A race track is in the shape of a ring whose inner and outer circumferences are 440 m and 506 m, respectively. What is the cost of levelling the track at ₹6/m2 ? (Take π = $\frac{22}{7}$)
Answer & explanation
Correct answer: option 4
We know that,
Internal circumference of the track = 2πr
2πr = 440
= 2 × \(\frac{22}{7}\)× r = 440
= r = 70 m
External circumference of the track = 2πR
2πR = 506
= 2 × \(\frac{22}{7}\)× R = 506
= R = 506 × \(\frac{7}{44}\) = 80.5 m
Area of track = π (R2 – r2) = \(\frac{22}{7}\)× (80.52 – 702) = \(\frac{22}{7}\)× 150.5 × 10.5 = 4966.5 m2
Cost of 1 m2 = Rs. 6
Cost of 4966.5 m2 = 4966.5 × 6 = Rs. 29799