 176 hasthe maximum quantity of divisors.
Q46. 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 means of first teach = 36x5/18x30 = 300m
distance travelled by using 2nd train = 48x5/18x30 = 400m
so distance between them =v( 90000 + 160000) = v250000 = 500mts.
Q47. Raghu Earns 25% On An Investment But Loses 10% On Another Investment. If The Ratio Of The Two Investment Is three:
Taking the two investments to be 3x and 5x respectively
Total profits of Raghu = (3x) x 1.25 + (5x) x zero.9 = eight.25
Therefore, Gain% = zero.25/8 x 100 = 3.One hundred twenty five %.