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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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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I am not sure if I understand correctly.

There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) so we need both k & n. Am I missing something here?

There are two examples given which give two different answers to the question whether p is a factor of n!.

(1) p is a factor of (n+2)!-n!

If \(n=2\) and \(p=2\) (notice that in this case \((n+2)!-n!=22\) and \(p=2\) is a factor of 22), then since \(p=2\) is a factor of \(n!=2!\) the answer to the question is YES;

If \(n=2\) and \(p=11\) (notice that in this case \((n+2)!-n!=22\) and \(p=11\) is a factor of 22), then since \(p=11\) is NOT a factor of \(n!=2!\) the answer to the question is NO.

Two different answers, hence not sufficient.
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Answer to the question is B, not E.

If there is some other snake, then group of {other, cobra} will be possible and the statement (2) will be violated (basically the possibility of other variety of snake is ruled out by the same logic as the possibility of second cobra).

Hope it's clear.
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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.
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Again: you count the probability of selecting 5 numbers out of 12 or 24, whereas the question asks about the probability of selecting some group of numbers in a certain ORDER. You just interpret the question in a wrong way.
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Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.

But:
Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.



Yes, first two terms in \(p^2=36n^2+12n+1\) are divisible by 12 so leave no remainder and 1 divided by 12 yields reminder of 1.

Hope it helps.
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Again, the answer would be the same 1/n!=1/3!=1/6:
1, 2, 3;
1, 2, 4;
1, 2, 5;
1, 2, 6;
1, 2, 7;
1, 3, 4;
1, 3, 5;
1, 3, 6;
1, 3, 7;
1, 4, 5;
1, 4, 6;
1, 4, 7;
1, 5, 6;
1, 5, 7;
1, 6, 7;
2, 3, 4;
2, 3, 5;
2, 3, 6;
2, 3, 7;
2, 4, 5;
2, 4, 6;
2, 4, 7;
2, 5, 6;
2, 5, 7;
2, 6, 7;
3, 4, 5;
3, 4, 6;
3, 4, 7;
3, 5, 6;
3, 5, 7;
3, 6, 7;
4, 5, 6;
4, 5, 7;
4, 6, 7;
5, 6, 7.

35 ways (naturally) to select 3 numbers out of 7 in ascending order. Total # of ways to select 3 numbers out of 7 when order matters is \(P^3_7=210\). Hence the probability is 35/210=1/6.

Hope it's clear.
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You cannot just substitute \(\leq\) sign with \(=\) sign and solve. Those are two very different signs.

To see that (2) is not sufficient, consider two cases:
A. Muffins and brownies are free - answer YES;
B. Muffins are free and each brownie costs $1.8 - answer NO.

Hope it's clear.
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