Question #78c95

Question

1011 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-24T06:28:28

The original number is $83$ $83$ .

Let the digit at tenth place be $x$ $x$ and at unit place be $y$ $y$ .

So, the number in algebraic form is $10 x + y$ $10x+y$ .

From the first condition

$y = \frac{x}{2} - 1 = \frac{x - 2}{2}$ $y=x/2-1=(x-2)/2$ .......(i)

From second condition

$10 y + x = 3 (x + y) + 5$ $10y+x=3(x+y)+5$

or, $10 y + x = 3 x + 3 y + 5$ $10y+x=3x+3y+5$

or, $7 y = 2 x + 5$ $7y=2x+5$

Putting the value of $y$ $y$ from (i) we get

$7 (\frac{x - 2}{2}) = 2 x + 5$ $7((x-2)/2)=2x+5$

or, $7 x - 14 = 4 x + 10$ $7x-14=4x+10$

or, $3 x = 24$ $3x=24$

or, $x = \frac{24}{3} = 8$ $x=24/3=8$

Putting the value of $x$ $x$ in (i) we get

$y = \frac{x - 2}{2} = \frac{8 - 2}{2} = 3$ $y=(x-2)/2=(8-2)/2=3$

Now $10 x + y = 10 \cdot 8 + 3 = 83$ $10x+y=10*8+3=83$

So the required number is $83$ $83$ .