Solving Rate and Mixture Problems with Systems | Grade 8 Math | LevelKite

Picture a 10-pound bag of nut mix. The nuts in it cost 60 dollars in all, but you don't know how many pounds of each nut are inside. Two numbers are unknown, so one fact is not enough to find them. You need two facts, and each fact becomes an equation.

The tool that connects the facts is a rate: a number that tells how much of one quantity goes with one unit of another, such as 4 dollars per pound or 60 miles per hour. When a rate stays the same, multiplying it by an amount gives a total. For example, 6 pounds at 4 dollars per pound cost $6 \times 4 = 24$ dollars.

Rate times amount gives a total when the rate stays the same

Situation	Amount	Rate	Rate times amount
Buying nuts by weight	weight in pounds	price in dollars per pound	cost in dollars
Driving at a steady speed	time in hours	speed in miles per hour	distance in miles
Dried fruit in granola	weight of granola in pounds	pounds of fruit per pound of granola (20% means $0.20$ )	weight of fruit in pounds

A mixture problem blends two ingredients that have different rates, such as different prices, into one batch. A rate problem often splits one total, such as a trip, into two parts done at different rates. The problems in this lesson all give you the same two facts:

the two unknown amounts add to a known total, and
the two rate-times-amount totals add to a known total.

You have already compared two plans with a system of equations, which is two equations that share the same unknowns. Here you write one equation for each fact. A solution of the system is a pair of values that makes both equations true at once.

Key concept

One equation for amounts, one for rate times amount

Let $x$ and $y$ be the two unknown amounts, measured in the same unit. Suppose they add to a known total $T$ . Suppose their rates are $r_1$ and $r_2$ , and the two rate-times-amount totals add to a known total $V$ . Then the system below models the problem.

When $r_1 \neq r_2$ , exactly one pair $(x, y)$ makes both equations true. That pair answers the problem only if it also fits the story. So check it in both original equations, and make sure $x$ and $y$ are permitted values, such as weights that are not negative.

x + y = T \quad\text{and}\quad r_1 x + r_2 y = V

Worked example

A nut mix priced by the pound

A shop mixes peanuts that cost 4 dollars per pound with cashews that cost 9 dollars per pound. It makes 10 pounds of mix, and the nuts in it cost 60 dollars in all. How many pounds of each nut are in the mix?

Name each unknown with its unit. Let $p$ be the pounds of peanuts and $c$ the pounds of cashews. Each is at least 0 and at most 10, because the whole mix weighs 10 pounds. The two weights make up the whole mix, so they add to 10.
$p + c = 10$
Use rate times amount. At 4 dollars per pound, $p$ pounds of peanuts cost $4p$ dollars. At 9 dollars per pound, $c$ pounds of cashews cost $9c$ dollars. The two costs add to the 60 dollars for the whole mix.
$4p + 9c = 60$
Subtract $c$ from both sides of the first equation. Subtracting the same quantity from both sides keeps exactly the same solutions, so you get $p = 10 - c$ . In any solution, $p$ and $10 - c$ are the same number, so you can replace $p$ with $10 - c$ in the second equation.
$4(10 - c) + 9c = 60$
Distribute, which means multiply 4 by each term in the parentheses: $4(10 - c) = 40 - 4c$ . Then combine the like terms, the ones with $c$ : $-4c + 9c = 5c$ . The expressions $4(10 - c) + 9c$ and $40 + 5c$ are equivalent, which means they have the same value for every value of $c$ . Swapping one for the other keeps the same solutions.
$40 + 5c = 60$
Subtract 40 from both sides to get $5c = 60 - 40 = 20$ . Then divide both sides by 5. Dividing both sides by the same nonzero number keeps the same solutions, so $c = 4$ . Put $c = 4$ into $p = 10 - c$ to find the pounds of peanuts.
$c = 20 \div 5 = 4 \quad\text{and}\quad p = 10 - 4 = 6$
Check both values in the two original equations, because those come straight from the story. Also check that the values are permitted: 6 and 4 are each between 0 and 10.
$6 + 4 = 10 \quad\text{and}\quad 4(6) + 9(4) = 24 + 36 = 60$

Answer

The mix has $6$ pounds of peanuts and $4$ pounds of cashews.

