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06 Feb 2018, 03:20
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A store sells five different kinds of nuts. If it is possible to buy x pounds of the most expensive nuts for $3.20 and x pounds of the cheapest nuts for $1.40, then which of the following could be the cost of purchasing a mixture containing x pounds of each type of nut?
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
06 Feb 2018, 03:58
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Bunuel
A store sells five different kinds of nuts. If it is possible to buy x pounds of the most expensive nuts for $3.20 and x pounds of the cheapest nuts for $1.40, then which of
the following could be the cost of purchasing a mixture containing x pounds of each type of nut?
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
the following could be the cost of purchasing a mixture containing x pounds of each type of nut?
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
From a logical standpoint, we can eliminate Options A,B, and C because
if the cheapest nut costs $1.4 and we have to buy x pounds of each of the 5 nuts.
Option E also will not work since the costliest nut will cost $3.2 and all the 5
varieties would need to cost $3.2 for the total cost to be $16.00.
We are left with Option D($13.60) which can be realized by having 3 boxes of the
costliest nut, 1 box of the cheapest nut and one box which costs $2.6.
rahul16singh28
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06 Feb 2018, 08:18
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Bunuel
A store sells five different kinds of nuts. If it is possible to buy x pounds of the most expensive nuts for $3.20 and x pounds of the cheapest nuts for $1.40, then which of
the following could be the cost of purchasing a mixture containing x pounds of each type of nut?
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
the following could be the cost of purchasing a mixture containing x pounds of each type of nut?
(A) $1.76
(B) $2.84
(C) $3.54
(D) $13.60
(E) $16.00
I think this is more of a Logical Question than algebraic.
By looking at the Options only D & E are left. Option E is multiple of most Expensive nuts (i.e. \(3.2 * 5\)), so this Option is ruled out.
Hence, D.
