## VMware Aptitude Interview Questions & Answers

1. Question 1. A 270 M Long Train Running At The Speed Of 120 Km/hr Crosses Another Train Running In Opposite Direction At The Speed Of 80 Km/hr In 9 Sec. What Is The Length Of The Other Train?

Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.

2. Question 2. Two Men Amar And Bhuvan Have The Ratio Of Their Monthly Incomes As 6 : 5. The Ratio Of Their Monthly Expenditures Is 3 : 2. If Bhuvan Saves One-fourth Of His Income, Find The Ratio Of Their Monthly Savings?

Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.

Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.

Savings of Bhuvan every month = 1/4(5x)

= (His income) – (His expenditure) = 5x – 2y.

=> 5x = 20x – 8y => y = 15x/8.

Ratio of savings of Amar and Bhuvan

= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4

= 3 : 10.

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4. Question 3. In 100 M Race, A Covers The Distance In 36 Seconds And B In 45 Seconds. In This Race A Beats B By?

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds.
Clearly, A beats B by (45-36)=9 seconds
Speed of B = Distance Time=10045 m/sDistance Covered by B in 9 seconds = Speed × Time = 10045×9 = 20 metre i.e., A beats B by 20 metre

5. Question 4. Three Persons A, B And C Divide A Certain Amount Of Money Such That A’s Share Is Rs. 4 Less Than Half Of The Total Amount, B’s Share Is Rs. 8 More Than Half Of What Is Left And Finally C Takes The Rest Which Is Rs. 14. Find The Total Amount They Initially Had With Them?

Let the total amount be Rs. p.

Let shares of A and B be Rs. x and Rs. y respectively.

C’s share was Rs. 14

we have, x + y + 14 = p —– (1)

From the given data, x = (p/2) – 4 —– (2)

Remaining amount = p – (p/2 – 4) => p/2 + 4.

y = 1/2(p/2 + 4) + 8 => p/4 + 10 —– (3)

From (1), (2) and (3)

p/2 – 4 + p/4 + 10 + 14 = p

3p/4 + 20 = p

p/4 = 20 => p = Rs. 80.

6. Question 5. The Mean Of Eight Numbers Is 25. If Five Is Subtracted From Each Number, What Will Be The New Mean?

Let the given numbers be x1, x2, . . ., x8.

Then, the mean of these numbers = (x1 + x2 + …+ x8)/8.

Therefore, (x1 + x2+…+x8)/8 = 25

⇒ (x1 + x2 + … + x8) = 200 ……. (A)

The new numbers are (x1 – 5), (x2 – 5), …… ,(x8 – 5)

Mean of the new numbers = {(x1 – 5) + (x1 – 5) + …… + (x8 – 5)}/8

= [(x1 + x2 + … + x8) – 40]/8

= (200 – 40)/8, [using (A)]

= 160/8

= 20

Hence, the new mean is 20.

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8. Question 6. In A Fort, There Are 1200 Soldiers. If Each Soldier Consumes 3 Kg Per Day, The Provisions Available In The Fort Will Last For 30 Days. If Some More Soldiers Join, The Provisions Available Will Last For 25 Days Given Each Soldier Consumes 2.5 Kg Per Day. Find The Number Of Soldiers Joining The Fort In That Case?

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)

= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available

=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = [(1200)(30)(3)] / (25)(2.5) – 1200 => x = 528.

9. Question 7. A Man, A Woman And A Boy Can Complete A Job In 3, 4 And 12 Days Respectively. How Many Boys Must Assist 1 Man And 1 Woman To Complete The Job In 1/4 Of A Day?

(1 man + 1 woman)’s 1 day work = (1/3 + 1/4) = 7/12 Work done by 1 man and 1 woman in 1/4 day = (7/12 * 1/4) = 7/48
Remaining work = (1 – 7/48) = 41/48
Work done by 1 boy in 1/4 day = ( 1/12 * 1/4) = 1/48
Number of boys required = 41/48 * 41 = 41

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11. Question 8. The Mean Of 14 Numbers Is 6. If 3 Is Added To Every Number, What Will Be The New Mean?

Let the given numbers be x1, x2, x3, ….. x14.

