Sec 1 Maths: Rate, Speed & Time — practice questions & worked solutions

(a)

$$\begin{aligned}\text{distance} &= \frac{15}{4}\,v \text{ km}\end{aligned}$$

(b)

New time $= 3\dfrac{3}{4} – \dfrac{1}{2} = \dfrac{13}{4}$ hours.

$$\begin{aligned}\text{distance} &= \frac{13}{4}(v + 8) \text{ km}\end{aligned}$$

(c)

$$\begin{aligned}\frac{15}{4}v &= \frac{13}{4}(v + 8)\\ 15v &= 13(v + 8)\\ 15v &= 13v + 104\\ 2v &= 104\\ v &= 52\end{aligned}$$

Rate of Pat:

$$\begin{aligned}\frac{15}{30} &= \frac{1}{2}\text{ cup per minute}\end{aligned}$$

Rate of Sally:

$$\begin{aligned}\frac{12}{36} &= \frac{1}{3}\text{ cup per minute}\end{aligned}$$

Combined rate:

$$\begin{aligned}\frac{1}{2} + \frac{1}{3} &= \frac{5}{6}\text{ cup per minute}\end{aligned}$$

Time for $100$ cups:

$$\begin{aligned}100 \div \frac{5}{6} &= 120\text{ minutes}\\ &= 2\text{ hours}\end{aligned}$$

Rate $B$ is better as it pays more than Rate $A$ (740 dollars $>$ 690 dollars).

(i) At the start ($t=0$), the man was already $6$ km away from town $P$.

(ii) Gradient $=\dfrac{20-6}{3.5-0}=4$ km/h, so distance at $t=2$ is $6+4(2)=14$ km from $P$. Distance from museum $=20-14=6$ km.

(iii) To travel $10$ km, the man’s distance from $P$ becomes $6+10=16$ km.

$$16=6+4t$$
$$t=2.5\ \text{hours}$$

(iv)(a)

$$\begin{aligned}\text{Gradient}&=\frac{20-6}{3.5-0}\\ &=4\end{aligned}$$

(b) The gradient represents the speed of the man, $4$ km/h.

(i) The flat-ground speed corresponds to $4$ parts $=0.2$ km/min, so $1$ part $=0.05$ km/min. Speed up the slope is $2$ parts:

$$\begin{aligned}\text{up speed} &= 2\times 0.05\\ &= 0.1 \text{ km/min}\end{aligned}$$

(ii) Distance up the slope:

$$\begin{aligned}0.1\times 10 &= 1 \text{ km}\end{aligned}$$

Speed down the slope $=5\times 0.05=0.25$ km/min, distance down:

$$\begin{aligned}0.25\times 8 &= 2 \text{ km}\end{aligned}$$

Total distance:

$$\begin{aligned}1+3+2 &= 6 \text{ km}\end{aligned}$$

Time on flat ground $=\dfrac{3}{0.2}=15$ min. Total time (including $2$ min rest):

$$\begin{aligned}10+15+8+2 &= 35 \text{ min}\end{aligned}$$

Average speed:

$$\begin{aligned}\frac{6}{35} &= 0.17 \text{ km/min}\end{aligned}$$

(i) Time for first part:

$$\begin{aligned}\frac{x}{96} \text{ hours}\end{aligned}$$

(ii) Total time $= 2\tfrac{5}{60} = \dfrac{25}{12}$ hours; remaining distance $=(176 – x)$ km.

$$\begin{aligned}\frac{x}{96} + \frac{176 – x}{80} &= \frac{25}{12}\end{aligned}$$

Multiply throughout by $480$:

$$\begin{aligned}5x + 6(176 – x) &= 1000 \\ 5x + 1056 – 6x &= 1000 \\ -x &= -56 \\ x &= 56\end{aligned}$$

(iii) Time for first part:

$$\begin{aligned}\frac{56}{96} \times 60 &= 35 \text{ minutes}\end{aligned}$$

$A=(1100,\,0)$ and $B=(1145,\,40)$, so the time taken is $45$ min $=\tfrac{3}{4}$ h.

$$\begin{aligned}\text{Gradient}&=\frac{40-0}{\tfrac{3}{4}}\\ &=53\frac{1}{3}\end{aligned}$$

The gradient represents the speed, in km/h, at which Penelope travels from her house to the café.

