Question

A number of two digits exceeds four times the sum of its digit by 6 and it is increased by 9 on reversing the digits. Find the number. ​

Answer 1

Then, the number is 10×x+y×1=10x+y

So, Number formed by reversing the digits =10×y+x×1=10y+y

According to the question,

10x+y=4(x+y)+6

⇒10x−4x+y−4y=6

⇒6x−3y=6

⇒2x−y=2 … (i)

And,

10x+y+9=10y+x

⇒10x−x+y−10y=−9

⇒9x−9y=−9

⇒x−y=−1 … (ii)

Now, subtracting eq. (ii) from eq. (i), we get

(2x−y)−(x−y=(−1)

⇒x=3

Substituting the value of x in eq. (i), we get

2(3)−y=2

⇒6−y=2

⇒y=6−2

⇒y=4

Therefore, the number is 10×3+4=30+4=34.

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