Interview Questions.

Top 100+ Cognizant Aptitude Interview Questions And Answers

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Top 100+ Cognizant Aptitude Interview Questions And Answers

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

Answer :

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

general no of high aspect =(five+1)*(five+1)*(2+1)*(1+1)=216.

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

Answer :

The take a look at for divisibility by using 8 is that the remaining 3 digits of the wide variety in question should be divisible by eight.

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

I recognize 512 is divisible via 8.

Also 592 is divisible by using 8.

So, 632 is divisible by means of 8.

So * is 3.

HR Management Interview Questions
Question 3. The Smallest Number, Which Is A Perfect Square And Contains 7936 As A Factor Is:

Answer :

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

To make it as a great square, we need 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?

Answer :

quantity=divisor*quotient+remainder

so 17*5+zero;

the rest is 0;

divisor is 17;

quotient is five.

HR Management Tutorial
Question five. 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?

Answer :

p-7= 11*a (as it's far more than one of 11)

p=11*(a+7)

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

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

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

=121*2

=242.

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

Answer :

The finest number = H.C.F of (138-sixty three), (228-138), (228-63)

H.C.F of seventy five, 90, a hundred sixty five = 15.

15 is the greatest number.

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

Answer :

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

BUt the no should depart remainder 2 in every case, so the no is of the shape: 420k + 2.

The smallest 4-digit no is one thousand. So preserving okay=0,1,2,three.

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

TCS Technical Interview Questions
Question eight. What Is The Highest Power Of 5 That Divides ninety X eighty X 70 X 60 X 50 X forty X 30 X 20 X 10?

Answer :

Take LCM of Each Number:

ninety/5=5*2*3*three——————>here we are able to get one five

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

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

60/five=five*2*2*three——————>here we can get one 5

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

40/5=five*2*2*2——————>right here we are able to get one five

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

20/five=5*2*2———————>here we are able to get one five

10/five=five*2————————>right here we are able to get one five

Here we are able to get one five in every range rather than 50(five*five*2)

So solution is 5^10.

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

Answer :

If a − b is divisible via 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)

= nine ok(3k+ab)

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

Infosys Technical Interview Questions
Question 10. What Is The Greatest Positive Power Of 5 That Divides 30! Exactly?

Answer :

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

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

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

Answer :

remainder while  m is divided via five  = 2

Smallest m is two.

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

HR Interview Questions
Question 12. P Is An Integer. P>883. If P-7 Is A Multiple Of eleven, Then The Largest Number That Will Always Divide (p+four) (p+15) Is?

Answer :

Given P is an integer>883.

P-7 is a multiple of eleven=>there exist a superb integer a such that

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

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

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

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

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

HR Management Interview Questions
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?

Answer :

c + n^2 is divisible by using five if and most effective if c and n^2 are both divisible through five.

But, if c is divisible by using five then c + 5 will not be divisible by way of 5.

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

Answer :

first we to find the L.C.M. Of 6, 7, eight and nine.

Prime factorization of 6 = 2*3

Prime factorization of seven = 7

Prime factorization of eight = 2*2*2

Prime factorization of nine = 3*3

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

= 504

The L.C.M. Of 6 seconds, 7 seconds, 8 seconds and nine seconds is 504

seconds.

Now, 1 hour = 3600 seconds

So, 2 hours = 3600*2 = 7200 seconds

The number of instances the 4 bells will toll together in the next 2 hour

= 7200/504

= 14.28 or 14 instances

They will toll collectively 14 instances within the subsequent 2 hours

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

Answer :

Number (Dividend) = Divisor * quotient + remainder.

Number = 999 * 377 + 105 = 3767.

ABB Group Aptitude Interview Questions
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?

Answer :

If a range of is Divisiable by using 6 , it need to be divisible via both 2 and 3

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

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

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

If placed x =2 , 12+2=14 now not a a couple of of 3

If put x =4 , 12+6=18  is a a couple of of three

If put x =6 , 12+2=14 not a more than one of 3

The fee of x is 6.

Question 17. In An Election Between Two Candidates, One Got fifty five% 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?

