Question #129600
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)
1
Expert's answer
2020-08-17T17:56:26-0400

1.2.1.1

Let s=s= distance in meters, v=v= velocity in meters per minute, t=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=vts=vt

Solve for t


t=svt={s\over v}

Given


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

Then


t=1800 m75 m/min=24mint=\dfrac{1800\ m}{75\ m/min}=24 \min

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


1.3.1. Let v1=v_1= the rate of cleaning a paving by Mpho, m2/hm^2/h .

Given v1=4 m2/h.v_1=4\ m^2/h. The area AA of the rectangular paving with the length ll and the width ww is


A=l×wA=l\times w

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


A=12 m×6 m=72 m2A=12\ m\times 6\ m=72\ m^2

If the rate of cleaning a paving is constant, then


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


Substitute


tI=Av1=72 m24 m2/h=18 ht_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 v2=v_2= the rate of cleaning a paving by Maria, m2/hm^2/h .

Given v2=4 m2/h=v1.v_2=4\ m^2/h=v_1. The area AA of the rectangular paving with the length ll and the width ww is


A=l×wA=l\times w

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


A=12 m×6 m=72 m2A=12\ m\times 6\ m=72\ m^2

If the rate of cleaning a paving is constant, then


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


Substitute


tII=Av1+v2=72 m24 m2/h+4 m2/h=9 ht_{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 v3=v_3= the rate of cleaning a paving by Lerato, m2/hm^2/h .

Given v3=4 m2/h=v1=v2.v_3=4\ m^2/h=v_1=v_2. The area AA of the rectangular paving with the length ll and the width ww is


A=l×wA=l\times w

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


A=12 m×6 m=72 m2A=12\ m\times 6\ m=72\ m^2

If the rate of cleaning a paving is constant, then


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


Substitute


tIII=Av1+v2+v3=72 m24 m2/h+4 m2/h+4 m2/h=6 ht_{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


p1=R452.5 kg=R18/kgp_1=\dfrac{R45}{2.5\ kg}=R18/kg

R55.50 for 3 kg

The price for washing powder


p2=R55.503 kg=R18.50/kgp_2=\dfrac{R55.50}{3\ kg}=R18.50/kg

R72 for 4 kg

The price for washing powder


p3=R724 kg=R18/kgp_3=\dfrac{R72}{4\ kg}=R18/kg

18<18.5018<18.50

The best price for washing powder is R18/kgR18/kg.



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