Travel problems have the same shape. The amounts are times, the rates are speeds, and the rate-times-amount totals are distances. One condition matters: distance equals speed times time for a stretch where the speed stays the same. When a trip has two steady speeds, each part of the trip gets its own term.

Worked example

A trip at two speeds

A family drives for 5 hours in all. They drive at a steady 30 miles per hour through towns and at a steady 60 miles per hour on the highway. The whole trip is 240 miles. How many hours do they drive at each speed?

Let $t$ be the hours in towns and $h$ the hours on the highway. Each is at least 0 and at most 5. The two times make up the whole trip, so they add to 5.
$t + h = 5$
Each speed is steady, so distance equals speed times time for each part. The town part is $30t$ miles and the highway part is $60h$ miles. The two distances add to 240 miles.
$30t + 60h = 240$
Subtract $h$ from both sides of the first equation, which keeps the same solutions, to get $t = 5 - h$ . Replace $t$ with $5 - h$ in the second equation, because the two are equal in any solution.
$30(5 - h) + 60h = 240$
Distribute: $30(5 - h) = 150 - 30h$ . Combine like terms: $-30h + 60h = 30h$ . The new left side is equivalent to the old one, so it has the same value for every value of $h$ .
$150 + 30h = 240$
Subtract 150 from both sides to get $30h = 240 - 150 = 90$ . Divide both sides by 30, which is allowed because 30 is not zero. Then use $t = 5 - h$ .
$h = 90 \div 30 = 3 \quad\text{and}\quad t = 5 - 3 = 2$
Check in the two original equations. Both values are permitted, because 2 and 3 are each between 0 and 5.
$2 + 3 = 5 \quad\text{and}\quad 30(2) + 60(3) = 60 + 180 = 240$

Answer

The family drives for $2$ hours through towns and $3$ hours on the highway.

A percent can act as a rate too. Granola that is 20% dried fruit by weight has $0.20$ pounds of fruit in each pound of granola. So the pounds of fruit equal the decimal rate times the pounds of granola. When you blend two granolas, the amounts that add are pounds of granola, and the rate-times-amount totals that add are pounds of fruit.

Worked example

A percent mixture

Granola A is 10% dried fruit by weight. Granola B is 40% dried fruit by weight. A camp cook wants 30 pounds of granola that is 20% dried fruit by weight. How many pounds of each granola should the cook mix?

Let $a$ be the pounds of Granola A and $b$ the pounds of Granola B. Each is at least 0 and at most 30. Together they make the 30-pound batch.
$a + b = 30$
Write each percent as a decimal rate in pounds of fruit per pound of granola: 10% is $0.10$ , 40% is $0.40$ and 20% is $0.20$ . The final batch must hold $0.20 \times 30 = 6$ pounds of fruit. The fruit from A plus the fruit from B makes that total.
$0.10a + 0.40b = 6$
Subtract $b$ from both sides of the first equation, which keeps the same solutions, to get $a = 30 - b$ . Replace $a$ with $30 - b$ in the second equation.
$0.10(30 - b) + 0.40b = 6$
Distribute: $0.10(30 - b) = 3 - 0.10b$ . Combine like terms: $-0.10b + 0.40b = 0.30b$ . The new left side is equivalent to the old one for every value of $b$ .
$3 + 0.30b = 6$
Subtract 3 from both sides to get $0.30b = 6 - 3 = 3$ . Divide both sides by $0.30$ , which is allowed because $0.30$ is not zero. Then use $a = 30 - b$ .
$b = 3 \div 0.30 = 10 \quad\text{and}\quad a = 30 - 10 = 20$
Check in the two original equations. Both values are permitted, because 20 and 10 are each between 0 and 30. The batch holds 6 pounds of fruit in 30 pounds of granola, and $6 \div 30 = 0.20$ , which is 20%.
$20 + 10 = 30 \quad\text{and}\quad 0.10(20) + 0.40(10) = 2 + 4 = 6$

Answer

The cook should mix $20$ pounds of Granola A with $10$ pounds of Granola B.

Many rate and mixture problems come down to two facts about two unknown amounts: the amounts add to a known total, and the rate-times-amount totals add to a known total. Define each variable with its unit and its permitted values, then write one equation for each fact. Solve by substituting and by doing the same valid operation on both sides of an equation: add or subtract the same quantity, or multiply or divide by the same nonzero number. Finish by checking your pair in both original equations and against the story.