Answer :

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

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

BHEL Aptitude Interview Questions
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?

Answer :

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

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

Thus, we get whilst 2n is divided by means of 4, the remainder is 2.

Aptitude Interview Questions
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?

Answer :

We need t first be counted the quantity of work done in 2 days by using Raju.

Raju can do a piece of labor in 20 days.

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

Amount of labor executed by using Raju, Ramu and Razi in 1 day = 1/20 + 1/30 + 1/60 = 1/10.

Amount of work executed in three days = 1/10 + 1/10 = 1/five.

So the work could be completed in three * 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?

Answer :

Time is taken by using one tap to fill half of the bucket = three hours.

So the element stuffed 4 faucets in one hour = four * (1/6) = 2/three of the bucket.

Therefore, the closing part is = (1 – half) = half of

Proportionally à 2/3: half of:: 1: x

=> x = three/four hours = 45 mins. So the total time = 3 hrs 45 minutes.

IBM Aptitude Interview Questions
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?

Answer :

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 sixteen = five.

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:three. How Many Black Balls Were In The Bag?

Answer :

Consider x black balls have been there.

After adding 9 gray balls the ratio is four/3.

That manner, x/9 = 4/three.

On fixing we can get x = 12.

Question 23. The Average Weight Of eight Person’s Increases By 2.Five Kg When A New Person Comes In Place Of One Of Them Weighing sixty five Kg. What Might Be The Weight Of The New Person?

Answer :

Total weight extended = (8 x 2.Five) kg = 20 kg.

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

Capgemini Aptitude Interview Questions
Question 24. A Shopkeeper Gives Two Successive Discounts Of 20 % And 10 % On Surplus Stock. Further, He Also Gives five % 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?

Answer :

Let the marked charge of the shirt be Rs. 1000

=> Price after first bargain = Rs. A thousand – 20 % of Rs. A thousand = Rs. A thousand – two hundred = Rs. 800

=> Price after 2nd bargain = Rs. 800 – 10 % of Rs. 800 = Rs. 800 – eighty = Rs. 720

=> Price after cash bargain = Rs. 720 – five % of Rs. 720 = Rs. 720 – 36 = Rs. 684

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

=> Overall cut price percent = (316 / 1000) x a hundred = 31.60 %.

TCS Technical Interview Questions
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?

Answer :

The hen flies for the equal time as both A and B take to satisfy.

Since the time taken via A and B together and the bird is equal, so the space blanketed can be inside the ratio in their speeds.

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

Hence, if A and B cover 800 m, the fowl will cover 800*four = 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?

Answer :

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

Distance = (35 x 4) m = a hundred and forty m.

Time taken = one hundred forty x (2/5) sec= fifty six 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 three Men Are There In The Committee. In How Many Ways Can It Be Done?

Answer :

From a group of seven men and six women, 5 humans are to be selected with at least 3 men.

Hence we've the subsequent 3 alternatives.

We can choose five guys à Number of ways to do this = 7C5

ii) We can pick four guys and 1 woman à Number of ways to do this = 7C4 × 6C1

iii)   We can choose three guys and a pair of women à Number of ways to do this = 7C3 × 6C2

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

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

= 21 + 210 + 525 = 756.

Infosys Technical Interview Questions
Question 28. What Is The Number Of Digits In 333? Given That Log3 = zero.47712?

Answer :

Let   Let x=(333) = (33)three 

 Then, log(x) = 33 log(3)  

= 27 x 0.47712 = 12.88224 

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

∴The quantity of digits in x is (12 + 1) = thirteen 

Hence the desired quantity of digits in 333is 13.

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

Answer :

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

The thickness = 1 cm. Therefore the internal radius = four – 1 = three cm

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

Therefore, the weight of iron = 462 * 8 gm = three.696 kg.

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

Answer :

Since the dice is melted so the volume of the new dice ought to be the same.

Volume of recent cube = Volume of dice 1 + dice 2 + dice three = 63 + 83 + 103 = 216 + 512 + one thousand

a^3 = 1728, a = (1728)^(1/3) = 12.




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