Question

the sum of digits of two digit number is 10 if 18 is subtracted from the original number the digits interchange places find the number

Answer 1

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Hey friends !!

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

And the unit's digit be y.

▶ Now,

A/Q,

=> x + y = 10................(1).

Original number = 10x + y.

And number obtained interchange = 10y + x.

A/Q,

=> 10x + y - 18 = 10y + x.

=> 10x - x + y - 10y = 18.

=> 9x - 9y = 18.

=> 9( x - y ) = 18.

=> x - y = $\bf{ \frac{18}{9} }$

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

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

x + y = 10.

x - y = 2.

(-)..(+)..(-)

________

=> 2y = 8.

=> y = $\bf{ \frac{8}{2} }$

=> y = 4.

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

=> x + 4 = 10.

=> x = 10 - 4.

=> x = 6.

↪ Original number = 10x + y.

=> 10 × 6 + 4.

$\huge \boxed{=> 64.}$

✔✔ Hence, the original number is founded ✅✅.

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THANKS

#BeBrainly.

Answer 2

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

x + y = 10............1

required number = 10x + y.

And number obtained interchanging = 10y + x.

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

Substracte eq 1 and 2, we get
2y = 8.
y = 4.

put the value of y in eq 1
x + 4 = 10.
x = 10 - 4.
x = 6.

Original number = 10x + y.
10 × 6 + 4=64.