Mixtures and their prices – Ratio and Proportions (Senior 1 Mathematics) | Ganzaa

Ratio and Proportions

Lessons

Senior 1 · Mathematics

Ratio and Proportions

5 lessons

Lesson 5 of 5

Mixtures and their prices

25 min

Let's begin

A bean trader at Kimironko has two kinds of beans: a cheaper variety and a better one that costs more. If she pours some of each into one sack, what price per kilogram should the mixture sell for so that she earns exactly what she would have earned selling them apart? Traders, cooperatives and cooks meet this question every day, and it combines everything in this unit: ratios, proportion and equations.

Definition

Mixture: a quantity made by combining two or more ingredients, each with its own price per unit. The average price of the mixture (or mean price) is the price per unit that makes the mixture worth exactly the same as its ingredients: $\text{average price} = \frac{\text{total cost of the ingredients}}{\text{total quantity}} = \frac{q_1 p_1 + q_2 p_2}{q_1 + q_2},$ where $q_{1}, q_{2}$ are the quantities and $p_{1}, p_{2}$ the prices per unit.

Example: 1 kg at 800 Frw mixed with 1 kg at 1,000 Frw is worth $\frac{1{,}800}{2} = 900$ Frw per kg.

Important

The key idea. Mixing changes nothing about value: the total cost of the ingredients equals the total value of the mixture, $q_1 p_1 + q_2 p_2 = (q_1 + q_2) \times p_m,$ where $p_{m}$ is the price of the mixture. Every mixture problem is this one equation, with a different letter unknown.

Why the average price is not just the midpoint. If you mix equal amounts, the mixture price is halfway between the two prices. If you use more of one ingredient, its price counts more. So the mixture price always lies between the two prices, nearer to the price of the ingredient you used more of. It is a weighted average, with the quantities as the weights.

Type of problem	Unknown	Section
Find the average price	$p_{m}$	Example 1
Find the quantity of one type	$q_{1}$ or $q_{2}$	Example 2
Find the price of one ingredient	$p_{1}$ or $p_{2}$	Example 3
Find both quantities	$q_{1}$ and $q_{2}$ , total known	Example 4

Worked example

Example 1 (average price). A trader mixes 30 kg of beans at 800 Frw per kg with 20 kg of beans at 1,000 Frw per kg. At what price per kg should she sell the mixture to receive the same total?

	Quantity (kg)	Price (Frw per kg)	Cost (Frw)
Beans 1	30	800	24,000
Beans 2	20	1,000	20,000
Mixture	50	?	44,000

p_m = \frac{44{,}000}{50} = 880 \text{ Frw per kg}.

880 lies between 800 and 1,000, and nearer to 800, because more of the cheaper beans was used.

Worked example

Example 2 (quantity of one type). How many kilograms of rice at 1,200 Frw per kg must be mixed with 40 kg of rice at 1,500 Frw per kg to make a mixture worth 1,300 Frw per kg?

Let $x$ be the mass of the cheaper rice, in kg. Total cost of the ingredients = total value of the mixture:

\begin{aligned} 1{,}200x + 40 \times 1{,}500 &= 1{,}300(x + 40) \\ 1{,}200x + 60{,}000 &= 1{,}300x + 52{,}000 \\ 8{,}000 &= 100x \\ x &= 80 \end{aligned}

80 kg of the cheaper rice. Check: $80 \times 1{,}200 + 60{,}000 = 156{,}000$ and $120 \times 1{,}300 = 156{,}000$ . ✓

Worked example

Example 3 (price of one ingredient). A miller mixes 15 kg of sorghum flour at 900 Frw per kg with 10 kg of maize flour. The mixture is worth 1,020 Frw per kg. What is the price of the maize flour?

\begin{aligned} 15 \times 900 + 10p &= 25 \times 1{,}020 \\ 13{,}500 + 10p &= 25{,}500 \\ 10p &= 12{,}000 \\ p &= 1{,}200 \text{ Frw per kg} \end{aligned}

Worked example

Example 4 (both quantities). A coffee cooperative wants 100 kg of a blend worth 5,200 Frw per kg, made from grade A coffee at 6,000 Frw per kg and grade B at 4,000 Frw per kg. How much of each grade should it use?

