# Elitmus Model Paper and solution

Posted by admin at November 6, 2019

What are the last two digits of the number 7^45

1.  07
2.  23
3.  49
4.  43

Sol: The last two digits of 7^1 are 07.
The last two digits of 7^2 are 49.
The last two digits of 7^3 are 43.
The last two digits of 7^4 are 01.
The last two digits of powers of 7 go in a cycle – 07, 49, 43, 01
So, the last two digits of 7^45 are 07.

The number of factors common to 30^11 and 20^13 is

1.  144
2.  156
3.  168
4.  136

Sol. The HCF of the two given numbers is 10^11.
∴ all the factors of 10^11 would be common to both the numbers 30^11 and 20^13.
10^11 = 2^11 × 5^11
Total number of common factors = Number of factors of 2^11 × 5^11 = (11 + 1)(11 + 1) = 12 × 12 = 144

A group of workers can do a piece of work in 24 days. However, as 7 of them were absent it took 30 days to complete the work. How any people actually worked on the job to complete it?

1.  35
2.  30
3.  28
4.  42

Sol. Let the original number of workers in the group be ‘x’
Therefore, the actual number of workers = x – 7.
We know that the number of manhours required to do the job is the same in both the cases.
Therefore, x (24) = (x-7).30
24x = 30x – 210
6x = 210
x = 35.
the actual number of workers who worked to complete the job = x – 7 = 35 – 7 = 28.

the actual number of workers who worked to complete the job = x – 7 = 35 – 7 = 28.

The sum of three numbers in A.P. is 45. If the sum of their squares is 683, what is the largest of the three numbers?

1.  16
2.  19
3.  17
4.  18

Sol. Average of the three numbers = 15.
Let the numbers be 15 – d, 15 and 15 + d.
⇒ (15 – d)2 + 152 + (15 + d)2 = 683
⇒ d = 2
The numbers are [15 – 2] =13,
15 and [15 + 2] = 17.

Two jars contain milk and water in the ratio 7:3 and 3:2 respectively. In what ratio should the contents of the two jars be mixed such that the final ratio of milk and water in the resultant solution becomes 23:17?

1.  1:3
2.  1:5
3.  3:5
4.  None of the above

Sol.  the concentration of milk in Jar 1 = 70%
The concentration of milk in Jar 2 = 60%.
The concentration of milk in any mixture of these two will lie between 60% and 70% depending on the actual ratio of the two.
the concentration of the resultant solution can never be less than 60%. So it is not possible.

A = k^2 – 1 and B = (k + 1)^2 – 1, where k is a natural number greater than 1. How many prime numbers are there by which both A and B are divisible for at least one value of k?

1.  0
2.  1
3.  2
4.  More than 2

Sol. A = (k – 1)(k + 1)
B = k(k + 2)
For all values of k greater than or equal to 2, the natural numbers ‘k – 1 ’ and ‘k + 1’ are co-prime with both ‘k’ and ‘k + 2’ except when ‘k – 1’ and ‘k + 2’ are both multiples of 3.
Note that (k + 2) – (k –1) = 3.
Here, the common factor of A and B is 3. Which is also a prime number,
E.g. when k – 1 = 3 or k = 4, A = 15 and B = 24.
The only common factor of A and B, in this case, is 3.

5765X4Y is divisible by 9. What is the maximum number of values that X can take for any particular value of Y?

1.   1
2.   2
3.   3
4.   4

Sol. Sum of the digits = 5 + 7 + 6 + 5 + X + 4 + Y = 27 + X + Y
27 is divisible by 9
∴ Checking divisibility of X and Y only
If the number is divisible then X + Y = 0 or 9 multiple
Since X + Y are the sum of single digit, the maximum sum can only be 9 + 9 = 18
If X + Y = 0
then X = 0, Y = 0
But we are looking for maximum value of X
If X + Y = 18
then X = 9, Y = 9

A hundred digit number is formed by writing the first 54 natural numbers in front of each other 12345678910111213… Find the remainder when this number is divided by 8.

1.   4
2.   7
3.   2
4.   0

Sol. Divisibility of 8 is checked by dividing the last 3 digits of a number.
Last three digits of the number: 12345678910111213……5354
Remainder[354/8] = 2

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)^3  z is even

Sol. (x – z)^2 y is even cannot be true.
x is odd and z is even.
∴ x – z is odd.
And y is odd.
∴ (x – z)^2 will be odd and (x – z)^2 y will be odd.

When the integer n is divided by 8, the remainder is 3. What is the remainder if 6n is divided by 8?

1.   0
2.   1
3.   2
4.   3

Sol. When n is divided by 8, the remainder is 3 may be written as
n = 8 k + 3
multiply all terms by 6
6 n = 6(8 k + 3) = 8(6k) + 18
Write 18 as 16 + 2 since 16 = 8 * 2.
= 8(6k) + 16 + 2
Factor 8 out.
= 8(6k + 2) + 2

The above indicates that if 6n is divided by 8, the remainder is 2.