(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?
(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \(\pi = \frac{22}{7}\)].
(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \(\pi = \frac{22}{7}\)].
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Correct Answer: Option n
Explanation:
(a) Let the cost of the car to the dealer be x.
Sales of car to a man with 15% profit : (115% of x = 1.15 x\)
Sale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)
\(0.95 \times 1.15x = N120,175.00\)
\(1.0925x = N120,175.00\)
\(x = \frac{120,175.00}{1.0925}\)
\(x = N110,000\)
The original cost of the car to the dealer = N110,000.00.
(b) Circumference of a circle = \(2\pi r\)
Distance covered by the wheel = 1.05 km
Radius of the wheel = 18cm
Circumference of the wheel = \(2 \times \frac{22}{7} \times 18\)
= \(113.14 cm\)
Note : 1.05 km = 105,000 cm
Number of revolution = \(\frac{105,000}{113.14}\)
= \(928.05 revs\)
\(\approxeq 928 revs\)
(a) Let the cost of the car to the dealer be x.
Sales of car to a man with 15% profit : (115% of x = 1.15 x\)
Sale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)
\(0.95 \times 1.15x = N120,175.00\)
\(1.0925x = N120,175.00\)
\(x = \frac{120,175.00}{1.0925}\)
\(x = N110,000\)
The original cost of the car to the dealer = N110,000.00.
(b) Circumference of a circle = \(2\pi r\)
Distance covered by the wheel = 1.05 km
Radius of the wheel = 18cm
Circumference of the wheel = \(2 \times \frac{22}{7} \times 18\)
= \(113.14 cm\)
Note : 1.05 km = 105,000 cm
Number of revolution = \(\frac{105,000}{113.14}\)
= \(928.05 revs\)
\(\approxeq 928 revs\)