Cognizant Aptitude Placement Papers - Cognizant Aptitude Interview Questions and Answers updated on 07.Jul.2022

Number (Dividend) = Divisor * quotient + remainder.

Number = 999 * 377 + 105 = 3767.

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.

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.

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.

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.

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.

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.

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.

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 %.

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.

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.

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.

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.

number=divisor*quotient+remainder

so 17*5+0;

remainder is 0;

divisor is 17;

quotient is 5.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.