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# M27-09

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Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135178 [0], given: 12671

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16 Sep 2014, 01:27
Expert's post
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00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:34) correct 70% (03:51) wrong based on 40 sessions

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Alice has $15, which is enough to buy 11 muffins and 7 brownies, is$45 enough to buy 27 muffins and 27 brownies?

(1) $15 is enough to buy 7 muffins and 11 brownies. (2)$15 is enough to buy 10 muffins and 8 brownies.
[Reveal] Spoiler: OA

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Kudos [?]: 135178 [0], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135178 [2], given: 12671

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16 Sep 2014, 01:27
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Official Solution:

This is a 700+ question.

Given: $$11m+7b \le 15$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.

Question: is $$27m+27b \le 45$$? Reduce by 3: $$9m+9b \le 15$$. The question basically asks whether we can substitute 2 muffins with 2 brownies.

Now, if $$m \gt b$$, we can easily substitute 2 muffins with 2 brownies (since $$2m$$ will be more than $$2b$$). But if $$m \lt b$$ we won't know this for sure.

But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases ($$m \gt b$$ or $$m \lt b$$) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies. $$7m+11b \le 15$$: we can substitute 4 muffins with 4 brownies, so according to the above, we can surely substitute 2 muffins with 2 brownies. Sufficient. (1)$15 is enough to buy 10 muffins and 8 brownies. $$10m+8b \le 15$$: we can substitute 1 muffin with 1 brownie, so according to the above, this does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

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Kudos [?]: 135178 [2], given: 12671

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Joined: 22 Jul 2013
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Location: United States
Concentration: Technology, Entrepreneurship
Schools: IIM A '15
GMAT 1: 650 Q46 V34
GMAT 2: 720 Q49 V38
GPA: 3.67
WE: Engineering (Non-Profit and Government)

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24 Oct 2014, 18:04
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I solved it as below,

Muffin : M; Brownie: B

11M + 7B <= 15

Question 27(M+B)<=45 or M+B <= 5/3

Stmt 1:

7M+11B <=15 --1

Given 11M + 7B <= 15 --2

Add equation 1 and 2 we get 18M+18B <=30 or M+B <= 5/3 (Enough)

Stmt 2:

10M+8B<=15 -- 3

Adding stmt 2 and 3 we get 21M+15B<=30; or
7M+5B <= 10; --- 4

Now if M+B<= 5/3
then 7M+ 7B <= 35/3 --5; subtract equation 4 from equation 5 we get 2B<= 5/3 which we do not know so Only A is the answer. Bunuel Am I correct? Please advice

Bunuel wrote:
Official Solution:

This is a 700+ question.

Given: $$11m+7b \le 15$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.

Question: is $$27m+27b \le 45$$? Reduce by 3: $$9m+9b \le 15$$. The question basically asks whether we can substitute 2 muffins with 2 brownies.

Now, if $$m \gt b$$, we can easily substitute 2 muffins with 2 brownies (since $$2m$$ will be more than $$2b$$). But if $$m \lt b$$ we won't know this for sure.

But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases ($$m \gt b$$ or $$m \lt b$$) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies. $$7m+11b \le 15$$: we can substitute 4 muffins with 4 brownies, so according to the above, we can surely substitute 2 muffins with 2 brownies. Sufficient. (1)$15 is enough to buy 10 muffins and 8 brownies. $$10m+8b \le 15$$: we can substitute 1 muffin with 1 brownie, so according to the above, this does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

Kudos [?]: 20 [2], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135178 [0], given: 12671

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25 Oct 2014, 06:09
amariappan wrote:
I solved it as below,

Muffin : M; Brownie: B

11M + 7B <= 15

Question 27(M+B)<=45 or M+B <= 5/3

Stmt 1:

7M+11B <=15 --1

Given 11M + 7B <= 15 --2

Add equation 1 and 2 we get 18M+18B <=30 or M+B <= 5/3 (Enough)

Stmt 2:

10M+8B<=15 -- 3

Adding stmt 2 and 3 we get 21M+15B<=30; or
7M+5B <= 10; --- 4

Now if M+B<= 5/3
then 7M+ 7B <= 35/3 --5; subtract equation 4 from equation 5 we get 2B<= 5/3 which we do not know so Only A is the answer. Bunuel Am I correct? Please advice

Bunuel wrote:
Official Solution:

This is a 700+ question.

Given: $$11m+7b \le 15$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.

Question: is $$27m+27b \le 45$$? Reduce by 3: $$9m+9b \le 15$$. The question basically asks whether we can substitute 2 muffins with 2 brownies.

