# E Maths Group Tuition Centre; Clementi Town Secondary School Prelim 2012 Solution

Q5) The speed of a boat in still water is 60 km/h.

On a particular day, the speed of the current is $x$ km/h.

(a) Find an expression for the speed of the boat

(I) against the current, [1]

Against the current, the boat would travel slower! This is related to the Chinese proverb, 逆水行舟，不进则退, which means “Like a boat sailing against the current, we must forge ahead or be swept downstream.”

Hence, the speed of the boat is $60-x$ km/h.

(ii) with the current. [1]

$60+x$ km/h

(b) Find an expression for the time required to travel a distance of 80km

(I) against the current,  [1]

Recall that $\displaystyle \text{Time}=\frac{\text{Distance}}{\text{Speed}}$

Hence, the time required is $\displaystyle \frac{80}{60-x}$ h

(ii) with the current. [1]

$\displaystyle \frac{80}{60+x}$ h

(c) If the boat takes 20 minutes longer to travel against the current than it takes to travel with the current, write down an equation in $x$ and show that it can be expressed as $x^2+480x-3600=0$   [2]

Note: We must change 20 minutes into 1/3 hours!

$\frac{80}{60-x}=\frac{1}{3}+\frac{80}{60+x}$

There are many ways to proceed from here, one way is to change the Right Hand Side into common denominator, and then cross-multiply.

$\displaystyle \frac{80}{60-x}=\frac{60+x}{3(60+x)}+\frac{240}{3(60+x)}=\frac{300+x}{3(60+x)}$

Cross-multiply,

$240(60+x)=(300+x)(60-x)$

$14400+240x=18000-300x+60x-x^2$

$x^2+480x-3600=0$ (shown)

(d) Solve this equation, giving your answers correct to 2 decimal places. [2]

$\displaystyle x=\frac{-480\pm\sqrt{480^2-4(1)(-3600)}}{2}=7.386 \text{ or } -487.386$

Answer to 2 d.p. is $x=7.39 \text{ or } -487.39$

(e) Hence, find the time taken, in hours, by the boat to complete a journey of 500 km against the current. [2]

Now we know that the speed of the current is 7.386 km/h.

Hence, the time taken is $\frac{500}{60-7.386}=9.50$ h

## Author: mathtuition88

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