Question

For the given problems in context,

(i) set up and solve the algebraic statement

(ii) evaluate the problem.

(a) There were 579 patients infected with COVID-19 at a hospital. Of these patients, 123 were children and there were twice as many men as women. How many women who have become infected with the coronavirus were in the hospital? (5)

(b) Phumza bought 12 metres of curtaining. Some of the material came from a roll with a slight flaw in it and cost R7,00 per metre instead of the normal R8,00 per metre. If Phumza paid R93,50 altogether, how many metres of the damaged material did she buy? (5)

(c) Nancy buys 15 meters of ribbon. Her sister buys 6 meters less ribbon but pays twice as much per metre. Between them they pay R6,60. What is the price per meter of the two different types of ribbon? (5) (d) You have twice as many 20 cent coins as 10 cents coins and half the number of 5 cent coins as 10 cent coins. If you have R4,20 altogether, find out how many of each coin denomination do you have?

Answer 1

(a) Total number of people infected =579

Number of children infected = 123

(i) Let $x$ x be number of men infected. Then number of women infected will be $2x.$

Complete the equation

$$123+x+2x=579$$

(ii) Solve for $x$

$$3x+123-123=579-123$$

$$3x=456$$

$$\dfrac{3x}{3}=\dfrac{456}{3}$$

$$x=152$$

Therefore 152 women who have become infected with the coronavirus were in the hospital.

(b)

(i) Let $x$ be amount of metres of the damage material, $y$ be amount of metres of the normal material.

Phumza bought 12 metres of curtaining

$$x+y=12$$

Phumza paid R93,50 altogether

$$7x+8y=93.50$$

Solve the following system

$$\begin{matrix} x+y=12\\ 7x+8y=93.50 \end{matrix}$$

(ii) Solve the first equation for $y$

$$y=12-x$$

Substitute in the second equation

$$7x+8(12-x)=93.50$$

Solve for $x$

$$7x+96-8x=93.50$$

$$-x+96-96=93.50-96$$

$$-x=-2.50$$

$$\dfrac{-x}{-1}=\dfrac{-2.50}{-1}$$

$$x=2.50$$

Find $y$

$$y=12-2.50$$

$$y=9.50$$

Phumza bought $2.50$ metres of the damaged material.

(c)

(i) Let $x$ be the price per metre of ribbbons, Nancy buys, then $2x$ is the price per metre of ribbons, her sister buys.

Complete the equation

$$15x-6(2x)=6.60$$

(ii) Solve the equation for $x$

$$15x-12x=6.60$$

$$3x=6.60$$

$$\dfrac{3x}{3}=\dfrac{6.60}{3}$$

$$x=2.20$$

$$2(2.20)=4.40$$

The price per metre of ribbons, Nancy buys is $R2.20$

The price per metre of ribbons, her sister buys $R4.40.$

(d)

(i) Let x be the number of 5 cent coins. Then the number of 10 cent coins will be $2x$ and the number of 20 cent coins will be $4x.$

Complete the equation

$$5(x)+10(2x)+20(4x)=420$$

(ii) Solve for $x$

$$5x+20x+80x=420$$

$$105x=420$$

$$\dfrac{105x}{105}=\dfrac{420}{105}$$

$$x=4$$

$$2x=8$$

$$4x=16$$

I have four 5 cent coins, eight 10 cent coins, and sixteen 20 cent coins.