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# M27-09

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Math Expert
Joined: 02 Sep 2009
Posts: 55732

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16 Sep 2014, 01:27
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Difficulty:

95% (hard)

Question Stats:

28% (01:35) correct 72% (02:10) wrong based on 239 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.

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16 Sep 2014, 01:27
4
4
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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24 Oct 2014, 18:04
5
2
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.

Math Expert
Joined: 02 Sep 2009
Posts: 55732

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 55732

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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 !
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18 Oct 2017, 07:25
1
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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25 Dec 2017, 05:02

we are given that 11M+7B<= 15.
We need to prove--> 27M+27B<= 45 --> 3M+3B<= 5 ?

Statement 1.
7M+11B<=15
+11M+7B<= 15
_______________
18M+18B<= 30 --> 3M+3B<= 5. Hence Sufficient.

I dont know how to go about statement 2. Below is how i tried.

Statement 2.
10M+8B<=15
+11M+7B<=15
______________
21M+15B<=30 --> 7M+5B<=10 --> 3.5M+2.5B<=5

I could only go this far. Can you please help me how further can i prove statement 2 is insufficient.

Regards
Math Expert
Joined: 02 Aug 2009
Posts: 7756

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25 Dec 2017, 05:50
sandysilva wrote:

we are given that 11M+7B<= 15.
We need to prove--> 27M+27B<= 45 --> 3M+3B<= 5 ?

Statement 1.
7M+11B<=15
+11M+7B<= 15
_______________
18M+18B<= 30 --> 3M+3B<= 5. Hence Sufficient.

I dont know how to go about statement 2. Below is how i tried.

Statement 2.
10M+8B<=15
+11M+7B<=15
______________
21M+15B<=30 --> 7M+5B<=10 --> 3.5M+2.5B<=5

I could only go this far. Can you please help me how further can i prove statement 2 is insufficient.

Regards

Hi..
Without breaking it down, I'll tell you Logic here..
10m+8b<=15
Also 11m+7b<=15
In both Number of m is more than b and total is 18
So if all 18 items are so cheap that they all are for $1, you can buy 15 times given number. Also if they just equal$15 and price of m is more than b..
So when you replace m with b in 10m+8b to make it 9 of each, it is possible

But if b is more than m and there combined price equals exactly 15, change of B to m will take it above 15, hence not possible..
So II is insufficient

In statement I, it gives you more b than m and main statement gives you more m than b, so equal number of both is possible
Hence sufficient

A
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04 Jun 2018, 04:57
perfect question!
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28 Jun 2018, 10:42
Bunuel wrote:

Yes, you can do this way too.

Hi Bunuel,

Was your above comment an oversight or is there anything peculiar going on in this question?

To amariappan, you were ok with the approach but to shohm you pointed out the mistake when both essentially are the same thing which subtracting inequalities when both signs points in same direction. Even I used to do it. Just bringing it up to know the correct approach or if there if any thing special in this question which allows the former to work out.

Needless to say, brilliant question!

Thanks as always.
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28 Jun 2018, 12:10
1
1
I got this problem.
Given: 11m + 7b <= 15
Is 9m + 9b <= 15?

(1) 7m + 11b <= 15
Subtract from Given: 4m - 4b <= 0
m <= b
The result is less than equal to original
Sufficient

(1) 10m + 8b <= 15
Subtract from Given: m - b <= 0; same m <= b
The result is greater than or equal to original
9m + 9b >= 15
Insufficient

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07 Oct 2018, 21:32

Hello experts,

As there are at least 2 cases for each option <15 and =15; so this one becomes quite complicated to understand.
Can anyone please provide any examples to explain the logic already shared multiple times?

I understood the pure algebra way as shown by amariappan but dont quite fully understand the logic.

Thank you!
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Posts: 9340
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08 Oct 2018, 01:28
1
Bunuel wrote:
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.

There are two variables here: cost of a muffin (M) and cost of a brownie (B)

Given:
11M + 7B <= 15 ........(I)

Is 27M + 27B <= 45 ?
Is M + B <= 5/3 ?

Often, in such cases, we try to algebraically manipulate the given equations/inequalities to get the asked equation/inequality.
Say multiply both sides of a given inequality by 3 to see if we get the asked inequality. Before we move on to the statements, note that the given inequality talks about 11 muffins and 7 brownies. But in the asked inequality number of muffins and brownies is the same. So what we need is information about the cost of same proportion of muffins and brownies as asked.

(1) $15 is enough to buy 7 muffins and 11 brownies. 7M + 11B <= 15 ...... (II) The interesting thing about this inequality is that it complements the given inequality. Together, they will give us cost info on equal number of muffins and brownies. (I) + (II) 18M + 18B <= 30 M + B <= 5/3 Answer "Yes". Sufficient (2)$15 is enough to buy 10 muffins and 8 brownies.
10M + 8B <= 15
When added to (I), it will give us
21M + 15B <= 30
We still cannot say anything about the price when muffins and brownies are in equal proportion.
Not sufficient.

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27 Nov 2018, 01:15
Thanks VeritasKarishma for the alternate solution. That's helpful. However, I'm still struggling to understand the proportion at play.

Official solution says:
Now, if m>b, we can easily substitute 2 muffins with 2 brownies (since 2m will be more than 2b). But if m<b we won't know this for sure.

The first part I clearly understand. What I dont know understand is the latter half : "But if m<b we won't know this for sure.".
If m = b or if m is > than b then of course, we can replace 2m with 2b. But if m is less than b, then isn't it obvious that can't replace 2m with 2b because 2m is lesser than 2b. I know I'm missing something but just can't understand what.

Re: M27-09   [#permalink] 27 Nov 2018, 01:15
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# M27-09

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