Segment $BC$ is horizontal (distance stays at $40$ km):

$$\begin{aligned}\text{Gradient}&=0\end{aligned}$$

The gradient represents the speed. In this case Penelope is having her lunch, so her speed is zero.

Speed from the café to Hudson’s house (segment $CD$): distance $=80-40=40$ km over $1300-1224=36$ min $=\tfrac{3}{5}$ h.

$$\begin{aligned}\text{Gradient of }CD&=\frac{40}{\tfrac{3}{5}}\\ &=66\frac{2}{3}\text{ km/h}\end{aligned}$$

Since $66\tfrac{2}{3}>53\tfrac{1}{3}$, Penelope travelled faster from the café to Hudson’s house than from her house to the café. Hudson’s claim is not accurate.

(i) Total time $= 1\frac{3}{4} \text{ h} = 105$ min. Swim time $= (x – 25)$ min, cycle time $= 1.25x$ min.

$$\begin{aligned}x + (x – 25) + 1.25x &= 105 \\ 3.25x – 25 &= 105 \\ 3.25x &= 130 \\ x &= 40\end{aligned}$$

Time taken to run $= 40$ min.

(ii) Swim time $= 40 – 25 = 15$ min, so swimming distance $= 50 \times 15 = 750$ m.

$$\begin{aligned}\text{1 ratio unit} &= \frac{750}{3} = 250 \text{ m} \\ \text{cycling distance} &= 80 \times 250 = 20\,000 \text{ m} \\ \text{running distance} &= 25\,750 – 750 – 20\,000 = 5000 \text{ m} \\ y &= \frac{5000}{250} = 20\end{aligned}$$

(iii) Cycle time $= 1.25 \times 40 = 50$ min.

$$\begin{aligned}\text{average cycling speed} &= \frac{20\,000}{50} \\ &= 400 \text{ m/min}\end{aligned}$$

(a)

$$\begin{aligned}20 \text{ m/s} &= 20 \times \frac{3600}{1000} \text{ km/h}\\ &= 72 \text{ km/h}\end{aligned}$$

(b)

From 2020 to 2023 is 3 years.

$$\begin{aligned}\text{Value} &= 3000 \times (1.05)^3\\ &= 3472.875\\ &= 3472.88 \text{ dollars}\end{aligned}$$

(c)

The square has side $2r$.

$$\begin{aligned}\text{Area of circle} &= \pi r^2\\ \text{Area of square} &= (2r)^2 = 4r^2\\ \text{Shaded area} &= 4r^2 – \pi r^2\end{aligned}$$

Ratio:

$$\begin{aligned}\pi r^2 : (4r^2 – \pi r^2) &= \pi : (4 – \pi)\end{aligned}$$

(i) Running speed $= x$, cycling speed $= x+15$.

$$\begin{aligned}\text{Running time} &= \frac{15}{x}\\ \text{Cycling time} &= \frac{40}{x+15}\end{aligned}$$

Cycling took $\frac{1}{x}$ hour longer:

$$\begin{aligned}\frac{40}{x+15} – \frac{15}{x} &= \frac{1}{x}\\ \frac{40}{x+15} &= \frac{16}{x}\\ 40x &= 16(x+15)\\ 40x &= 16x + 240\\ 24x &= 240\\ x &= 10\end{aligned}$$

(ii) Total time:

$$\begin{aligned}&= \frac{15}{10} + \frac{40}{25}\\ &= 1.5 + 1.6\\ &= 3.1 \text{ h}\end{aligned}$$

Marking rates:

$$\begin{aligned}\text{Mr Tan} &= \frac{30}{2.5}=12\text{ scripts/h}\\ \text{Mrs Lim} &= \frac{25}{1.25}=20\text{ scripts/h}\end{aligned}$$

