Question

A two digit number is 3 more than 4 times the sum of its digit if 18 is addard to the number its digits are reversed find the number

Answer 1

35

Step-by-step explanation:

${\underline{{\text{To Find:}}}}$

The two digit no.

${\underline{{\text{Solution:}}}}$

let, the one digit be x and ten digit be y.

So, The no. is (10y + x)

According to given,

(10y + x) = 4(x+y) + 3 .... (1)

(10y + x) + 18 = (10x + y) ....(2)

Solving 1st equation....

$10y + x = 4(x + y) + 3 \\ \\ 10y + x = 4x + 4y + 3 \\ \\ 10y - 4y = 4x - x + 3 \\ \\ 6y = 3x + 3 \\ \\ 6y = 3(x + 1) \\ \\ 2y = x + 1 \\ \\ 2y - 1 = x$

Putting the value of x in 2nd equation...

$10y + x + 18 = 10x + y \\ \\ 10y - y + 18 = 10x - x \\ \\ 9y + 18 = 9x \\ \\ 9(y + 2) = 9x \\ \\ y + 2 = x .....(a)\\ \\ y + 2 = 2y - 1 \\ \\ 2 + 1 = 2y - y \\ \\ 3 = y \\ \\ x = 5 \: \: \: from \: (a)$

So, no. is (10y + x) = 30 + 5 = 35

Therefore the no. is 35.

Answer 2

${\purple{\underline{\underline{\large{\mathtt{Answer:}}}}}}$

Given:

We have been given that a two digit number is 3 more than 4 times the sum of its digit and when 18 is addard to the number its digits are reversed

To Find:

We need to find the number.

Solution:

Let the tens digit be x and ones digit be y.

So the number formed is (10x + y)

Now, according to the question, we have

10x + y = 4(x + y) + 3

=> 10x + y = 4x + 4y + 3

=> 10x - 4x + y = 4y + 3

=> 6x + y = 4y + 3

=> 6x - 3 = 4y - y

=> 6x - 3 = 3y

=> 6x = 3y + 3

=> 6x - 3y = 3

=> 2x - y = 1 _____(1)

Also, 10x + y + 18 = 10y + x

=> 9x - 9y = -18

=> x - y = -2______(2)

On subtracting equation 2 from equation 1, we get

x = 3____________(3)

Now, substituting the value of x from equation 3 in equation 1, we have

2x - y = 1

=> 2(3) - y = 1

=> 6 - y = 1

=> 6 - 1 = y

=> 5 = y

=> y = 5

So, number = (10x + y)

= 10(3) + 5

= 30 + 5

= 35

Hence, the required number is 35.