question_answer

(a) The sum of the digits of a two-digit number is 15. If the number formed by reversing the digits is less than the original number by 27, find the original number.

(b) Verify that \[x=2\]is a solution of the equation

\[2\left( x+1 \right)=3\left( x+1 \right)-3.\]

Answer:

(a) Let the unit place = x

Then the tens place \[=\left( 15x \right)\]

Therefore, original number\[=10\left( 15x \right)+x\]

\[=\left( 1509x \right)\]

By reversing the digits, we get

New number \[=10x+(15-x)\]

\[=\text{ }9x\text{ }+\text{ }15\]

According to question,

(original number) - (new number) = 27

\[\left( 1509x \right)\left( 9x+15 \right)=27\]

or         \[13518x=27\]

or         \[18x=13527\]

or         \[18x=108\]

or         \[x\text{ }=\text{ }\frac{108}{18}\]

or         x = 6

Hence, original number \[=1509x\]

\[=1509\times 6\]

\[=15054=96\]

(b) Verification: Since \[2\left( x+1 \right)=3\left( x+1 \right)\text{3}\]

The putting x = 2, then

LHS \[=\text{ }2\left( x\text{ }+\text{ }1 \right)\]

= 2 (2 + 1)

\[=2\times 3=6\]

and      RHS \[=3\left( x+1 \right)3\]

\[=3\left( 2+1 \right)3\]

\[=3\times 33\]

\[=93=6\]

Hence, LHS =   RHS = 6     [Hence, Verified]