Proportion Word Problems

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Question 1 of 5

1. Question

A cup is half full. After emptying 55mL, the cup is now one third full. What is the capacity of the cup?

(330)mL

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A word problem can be drawn as a diagram and then translated into an equation for easier solving.

First, draw a diagram of the problem

capacity of the cup:

x

$x$

Translate the problem into an equation with the help of the diagram above

A half-full cup, emptied $55$ $55$ mL,	is	now one third full
$\frac{1}{2} x - 55$ $1/2x-55$	$=$ $=$	$\frac{1}{3} x$ $1/3x$

Solve for

x

$x$

$\frac{1}{2} x - 55$ $1/2x-55$	$=$ $=$	$\frac{1}{3} x$ $1/3x$

$(\frac{1}{2} x - 55)$ $(1/2x-55)$ $\times 6$ $xx6$	$=$ $=$	$\frac{1}{3} x$ $1/3x$ $\times 6$ $xx6$	Multiply $6$ $6$ to both sides

$3 x - 330$ $3x-330$	$=$ $=$	$2 x$ $2x$
$3 x - 330$ $3x-330$ $- 2 x$ $-2x$	$=$ $=$	$2 x$ $2x$ $- 2 x$ $-2x$	Subtract $2 x$ $2x$ from both sides
$x - 330$ $x-330$	$=$ $=$	$0$ $0$
$x - 330$ $x-330$ $+ 330$ $+330$	$=$ $=$	$0$ $0$ $+ 330$ $+330$	Add $330$ $330$ to both sides
$x$ $x$	$=$ $=$	$330$ $330$ mL

330

$330$ mL

Question 2 of 5

2. Question

Greg spent a quarter of his money. He then spent half of the remainder. Altogether he spent

$ 50

$$50$ . How much did he have to start with?

$$$ $$$ (80)

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A word problem can be drawn as a diagram and then translated into an equation for easier solving.

First, draw a diagram of the problem

Amount of money that Greg started with:

x

$x$

Translate the problem into an equation with the help of the diagram above

He spent a quarter of his money then half of the remainder	which is	$$ 50$ $$50$ in total
$\frac{1}{4} x + \frac{3}{8} x$ $1/4x+3/8x$	$=$ $=$	$$ 50$ $$50$

Solve for

x

$x$

$\frac{1}{4} x + \frac{3}{8} x$ $1/4x+3/8x$	$=$ $=$	$$ 50$ $$50$

$(\frac{1}{4} x + \frac{3}{8} x)$ $(1/4x+3/8x)$ $\times 8$ $xx8$	$=$ $=$	$50$ $50$ $\times 8$ $xx8$	Multiply $8$ $8$ to both sides

$2 x + 3 x$ $2x+3x$	$=$ $=$	$400$ $400$
$5 x$ $5x$	$=$ $=$	$400$ $400$
$5 x$ $5x$ $\div 5$ $-:5$	$=$ $=$	$400$ $400$ $\div 5$ $-:5$	Divide both sides by $5$ $5$
$x$ $x$	$=$ $=$	$$ 80$ $$80$

$ 80

$$80$

Question 3 of 5

3. Question

A special concrete mix is made by mixing

3

$3$ tonnes (tons) of sand with

4

$4$ tonnes (tons) of cement. Per tonne, the cement costs

$ 180

$$180$ more than the sand. The total cost is

$ 1392

$$1392$ . What is the cost of each component?

Sand: $$$ $$$ (288)

Cement: $$$ $$$ (1104)

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A word problem can be drawn as a diagram and then translated into an equation for easier solving.

