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# If a certain toy store's revenue in November was 2/5 of its

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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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45% (medium)

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60% (02:27) correct 40% (01:28) wrong based on 933 sessions

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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
[Reveal] Spoiler: OA

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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, 02:36
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SOLUTION

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

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.

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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
1
KUDOS
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
1
KUDOS
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
Expert's post
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SOLUTION

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

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.

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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]

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11 Jul 2012, 22:00
1
KUDOS
Bunuel,

The answer is perfect, but I think there is some confusion:-

You have written:-
Revenue in December = 20;
Revenue in November = 20*2/5 = 8;
Revenue in December = 8*1/4 = 2;

I think it should be:-
Revenue in December = 20;
Revenue in November = 20*2/5 = 8;
Revenue in January = 8*1/4 = 2;

Rgds
Rahul Goel
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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, 01:04
rggoel9 wrote:
Bunuel,

The answer is perfect, but I think there is some confusion:-

You have written:-
Revenue in December = 20;
Revenue in November = 20*2/5 = 8;
Revenue in December = 8*1/4 = 2;

I think it should be:-
Revenue in December = 20;
Revenue in November = 20*2/5 = 8;
Revenue in January = 8*1/4 = 2;

Rgds
Rahul Goel

Yes, I had two Decembers there. Edited. Thank you. +1.
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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
600 level.

if we translate can write: N= 2/5 D and J = 1/4 N

Now observing and substituting we obtaine N = 4J .......ravenues of November and January is 4. E is the answer.
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Re: If a certain toy store's revenue in November was 2/5 of its [#permalink]

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16 Jul 2014, 19:48
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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, 06:24
Hello from the GMAT Club BumpBot!

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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 08 Aug 2015 Posts: 8 GMAT 1: 770 Q49 V47 Followers: 0 Kudos [?]: 4 [0], given: 16 If a certain toy store's revenue in November was 2/5 of its [#permalink] ### Show Tags 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). EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 7727 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 343 Kudos [?]: 2288 [0], given: 162 Re: If a certain toy store's revenue in November was 2/5 of its [#permalink] ### Show Tags 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.

[Reveal] Spoiler:
E

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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 up-front 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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If a certain toy store's revenue in November was 2/5 of its   [#permalink] 28 Mar 2016, 17:57
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# If a certain toy store's revenue in November was 2/5 of its

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