Question

Please help with me Question-3rd
of exercise- '6.2'​

Answer 1

Answer:

Let the two digit number be denoted by pq. Since q is in unit place and p is in ten place, in the decimal system where the base is 10,

The value of pq = 10¹. p + 10⁰ .q = 10p + q ……………………………………….…….(1)

After reversing the digits pq→qp and by the same argument as above,

The value of the reversed number qp = 10q + p ……………………………….……(2)

By hypothesis difference between pq and qp = 45.

∴ From (1) and (2),

10p + q - (10q + p) = 45 Or, 10p + q - 10q - p = 45

Or, 10p - p + q - 10q = 45

Or, 9p - 9q = 45 Dividing both sides by 9,

9p/9 - 9q/9 = 45/9 = 5.9/9

⇒ p -q = 5

∴ The difference between the two digits of the number = 5 (Proved)

Step-by-step explanation:

hope its help you ..❤

Answer 2

Answer:

Let the ten's digit and the one's digit be x and y respectively. (x > y)

Given

The difference between a 2-digit number and the number formed by reversing its digits is 45.

\begin{gathered}10x+y-10y-x=45\\\implies 9x-9y=45\\\implies x-y=5--(1)\end{gathered}

10x+y−10y−x=45

⟹9x−9y=45

⟹x−y=5−−(1)

The sum of the digits of the original number is 13.

\implies x+y=13--(2)⟹x+y=13−−(2)

Solving (1) and (2), we get,

\begin{gathered}x-y=5\\\underline{x+y=13}\\\underline{\underline{2x=18}}\\\implies x = 9\\\implies y = 4\end{gathered}

x−y=5

x+y=13

2x=18

⟹x=9

⟹y=4