Question

A tea shop owner is mixing a blend of two teas, one of which costs $6.50 per pound, the other costing $4.00 per pound. The owner wants to have 20 pounds of a mixture that will sell for $5.50 per pound. How much of each type of tea should be used?

Answer 1

12 pounds of tea-1 and 8 pounds of tea-2 should be used.

Explanation:

Let the mass of the tea-1 be x and tea-2 be y.

Total mass of mixture tea owner wants to make = 20 pounds

x + y = 20 ...[1]

Cost of 1 pound of tea-1 = $6.50

Cost of 1 pound of tea-2 = $4.00

Cost of the 1 pounds mixture = $5.50

`x* \$6.50+y* \$4.00=20* \$5.50`

6.5x + 4.00y = 110...[2]

Solving equation [1] and [2] by substitution method, putting value of x from [1] in [2].

x = 20 - y

6.5(20 - y) + 4.00y = 110

130 - 6.50y + 4.00y = 110

-2.5 y=-20

y = 8

x = 20 - y = 20 -8 = 12

12 pounds of tea-1 and 8 pounds of tea-2 should be used.

Answer 2

`\bf \begin{array}{lccclll} &amount&\$&cost\\ &-----&-----&-----\\ \textit{first tea}&x&6.50&(6.50)(x)\\ \textit{second tea}&y&4&(4)(y)\\ -----&-----&-----&-----\\ mixture&20&5.50&(20)(5.50) \end{array}`

whatever the amounts of "x" and "y" are, they have to sum up to 20 for the mixture
thus x + y = 20

whatever 6.50x and 4y prices are, they must add up to (20)(5.50)
thus
`\bf \begin{cases} x+y=20\to \boxed{y}=20-x\\ 6.50x+4y=20\cdot 5.50\\ ----------\\ 6.50x+4\left( \boxed{20-x} \right)=20\cdot 5.50 \end{cases}`

solve for "x", to see how much will be needed for the first tea type

what about the second tea type? well y = 20 - x