Method 1, an equation. Let $x$ kg be grade A; then grade B is $(100 - x)$ kg.

\begin{aligned} 6{,}000x + 4{,}000(100 - x) &= 100 \times 5{,}200 \\ 2{,}000x + 400{,}000 &= 520{,}000 \\ x &= 60 \end{aligned}

60 kg of grade A and 40 kg of grade B.

Method 2, a ratio. The blend price is 1,200 above grade B and 800 below grade A. The amounts are in the ratio of these distances the other way round: A : B $= 1,200 : 800 = 3 : 2$ . Sharing 100 kg in the ratio $3 : 2$ gives 60 kg and 40 kg, as before.

Why the distances swap. Each kg of A adds 800 Frw more than the blend price, and each kg of B brings in 1,200 Frw less. The extra must cancel the shortfall: $800 \times 60 = 1{,}200 \times 40 = 48{,}000$ . That only works if the amounts are in the ratio $1,200 : 800$ .

Warning

Don't average the prices. $\frac{800 + 1{,}000}{2} = 900$ is the mixture price only when the quantities are equal. In Example 1 it would overcharge the customers: the right price is 880 Frw per kg. Always divide the total cost by the total quantity.

Warning

The mixture price must lie between the two prices. If your answer is cheaper than the cheapest ingredient or dearer than the dearest, there is an error. Check the equation before you do any more arithmetic.

Tip

Set out every mixture problem in a table with the columns quantity, price and cost, and one row per ingredient plus a row for the mixture. The cost column always adds up. That sum is your equation.

Mathispower4u: "Mixture App: Coffee Blend Given Mix Total Weight". Watch how the table of amount, price and value turns into one equation, as in Example 4 (the prices there are in dollars per pound).

✦

Practice

Use a quantity, price and cost table for each problem.

10 kg of sugar at 1,500 Frw per kg is mixed with 10 kg at 1,700 Frw per kg. Find the price of the mixture.
25 kg of rice at 1,000 Frw per kg is mixed with 15 kg at 1,400 Frw per kg. Find the price per kg of the mixture.
How many kg of tea at 3,000 Frw per kg must be mixed with 20 kg at 4,500 Frw per kg to give a blend worth 3,500 Frw per kg?
12 kg of flour at 800 Frw per kg is mixed with 8 kg of another flour. The mixture is worth 920 Frw per kg. Find the price of the second flour.
A shop wants 60 kg of beans worth 950 Frw per kg by mixing beans at 800 and 1,100 Frw per kg. How much of each should it use?
In what ratio must coffee at 4,500 Frw per kg be mixed with coffee at 7,000 Frw per kg to give a blend worth 5,500 Frw per kg?
Mutoni says that mixing 5 kg at 600 Frw per kg with 1 kg at 1,200 Frw per kg gives a mixture worth 900 Frw per kg. Is she right? Explain.

Show answersHide answers

$\frac{15{,}000 + 17{,}000}{20} = 1{,}600$ Frw per kg (equal amounts, so it is the midpoint).
$\frac{25{,}000 + 21{,}000}{40} = 1{,}150$ Frw per kg.
$3,000 x + 90,000 = 3,500 (x + 20)$ , so $500 x = 20,000$ and $x = 40$ kg.
$9,600 + 8 p = 18,400$ , so $p = 1,100$ Frw per kg.
Distances: 150 and 150, so the ratio is $1 : 1$ : 30 kg of each. Check: $24{,}000 + 33{,}000 = 57{,}000 = 60 \times 950$ .
Cheaper : dearer $= (7,000 - 5,500) : (5,500 - 4,500) = 1,500 : 1,000 = 3 : 2$ .
No. $\frac{3{,}000 + 1{,}200}{6} = 700$ Frw per kg. She averaged the prices, but she used five times as much of the cheaper kind.