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A survey of n people found that 60 percent preferred brand A. An addit

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A survey of n people found that 60 percent preferred brand A. An addit  [#permalink]

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New post 05 Nov 2017, 00:29
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A survey of n people found that 60 percent preferred brand A. An additional x people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find x in terms of n.

(A) n/6
(B) n/3
(C) n/2
(D) n
(E) 3n

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Re: A survey of n people found that 60 percent preferred brand A. An addit  [#permalink]

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New post 05 Nov 2017, 00:58
Bunuel wrote:
A survey of n people found that 60 percent preferred brand A. An additional x people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find x in terms of n.

(A) n/6
(B) n/3
(C) n/2
(D) n
(E) 3n



algebraic method..

A survey of n people found that 60 percent preferred brand A means \(\frac{60}{100}*n\)
add x people to it...\(\frac{60}{100}*n+x\)
IMPORTANT... total becomes = n+x

"Seventy percent of all the people surveyed preferred brand A" means \(\frac{60}{100}*n+x=\frac{70}{100}*(n+x)\)...
\(\frac{60}{100}*n+x=\frac{70}{100}*(n+x).........\frac{10n}{100}=\frac{30x}{100}...........10n=30x......x=n/3\)
B
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A survey of n people found that 60 percent preferred brand A. An addit  [#permalink]

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New post 05 Nov 2017, 14:32
Bunuel wrote:
A survey of n people found that 60 percent preferred brand A. An additional x people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find x in terms of n.

(A) n/6
(B) n/3
(C) n/2
(D) n
(E) 3n

This question can be approached as if it were a mixture problem, with weighted average.

Two different populations, with two different percentages of preference for A, contribute to an overall population with yet a third percentage of preference for A.

You can use decimals for percentages (e.g., 60 percent of n = .60n), or multiply all by 100 and use integers; it doesn't affect the weighting of x and n.

60(n) + 100(x) = 70(n + x)
60n + x = 70n + 70x
30x = 10n
x = \(\frac{10}{30}\)n
x = n/3

Answer B
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Re: A survey of n people found that 60 percent preferred brand A. An addit  [#permalink]

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New post 06 Nov 2017, 08:35
Bunuel wrote:
A survey of n people found that 60 percent preferred brand A. An additional x people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find x in terms of n.

(A) n/6
(B) n/3
(C) n/2
(D) n
(E) 3n



0.6n+x=0.7(n+x)
0.6n+x=0.7n+0.7x
x-0.7x=0.7n-0.6n
0.3x=0.1n
n=1/3n
Ans is B
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Re: A survey of n people found that 60 percent preferred brand A. An addit  [#permalink]

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New post 08 Nov 2017, 16:31
Bunuel wrote:
A survey of n people found that 60 percent preferred brand A. An additional x people were surveyed who all preferred brand A. Seventy percent of all the people surveyed preferred brand A. Find x in terms of n.

(A) n/6
(B) n/3
(C) n/2
(D) n
(E) 3n


Since 60 percent of the n people who were initially surveyed preferred brand A, the number of initially surveyed people who preferred brand A is 0.6n. Since all of the additional x people who were surveyed preferred brand A, the total number of people who prefer brand A is 0.6n + x. We are given that 70% of all the surveyed people prefer brand A and the total number of people who were surveyed, together with the x additional people, is n + x. Therefore, 0.6n + x must equal 70% of n + x, which gives us the following equation:

0.6n + x = 0.7(x + n)

0.6n + x = 0.7x + 0.7n

0.3x = 0.1n

3x = n

x = n/3

Answer: B
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Re: A survey of n people found that 60 percent preferred brand A. An addit &nbs [#permalink] 08 Nov 2017, 16:31
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