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Re M2709 [#permalink]
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16 Sep 2014, 01:27
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. Answer: A
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Re: M2709 [#permalink]
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24 Oct 2014, 18:04
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.
Answer: A



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Re: M2709 [#permalink]
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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.
Answer: A Yes, you can do this way too.
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Re: M2709 [#permalink]
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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 2M2B=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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Re: M2709 [#permalink]
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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 mb≤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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Re: M2709 [#permalink]
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20 Dec 2016, 00:47



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Re: M2709 [#permalink]
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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.
Answer: A 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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Re M2709 [#permalink]
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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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Re: M2709 [#permalink]
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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. ADDING/SUBTRACTING INEQUALITIES1. 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Check for more the links below: Inequalities Made Easy!
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Hi Bunuel. Can you please help me know where am I falling short ?
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



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Re: M2709 [#permalink]
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25 Dec 2017, 05:50
sandysilva wrote: Hi Bunuel. Can you please help me know where am I falling short ?
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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Re: M2709 [#permalink]
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04 Jun 2018, 04:57
perfect question!



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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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Re: M2709 [#permalink]
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28 Jun 2018, 12:10
I got this problem. Given: 11m + 7b <= 15 Is 9m + 9b <= 15?
(1) 7m + 11b <= 15 Subtract from Given: 4m  4b <= 0 m <= b Subtract 2b and add 2m The result is less than equal to original Sufficient
(1) 10m + 8b <= 15 Subtract from Given: m  b <= 0; same m <= b Subtract m and add b The result is greater than or equal to original 9m + 9b >= 15 Insufficient
Answer A










