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
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 in total
14x+38x
=
$50
Solve for x
14x+38x
=
$50
(14x+38x)×8
=
50×8
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?
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
Cost of the cement per ton: x+180
Translate the problem into an equation with the help of the diagram above
3tons(x)+4tons(x+180)
=
$1392
Solve for x
3tons(x)+4tons(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 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 8kg of hard rubber to sell for $9 per kg?
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
=
7kg
7kg
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 100g candy bar that contains 35% chocolate?