Now, if $$m \gt b$$, we can easily substitute 2 muffins with 2 brownies (since $$2m$$ will be more than $$2b$$). But if $$m \lt b$$ we won't know this for sure.

But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases ($$m \gt b$$ or $$m \lt b$$) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies. $$7m+11b \le 15$$: we can substitute 4 muffins with 4 brownies, so according to the above, we can surely substitute 2 muffins with 2 brownies. Sufficient. (1)$15 is enough to buy 10 muffins and 8 brownies. $$10m+8b \le 15$$: we can substitute 1 muffin with 1 brownie, so according to the above, this does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

Yes, you can do this way too.
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22 Jul 2016, 11:28
I still cant figure this question out

2M can be replaced by 2B =Sufficient
1M can be replaced by 1B =insufficient

The question seems to not have asked for either but for sufficiency. I don't know what am missing

9M+ 9B = 15
7M+11B = 15 2M-2B=0

9M+ 9B = 15
10M+8B = 15 -M+B=0

I still see both being independently able to provide a yes/no answer
Can someone further explain
Thanks

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20 Dec 2016, 00:43
The explanation makes no sense to me.

Out of:
11m+7b≤15
10m+8b≤15
we get m-b≤0, or m≤b

Therefore 11m+7b≤18b≤15
Or, m≤b≤5/6
Then, 9m+9b≤9*5/6*2, or 9m+9b≤15

Therefore, statement 2 is sufficient alone and answer should be D.
Looks like there's a mistake in official solution.

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Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135178 [0], given: 12671

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20 Dec 2016, 00:47
Nikotino wrote:
The explanation makes no sense to me.

Out of:
11m+7b≤15
10m+8b≤15
we get m-b≤0, or m≤b

Therefore 11m+7b≤18b≤15
Or, m≤b≤5/6
Then, 9m+9b≤9*5/6*2, or 9m+9b≤15

Therefore, statement 2 is sufficient alone and answer should be D.
Looks like there's a mistake in official solution.

No. There is no mistake. The question is quite hard and it seems that you did not understand it. Please check the discussion of this question here: devil-s-dozen-129312-20.html

Hope it helps.
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21 Dec 2016, 06:35
Bunuel wrote:

Stmt 2:

10M+8B<=15 -- 3

Adding stmt 2 and 3 we get 21M+15B<=30; or
7M+5B <= 10; --- 4

Now if M+B<= 5/3
then 7M+ 7B <= 35/3 --5; subtract equation 4 from equation 5 we get 2B<= 5/3 which we do not know so Only A is the answer. Bunuel Am I correct? Please advice

Bunuel wrote:
Official Solution:

This is a 700+ question.

Given: $$11m+7b \le 15$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.

Question: is $$27m+27b \le 45$$? Reduce by 3: $$9m+9b \le 15$$. The question basically asks whether we can substitute 2 muffins with 2 brownies.

Now, if $$m \gt b$$, we can easily substitute 2 muffins with 2 brownies (since $$2m$$ will be more than $$2b$$). But if $$m \lt b$$ we won't know this for sure.

But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases ($$m \gt b$$ or $$m \lt b$$) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies. $$7m+11b \le 15$$: we can substitute 4 muffins with 4 brownies, so according to the above, we can surely substitute 2 muffins with 2 brownies. Sufficient. (1)$15 is enough to buy 10 muffins and 8 brownies. $$10m+8b \le 15$$: we can substitute 1 muffin with 1 brownie, so according to the above, this does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

Yes, you can do this way too.[/quote]

Dear Bunuel,

I think there is a mistake. we can't subtract 2 inequalities in same direction. Am I right or miss something?

Thanks

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18 Oct 2017, 07:09
Wondering why statement 2 is not sufficient :
(1) Original condition : 11m + 7b <= 15
(2) Statement 2: 10m + 8b <= 15

(1) + (2) gives 21m + 15 b <=30 OR 7m +5b <=10

Now, (2) - [(1) + (2)] gives 3m + 3b <=5 OR 27m + 27b <=45. Sufficient !

Kudos [?]: [0], given: 66

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135178 [0], given: 12671

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18 Oct 2017, 07:25
shohm wrote:
Wondering why statement 2 is not sufficient :
(1) Original condition : 11m + 7b <= 15
(2) Statement 2: 10m + 8b <= 15

(1) + (2) gives 21m + 15 b <=30 OR 7m +5b <=10

Now,(2) - [(1) + (2)] gives 3m + 3b <=5 OR 27m + 27b <=45. Sufficient !

You cannot subtract inequalities that way.

1. You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

2. You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Check for more the links below:
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Re: M27-09   [#permalink] 18 Oct 2017, 07:25
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# M27-09

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