Then, the mean of these numbers = x1 + x2 + x3+ ….. x14/14

Therefore, (x1 + x2 + x3 + ….. x14)/14 = 6

⇒ (x1 + x2 + x3 + ….. x14) = 84 ………………. (A)

The new numbers are (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)

Mean of the new numbers

= (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)/14

= (x1 + x2 + x3 + ….. x14) + 42

= (84 + 42)/14, [Using (A)]

= 126/14

= 9

Hence, the new mean is 9.

12. Question 9. The Aggregate Monthly Expenditure Of A Family Was \$ 6240 During The First 3 Months, \$ 6780 During The Next 4 Months And \$ 7236 During The Last 5 Months Of A Year. If The Total Saving During The Year Is \$ 7080, Find The Average Monthly Income Of The Family?

Total expenditure during the year

= \$[6240 × 3 + 6780 × 4 + 7236 × 5]

= \$ [18720 + 27120 + 36180]

= \$ 82020.

Total income during the year = \$ (82020 + 7080) = \$ 89100.

Average monthly income = (89100/12) = \$7425.

Hence, the average monthly income of the family is \$ 7425.

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14. Question 10. Which Of The Following Symbol Should Replace Question Mark (?) In The Given Expression In Order To Make The Statements ‘x > V’ And ‘z < T’ Definitely Follow? T = X ? Y = Z > V = W?

By replacing question mark with ‘>’ symbol we get,
T = X > Y = Z > V = W.
From the above ‘X > V’ and ‘Z < T’ definitely follow.

15. Question 11. 24, 60, 120, 210, ?

The pattern is + 36, + 60, + 90,…..i.e. + [6 x (6 + 0)], + [6 x (6 + 4)], + [6 x (6 + 9)],…
So, missing term = 210 + [6 x (6 + 15)] = 210 + 126 = 336.

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17. Question 12. Find Remainder Of (9^1+9^2+………+9^n)/6 N Is Multiple Of 11?

9/6 remainder is 3 9^2/6 remainder is 3 9^3/6 remainder is 3 9 to the power of any number when divided by 6 ,

the remainder will always be 3.

Now, ( 3+3+3 ……11 times)/6 =(3*11)/6;

therefore the remainder will be 3.

If we take the even multiple of 11, then remainder will be zero.

Therefore answer is cannot be determine

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19. Question 13. 12l : 24x :: 5e : __?

12L : 24X :: 5E : __
L is 12th letter and 12 * 2 = 24
The 24th letter is X.
Similarly, E is the 5th letter and 5 * 2 = 10
The 10th letter is J.

20. Question 14. The Mean Of 25 Observations Is 36. If The Mean Of The First Observations Is 32 And That Of The Last 13 Observations Is 39, Find The 13th Observation?

Mean of the first 13 observations = 32.

Sum of the first 13 observations = (32 × 13) = 416.

Mean of the last 13 observations = 39.

Sum of the last 13 observations = (39 × 13) = 507.

Mean of 25 observations = 36.

Sum of all the 25 observations = (36 × 25) = 900.

Therefore, the 13th observation = (416 + 507 – 900) = 23.

Hence, the 13th observation is 23.

21. Question 15. The Mean Of 16 Items Was Found To Be 30. On Rechecking, It Was Found That Two Items Were Wrongly Taken As 22 And 18 Instead Of 32 And 28 Respectively. Find The Correct Mean?

Calculated mean of 16 items = 30.

Incorrect sum of these 16 items = (30 × 16) = 480.

Correct sum of these 16 items

= (incorrect sum) – (sum of incorrect items) + (sum of actual items)

= [480 – (22 + 18) + (32 + 28)]

= 500.

Therefore, correct mean = 500/16 = 31.25.

Hence, the correct mean is 31.25.

22. Question 16. The Average Height Of 30 Boys Was Calculated To Be 150 Cm. It Was Detected Later That One Value Of 165 Cm Was Wrongly Copied As 135 Cm For The Computation Of The Mean. Find The Correct Mean?

Calculated average height of 30 boys = 150 cm.

Incorrect sum of the heights of 30 boys

= (150 × 30)cm

= 4500 cm.

Correct sum of the heights of 30 boys

= (incorrect sum) – (wrongly copied item) + (actual item)

= (4500 – 135 + 165) cm

= 4530 cm.

Correct mean = correct sum/number of boys

= (4530/30) cm

= 151 cm.

Hence, the correct mean height is 151 cm.

