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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
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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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
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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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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shohm
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 INEQUALITIES

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:
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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sandysilva
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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perfect question!
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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
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Can someone explain why?

"However, consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (m>b or m<b), it would mean that we can substitute 2 (fewer than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (more than 3) muffins with 4 brownies."


Bunuel
Official Solution:


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

This is a 700+ level question.

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

Question: Is \(27m+27b \le 45\)? Reduce by 3: \(9m+9b \le 15\). The question essentially asks whether we can substitute 2 muffins with 2 brownies.

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.

However, 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 (fewer than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (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 certainly 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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Shubham2599jain
Can someone explain why?

"However, consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (m>b or m<b), it would mean that we can substitute 2 (fewer than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (more than 3) muffins with 4 brownies."




If we can substitute 3 muffins for 3 brownies, then if muffins cost more, we can obviously substitute 2 muffins for 2 brownies because 2 brownies will cost less than 2 muffins. If muffins cost less, the difference between 3 muffins and 3 brownies will be greater than the difference between 2 muffins and 2 brownies, so if we can substitute 3 for 3, we can also substitute 2 for 2.

However, if we can substitute 3 muffins for 3 brownies, and muffins cost more, then we can also substitute 4 muffins for 4 brownies because 4 brownies will cost less than 4 muffins. But if muffins cost less, the difference between 3 muffins and 3 brownies will be smaller than the difference between 4 muffins and 4 brownies, so we cannot be sure that we can substitute 4 for 4.

Hope it's clear.
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The biggest trick here is to realise the inequality and that 2nd equation gives M<=B. I made the mistake assuming M=B here and thus 2nd being solvable because of it.

I think it is a good practise to maintain the inequality from the start to the very end so one does not mistake the value assumptions towards the end if they have to make it.
Bunuel
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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