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30 Jun 2021, 08:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
75% (02:46) correct
25%
(03:05)
wrong
based on 648
sessions
History
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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
(A) 16
(B) 20
(C) 32
(D) 44
(E) 48
08 Mar 2022, 07:31
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Let the quantity of brand Q = x
Let the quantity of brand P = (64-x)
0.2(64-x)+0.25x = 15
12.8-0.2x+0.25x = 15
0.05x = 2.2
x = 44 (D)
Let the quantity of brand P = (64-x)
0.2(64-x)+0.25x = 15
12.8-0.2x+0.25x = 15
0.05x = 2.2
x = 44 (D)
General Discussion
30 Jun 2021, 12:24
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Bunuel
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
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.
