# Dell Emc Aptitude Placement Papers - Dell Emc Aptitude Interview Questions and Answers updated on 14.Jun.2024

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.

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.

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

Me, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.

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 %.

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.

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.

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.

Amount

= Rs.(25000x(1+12/100)³

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

= Rs. 35123.20.

C.I = Rs(35123.20 -25000)

= Rs.10123.20.

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

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms.

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.

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.

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

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

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 @p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph.

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.

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.

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.

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.

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.

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.

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

=Rs.840

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

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

Number of c of Maaza = 368/16 = 23

Number of c of Pepsi = 80/16 = 5

Number of c of Sprite = 144/16 = 9

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

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

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%.

Initial speed = 80km/hr

Total distance = 80 x 10 = 800km

New speed = 800/4 =200km/hr

Increase in speed = 200 - 80 = 120km/hr.

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.

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.

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.

(x - 4) / 6 = 9

Multiply both sides by 6:

x - 4 = 54

x = 58

(58 - 3) / 5 = 55 / 5 = 11.

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.

Let the actual distance travelled be x km.

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

=> 12x = 8x + 160

=> 4x = 160

=> x = 40 km.

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.

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.

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(8000x21/20x21/20x21/20)

= Rs.9261

C.I

=Rs.(9261-8000)

=Rs.1261.

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

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.

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.

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).

No. of letters in the word = 6

No. of 'E' repeated = 2

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

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.

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.

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.

Principal = Rs.16,000;

Time=9 months = 3 quarters;

Amount

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

= Rs.18522.

C.I

= Rs.(18522 - 16000)

= Rs.2522.

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.

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.

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.

Using pythagarous theorem,

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

distance travelled by second train = 48x5/18x30 = 400m

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

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 %.