Answer to Question #129600 in Math for Sphesihle Zuma

Question

Ntsako walks to school each day at a constant pace of 75 m per minute.
The school is 1.8 km from his house.
1.2.1.1. Use a sketch or diagram to represent the problem. (2)
1.2.1.2. Solve the problem (3)
1.3. Use the concept of ratio to answer the following questions.
1.3.1. In week 1, Mpho deep cleans a paving at 4 m2 per hour. How long will it take her to deep clean
an area of 12 m long and 6 m wide? (3)
1.3.2. In week 2, Maria joins and each of them cleans 4 m2 per hour. How long will it take the two of
them to deep clean an area of 12 m long and 6m wide? (4)
1.3.3. In week 3, Lerato joins and each of them cleans 4 m2 per hour. How long will it take the three of
them to deep clean an area of 12 m long and 6 m wide? (4)
1.4. Calculate the following:
1.4.1. Which is the best price for washing powder? Motivate your answer.
 R45 for 2.5 kg
 R55.50 for 3 kg
 R72 for 4 kg (3)

Answer 1

1.2.1.1

Let $s=$ distance in meters, $v=$ velocity in meters per minute, $t=$ time in minutes.

Distance Time Graph of an object moving along line with a uniform speed 

1.2.1.2. The equation of the uniform line motion

$$s=vt$$

Solve for t

$$t={s\over v}$$

Given

$$s=1.8 \ km=1800\ m, v=75\ m/min$$

Then

$$t=\dfrac{1800\ m}{75\ m/min}=24 \min$$

A boy takes 24 minutes to walk from home to school.

1.3.1. Let $v_1=$ the rate of cleaning a paving by Mpho, $m^2/h$ .

Given $v_1=4\ m^2/h.$ The area $A$ of the rectangular paving with the length $l$ and the width $w$ is

$$A=l\times w$$

Given $l=12\ m, w=6\ m.$ Then

$$A=12\ m\times 6\ m=72\ m^2$$

If the rate of cleaning a paving is constant, then

$$t=\dfrac{A}{v}$$

Substitute

$$t_I=\dfrac{A}{v_1}=\dfrac{72\ m^2}{4\ m^2/h}=18\ h$$

It will take her18 hours to clean a paving.

1.3.2.Let now $v_2=$ the rate of cleaning a paving by Maria, $m^2/h$ .

Given $v_2=4\ m^2/h=v_1.$ The area $A$ of the rectangular paving with the length $l$ and the width $w$ is

$$A=l\times w$$

Given $l=12\ m, w=6\ m.$ Then

$$A=12\ m\times 6\ m=72\ m^2$$

If the rate of cleaning a paving is constant, then

$$t=\dfrac{A}{v}$$

Substitute

$$t_{II}=\dfrac{A}{v_1+v_2}=\dfrac{72\ m^2}{4\ m^2/h+4\ m^2/h}=9\ h$$

It will take them 9 hours to clean a paving.

1.3.3. Let now $v_3=$ the rate of cleaning a paving by Lerato, $m^2/h$ .

Given $v_3=4\ m^2/h=v_1=v_2.$ The area $A$ of the rectangular paving with the length $l$ and the width $w$ is

$$A=l\times w$$

Given $l=12\ m, w=6\ m.$ Then

$$A=12\ m\times 6\ m=72\ m^2$$

If the rate of cleaning a paving is constant, then

$$t=\dfrac{A}{v}$$

Substitute

$$t_{III}=\dfrac{A}{v_1+v_2+v_3}=\dfrac{72\ m^2}{4\ m^2/h+4\ m^2/h+4\ m^2/h}=6\ h$$

It will take the three of them 6 hours to clean a paving.

1.4.1

R45 for 2.5 kg

The price for washing powder

$$p_1=\dfrac{R45}{2.5\ kg}=R18/kg$$

R55.50 for 3 kg

The price for washing powder

$$p_2=\dfrac{R55.50}{3\ kg}=R18.50/kg$$

R72 for 4 kg

The price for washing powder

$$p_3=\dfrac{R72}{4\ kg}=R18/kg$$

$18<18.50$

The best price for washing powder is $R18/kg$.