First, draw a diagram of the problem

Cost of the sand per ton:

x

$x$

Cost of the cement per ton:

x + 180

$x+180$

Translate the problem into an equation with the help of the diagram above

3

$3$ tons

(x) + 4

$(x)+4$ tons

(x + 180)

$(x+180)$

=

$=$

$ 1392

$$1392$

Solve for

x

$x$

$3$ $3$ tons $(x) + 4$ $(x)+4$ tons $(x + 180)$ $(x+180)$	$=$ $=$	$$ 1392$ $$1392$
$3 x + 4 x + 720$ $3x+4x+720$	$=$ $=$	$1392$ $1392$
$7 x + 720$ $7x+720$	$=$ $=$	$1392$ $1392$
$7 x + 720$ $7x+720$ $- 720$ $-720$	$=$ $=$	$1392$ $1392$ $- 720$ $-720$	Subtract $720$ $720$ from both sides
$7 x$ $7x$	$=$	$672$
$7x$ $-:7$	$=$	$672$ $-:7$	Divide both sides by $7$
$x$	$=$	$$96$	cost of sand per ton

Solve for

$x+180$

	$x+180$
$=$	$96+180$	Substitute $x$
$=$	$$276$	cost of cement per ton

Finally, compute for the total cost of each component

Remember that there are

$3$ tons of sand and

$4$ tons of cement

Sand:

		$3($96)$
	$=$	$$288$	cost of $3$ tons sand

Cement:

		$4($276)$
	$=$	$$1104$	cost of $4$ tons cement

Sand:

$$288$

Cement:

$$1104$

Question 4 of 5

4. Question

A tyre manufacturer wants to create a special mix of

$2$ rubber types. Hard rubber is priced at

$$7.25$ /kg and soft rubber is priced at

$$11$ /kg. How many kilograms of soft rubber should be mixed with

$8$ kg of hard rubber to sell for

$$9$ per kg?

(7)kg

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A word problem can be drawn as a diagram and then translated into an equation for easier solving.

First, draw a diagram of the problem

kilos of soft rubber:

$x$

total kilos of the mix:

$x+8$

Translate the problem into an equation with the help of the diagram above

$11(x)+7.25(8)$

$=$

$9(x+8)$

Solve for

$x$

$11(x)+7.25(8)$	$=$	$9(x+8)$
$11x+58$	$=$	$9x+72$
$11x+58$ $-9x$	$=$	$9x+72$ $-9x$	Subtract $9x$ from both sides
$2x+58$	$=$	$72$
$2x+58$ $-58$	$=$	$72$ $-58$	Subtract $58$ from both sides
$2x$	$=$	$14$
$2x$ $-:2$	$=$	$14$ $-:2$	Divide both sides by $2$
$x$	$=$	$7$ kg

$7$ kg

Question 5 of 5

5. Question

A special candy bar contains

$30%$ chocolate. Another candy bar contains

$70%$ chocolate. When melted together, how much of each candy bar should be used to make a

$100$ g candy bar that contains

$35%$ chocolate?

Candy bar with $30%$ chocolate: (87.5, 87.50)g

Candy bar with $70%$ chocolate: (12.5, 12.50)g

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A word problem can be drawn as a diagram and then translated into an equation for easier solving.

First, draw a diagram of the problem

grams of candy bar with

$30%$ chocolate:

$x$

grams of candy bar with

$70%$ chocolate:

$100-x$

Translate the problem into an equation with the help of the diagram above. Convert percentages to decimals for easier solving

$30%(x)+70%(100-x)$	$=$	$35%(100$ g $)$
$0.3(x)+0.7(100-x)$	$=$	$0.35(100)$

Solve for

$x$

$0.3(x)+0.7(100-x)$	$=$	$0.35(100)$
$0.3x+70-0.7x$	$=$	$35$
$70-0.4x$	$=$	$35$
$70-0.4x$ $-70$	$=$	$35$ $-70$	Subtract $70$ from both sides
$-0.4x$	$=$	$-35$
$-0.4x$ $-:(-0.4)$	$=$	$-35$ $-:(-0.4)$	Divide both sides by $-0.4$
$x$	$=$	$87.5$ g	grams of candy bar with $30%$ chocolate

Solve for

$100-x$

	$100-x$
$=$	$100-87.5$	Substitute $x$
$=$	$12.5$ g	grams of candy bar with $70%$ chocolate

Candy bar with

$30%$ chocolate:

$87.5$ g

Candy bar with

$70%$ chocolate:

$12.5$ g

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