Question

the sum of a two digit number is 15 the number is increased by 27 if the digits are reversed find the number

Answer 1

Hey there !!


→ Let the ten's digit of the original number be x.

→ And, the unit's digit of the original number be y.


▶Now,

A/Q,

=> x + y = 15..............(1).


➡ Original number = 10x + y.

➡ Number obtained by reversing the digits = 10y + x.


A/Q,

↪ If the number is increased by 27, the digits are reversed.

=> 10x + y + 27 = 10y + x.

=> 10x - x + y - 10y = -27.

=> 9x - 9y = -27.

=> 9( x - y ) = -27.

=> x - y = $\frac{ - 27}{9}$

=> x - y = -3..............(2).


▶Now, Substracte in equation (1) and (2), we get

x + y = 15.
x - y = -3.
(-)..(+)...(+)
________

=> 2y = 18.

=> y = $\frac{18}{2}$

=> y = 9.

▶Now, put the value of ‘y’ in equation (1), we get

=> x + 9 = 15.

=> x = 15 - 9.

=> x = 6.


↪ Therefore, original number = 10x + y.

= 10 × 6 + 9.

= 60 + 9.

$\huge \boxed{ \boxed{ \bf = 69. }}$


✔✔ Hence, it is solved ✅✅.

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$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

Let the ten's digit of the required number be x.
the unit's digit be y.

A/Q,
x + y = 15............1


Required number = 10x + y.
Number obtained by reversing the digits = 10y + x.


A/Q,

10x + y + 27 = 10y + x.
10x - x + y - 10y = -27.
9x - 9y = -27.

x - y = -3...........2


Substracte in eq 1 and 2,we get

2y = 18
y = 9.
put the value of ‘y’ in eq 1
x + 9 = 15.
x = 15 - 9.
x = 6.

required number = 10x + y.
= 10 × 6 + 9.
= 69.