Question

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

Answer 1

Let the ones place digit be a

And Unit place digit be b

According to first condition,

10a + b = 4 ( a+ b) + 3

=> 10a + b = 4a + 4b + 3

=> 10a - 4a + b - 4b = 3

=> 6a - 3b = 3

=> 2a - b = 1 --------(1)

According to second condition,

10a +b + 18 = 10b + a

=> 10a + b - 10b - a = - 18

=> 9a - 9b = - 18

=> a - b = - 2 ------(2)

On subtracting equation 2 from 1, we get

2a - b - a + b = 1 - (-2)

=> a = 3

On Substituting the value of a in equation (2), we get

3 - b = - 2

=> b = 5

Required number = 10a + b

= 10 × 3 + 5

= 30 + 5

= 35

Answer 2

Heya !!

Here's your answer.. ⬇⬇
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Let the ten's place = x
The unit place = y

Original no. = 10x + y

➡ Given that a two digit number is 3 more than 4 times the sum of its digits.

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

➡ If 18 is added to the number the digits are reversed.

Reversed no. = 10y + x

10x + y + 18 = 10y + x
10y - y + x - 10x = 18
9y - 9x = 18
9 ( y - x ) = 18
y - x = 2 -------(2 )

Add both eq. (1) and (2)

2x - y = 1
-x + y = 2
-----–-----–-–
x = 3

Put value of ( x ) in eq. ( 1 )

2x - y = 1
2 ( 3 ) - y = 1
y = 6 - 1
y = 5

The original no. = 10x + y

= 10( 3 ) + 5
= 30 + 5
= 35
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Hope it helps..

Thanks :))