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 problemcapacity of the cup: `x`
Translate the problem into an equation with the help of the diagram aboveA half-full cup, emptied `55`mL, is now one third full `1/2x-55` `=` `1/3x` Solve for `x``1/2x-55` `=` `1/3x` `(1/2x-55)``xx6` `=` `1/3x``xx6` Multiply `6` to both sides `3x-330` `=` `2x` `3x-330` `-2x` `=` `2x` `-2x` Subtract `2x` from both sides `x-330` `=` `0` `x-330` `+330` `=` `0` `+330` Add `330` to both sides `x` `=` `330`mL `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 `$50`. How much did he have to start with?- `$` (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 problemAmount of money that Greg started with: `x`
Translate the problem into an equation with the help of the diagram aboveHe spent a quarter of his money then half of the remainder which is `$50` in total `1/4x+3/8x` `=` `$50` Solve for `x``1/4x+3/8x` `=` `$50` `(1/4x+3/8x)``xx8` `=` `50``xx8` Multiply `8` to both sides `2x+3x` `=` `400` `5x` `=` `400` `5x``-:5` `=` `400``-:5` Divide both sides by `5` `x` `=` `$80` `$80` -
Question 3 of 5
3. Question
A special concrete mix is made by mixing `3` tonnes (tons) of sand with `4` tonnes (tons) of cement. Per tonne, the cement costs `$180` more than the sand. The total cost is `$1392`. What is the cost of each component?-
Sand: `$` (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 problemCost of the sand per ton: `x`Cost of the cement per ton: `x+180`
Translate the problem into an equation with the help of the diagram above`3`tons`(x)+4`tons`(x+180)` `=` `$1392` Solve for `x``3`tons`(x)+4`tons`(x+180)` `=` `$1392` `3x+4x+720` `=` `1392` `7x+720` `=` `1392` `7x+720` `-720` `=` `1392` `-720` Subtract `720` from both sides `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 componentRemember that there are `3` tons of sand and `4` tons of cementSand:`3($96)` `=` `$288` cost of `3` tons sand Cement:`4($276)` `=` `$1104` cost of `4` tons cement Sand: `$288`Cement: `$1104` -
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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 problemkilos 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)gCandy 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 problemgrams 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`gCandy bar with `70%` chocolate: `12.5`g -