A wholesaler of nuts sells two mixtures of peanuts and cashews. The first mixture is 3/4 peanuts by weight. The second mixture has the same number of pounds of cashews as the first mixture but 8 fewer pounds of peanuts. If 7/10 of the weight of the second mixture is peanuts, how many pounds of nuts are in the first mixture?One thing that jumps out about this question is that the quantity of cashews is the same in both cases.
A second thing that jumps out is that, if we know the proportion of peanuts in each mixture, we also know the proportion of cashews and the relative proportions of peanuts and cashews in each mixtures.
So, given those two facts, we can base all of our work on the quantity of cashews.
Let \(C\) be the quantity of cashews in both mixtures.
Then, since the first mixture is \(3/4\) peanuts by weight, \(C = 1/4\) of the first mixture. So, the weight of the first mixture is \(4C\).
In the second mixture, peanuts represent \(7/10\) and \(C\) represents \(3/10\). So, the total weight of the second mixture is the following:
\(\frac{7}{3} C + C = \frac{10}{3} C\)
Since the difference between the weights of the mixture is 8 pounds, we have the following:
\(4C - 8 = \frac{10}{3} C\)
\(12C - 24 = 10C\)
\(2C = 24\)
\(C = 12\)
So, the total weight of the first mixture is \(4C = 48\).
Correct Answer