Combined rate:

$$\begin{aligned}12+20 &= 32\text{ scripts/h}\end{aligned}$$

Time for $42$ scripts:

$$\begin{aligned}\frac{42}{32} &= 1.3125\text{ h}\\ &= 1\text{ h }19\text{ min}\end{aligned}$$

(a) SUV in city uses 9.2 litres/100 km.

$$\begin{aligned}\frac{76}{100} \times 9.2 &= 6.992 \\ &\approx 7.0 \text{ litres}\end{aligned}$$

(b) Distance on expressway:

$$\begin{aligned}\frac{50}{60} \times 90 &= 75 \text{ km}\end{aligned}$$

SUV on expressway uses 6.6 litres/100 km.

$$\begin{aligned}\frac{75}{100} \times 6.6 &= 4.95 \text{ litres}\end{aligned}$$

(c)(i) Total time (in hours) $= 2$ h $36$ min $= 2.6$ h. Total distance $= x + (x + 23)$.

$$\begin{aligned}\frac{x}{105} + \frac{x + 23}{105} &= 2.6 \\ x + x + 23 &= 2.6 \times 105 \\ 2x + 23 &= 273 \\ 2x &= 250 \\ x &= 125\end{aligned}$$

(c)(ii) Total distance $= 125 + 148 = 273$ km (combination of city and expressway, SUV $= 7.6$ litres/100 km).

$$\begin{aligned}\text{Fuel used} &= \frac{273}{100} \times 7.6 = 20.748 \text{ litres}\end{aligned}$$

Using regular fuel at RM 2.05 with 10% loyalty discount:

$$\begin{aligned}\text{Fuel cost} &= 20.748 \times 2.05 \times 0.9 \\ &= \text{RM } 38.28\end{aligned}$$

Half the cost $= 38.28 \div 2 = \text{RM } 19.14$, so a suitable amount for Hamid to pay is about RM 19.14 (or rounded to RM 19.10 / RM 19.20).

(a) Company $X$, because its line has a steeper gradient (a greater rate of increase in charges per hour).

(b) Company $Y$. For a budget of 60 dollars, Company $Y$ allows about $4$ hours of rental compared with about $3$ hours for Company $X$, so Company $Y$ gives better value for money (an extra hour of riding for the same 60 dollars).

(a)

$$\begin{aligned}5 \text{ km}\end{aligned}$$

(b)

$$\begin{aligned}1.5 \text{ hours}\end{aligned}$$

(c)

$$\begin{aligned}\text{Gradient} &= \frac{\text{Rise}}{\text{Run}} \\ &= 20\end{aligned}$$

$20$ km/h represents the speed of Daniel travelling from home to the supermarket.

(a)

$$\begin{aligned}9 \text{ km/h} &= \frac{9\times 1000}{60} \text{ m/min} \\ &= 150 \text{ m/min}\end{aligned}$$

(b) Distance in $40$ min $=150\times 40=6000$ m. Circumference $=2\pi(0.2)=0.4\pi$ m.

$$\begin{aligned}\text{Revolutions} &= \frac{6000}{0.4\pi} \\ &= 4774.6\ldots \\ &= 4775\end{aligned}$$

(a)

$$120 \text{ m/min} = \frac{120 \times 60}{1000} \text{ km/h}$$
$$= 7.2 \text{ km/h}$$

(b) New speed $=7.2+0.8=8$ km/h. Time from $9.15$ am to $10.45$ am $=1.5$ h:

$$\text{distance} = 8 \times 1.5$$
$$= 12 \text{ km}$$

(a) Cycling time $+$ running time $=4$:

$$\frac{x}{36} + \frac{102 – x}{9} = 4$$

Multiply by $36$:

$$x + 4(102 – x) = 144$$
$$x + 408 – 4x = 144$$
$$-3x = -264$$
$$x = 88$$

(b) Running distance $=102 – 88 = 14$ km:

$$\text{time} = \frac{14}{9}$$
$$= 1\tfrac{5}{9} \text{ hours}$$