(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5. (b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \(\pi\) = 3.142].

Saturday, 05 April 2025

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(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R ...

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(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.
(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and

π

= 3.142].

Explanation

(a)

P \propto Q

and

P \propto \frac{1}{R^{2}}

P \propto \frac{Q}{R^{2}} ⟹ P = \frac{k Q}{R^{2}}

When P = 1, Q = 8, R = 2

1 = \frac{8 k}{2^{2}} ⟹ 8 k = 4

k = \frac{4}{8} = 0.5

P = \frac{Q}{2 R^{2}}

When P = 3, R = 5, Q = ?

3 = \frac{Q}{2 (5^{2})}

3 = \frac{Q}{50} ⟹ Q = 150

.
(b)
Angular difference : 60° - 20° = 40°

\frac{θ}{360} \times 2 π r

r = R \cos θ

\frac{40}{360} \times 2 \times 3.142 \times 6400 \cos 20 = \frac{1511687.278}{360}

4199.13 k m

S p e e d = \frac{D i s t a n c e}{T i m e}

\frac{4199.13}{6}

699.85 k m / h r

≊ 700 k m / h r

(to 3 significant figures)
(ii)

D = \frac{θ}{360} \times 2 π r

420 = \frac{θ}{360} \times 2 \times 3.142 \times 6400

θ = \frac{420 \times 360}{2 \times 3.142 \times 6400}

θ = \frac{151200}{40217.6} = 3.76 °

Latitude of C : 20° + 3.76° = 23.76°
Note that due North implies same longitude but different latitude.

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