## Cognizant Aptitude Interview Questions & Answers

• Question 1. The Number Of Prime Factors Of (3 X 5)12 (2 X 7)10 (10)25 Is?

The equation can be facorize as 3*5*3*2*2*2*7*2*5*2*5*5*5 or 2^5*3^2*5^5*7^1

total no of prime factor =(5+1)*(5+1)*(2+1)*(1+1)=216.

• Question 2. What Least Value Must Be Assigned To * So That The Number 63576*2 Is Divisible By 8?

The test for divisibility by 8 is that the last 3 digits of the number in question have to be divisible by 8.

So, 6*2 has to be divisibile by 8.

I know 512 is divisible by 8.

Also 592 is divisible by 8.

So, 632 is divisible by 8.

So * is 3.

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• Question 3. The Smallest Number, Which Is A Perfect Square And Contains 7936 As A Factor Is:

7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1

To make it as a perfect square, we have to multiply 7936 with 31.

Hence the reqd no. is 7936*31 = 246016.

• Question 4. If A Number Is Exactly Divisible By 85, Then What Will Be The Remainder When The Same Number Is Divided By 17?

number=divisor*quotient+remainder

so 17*5+0;

remainder is 0;

divisor is 17;

quotient is 5.

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• Question 5. P Is An Integer. P Is Greater Than 883.if P -7 Is A Multiple Of 11, Then The Largest Number That Will Always Divide (p+4)(p+15) Is?

p-7= 11*a (as it is multiple of 11)

p=11*(a+7)

so (p+4)(p+15)= (11a+7+4)(11a+7+15);

= (11a+11)(11a+22);

=11*11(a+1)(a+2);

=121*2

=242.

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• Question 6. The Greatest Number That Will Divide 63, 138 And 228 So As To Leave The Same Remainder In Each Case?

The greatest number = H.C.F of (138-63), (228-138), (228-63)

H.C.F of 75, 90, 165 = 15.

15 is the greatest number.

• Question 7. Find The Largest Number, Smaller Than The Smallest Four-digit Number, Which When Divided By 4,5,6and 7 Leaves A Remainder 2 In Each Case?

Take LCM of 4,5,6,7. It is 420

BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2.

The smallest 4-digit no is 1000. So keeping k=0,1,2,3.

We get that the largest no smaller than the smallest 4 -digit no is 842.

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• Question 8. What Is The Highest Power Of 5 That Divides 90 X 80 X 70 X 60 X 50 X 40 X 30 X 20 X 10?

Take LCM of Each Number:

90/5=5*2*3*3——————>here we will get one 5

80/5=5*2*2*2*2—————>here we will get one 5

70/5=5*2*7————–___—->here we will get one 5

60/5=5*2*2*3——————>here we will get one 5

50/5=5*5*2___——————>here we will get Two 5^2

40/5=5*2*2*2——————>here we will get one 5

30/5=5*2*3———————>here we will get one 5

20/5=5*2*2———————>here we will get one 5

10/5=5*2————————>here we will get one 5

Here we will get one 5 in each number instead of 50(5*5*2)

• Question 9. If A And B Are Natural Numbers And A-b Is Divisible By 3, Then A3-b3 Is Divisible By?

If a − b is divisible by 3, then a − b = 3k, for some integer k

(a − b)² = (3k)²

a² − 2ab + b² = 9k²

a³ − b³ = (a−b) (a² + ab + b²)

= (a−b) (a² − 2ab + b² + 3ab)

= 3k (9k + 3ab)

= 3k * 3 (3k + ab)

= 9 k(3k+ab)

Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9.

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• Question 10. What Is The Greatest Positive Power Of 5 That Divides 30! Exactly?

Only the numbers 5, 10, 15, 20, 25, and 30 have divisors of 5. And 25 is divisible by 5^2.

So the answer is 5*5*5*5*(5^2)*5 = 5^7.

• Question 11. What Is The Smallest Four-digit Number Which When Divided By 6, Leaves A Remainder Of 5 And When Divided By 5 Leaves A Remainder Of 3?

remainder when  m is divided by 5  = 2

Smallest m is 2.

Hence, N = 1001 + 6 * 2 = 1013.

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• Question 12. P Is An Integer. P>883. If P-7 Is A Multiple Of 11, Then The Largest Number That Will Always Divide (p+4) (p+15) Is?

Given P is an integer>883.

P-7 is a multiple of 11=>there exist a positive integer a such that

P-7=11 a=>P=11 a+7

(P+4)(P+15)=(11 a+7+4)(11 a+7+15)

=(11 a+11)(11 a+22)

=121(a+1)(a+2)

As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242.

