Question

8. A two-digit number can be expressed as
8 times the sum of its digits increased by
1, as well as, 13 times the difference of its
digits increased by 2. Find that number.
(Hint: Following the second condition,
Number = 13 (Units digit - Tens
digit) + 2 will give digits to be
fractional and that is not possible.​

Answer 1

Answer:

x = 4 , y = 1

Step-by-step explanation:

Let the two digit of the no. be x and y.

given,

10x + y = 8(x + y) + 1

=> 10x - 8x + y - 8y = 1

=> 2x - 7y = 1 ..........(i)

also,

given 10x + y = 13(x-y) + 2

=> 10x - 13x + y - 13y = 2

=> -3x + 14y = 2 ........(ii)

multiplying eq (i) with 2

4x - 14y = 2 .....(iii)

solving equation (iii) - (ii)

=> x = 4

put the value in equation (i)

2x - 7y = 1

=> 8 - 1 = 7 y

=> y = 1

Answer 2

SOLUTION :-

Let,

Digit in ten's place = x

Digit in unit's place = y

Two digit number = 10x + y

According to the first condition,

$\longrightarrow\sf 10x+y = 8(x+y)+1\\\\\longrightarrow 10x+y=8x+8y+1\\\\\longrightarrow 10x-8x+y-8y=1\\\\\longrightarrow 2x-7y = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............................(1)$

According to the second condition,

$\longrightarrow\sf 10x+y = 13(x-y)+2\\\\\longrightarrow 10x+y = 13x-13y+2\\\\\longrightarrow 10x-13x+y+13y=2 \\\\\longrightarrow -3x+14y = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............................(2)$

Equation (1) × 2,

$\longrightarrow\sf 4x-14y = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............................(3)$

Equation (2) + Equation (3),

$\longrightarrow \bf x = 4$

Substitute x = 4 in equation (1),

$\longrightarrow\sf 2\times 4- 7y = 1 \\\\\longrightarrow 8-7y = 1 \\\\\longrightarrow -7y = -7 \\\\\longrightarrow \bf y = 1$

TWO DIGIT NUMBER = 41