16-08-2021 Answers

1. A two-digit number when read from left to right, is 4.5 times less than the same number read from right to left. What is the first digit of the number?

Answer: 1

Solution: Let the number be xy.

4.5 * (10x+y) = 10y + x,

9/2 * (10x+y) = 10y + x,

90x+9y = 20y + 2x,

88x = 11y, 8x=y.

As x has to be chose from 1 to 9, (zero cannot be in most significant digit), the possible combinations are (1,8). Thus the number is 18 and the first digit is 1.

2. Find the highest power of 10 in 120!

Answer: 28

Solution: Using formula to find num of zeros,

highest power = 120/5 + 120/25 = 24+4 =28

3. What is the value of the below expression?

(552*552 + 1600 - 220800) / (552*552*552 + 64000000)

Answer: 1/952

Solution: The expression can be written as (552*552 + 400*400 - (552 * 400)) / (552*552*552 + 400*400*400).

Using the formula,  (a2 + b2 - ab)(a+b) = (a3 + b3) the expression becomes,

1/(552+400) = 1/952

4. If a number 774958A96B is to be divisible by 8 and 9, the values of A and B, respectively, will be:

Answer: 8, 0

Solution: If the number is divisible by 8, it's last 3 digits are divisible by 8. 960 or 968 are the two possibilities.

If the number is divisible by 9, the sum of its digits is divisible by 9. Adding the other digits, we get (55+ A + B) is a multiple of 9.

That is (1+A+B) is a multiple of 9. 

Among the given options only 8,0 meets both the conditions.

5. The highest prime number that can be stored in a 8-bit microprocessor is

Answer: 251

Solution: The max address (highest number) that can be denoted is 2^8 - 1 = 255.

255 is not prime (divisible by 5). 254 and 252 are not prime (even numbers). 253 is not prime (divisible by 11).

Hence the highest prime number that can be stored in a 8-bit microprocessor is 251.

6. ABCD is a popular software company and hence for the hiring process 2557 applicants were standing in the queue. Between every two females there were five males in the queue. The maximum number of females could be

Answer: 427

Solution: The arrangement of 2557 people in the queue is like F M M M M M F M M M M M F......F.

Hence max number of females = (2557-1)/6 + 1 = 426+1 = 427

7. On dividing a number by 209, we get 50 as remainder. What will be the remainder when dividing the same number by 19?

Answer: 12

Solution: Let the number be x and k be the quotient when dividing by 209

x=209k + 50

Dividing by 19,

(209k+50)/19 = 19(11k+2)/19 + 12/19.

Hence remainder is 12.

8. Determine the digit in the unit position of 11^21 * 17^17 * 21^21

Answer: 7

Solution: The result of all powers of 11 ends in 1.

17^4 ends in 1 so 17^17 = (17^4)^4*17 will end in 7.

The result of all powers of 21 ends in 1.

Hence the unit digit of the given expression will be unit digit of 1*7*1 = 7.

9. If m is an odd integer and n an even integer, which of the following is odd?

A) (2m+n)(m-n)

B) (m+n^2)(m-n^2)

C) m^2+mn+n^2

D) m+n  

 Answer: D

Solution: Always sum of an odd and even is an odd integer.

10. What are the last two digits of 7^2008 ?

Answer: 01

Solution: 7^2008 = (7^4)^502

7^4 = 2401

As 01 are last two digits, when multiplied any number of times 01 will be the last two digits.

11. The difference between the squares of two consecutive odd integers is always divisible by:

Answer: 8

Solution: Let the consecutive odd integers be x and x+2.

Diff between the squares is = x^2 + 4 +4x - x^2 = 4 + 4x = 4(1+x).

Hence 4*(1+x) will be divisible by both 4 and 2 (As x is odd, 1+x is always even and hence will be divisible by 2.).

So the difference will always divisible by 8.

12. What is the remainder when 2^256 is divided by 17?

Answer: 1

Solution: 2^256 = (2^8)^32

2^8 mod 17 =1, So (2^8)^32 mod 17 = 1^32 mod 17 = 1

13. If x is an integer, which of the following CANNOT be an even integer?

Answer: 2x + 3

Solution: x can be even or odd.

