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# Devil's Dozen!!!

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Joined: 05 May 2012
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14 Jul 2012, 21:30
1
Bunuel wrote:
3. If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: $$3x^2-x^4=y^4-4y^2$$ --> $$x^2(3-x^2)=y^2(y^2-4)$$. Notice that LHS is even for any value of $$x$$: if $$x$$ is odd then $$3-x^2=odd-odd=even$$ and if $$x$$ is even then the product is naturally even. So, $$y^2(y^2-4y)$$ is also even, but in order it to be even $$y$$ must be even, since if $$y$$ is odd then $$y^2(y^2-4y)=odd*(odd-even)=odd*odd=odd$$. Sufficient.

(2) y=4-x^2 --> if $$x=odd$$ then $$y=even-odd=odd$$ but if $$x=even$$ then $$y=even-even=even$$. Not sufficient.

Regarding statement (1), I understood how we got $$x^2(3-x^2)=y^2(y^2-4)$$. Later, we are examining $$y^2(y^2-4y)$$ - question: how did we get this extra 'y' here? Am getting a lot of DS questions wrong, would appreciate if you could explain. Thanks.
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15 Jul 2012, 05:41
lateapp wrote:
Bunuel wrote:
3. If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: $$3x^2-x^4=y^4-4y^2$$ --> $$x^2(3-x^2)=y^2(y^2-4)$$. Notice that LHS is even for any value of $$x$$: if $$x$$ is odd then $$3-x^2=odd-odd=even$$ and if $$x$$ is even then the product is naturally even. So, $$y^2(y^2-4y)$$ is also even, but in order it to be even $$y$$ must be even, since if $$y$$ is odd then $$y^2(y^2-4y)=odd*(odd-even)=odd*odd=odd$$. Sufficient.

(2) y=4-x^2 --> if $$x=odd$$ then $$y=even-odd=odd$$ but if $$x=even$$ then $$y=even-even=even$$. Not sufficient.

Regarding statement (1), I understood how we got $$x^2(3-x^2)=y^2(y^2-4)$$. Later, we are examining $$y^2(y^2-4y)$$ - question: how did we get this extra 'y' here? Am getting a lot of DS questions wrong, would appreciate if you could explain. Thanks.

Extra $$y$$ there is clearly a typo, it should be $$y^2(y^2-4)$$. Edited. Thank you.
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15 Dec 2012, 21:22
Bunnel can you please elaborate Q 5....Why x cannot take a value of 50

14x/25-14*2=28 less than 35...
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16 Dec 2012, 08:26
apoorvarora wrote:
Bunnel can you please elaborate Q 5....Why x cannot take a value of 50

14x/25-14*2=28 less than 35...

x is # of astronauts who do NOT listen to Bach, there are total of 35 astronauts, so how can x be 50>35?
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11 Feb 2013, 10:27
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

we could also conclude that there is also one viper snake??
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13 Feb 2013, 01:52
FTG wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

we could also conclude that there is also one viper snake??

No, that's not true. From the second statement we have that there must be one cobra in the box, but all we can say about vipers is that there must be at least one.

For example, there can be 1 cobra and 1 viper, or 1 cobra and 2 vipers, 1 cobra and 55 vipers... In all these cases from any two snakes from the box at least one will be viper.

Hope it's clear.
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13 Feb 2013, 21:08
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ then answer will be YES but for $$p=11$$ the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$ --> if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ then answer will be YES but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it can not be a factor of $$(n+1)(n+2)-1$$, thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Bunuel,
I did not understand the below:
thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$

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14 Feb 2013, 02:59
1
Sachin9 wrote:
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ then answer will be YES but for $$p=11$$ the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$ --> if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ then answer will be YES but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it can not be a factor of $$(n+1)(n+2)-1$$, thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Bunuel,
I did not understand the below:
thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$

We got that prime number p is NOT a factor of $$(n+1)(n+2)-1$$ but it IS a factor of $$n!*((n+1)(n+2)-1)$$, thus it must be a factor of n!.

For example, if we are told that 3 IS a factor of xy (where x and y are positive integers) and is NOT a factor of y, then it wold mean that 3 IS a factor of x.

Hope it's clear.
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09 Apr 2013, 09:46
1
Bunuel wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:
Attachment:
Vertigo.png
As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Sorry to disturb you Bunuel, but why haven't you considered in your matrix the case where people can have acrofobia but not arachnofobia? If you add a new incognite there (let's say "y"), first statement is insufficient and you would need the second one to solve the problem. Where is my argument failing?

