Question

the sum of the digits of a 2 digit number is 9. On reversing its digits, the new number obtained is 45 more than the original number. Find the number

Answer 1

27 is the answer......

Answer 2

Concept:

An algebraic equation may be described as a mathematical assertion wherein expressions are set identical to every other. The algebraic equation typically includes a variable, coefficients and constants.

Given:

We are given that the sum of the digits of a $2$ digit number is $9$ and reversing its digits, the new number obtained is $45$ more than the original number.

Find:

We have to find the number.

Solution:

Let the digit at tens place be $x$ then digit at ones place will be $9-x$.

Original two digit number = $10x+(9-x)$

After interchanging the digits, the new number $=10(9-x)+x$

According to the given question,

$\begin{aligned}10x + (9-x) + 45 &= 10(9-x) + x\\9x+54&=90-9x\\ 9x+9x&=90-54\\18x&=36\\x&=2\end$

Substitute the value of $x$ in the original two-digit number, we get

$\begin{aligned}10x+(9-x)&=10\times 2+(9-2)\\ &=20+7\\ &=27\end$

Hence, the number when the sum of the digits of a two digit number is $9$ is $27.$

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