2x+2 is always even.

x-5 can be even when x is odd (example x is 7 or 9)

2x+3 is always odd (As 2x is always even) Hence 2x+3 CANNOT BE AN EVEN INTEGER.

5x+2 can be even for even values of x (x is 4 or 6)

14. The sum of four consecutive odd numbers is always divisible by 

Answer: 4

Solution: Let the numbers be x,x+2,x+4,x+6. 

The sum = 4x+12 = 4(x+3). 

Hence they are always divisible by 4.

15. What is the remainder when 9^113 * 7^110 is divided by 31 ?

Answer: 16

Solution: 9^113 * 7^110 = 63^110 * 9^3

63^110 * 9^3 mod 31 = 1^110 mod 31 * 9^3 mod 31

= (1*16) mod 31 = 16

16. 16^4 + 2^20 is divisible by

Answer: 17

Solution: 16^4 = (2^4)^4 = 2^16

Hence the expression becomes 2^16 (1+2^4)

= 2^16 * 17

Hence the expression is divisible by 17

17. Instead of multiplying a number by 53, Suresh multiplied it by 35 and got the answer which was 1206 less than the expected answer. What is the number?

Answer: 67

Solution: Let the number be x. 

As it was 1206 less,  x(53-35) = 1206,

x=1206/18 = 67.

18. What will be the last digit of the multiplication  22^22 * 33^33 * 44^44 ?

Answer: 2

Solution: Last digit of 22^22 = Last digit of 2^22 = 4  (Cyclicity of power of 2 is 4)

Similarly for 33^33 it is 3,

44^44 it is 6

Hence the last digit of the entire multiplication is last digit of 4*3*6 (72) = 2

19. One-fourth of one-third of two-fifth of a number is 25. What will be 60% of the number?

Answer: 450

Solution: x * 1/4 * 1/3 * 2/5 = 25, x = 25 * 30 

60% of x = 60/100 * 25 * 30 = 450.

20. Which one of the following numbers is divisible by 99?

Answer: 342342 

Solution: To be divisible by 99, the number should be divisible by both 9 and 11.

Applying related divisibility rules we find 342342 being divisible by both 9 and 11 and hence by 99.

21. What is the remainder when 50! is divided by 16 ^ 8?

Answer: None of these

Solution: 50! = 1*2*3*......*50. 

16^8 = 2^32.

In 50! we need to calculate the number of 2s.

Number of even numbers = 25  (so 25 twos in them)

Multiples of 4 = 4,8,12,16,20,24,28,32,36,40,44,48 - Hence 12 additional twos.

Multiples of 8 = 8,16,24,32,40,48 - 6 additional twos.

Mutiples of 16 - 16,32,48 - 3 additional twos.

Multiples of 32 - 1 additional twos.

So total number of twos = 25+12+6+3+1 = 47.

 

As 2^32 divides 2^47 without leaving a remainder, the answer is zero.

22. How many positive numbers less that 50000 exist which are both perfect squares and perfect cubes?

Answer: 6

Solution: Numbers which are perfect square and cubes will have a number raised to the power 6 (LCM of 2 and 3).

Such numbers are 1^6, 2^6, 3^6, 4^6, 5^6, 6^6  (7^6 is 117649 which is more than 50000).

23. In a number system with base b, 12*25 = 333, the value of b is 

Answer: 7

Solution: (b + 2)(2b+5) = 3b^2 + 3b + 3,

2b^2 + 9b + 10 = 3b^2 + 3b + 3,

b^2 -6b -7 = 0,  (b-7)(b+1)=0

Solving b=7 or b=-1,

Among the given options, 7 is the answer.

24. Find the sum up to 20 terms in the below series

1 + (1+3) + (1+3+5) + (1+3+5+7) + ....

Answer: 2870

Solution: Sum = 1+4+9+16+25+...400 which is nothing but the sum of squares of numbers from 1 to 20.

So sum = n*(n+1)*(2n+1)/6,

Here as n=20, sum = 20*21*41/6 = 2870

25. A boy counted in the following way on the fingers in his hand. He started numbering the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5. Then he reversed the direction calling ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to index finger as 10, middle finger as 11 and so on. If he counter upto 1985, on which finger did he end with?