23. Question 17. Ax^2+bx+c = 0 Has Two Roots X1 And X2. If Mode X1 = Mode X2, Then?

mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.

in this case b=0 eqn is ax^2+c = 0 => x=sqrt(-c/a) for x to be real (-c/a) should +ve => c & a are of opposite sign so c>a or a>c for ex x^2-4=0 => x1=-2,

x2=2 & modx1=modx2=2 [a>c] or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 (d) none

24. Question 18. How Many Values Of C In The Equation X^3-5x+c Result In Rational
Roots Which Are Integers?

x^3-5x+c=0 if x=1,

1-5+c=0 or c=4 if x=-1,

-1+5+c=0 or c=-4 if x=2, 8-10+c=0 or

c=2 if x=-2, -8+10+c=0 or c=-2 we see that,

for c=-2,2,-4,4

we get x=-2,2,-1,1 etc i.e we get integral roots of x for infinite values of c. d)infinite

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26. Question 19. What Is The Least Number To Be Subtracted From 11, 15, 21 And 30 Each So That Resultant Numbers Become Proportional?

Let the least number to be subtracted be x, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.

<=> (11 – x) : (15 – x) = (21 – x) : (30 – x)

=> (11 – x)(30 – x) = (15 – x)(21 – x)

From the options, when x = 3

=> 8 * 27 = 12 * 18

27. Question 20. The Mean Weight Of A Class Of 35 Students Is 45 Kg. If The Weight Of The Teacher Be Included, The Mean Weight Increases By 500 G. Find The Weight Of The Teacher?

Mean weight of 35 students = 45 kg.

Total weight of 35 students = (45 × 35) kg = 1575 kg.

Mean weight of 35 students and the teacher (45 + 0.5) kg = 45.5 kg.

Total weight of 35 students and the teacher = (45.5 × 36) kg = 1638 kg.

Weight of the teacher = (1638 – 1575) kg = 63 kg.

Hence, the weight of the teacher is 63 kg.

28. Question 21. If P, Q And R Are Positive Integers And Satisfy X = (p + Q -r)/r = (p – Q + R)/q = (q + R – P)/p, Then The Value Of X Is?

When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.

Hence, x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p

=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)

=> x = (r + q + p) / (r + q + p) = 1

p + q + r is non-zero.

29. Question 22. How Many Values Of C In Equation X^2-5x+c Result In Rational Roots Which Are Integers?

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg = 338 kg.

Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys) = (392 – 338) kg = 54 kg.

30. Question 23. The Weights Of Three Boys Are In The Ratio 4 : 5 : 6. If The Sum Of The Weights Of The Heaviest And The Lightest Boy Is 45 Kg More Than The Weight Of The Third Boy, What Is The Weight Of The Lightest Boy?

Let the weights of the three boys be 4k, 5k and 6k respectively.

4k + 6k = 5k + 45

=> 5k = 45 => k = 9

Therefore the weight of the lightest boy

= 4k = 4(9) = 36 kg.

31. Question 24. The Mean Of Five Numbers Is 28. If One Of The Numbers Is Excluded, The Mean Gets Reduced By 2. Find The Excluded Number?

Mean of 5 numbers = 28.

Sum of these 5 numbers = (28 x 5) = 140.

Mean of the remaining 4 numbers = (28 – 2) =26.

Sum of these remaining 4 numbers = (26 × 4) = 104.

Excluded number

= (sum of the given 5 numbers) – (sum of the remaining 4 numbers)

= (140 – 104)

= 36.

Hence, the excluded number is 36.

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33. Question 25. A Cricketer Has A Mean Score Of 58 Runs In Nine Innings. Find Out How Many Runs Are To Be Scored By Him In The Tenth Innings To Raise The Mean Score To 61?

Mean score of 9 innings = 58 runs.

Total score of 9 innings = (58 x 9) runs = 522 runs.

Required mean score of 10 innings = 61 runs.

Required total score of 10 innings = (61 x 10) runs = 610 runs.

Number of runs to be scored in the 10th innings

= (total score of 10 innings) – (total score of 9 innings)

= (610 -522) = 88.

Hence, the number of runs to be scored in the 10th innings = 88.

34. Question 26. The Mean Weight Of A Group Of Seven Boys Is 56 Kg. The Individual Weights (in Kg) Of Six Of Them Are 52, 57, 55, 60, 59 And 55. Find The Weight Of The Seventh Boy?

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg

= 338 kg.

Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys)

= (392 – 338) kg

= 54 kg.

Hence, the weight of the seventh boy is 54 kg.