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When 9 is subtracted from a two digit number, the number so formed is reverse of the original number. Also, the average of the digits of the original number is 7.5. What is definitely the original number ?
- A87
- B92
- C90
- D69
Answer not set
Discussion 1
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Solution
Let the ten's digit and unit's digit of the original number be $x$ and $y$ respectively.
=> original number = $10x + y$
Average of digits = $\frac{x + y}{2} = 7.5$
=> $x + y = 7.5 \times 2 = 15$ -------------(i)
When 9 is subtracted from it, => Reverse number = $10y + x$
=> $(10x + y) - 9 = 10y + x$
=> $9x - 9y = 9$
=> $x - y = \frac{9}{9} = 1$ ----------------(ii)
Adding equations (i) & (ii), we get :
=> $2x = 16$ => $x = \frac{16}{2} = 8$
Putting it in eqn(i), => $y = 15 - 8 = 7$
$\therefore$ Original number = $87$
Let the ten's digit and unit's digit of the original number be $x$ and $y$ respectively.
=> original number = $10x + y$
Average of digits = $\frac{x + y}{2} = 7.5$
=> $x + y = 7.5 \times 2 = 15$ -------------(i)
When 9 is subtracted from it, => Reverse number = $10y + x$
=> $(10x + y) - 9 = 10y + x$
=> $9x - 9y = 9$
=> $x - y = \frac{9}{9} = 1$ ----------------(ii)
Adding equations (i) & (ii), we get :
=> $2x = 16$ => $x = \frac{16}{2} = 8$
Putting it in eqn(i), => $y = 15 - 8 = 7$
$\therefore$ Original number = $87$