Answer: Thumb

Solution: After 9, it is multiple of 8 to end on thumb and hence 17,25,33...(8n+1) will end on the thumb.

As 1985 = 8n + 1, he ended up on thumb

26. If 50% of the number is added to 50, the result is number itself, then the number is

Answer: 100

Solution: Let the number be x.

0.5x + 50 = x,  Hence x = 50/0.5 = 100

27. A number is divided by 5, 2 and 3 successively to get remainder of 0, 1 and 2 respectively. What will be the remainder if the same number is divided by 2,3,5 successively.

Answer: 1,0,4

Solution: Let the final quotient be q when divided by 3.

So the number is    5*(2*(3q+2)+1) = 5*(6q+5) = 30q+25

Now when the number is divided by 2, the quotient is 15q+12 leaving a remainder of 1.

Now when the number is divided by 3, the quotient is 5q+4 leaving a remainder of 0

Now ewhen divided by 5, the quotient is q leaving a remainder of 4.

As the number is divisible by 5 it's unit digit is either 5 or zero. 

Then when divided by 2 it gives a remainder 1 which implies the quotient is odd. So the unit digit of the number is 5.

Then when divided by 3 it gives a remainder of 2 which implies the quotient after division by 2 is = 3k + 2

 

Now as the number's unit digit is odd, when we divide by 2 the remainder is 1.

28. How many times does the digit 6 appear from 11 to 100?

Answer: 19

Solution: From 16 to 96 (Excluding 66) it appears 8 times

From 60 to 69 - 11 times (66 has two sixes)

Total 8+11 = 19 times

29. Find the last two digits in 2591^66

Answer: 41

Solution: The unit digit will be 1.

The tens digit will be the unit digit in the product of 9*6(the unit digit in power and the tens digit of the number)

= unit digit in 54 = 4

Hence last two digits will be 41

30. There are two numbers, one of which is twice the other. Both had the same number of prime factors while the larger number had four factors more than the smaller one. Choose the pair containing those two numbers.

Answer: 30,60

Solution: 40,80 and 20,40 - Here larger number has only two factors more and hence not the desired pair.

50 and 100 - Here larger number has only three factors more.

30,60 - 30 has 8 factors and 60 has 12  and both have same prime factors and hence the desired pair.

31. If all the numbers between 7 and 100 are written on a paper, how many times will the number "4" be used?

Answer: 19

Solution: 4 is used for 14,24,34,54,64,74,84,94 = 8 times.

In 40,41,42,43,44,45,46,47,48,49 it is used = 11 times (Note two fours in 44)

Total = 11+8 = 19 times.

 

32. Which one of the following cannot be the square of a natural number?

Answer: 42437

Solution: When a number is squared, the last digit will not be 7 (If the number ends with 1 it is 1, 2 it is 4, 3 it is 9 and so on). 

Hence 42437 is the answer.

 

33. What is the ten's digit in 2^9999 ?

Answer: 8

Solution: Tens's digit is nothing but remainder when divided by 100.

2^9999 mod 100

= (2^10)^999 * 2^9 mod 100

= (2^10)^999 mod 100  * 512 mod 100

Note: (2^10) raised to even power has last two digits as 76 and when raised to odd power has last two digits as 24.

= 24 * 12 mod 100

= 288 mod 100

= 88

Hence last two digits are 88 and ten's digit is 8

34. If X + Y = 6, then XY = ?

Answer: 9

Solution: If X+Y = 6, the possible value pairs are

0,6  1,5  2,4,  3,3.

So the product XY can be 0, 5, 8 or 9.

Among the options, 9 is the right choice.

35. What is the unit digit in the product (3^65 x 6^59 x 7^71)?

Answer: 4

Solution: Unit digit in 3^65 = (3^4)^16 * 3  = 3 (as always 3^4 has 1 as unit digit)

Unit digit in 6^59 = 6 (Any multiplication of 6 by 6 is always 6)

Unit digit in 7^71 = (7^4)^17 * 7^3  = 3 (as always 7^4 has 1 as unit digit)

Thus the unit digit is the unit digit in 3*6*3 which is 4.