Question 1

Q1. An old man has Rs (1! + 2! + 3! + ...+ 50!), all of which he wants to divide equally (without fractions) among his n children. Then, n may be

(a)5

(b)7

(c)9

(d)11

Solution:

If the man is dividing the amount (1! + 2! + 3! + … + 50!) among n children, then this sum should be divisible by n.

Hence in this problem, we work with options.

Lets start with option (a). So, now, let us suppose that n = 5. Now we find the remainder when (1! + 2! + 3! + … + 50!) is divided by 5. If the remainder is 0, that means the amount (1! + 2! + 3! + … + 50!) is perfectly divisible by 5 and hence a possible value of n can be 5 and if we get any other non – zero remainder, then the amount is not divisible by 5 and hence n cannot be equal to 5 in that case.

Now, note that of the given terms, in the amount, all the terms from 5! onwards are divisible by 5 and hence we have to find the remainder when (1! + 2! + 3! + 4!) is divided by 5. Now we can check that 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 and this is not divisible by 5 and hence n cannot be equal to 5.

Next we move on to option (b). Let us suppose that n = 7. Note that of the given terms, in the amount, all the terms from 7! onwards are divisible by 7 and hence we have to find the remainder when (1! + 2! + 3! + 4! + 5! + 6!) is divided by 7. Now we can check that 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 and this is not divisible by 7 and hence n cannot be equal to 7.

Next we move on to option (c). Let us suppose that n = 9. Note that of the given terms, in the amount, all the terms from 6! onwards are divisible by 9 and hence we have to find the remainder when (1! + 2! + 3! + 4! + 5!) is divided by 9. Now we can check that 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 and this is divisible by 9 and hence n can be equal to 9.

Therefore the possible number of children can be 9.

Thank You.

Ravi Raja

Question 3

Q3. What is the last digit of the number you get by multiplying the first 2002 odd prime numbers together?

a. 1

b. 3

c. 5

d. 7

e. 9

Solution:

Since one of the prime numbers is 5 and the rest of them are all odd primes, hence the last digit of the product of these numbers has to be 5. Since any odd number when multiplied by 5, the last digit of the result will always be 5.

Thank You.

Ravi Raja

Question 5

Q5.A tennis tournament is held in a school where every student plays 1 game against every other student. There are 210 boys vs. boys games held and 300 girls vs. girl games. Q19. How many boy vs. girl games were held?
a. 40
b. 225
c. 525
d. 810

Solution:

Let us suppose that the number of boys is x and the number of girls is y.

Now, the first boy plays with the remaining (x – 1) boys, the second boy plays with the remaining (x – 2) boys, the third plays with (x – 3), … and so on. Hence the total number of boys vs. boys games will be: 1 + 2 + 3 + … + (x – 2) + (x – 1) = x(x – 1)/2

Similarly, the total number of girls vs. girl games will be y(y – 1)/2

Now, it is given that:
x(x – 1)/2 = 210
or, x(x – 1) = 420
or, x^2 – x – 420 = 0
or, (x – 21)(x + 20) = 0
or, x = 21 (as x cannot be negative)

and y(y – 1)/2 = 300
or, y(y – 1) = 600
or, y^2 – y – 600 = 0
or, (y – 25)(y + 24) = 0
or, y = 25 (as y cannot be negative)

Now, if there are x boys and y girls, then the number of boys vs. girls games will be (x)(y) = (21)(25) = 525, which is the required answer.

Thank You.

Ravi Raja

Question 7

Q7. A very fast growing sunflower grows to a height of 12 feet in 12 weeks by doubling its height every week. If you only want your sunflower to be 6 feet tall, after how many weeks should you stop it growing?

Solution:

Since the sunflower doubles its height every week and it was 12 feet in 12 weeks, then clearly in the previous week, its height was half of 12 feet.

Hence if I want the sunflower to be 6 feet tall, then I should stop its growth at the end of 11 weeks.

Thank You.

Ravi Raja

Question 9

Q9. Let x, y and z be distinct integers x and y are odd and positive and z is even and positive. Which one of the following statements cannot be true?

(1) {(x – z)^2}(y) is even

(2) (x – z)(y^2) is odd

(3) (x – z)y is odd

(4) {(x – y)^2}(z) is even

Solution:

x and y being odd and z being even, we can check that (x – z) = odd – even = odd and hence (x – z)^2 will be odd and since (odd) x (odd) = odd, therefore, {9x – z)^2}(y) will always be odd and can never be even. So, option (a) is our required answer.

Thank You.

Ravi Raja