Posted by admin at November 6, 2019

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

**07****23****49****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**

**144****156****168****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?**

**35****30****28****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?**

**16****19****17****18**

Sol. Average of the three numbers = 15.

Let the numbers be 15 – d, 15 and 15 + d.

⇒ (15 – d)^{2} + 15^{2} + (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?**

**1:3****1:5****3:5****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?**

**0****1****2****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****2****3****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.**

**4****7****2****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?**

**(x – z)^2 y is even****(x – z) y^2 is odd****(x – z) y is odd****(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?**

**0****1****2****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.

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