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

Let Rashmi's present age = $x$ years and Nazma's present age = $y$ years.

Three years ago:
$x - 3 = 3(y - 3)$
$x - 3 = 3y - 9$
$x - 3y = -6$ … (1)

Ten years later:
$x + 10 = 2(y + 10)$
$x + 10 = 2y + 20$
$x - 2y = 10$ … (2)

Subtracting (1) from (2):
$(x - 2y) - (x - 3y) = 10 - (-6)$
$y = 16$

Substituting in (2):
$x - 2(16) = 10$
$x = 42$

∴ Rashmi is 42 years old and Nazma is 16 years old.

Verification: 3 years ago — Rashmi: 39, Nazma: 13 → 39 = 3 × 13 ✓
10 years later — Rashmi: 52, Nazma: 26 → 52 = 2 × 26 ✓

Source: Chapter 3, Exercise 3.3 (Q.2, similar to Q.2(ii))

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• This is identical in structure to the textbook's Nuri–Sonu problem (Ex 3.3, Q.2(ii)) — just different names and slightly different numbers.
• Always form two equations clearly labelled, then solve by elimination or substitution — either is acceptable.
• Write a verification at the end for full marks; CBSE awards 1 mark for checking.
• Common error: forgetting the "three years ago" means subtract 3 from both ages, not just one.