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• Question 13. Let C Be A Positive Integer Such That C + 7 Is Divisible By 5. The Smallest Positive Integer N (>2) Such That C + N2 Is Divisible By 5 Is?

c + n^2 is divisible by 5 if and only if c and n^2 are both divisible by 5.

But, if c is divisible by 5 then c + 5 will not be divisible by 5.

• Question 14. Four Bells Begin To Toll Together And Then Each One At Intervals Of 6 S, 7 S, 8 S And 9 S Respectively.the Number Of Times They Will Toll Together In The Next 2 Hr Is?

first we to find the L.C.M. of 6, 7, 8 and 9.

Prime factorization of 6 = 2*3

Prime factorization of 7 = 7

Prime factorization of 8 = 2*2*2

Prime factorization of 9 = 3*3

L.C.M. = 2*2*2*3*3*7

= 504

The L.C.M. of 6 seconds, 7 seconds, 8 seconds and 9 seconds is 504

seconds.

Now, 1 hour = 3600 seconds

So, 2 hours = 3600*2 = 7200 seconds

The number of times the four bells will toll together in the next 2 hour

= 7200/504

= 14.28 or 14 times

They will toll together 14 times in the next 2 hours

• Question 15. On Dividing A Number By 999,the Quotient Is 366 And The Remainder Is 103.the Number Is?

Number (Dividend) = Divisor * quotient + remainder.

Number = 999 * 377 + 105 = 3767.

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• Question 16. If 522x Is A Three Digit Number With As A Digit X . If The Number Is Divisible By 6, What Is The Value Of The Digit X Is?

If a number is Divisiable by 6 , it must be divisible by both 2 and 3

In 522x, to this number be divisible by 2, the value of x must be even. So it n be 2,4 or 6 from given options

552x is divisible by 3, If sum of its digits is a multiple of 3.

5+5+2+x =12+x ,

If put x =2 , 12+2=14 not a multiple of 3

If put x =4 , 12+6=18  is a multiple of 3

If put x =6 , 12+2=14 not a multiple of 3

The value of x is 6.

• Question 17. In An Election Between Two Candidates, One Got 55% Of The Total Valid Votes And Got 20% Invalid Votes. At The End Of The Day When The Total Number Of Votes Were Counted, The Total Number Was Found To Be 7500. So What Was The Total Number Of Valid Votes That The Winning Candidate Got, Was?

Since 20% of the votes were invalid, 80% of the votes were valid = 80% of 7500 = 6000 votes were valid.

One candidate got 55% of the total valid votes, then the second candidate must have 45% of the votes = 0.45 * 6000 = 2700 votes.

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• Question 18. A Whole Number N Which When Divided By 4 Gives 3 As Remainder. What Will Be The Remainder When 2n Is Divided By 4?

According to the question, n = 4q + 3.

Therefore, 2n = 8q + 6 or 2n = 4(2q + 1) + 2.

Thus, we get when 2n is divided by 4, the remainder is 2.

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• Question 19. Raju, Ramu And Razi Can Do A Piece Of Work In 20, 30 And 60 Days Respectively Depending On Their Capacity Of Doing Work. If Raju Is Assisted By Ramu And Razi On Every Third Day, Then In How Raju Will Complete The Work?

We need t first count the amount of work done in 2 days by Raju.

Raju can do a piece of work in 20 days.

So, in 2 days he can do = 1/20 * 2 = 1/10.

Amount of work done by Raju, Ramu and Razi in 1 day = 1/20 + 1/30 + 1/60 = 1/10.

Amount of work done in 3 days = 1/10 + 1/10 = 1/5.

So the work will be completed in 3 * 5 = 15 days.

• Question 20. A Tap Can Fill A Bucket In 6 Hours. After Half The Bucket Is Filled, Three More Similar Taps Are Opened. What Is The Total Time Taken To Fill The Bucket Completely?

Time is taken by one tap to fill half the bucket = 3 hours.

So the part filled 4 taps in one hour = 4 * (1/6) = 2/3 of the bucket.

Therefore, the remaining part is = (1 – 1/2) = 1/2

Proportionally à 2/3: 1/2:: 1: x

=> x = 3/4 hours = 45 minutes. So the total time = 3 hrs 45 minutes.

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• Question 21. A Reduction Of 20% In The Price Of Strawberries Enables A Person To Purchase 12 More For Rs. 15. What Was The Price Of 16 Strawberries Before Reduction Of Price?

