# 300+ TOP Cognizant Aptitude Interview Questions and Answers

Q1. 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.

Q2. 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.

Q3. A Reduction Of 20% In The Price Of Strawberries Enables A Person To Purchase 12 More For Rs.

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.

Q4. 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.

Q5. 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.

Q6. 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?

Since the cube is melted so the volume of the new cube must be the same.

Volume of new cube = Volume of cube 1 + cube 2 + cube 3 = 63 + 83 + 103 = 216 + 512 + 1000

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

Q7. 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.

Q8. 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.

Q9. 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 Ca

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 %.

Q10. 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.

Q11. 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.

Q12. 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.

Q13. 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.

Q14. 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.

Q15. 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.

Q16. Let C Be A Positive Integer Such That C + 7 Is Divisible By

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.

Q17. 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

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.

Q18. P Is An Integer. P>88

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.

Q19. 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)

So wer is 5^10.

Q20. 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

Q21. The Ratio Of The No. Of White Balls In A Bag To That Of Black Balls Is 1:

Consider x black balls were there.

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

That me, x/9 = 4/3.

On solving we will get x = 12.

Q22. 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.

Q23. 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, Til

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.

Q24. 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.

Q25. 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 @And 25 is divisible by 5^2.

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

Q26. 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.

Q27. 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

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.

Q28. 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.

Q29. 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,@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 10@So keeping k=0,1,2,3.

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