Proportion Word Problems

Question 1 of 5

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

(330)mL

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

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)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

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)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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
$7 x$ $7x$	$=$ $=$	$672$ $672$
$7 x$ $7x$ $\div 7$ $-:7$	$=$ $=$	$672$ $672$ $\div 7$ $-:7$	Divide both sides by $7$ $7$
$x$ $x$	$=$ $=$	$$ 96$ $$96$	cost of sand per ton

Solve for

x + 180

$x+180$

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

Finally, compute for the total cost of each component

Remember that there are

3

$3$ tons of sand and

4

$4$ tons of cement

Sand:

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

Cement:

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

Sand:

$ 288

$$288$

Cement:

$ 1104

$$1104$

Question 4 of 5

A tyre manufacturer wants to create a special mix of

2

$2$ rubber types. Hard rubber is priced at

$ 7.25

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

$ 11

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

8

$8$ kg of hard rubber to sell for

$ 9

$$9$ per kg?

(7)kg

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

$x$

total kilos of the mix:

x + 8

$x+8$

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

11 (x) + 7.25 (8)

$11(x)+7.25(8)$

=

$=$

9 (x + 8)

$9(x+8)$

Solve for

x

$x$

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

7

$7$ kg

Question 5 of 5

A special candy bar contains

30 %

$30%$ chocolate. Another candy bar contains

70 %

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

100

$100$ g candy bar that contains

35 %

$35%$ chocolate?

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

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 %

$30%$ chocolate:

x

$x$

grams of candy bar with

70 %

$70%$ chocolate:

100 - x

$100-x$

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

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

Solve for

x

$x$

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

Solve for

100 - x

$100-x$

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

Candy bar with

30 %

$30%$ chocolate:

87.5

$87.5$ g

Candy bar with

70 %

$70%$ chocolate:

12.5

$12.5$ g

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Correct!

3. Question

Hint

Keep Going!

4. Question

Hint

Fantastic!

5. Question

Hint

Excellent!

Quizzes