Two-Digit Number Reversal Problems | Class 10 Maths

Quick answer: Write a two-digit number with tens digit $x$ and units digit $y$ as $10x + y$; its reversal is $10y + x$. If the number plus its reversal is 66 and the digits differ by 2, then $11(x + y) = 66$ gives $x + y = 6$, and $x - y = \pm 2$ gives 42 or 24, so there are two such numbers.

Writing a two-digit number

A number like 57 means $5 \times 10 + 7$. So if the tens digit is $x$ and the units digit is $y$:

Key idea: Do not write the number as $xy$; in algebra that means $x \times y$. For digits 3 and 4, $xy = 12$, but the number is 34.

The last two rows are useful shortcuts: the sum of a number and its reversal is always a multiple of 11, and the difference is always a multiple of 9.

Problem 1: sum 66, digits differ by 2

The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers are there?

Step 1: Form the first equation.

$(10x + y) + (10y + x) = 66$

$11x + 11y = 66$

$x + y = 6 \qquad \text{(1)}$

Step 2: Form the second equation. "The digits differ by 2" does not say which digit is larger, so there are two cases.

Case 1: $x - y = 2$.

Adding to (1): $2x = 8$, so $x = 4$ and $y = 2$. The number is 42.

Case 2: $y - x = 2$.

Adding to (1): $2y = 8$, so $y = 4$ and $x = 2$. The number is 24.

Step 3: Check. $42 + 24 = 66$ ✓, and in each case the digits 4 and 2 differ by 2 ✓.

Answer: The number is 42 or 24. There are two such numbers.

Problem 2: nine times the number

The sum of the digits of a two-digit number is 9. Nine times the number is twice the number obtained by reversing its digits. Find the number.

Step 1: First equation.

$x + y = 9 \qquad \text{(1)}$

Step 2: Second equation.

$9(10x + y) = 2(10y + x)$

$90x + 9y = 20y + 2x$

$88x - 11y = 0$

$8x - y = 0 \qquad \text{(2)}$

Step 3: Solve. Adding (1) and (2):

$9x = 9 \quad \Rightarrow \quad x = 1$

From (1), $y = 8$. The number is $10 \times 1 + 8 = 18$.

Step 4: Check. $1 + 8 = 9$ ✓. $9 \times 18 = 162$ and $2 \times 81 = 162$ ✓.

Answer: The number is 18.

Summary table

Exam tips

• Start with "Let the tens digit be $x$ and the units digit be $y$" and write both the number and its reversal.
• Divide by common factors (11 or 9) early to keep the equations small.
• When digits "differ by" a value, solve both cases unless the question says which digit is bigger.
• Always check your answer against every condition in the question.

Common mistakes

• Writing the number as $x + y$ or $xy$ instead of $10x + y$.
• Confusing "sum of the digits" ($x + y$) with "sum of the number and its reversal" ($11(x + y)$).
• Giving only one answer when the digits "differ by" a value.
• Swapping $x$ and $y$ when writing the reversed number.

Frequently Asked Questions

How do you write a two-digit number in algebra?

If the tens digit is x and the units digit is y, the number is 10x + y. Reversing the digits gives 10y + x.

What two-digit numbers have a sum of 66 with their reversal and digits differing by 2?

The numbers are 42 and 24. Both satisfy x + y = 6 with the digits differing by 2.

Why is the sum of a number and its reversal a multiple of 11?

(10x + y) + (10y + x) = 11x + 11y = 11(x + y), which is always a multiple of 11.

Why must we consider two cases when digits "differ by 2"?

The question does not say which digit is larger, so both x − y = 2 and y − x = 2 are possible and may give different numbers.

What is the number whose digits add to 9 and nine times which equals twice its reversal?

The number is 18, because 9 × 18 = 162 = 2 × 81.