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12 Apr 2013, 05:49
Recobita wrote:
Bunuel wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:
Attachment:
The attachment Vertigo.png is no longer available
As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Sorry to disturb you Bunuel, but why haven't you considered in your matrix the case where people can have acrofobia but not arachnofobia? If you add a new incognite there (let's say "y"), first statement is insufficient and you would need the second one to solve the problem. Where is my argument failing?

Double-set matrix has all cases possible. The case you are talking about is red box below:
Attachment:

Untitled.png [ 6.46 KiB | Viewed 2441 times ]
But we don't need it to get the answer.

Hope it's clear.
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11 May 2013, 15:02
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ then answer will be YES but for $$p=11$$ the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$ --> if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ then answer will be YES but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it can not be a factor of $$(n+1)(n+2)-1$$, thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Hi Bunuel,
I have a question.From 1),cant we have : as p is a factor of (n+2)!-n! so, n![(n+2)(n+1)-1].

.so p is a factor of n! as it is some value * n! ??

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12 May 2013, 02:58
up4gmat wrote:
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ then answer will be YES but for $$p=11$$ the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$ --> if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ then answer will be YES but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it can not be a factor of $$(n+1)(n+2)-1$$, thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Hi Bunuel,
I have a question.From 1),cant we have : as p is a factor of (n+2)!-n! so, n![(n+2)(n+1)-1].

.so p is a factor of n! as it is some value * n! ??

p is a factor of $$(n+2)!-n!=n!((n+1)(n+2)-1)$$ does not necessarily means that p is a factor of n! it could be a factor of another multiple: ((n+1)(n+2)-1). Check examples in (1) to verify.

Hope it helps.
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21 May 2013, 05:46
Bunuel wrote:
13. Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?

(1) Charlie gets to the trailer in 55 minutes. No info about Buster. Not sufficient.

(2) Buster gets to the studio at the same time as Charlie gets to the trailer --> Charlie needed 20 minutes less than Buster to cover the same distance, which means that the rate of Charlie is higher than that of Buster. Since after they pass each other they need the same time to get to their respective destinations (they get at the same time to their respective destinations) then Buster had less distance to cover ahead (at lower rate) than he had already covered (which would be covered by Charlie at higher rate). Sufficient.

Responding to a pm:

This is not a weighted average question since there is no 'average' speed to be considered. This is more apt for relative speed concepts though I would think in terms of ratio of speeds (who is faster and who is slower) since we don't need to give values - only whether he is closer to trailer or studio. So we only need to figure out their speeds relative to each other (more/less).
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24 May 2013, 02:38
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is$45 enough to buy 27 muffins and 27 brownies?

700+ question.

Given: $$11m+7b\leq{15}$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.
Question: is $$27m+27b\leq{45}$$? --> $$9m+9b\leq{15}$$. Question basically asks whether we can substitute 2 muffins with 2 brownies.

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.

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

Hi Bunnel,

I did it as below, please confirm if this method can be used in all the cases.
is 45=27m+27b
i.e. 5=3m+3b
1.
15>=11m+7b (given in question stem)
15>=7m+11b
30>=18m+18b
5>=6m+6b
sufficient

2.
15>=10m+8b
=> 8/3*(5=3m+3b)
=> 13.33=8m+8b
here we don't know for sure whether 1.7 will be sufficient to buy 2 muffins hence insufficient
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24 May 2013, 02:53
1
cumulonimbus wrote:
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is$45 enough to buy 27 muffins and 27 brownies?

700+ question.

Given: $$11m+7b\leq{15}$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.
Question: is $$27m+27b\leq{45}$$? --> $$9m+9b\leq{15}$$. Question basically asks whether we can substitute 2 muffins with 2 brownies.

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.

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

Hi Bunnel,

I did it as below, please confirm if this method can be used in all the cases.
is 45=27m+27b
i.e. 5=3m+3b
1.
15>=11m+7b (given in question stem)
15>=7m+11b
30>=18m+18b
5>=6m+6b
sufficient

2.
15>=10m+8b
=> 8/3*(5=3m+3b)
=> 13.33=8m+8b
here we don't know for sure whether 1.7 will be sufficient to buy 2 muffins hence insufficient

Yes, it's a valid approach, though the red parts are not correct.

The question asks whether $$27m+27b\leq{45}$$ ($$3m+3b\leq{5}$$) not whether $$27m+27b={45}$$ (should be $$\leq$$ instead of =).

Next, when reducing $$18m+18b\leq{30}$$ you get $$3m+3b\leq{5}$$ or $$6m+6b\leq{10}$$ not $$6m+6b\leq{5}$$.

Hope it helps.
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23 Jun 2013, 21:16
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order?

(1) Set A consists of 12 even consecutive integers;
(2) n=5.

We should understand following two things:
1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.

