Mathematics — Standard (041) — CBSE Class 10 board question

Let the tens digit be $x$ and units digit be $y$.

So the two-digit number = $10x + y$.

Forming the equations:

Condition 1: The number is seven times the sum of its digits.
$$10x + y = 7(x + y)$$
$$10x + y = 7x + 7y$$
$$3x - 6y = 0 \implies x = 2y \tag{1}$$

Condition 2: The number is 2 more than 5 times the product of its digits.
$$10x + y = 5xy + 2 \tag{2}$$

Solving:

Substitute $x = 2y$ from (1) into (2):
$$10(2y) + y = 5(2y)(y) + 2$$
$$21y = 10y^2 + 2$$
$$10y^2 - 21y + 2 = 0$$
$$10y^2 - 20y - y + 2 = 0$$
$$(10y - 1)(y - 2) = 0$$

So $y = 2$ (since $y = \frac{1}{10}$ is not a digit).

From (1): $x = 2(2) = 4$

The required number = $10(4) + 2 = \mathbf{42}$

Verification: Sum of digits = 6; $7 \times 6 = 42$ ✓. Product = 8; $5 \times 8 + 2 = 42$ ✓

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• Represent the number as $10x + y$ (standard form for 2-digit numbers, as shown in Example 10 of the chapter).
• Condition 1 gives a simple linear equation; condition 2 gives a quadratic in one variable after substitution.
• Factorise the quadratic and reject the non-integer/non-digit root.
• Always verify both conditions — examiners award a mark for it.
• The key step students miss: correctly substituting $x = 2y$ into the product term $5xy$ to get $10y^2$.