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