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Bunuel
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Devil's Dozen!!! 14 Feb 2013, 02:59
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Sachin9
Bunuel
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.

Answer: C.

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: \(n!\)

Please clarify. .
Show more

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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Devil's Dozen!!! 09 Apr 2013, 09:46
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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.

Answer: A.
Show more


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?

Thank you in advance
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Devil's Dozen!!! 24 May 2013, 02:53
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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.

Answer: A.


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
add both equations:
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
Show more

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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Devil's Dozen!!! 07 Nov 2014, 04:28
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Dear Bunuel,

I have a question to your solution on the following problem:

2. If n is a positive integer and p is a prime number, is p a factor of n!?

you specify for statement 2 the following sentence:

(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.

now consider (n+2)!/n!, this can be reduced to (n+1)(n+2), for n is odd, the expression is even, further for n is even, the expression is even again.
the statement specifies that p is a factor of an even number and is prime, thus p should be even and the only even prime number is 2. How did you consider p= 3? in your solution.

Thanks in advance,
Sant
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The red part is not correct. An even number can have an odd factor. For example, 12 is even and it has an odd factor 3.
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Devil's Dozen!!! 27 Oct 2018, 14:08
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Hi Bunuel! I am a bit confused for the case n=1.

If n=1 then statement one equates to 5 and statement 2 equates to 6. Hence the possible values of P are 5,2,3. none of which is a factor of 1?

If each statement is true, then P should then be a prime factor of 5 and either 2 or 3, but that's impossible?

Will be grateful if you could clarify. Thank you.

Bunuel
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.

Answer: C.
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Devil's Dozen!!! 27 Jun 2020, 11:43
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Bunuel
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.

Answer: C.



Hi Bunuel,
As per 1, the following would be true
n+2!-n!= P*(some number)

Simplifying this we get-

n!(n2+3n+1)=p*(some number). The expression n2+3n+1 is an integer, so isn't p a factor of n!

Thanks,
Show more

Why can't p be a factor so n^2 + 3n + 1 instead ? There are two cases given in the solution, in one p is a factor of n! and in another p is not a factor of n! (it's a factor of n^2 + 3n + 1).
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Devil's Dozen!!! 23 Mar 2012, 18:44
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Bunuel
10. There is at least one viper and at least one cobra in a 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 the a 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.

Answer: B.
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How do we know that there are no other varieties of snakes other than cobra and viper. This is precisely why I chose E.
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Devil's Dozen!!! 24 Mar 2012, 08:41
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Bunuel
mithun2vrs
I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls

I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order.
Show more

Sorry to keep on with the problem,

Probability is defined as No of fav outcomes / No of Total possible outcomes

With your example,
in both cases No of favorable outcomes is 1 and in 1st case No of total possible outcomes is 12c5 and in the second case No of total possible outcomes is 24c5. Pls correct me where I am going wrong.
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Devil's Dozen!!! 15 Jul 2012, 05:41
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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.

Answer: A.

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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Extra \(y\) there is clearly a typo, it should be \(y^2(y^2-4)\). Edited. Thank you.
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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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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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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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Bunuel
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.

Answer: B.
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we could also conclude that there is also one viper snake??
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Devil's Dozen!!! 13 Feb 2013, 01:52
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Bunuel
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.

Answer: B.


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

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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Bunuel
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.

Answer: C.
Show more

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: \(n!\)

Please clarify. .
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Bunuel
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.

Answer: A.


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?

Thank you in advance
Show more

Double-set matrix has all cases possible. The case you are talking about is red box below:
Attachment:
Untitled.png
Untitled.png [ 6.46 KiB | Viewed 5719 times ]
But we don't need it to get the answer.

Hope it's clear.
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Bunuel
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.

Answer: C.
Show more

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! ??

Please clarify.
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Bunuel
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.

Answer: C.

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! ??

Please clarify.
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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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Devil's Dozen!!! 21 May 2013, 05:46
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Bunuel
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.

Answer: B.
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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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Bunuel
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.

Answer: A.
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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
add both equations:
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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Bunuel
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.

Answer: B.
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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
Please clear.
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