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?
\(\bf \begin{array}{lccclll}
&amount&\@@MATH_TOKEN_0@@cost\\
&-----&-----&-----\\
\textit{first tea}&x&6.50&(6.50)(x)\\
\textit{second tea}&y&4&(4)(y)\\
-----&-----&-----&-----\\
mixture&20&5.50&(20)(5.50)
\end{array}\)
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
Answer:
12 pounds of tea-1 and 8 pounds of tea-2 should be used.
Step-by-step 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
\(x\times \$6.50+y\times \$4.00=20\times \$5.50\)
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.