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If a certain toy store's revenue in November was 2/5 of its [#permalink]
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25 Jun 2012, 02:36
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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SOLUTIONIf a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?(A) 1/4 (B) 1/2 (C) 2/3 (D) 2 (E) 4 Probably, for most people it would be easier to solve such kind of questions by picking numbers. Notice that January is linked with November and November is linked with December. So, we should pick some smart number for December's revenue. Since the denominators in the question are 5 and 4, the let say December's revenue was 20 (the least common multiple of 4 and 5). So, we have that: Revenue in December = 20; Revenue in November = 20*2/5 = 8; Revenue in January = 8*1/4 = 2; The average of 8 and 2 is 5, so the store's revenue in December was 20/5=4 times that value. Answer: E.
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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25 Jun 2012, 04:29
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Hi,
Lets say, revenue in November = N was 2/5 of its revenue in December = D revenue in January = J As per given data, N = (2/5)D J = (1/4)N = 1/4 * 2/5 *D = (1/10)D
\(D = \frac {x(N+J)}2\), we need to find x,
\(D = \frac {x((2/5)D+(1/10)D}2\) x = 4,
Thus, Answer is (E).
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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26 Jun 2012, 10:11
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Use plug in value. Let Dec rev =100 Then Nov rev is 2/5 (100) => 40 Therefore Jan rev = 1/4(Nov rev) = 1/4(40) => 10
Hence Dec rev = x*( Nov rev+Jan rev)/2 100 = x* (40+10)/2 x = 100/25 => 4
Ans) E



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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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26 Jun 2012, 14:09
N = 2/5*D or D = 5/2N J= 1/4*N
avg (N+J) =( N+1/4*N)/ 2 =5/8*N so, avg(N+J)/D = 4
E



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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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29 Jun 2012, 03:05
SOLUTIONIf a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?(A) 1/4 (B) 1/2 (C) 2/3 (D) 2 (E) 4 Probably, for most people it would be easier to solve such kind of questions by picking numbers. Notice that January is linked with November and November is linked with December. So, we should pick some smart number for December's revenue. Since the denominators in the question are 5 and 4, then let say December's revenue was 20 (the least common multiple of 4 and 5). So, we have that: Revenue in December = 20; Revenue in November = 20*2/5 = 8; Revenue in January = 8*1/4 = 2; The average of 8 and 2 is 5, so the store's revenue in December was 20/5=4 times that value. Answer: E.
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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12 Jul 2012, 01:24



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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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06 Aug 2015, 07:16
Bunuel wrote: If a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January? (A) 1/4 (B) 1/2 (C) 2/3 (D) 2 (E) 4 Diagnostic Test Question: 8 Page: 21 Difficulty: 600 If N = 2/5 D and J = 1/4N, we should put all expressions into a common variable. Let's use D. D =D; N =2/5D; and J = 1/10D. Assuming D =10, we now have N = 4 and J = 1. The average of 4 and 1 is 2.5 10 is 4 times greater than 2.5 Answer E is correct
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If a certain toy store's revenue in November was 2/5 of its [#permalink]
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20 Nov 2015, 06:19
This can be solved even faster:
The revenue in November (N) is less than half the revenue in December (D). Therefore, December revenue is more than twice the November revenue: \(D>2N\)
The revenue in January (J) is even smaller than revenue in November. Consequently, the average of the two will be less than N: \(Average = (J+N)/2 < N\)
Therefore, December revenue will be far more than twice the average: \(D>2N>2*Average\)
The only option left is (E).



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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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29 Nov 2015, 11:50
Hi All, This question can be solved by TESTing VALUES. We're given information comparing revenue for 3 months: 1) November revenue = 2/5 of December revenue 2) January revenue = 1/4 of November revenue We're asked to compare the December revenue to the AVERAGE of November's and January's revenues (and we're asked how many times greater the December revenue is). With the two fractions involved, the common denominator is 20, but you can use any number you like. For example... December revenue = $100 November revenue = (2/5)(100) = $40 January revenue = (1/4)(40) = $10 The average of November and January is (40+10)/2 = 25 Since December revenue is 100, that is 4 TIMES the average of the other two months. Final Answer: GMAT assassins aren't born, they're made, Rich
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If a certain toy store's revenue in November was 2/5 of its [#permalink]
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28 Mar 2016, 17:57
Here is a visual that should help. Notice that I simplified the equation using algebra, and then I substituted numbers once the math was simplified. Trying to substitute numbers upfront can also work, but it often involves too much trial and error. Hence the hybrid technique. Please note that "(2d /10)" should have been written as "(2d / 20)."
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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25 Dec 2016, 22:35
Nice Official Question. Here is what i did on tis one =>
Method 1> December = x November =2x/5 January =x/10
Mean (J+N)= {2x/5+x/10}/2 => x/4
Hence x=p*x/4 p=4
Hence E
Method 2=> December = 10 November = 4 January => 1
Mean(J+N) = 2.5
Hence 10 = 4 times 2.5
Hence E
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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20 Dec 2017, 12:31
Bunuel wrote: If a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?
(A) 1/4 (B) 1/2 (C) 2/3 (D) 2 (E) 4 We can let n = revenue in November, d = revenue in December, and j = revenue in January. We can create the following equations: n = (2/5)d j = (1/4)n So, j = (1/4)(2/5)d = (1/10)d We now determine that the average revenue for November and January is: [(2/5)d + (1/10)d]/2 [(4/10)d + (1/10)d]/2 (5/10)d/2 = (1/4)d Thus, the revenue in December is d/[(1/4)d] = 4 times the average revenue for November and January. Answer: E
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]
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04 Jan 2018, 19:00
Kind of tricky... I glanced over the question and found the average for NOV, DEC and JAN instead of just NOV and JAN. I need to read all the way through the problem instead of taking a quick glance. Other than that little issue, the problem isn't difficult at all to solve, if you pick smart numbers.



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If a certain toy store's revenue in November was 2/5 of its [#permalink]
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17 Apr 2018, 11:39
Bunuel wrote: If a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January? (A) 1/4 (B) 1/2 (C) 2/3 (D) 2 (E) 4 Diagnostic Test Question: 8 Page: 21 Difficulty: 600 ok I got it revenue in december 100 revenue in november is 100 November was 2/5 of its revenue in December > N = 2/5*100 = 40 January was 1/4 of its revenue in November > J = 1/4 * 40 = 10 then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January? average (arithmetic mean) of its revenues in November and January > \(\frac{40+10}{2}\)\(= 25\) Average revenue 25, hence \(100/25\)\(= 4\) finally !




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