2. Now, consider that the subset selected is $$\{x_1, \ x_2, \ ..., \ x_n\}$$, where $$x_1<x_2<...<x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these n! ways only one, namely $$\{x_1, \ x_2, \ ..., \ x_n\}$$ will be in ascending order. So 1 out of n!. $$P=\frac{1}{n!}$$.

Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.

One doubt dear
Lets assume k=1 2 3 4 5 6 7 8 9 10
Now if first select the number 4 then the total number i.e. k is definitely going to affect the total probability
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23 Jun 2013, 22:04
Vinayprajapati wrote:
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order?

(1) Set A consists of 12 even consecutive integers;
(2) n=5.

We should understand following two things:
1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.

2. Now, consider that the subset selected is $$\{x_1, \ x_2, \ ..., \ x_n\}$$, where $$x_1<x_2<...<x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these n! ways only one, namely $$\{x_1, \ x_2, \ ..., \ x_n\}$$ will be in ascending order. So 1 out of n!. $$P=\frac{1}{n!}$$.

Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.

One doubt dear
Lets assume k=1 2 3 4 5 6 7 8 9 10
Now if first select the number 4 then the total number i.e. k is definitely going to affect the total probability

Great Question Bunuel!

As for your doubt Vinay, let me try to explain it. The question does not care about the number of ways of selecting n numbers from k numbers. It is only asking for the probability of selecting numbers in increasing order.
Probability of selecting numbers in increasing order + Probability of selecting numbers in some other order = 1

Say out of the 10 numbers, you select 3 numbers. 1, 5, and 7

You can select them in increasing order = 1, 5, 7 - total 1 way
You can select them in some other order = 1, 7, 5/ 5, 1, 7/5, 7, 1/7, 5, 1/7, 1, 5 - total 5 ways

Probability of selecting numbers in increasing order = 1/6
Probability of selecting numbers in some other order = 5/6

Notice that it doesn't matter how many numbers there are in the original list. All we care about is arranging the n selected numbers. One arrangement will be in increasing order and the others will be non-increasing.

A good tricky question!
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15 Jul 2013, 09:32
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is$45 enough to buy 27 muffins and 27 brownies?

700+ question.

Given: $$11m+7b\leq{15}$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.
Question: is $$27m+27b\leq{45}$$? --> $$9m+9b\leq{15}$$. Question basically asks whether we can substitute 2 muffins with 2 brownies.

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.

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

Hi Bunnel,

Thanks for updating such elaborate answers ,I have one confusion regarding the answer posted above,which I have marked in red.

If we can substitute one muffin with one brownie why can't we be sure to replace 2 brw with 2 muffins.

Regards,
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15 Jul 2013, 10:00
Countdown wrote:
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is$45 enough to buy 27 muffins and 27 brownies?

700+ question.

Given: $$11m+7b\leq{15}$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively.
Question: is $$27m+27b\leq{45}$$? --> $$9m+9b\leq{15}$$. Question basically asks whether we can substitute 2 muffins with 2 brownies.

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.

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

Hi Bunnel,

Thanks for updating such elaborate answers ,I have one confusion regarding the answer posted above,which I have marked in red.

If we can substitute one muffin with one brownie why can't we be sure to replace 2 brw with 2 muffins.

Regards,
Countdown

Say 11 muffins and 7 brownies cost $14 and the price of a muffin is less than the price of a brownie. 10 muffins and 8 brownies will cost more than$14 (since m<b).
9 muffins and 9 brownies will cost even more than that, so it could be more than $15. Hope it's clear. _________________ Manager Joined: 10 Mar 2014 Posts: 180 Re: Devil's Dozen!!! [#permalink] ### Show Tags 23 May 2014, 04:54 Bunuel wrote: 11. Alice has$15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies? 700+ question. Given: $$11m+7b\leq{15}$$, where $$m$$ and $$b$$ are prices of one muffin and one brownie respectively. Question: is $$27m+27b\leq{45}$$? --> $$9m+9b\leq{15}$$. Question basically asks whether we can substitute 2 muffins with 2 brownies. 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. But 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 (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 --> $$7m+11b\leq{15}$$: we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient.
(1) \$15 is enough to buy 10 muffins and 8 brownies --> $$10m+8b\leq{15}$$: we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

Hi Bunnel,

I just want to know why statement 2 is not sufficient. following is my logic

from question 11m+7b =15

st1- 7m+11b = 15

subtract st2 from 1

4m-4b = 0
m=b

so in question we can say 18b = 15
b=15/18=5/6. using this we can get the price for 27m and 27b

now same i can get from st2 then why this is not sufficient?

Thanks.
Re: Devil's Dozen!!!   [#permalink] 23 May 2014, 04:54

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