In what ratio must a person mix 3 kinds of tea costing $60 per kg, $75

The resultant value of \(z\) can be found by adding the value of \(x:y\) and \(y:z\). Hence, we are the value in \(z\) two time. Hope this helps.

Cheers meanup!

meanup, I still didn't understand the addition part. Could you please elaborate?

Thanks!

ankitapugalia

Here the common value is \(z\), and since we are averaging it, we do not standardize as per the normal process, but we add the values of the common factors.

\(x : z = 1 : 1\) and \(z : y = 1 : 4\)

Therefore, we may rewrite;
\(x : z : y = 1 : (1 + 1) : 4 = 1 : 2 : 4\)

\(x : y : z = 1 : 4 : 2\)

I set up a weighted-average formula and then worked from there.

\(\frac{(60*x + 75*y + 100*z)}{(x+y+z)}\)=80

60*x + 75*y + 100*z = 80*x + 80*y + 100*z

20*z = 20*x + 5*y

4*z = 4*x + y

Then from the last equation, the lowest values in which the equation holds true would be the ratio of values reduced to lowest form:

x=1, y=4, z=2

For a selling price of $96 per kilogram to yield a profit of 20%, the average cost per kilogram of the mixture must be $80.

From cheapest to most expensive, let the three teas be x, y and z.
Since the three costs for x, y and z -- $60 per kilogram, $75 per kilogram, and $100 per kilogram -- must be combined to yield a MIXTURE of x+y+z with an average cost of $80 per kilogram, the following equation is implied:
60x + 75y + 100z = 80(x+y+z)
60x + 75y + 100z = 80x + 80y + 80z
20z = 20x + 5y
4z = 4x + y

The resulting blue equation has an INFINITE number of solutions.

Option D is one solution for 4z=4x+y --> x=1, y=4, z=2
If 1 kilogram of x, 4 kilograms of y, and 2 kilograms of z are combined to form a 7-kilogram mixture, the average cost per kilogram \(= \frac{1*60 + 4*75 + 2*100}{7} = \frac{560}{7} = 80\)

Another solution for 4z=4x+y --> x=2, y=4, z=3
If 2 kilograms of x, 4 kilograms of y, and 3 kilograms of z are combined to form a 9-kilogram mixture, the average cost per kilogram \(= \frac{2*60 + 4*75 + 3*100}{9} = \frac{720}{9} = 80\)

Another solution for 4z=4x+y --> x=3, y=4, z=4
If 3 kilograms of x, 4 kilograms of y, and 4 kilograms of z are combined to form an 11-kilogram mixture, the average cost per kilogram \(= \frac{3*60 + 4*75 + 4*100}{11} = \frac{880}{11} = 80\)

Another solution for 4z=4x+y --> x=4, y=20, z=9
If 4 kilograms of x, 20 kilograms of y, and 9 kilograms of z are combined to form a 33-kilogram mixture, the average cost per kilogram \(= \frac{4*60 + 20*75 + 9*100}{33} = \frac{2640}{33} = 80\)

As the examples above illustrate, there are an infinite number of ways to combine the three teas to yield a mixture with an average cost of $80 per kilogram.
Since there is no unique ratio for the three teas, the problem is flawed.

meanup do you think is better your approach or just to find the cost ---> 80$ and then working backward? The second is much more intuitive but probably a little time consuming

­It depends upon the answer choice available. For example, if the answer contains only prime number and divisible by itself, then backtrack may be tricker at times. However, this problem can be solved in multiple ways. We can solve it using percentage, ration and proportions, and allegations and mixtures as well. Also, performing backtrack (in this problem statement) will result in even or odd factor (or equivalent) which may be optimized accordingly.­

PS: I know this reply has been delayed.­