Bunuel
A big bowl of nuts is prepared for a party. Brand P’s mixed nuts are 20% almonds and brand Q’s deluxe mixed nuts are 25% almonds. If the bowl contains a total of 64 ounces of nuts, representing a mixture of both brands, and 15 ounces of the mixture are almonds, how many ounces of brand Q’s deluxe mixed nuts are used?
(A) 16
(B) 20
(C) 32
(D) 44
(E) 48
Solution :
We are given,
• There are two types of nut brands : P and Q
• Brand P consists of 20% almonds
• Brand Q consists of 25% almonds
• Total ounces of almonds = 15
• Total ounces of the nuts = 64
We need to find the total ounces of Brand Q's nuts.
Let the total ounces of Brand P's nuts be x
and the total ounces of brand Q's nuts be y.
Then, \(x + y = 64\) ....... (1)
Total ounces of almonds present in Brand P
= 20% of \(x\)
= \(\frac{20}{100} * x\)
= \(\frac{x}{5}\)
Total ounces of almonds present in Brand Q
= 25% of \(y\)
= \(\frac{25}{100} * y\)
= \(\frac{y}{4}\)
Then, \(\frac{x}{5} + \frac{y}{4} = 15\\
⟹ 4x + 5y = 300\) ....... (2)
Multiplying (1) by 4, we get :
\(4x + 4y = 256\) ........ (3)
Solving (2) & (3) simultaneously, we have :
\(4x + 5y = 300\) ........ (2)
\(4x + 4y = 256\) ......... (3)
Subtracting (3) from (2), we get :
\((4x - 4x) + (5y - 4y) = 300 - 256\)
⟹ \(0 + y = 44\)
⟹ \(y = 44\)
Hence, the correct answer is
Option D.