The inner and outer radii of a circular track are 28 m and 25 m, respectively. The cost (in ₹) of levelling the track at ₹28 per m² is: [Use π = 22⁄7]

Correct — B. Factorise the difference of squares before touching π — the numbers stay small and nothing has to be rounded.

Track area = π(R² − r²)
28² − 25² = (28 + 25)(28 − 25) = 53 × 3 = 159
Area = 22/7 × 159 = 3,498/7 m²

Cost = area × ₹28 = (3,498/7) × 28 = 3,498 × 4 = ₹13,992 → option (b).

Why the others are wrong

• (a)
  17,355 — ₹17,355 implies 619.8 m² of track, which would need R² − r² near 197. The stated radii give 53 × 3 = 159, so this bill is far too large.
• (c)
  15,421 — ₹15,421 works back to 550.75 m², i.e. R² − r² near 175. The ring between radii 25 m and 28 m measures 499.71 m².
• (d)
  11,229 — ₹11,229 corresponds to just 401 m², about a fifth short of the ring's 499.71 m². No pairing of 28 m and 25 m produces that area.

A track between two circles is an annulus, and its area is the larger disc minus the smaller one: π(R² − r²).

Do not square and subtract separately. R² − r² factorises to (R + r)(R − r), which turns 784 − 625 into 53 × 3 = 159 in a single line and removes the chance of a squaring slip.

Cost questions then multiply area by rate. Here the rate is ₹28 per m² and π is 22/7, so the 7 in the denominator cancels against the 28 and the bill comes out whole.

Key facts

• Area of a circular ring = π(R² − r²) = π(R + r)(R − r).
• With the radii 28 m and 25 m, R² − r² = 53 × 3 = 159, so the ring is 22/7 × 159 = 499.71 m².
• At ₹28 per m² the bill is (22/7 × 159) × 28 = ₹13,992.
• The full 28 m disc would cost ₹68,992 and the 25 m disc ₹55,000, and the difference is ₹13,992.

Common traps

• Using (R − r)² = 9 in place of R² − r² = 159
• Computing 2π(R − r) and pricing a boundary when the question asks for a surface
• Rounding 22/7 × 159 to 499.7 before multiplying, which loses the exact ₹13,992

No directly related past PYQ was found.