Price x Consumption = Expenditure

(15 / 8x) – (15 / x) = 12

x = (15 x 2) / (12 x 8)

For 16 Strawberries = [(15 x 2) / (12 x 8)] x 16 = 5.

• Question 22. The Ratio Of The No. Of White Balls In A Bag To That Of Black Balls Is 1:2. If 9 Grey Balls Are Added The Ratio Of Nos. Of White, Black And Grey Become 2:4:3. How Many Black Balls Were In The Bag?

Consider x black balls were there.

After adding 9 grey balls the ratio is 4/3.

That means, x/9 = 4/3.

On solving we will get x = 12.

• Question 23. The Average Weight Of 8 Person’s Increases By 2.5 Kg When A New Person Comes In Place Of One Of Them Weighing 65 Kg. What Might Be The Weight Of The New Person?

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

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• Question 24. A Shopkeeper Gives Two Successive Discounts Of 20 % And 10 % On Surplus Stock. Further, He Also Gives 5 % Extra Discount On Cash Payment. If A Person Buys A Shirt From The Surplus Stock And Pays In Cash, What Overall Discount Percent Will He Get On The Shirt?

Let the marked price of the shirt be Rs. 1000

=> Price after first discount = Rs. 1000 – 20 % of Rs. 1000 = Rs. 1000 – 200 = Rs. 800

=> Price after second discount = Rs. 800 – 10 % of Rs. 800 = Rs. 800 – 80 = Rs. 720

=> Price after cash discount = Rs. 720 – 5 % of Rs. 720 = Rs. 720 – 36 = Rs. 684

Therefore, total discount = Rs. 1000 – 684 = Rs. 316

=> Overall discount percent = (316 / 1000) x 100 = 31.60 %.

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• Question 25. A & B Are At A Distance Of 800 M. They Start Towards Each Other At 20 & 24 Kmph. As They Start, A Bird Sitting On The Cap Of A, Starts Flying Towards B, Touches B & Then Returns Towards A & So On, Till They Meet. What Is The Distance Traveled By The Bird, If Its Speed Is 176 Kmph?

The bird flies for the same time as both A and B take to meet.

Since the time taken by A and B together and the bird is same, so the distance covered will be in the ratio of their speeds.

The ratio of the speeds is 44: 176 or 1: 4.

Hence, if A and B cover 800 m, the bird will cover 800*4 = 3200 m.

• Question 26. How Long Will A Boy Take To Run Round A Square Field Of Side 35 Meters, If He Runs At The Rate Of 9 Km/hr?

Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec

Distance = (35 x 4) m = 140 m.

Time taken = 140 x (2/5) sec= 56 sec.

• Question 27. From A Group Of 7 Men And 6 Women, Five Persons Are To Be Selected To Form A Committee So That At Least 3 Men Are There In The Committee. In How Many Ways Can It Be Done?

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.

Hence we have the following 3 options.

We can select 5 men à Number of ways to do this = 7C5

ii) We can select 4 men and 1 woman à Number of ways to do this = 7C4 × 6C1

iii)   We can select 3 men and 2 women à Number of ways to do this = 7C3 × 6C2

Total number of ways = 7C5+ (7C4 × 6C1) + (7C3× 6C2)

= 7C2+ (7C3× 6C1) + (7C3×6C2) —-     Expand this using nCr = nC (n – r)

= 21 + 210 + 525 = 756.

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• Question 28. What Is The Number Of Digits In 333? Given That Log3 = 0.47712?

Let   Let x=(333) = (33)3

Then, log(x) = 33 log(3)

= 27 x 0.47712 = 12.88224

Since the characteristic in the resultant value of log x is 12

∴The number of digits in x is (12 + 1) = 13

Hence the required number of digits in 333is 13.

• Question 29. A Hollow Iron Pipe Is 21 Cm Long And Its External Diameter Is 8 Cm. If The Thickness Of The Pipe Is 1 Cm And Iron Weighs 8 G/cm3, Then The Weight Of The Pipe Is?

Given the external diameter = 8 cm. Therefore, the radius = 4 cm.

The thickness = 1 cm. Therefore the internal radius = 4 – 1 = 3 cm

The volume of the iron = pi *(R^2 – r^2)*length = 22/7 *[(4^2) – (3^2)] *21 = 462 cm3>.

Therefore, the weight of iron = 462 * 8 gm = 3.696 kg.

• Question 30. Three Cubes Of Edges 6 Cm, 8 Cm And 10 Cm Are Meted Without Loss Of Metal Into A Single Cube. The Edge Of The New Cube Will Be?