Q9

Hi

can anyone solve 9th ques

thanks

Question 2

Q2. The number of 1's in the binary notation of (2^89) – 1 is
(a) 89
(b) 88
(c) 90
(d) 1

Solution:

(2^1) – 1 when written in the binary notation is written as 1 and thus contains one 1.

(2^2) – 1 when written in the binary notation is written as 11 and thus contains two 1s.

(2^3) – 1 when written in the binary notation is written as 111 and thus contains three 1s.

(2^4) – 1 when written in the binary notation is written as 1111 and thus contains four 1s.

……………………
……………………
……………………
……………………
and so on.

Now, make a note of the pattern that is being followed and we can conclude that (2^89)- 1 when written in the binary notation, will contain eighty – nine 1s.

Thank You.

Ravi Raja

Question 4

Q4. How many natural numbers between 1 and 900 are NOT multiples of any of the numbers 2, 3, or 5?

a. 650

b. 660

c. 240

d. 250

Solution:

The natural numbers between 1 and 900 that are divisible by 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, … , 900 and there are altogether 450 such numbers.

The natural numbers between 1 and 900 that are divisible by 3 are: 3, 6, 9, 12, 15, 18,  …, 900 and there are altogether 300 such numbers.

The natural numbers between 1 and 900 that are divisible by 5 are: 5, 10, 15, 20, 25, 30, 35, 40, …, 900 and there are altogether 180 such numbers.

The natural numbers between 1 and 900 that are divisible by both 2 and 3 are: 6, 12, 18, 24, 30, 36, …, 900 and there are altogether 150 such numbers.

The natural numbers between 1 and 900 that are divisible by both 3 and 5 are: 15, 30, 45, 60, 75, 90, …, 900 and there are altogether 60 such numbers.

The natural numbers between 1 and 900 that are divisible by both 5 and 2 are: 10, 20, 30, 40, 50, 60, …, 900 and there are altogether 90 such numbers.

The natural numbers between 1 and 900 that are divisible by 2, 3 and 5 are: 30, 60, 90, 120, 150, 180, …, 900 and there are altogether 30 such numbers.

So the number of numbers between 1 and 900 (both inclusive), that are divisible by at least one of 2, 3 or 5 is given by: The number of numbers divisible by 2 + The number of numbers divisible by 3 + The number of numbers divisible by 5 – The number of numbers divisible by both 2 and 3 – The number of numbers divisible by both 3 and 5 – The number of numbers divisible by both 5 and 2 + The number of numbers divisible by 2, 3 and 5 = 450 + 300 + 180 – 150 – 90 – 60 + 30 = 660.

Hence the number of natural numbers between 1and 900 that are NOT divisible by any of 2, 3 and 5 are: 900 – 660 = 240, which is the required answer.

Thank You.

Ravi Raja

Question 6

Q6. Assume that all bricklayers work at the same rate of speed. If it takes nine bricklayers (all working at the same time) fourteen days to do a job, how long would it take for the job to be done by
a) Seven bricklayers
b) Three bricklayers.

Solution:

Here we will be using the result (M1 x D1)/(W1) = (M2 x D2)/(W2)
Where M denotes the number of men involved in doing the work, D denotes the number of days required to complete the work and W denotes the quantity of work done or to be done (and is generally taken as 1, unless the exact work is specified)

Solution (a):

M1 = 9, D1 = 14, W1 = 1, M2 = 7, W2 = 1 (as the same work is to be done)

(M1 x D1)/(W1) = (M2 x D2)/(W2)
or, (9 x 14)/(1) = (7 X D2)/(1)
or. D2 = (9 x 14)/(7) = 9 x 2 = 18 days, which is the required answer.

Solution (b):

M1 = 9, D1 = 14, W1 = 1, M2 = 3, W2 = 1 (as the same work is to be done)

(M1 x D1)/(W1) = (M2 x D2)/(W2)
or, (9 x 14)/(1) = (3 X D2)/(1)
or. D2 = (9 x 14)/(3) = 3 x 14 = 42 days, which is the required answer.

Thank You.

Ravi Raja

Question 8

Q8. What is the remainder when 6^2002 is divided by 11?

Solution:

Fermat’s Theorem states that: “If p be a prime number and ‘a’ be a number not divisible by p, then (a)^(p – 1), when divided by p, the remainder is always 1”.

In the given problem, we see that 11 is a prime number and 6 is not divisible by 11 and so we can use Fermat’s Theorem in the given problem.

So, (6)^10 when divided by 11, the remainder is 1.
Therefore, 6^2000 = {(6)^10}^200 when divided by 11, the remainder will be (1)^200 = 1.

Now, 6^2002 = (6^2000)*(6^2), and when 6^2 is divided by 11, the remainder is 3.

So the remainder when 6^2002 is divided by 11, the remainder is (1)*(3) = 3.

Thank You.

Ravi Raja