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Math Expert V
Joined: 02 Sep 2009
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12. If x>0 and xy=z, what is the value of yz?

(1) $$x^2*y=3$$. If $$x=1$$ then $$y=z=3$$ and $$yz=9$$ but if $$x=3$$ then $$y=\frac{1}{3}$$, $$z=1$$ and $$yz=\frac{1}{3}$$. Not sufficient.

(2) $$\sqrt{x*y^2}=3$$ --> $$x*y^2=9$$ --> $$(xy)*y=9$$ --> since $$xy=z$$ then: $$z*y=9$$. Sufficient.

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Math Expert V
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shivamayam 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 bunuel - i am unable to find the gap in my logic. Can you pls help?

Given: 11m+7b=15 -- (1) (let's disregard the inequality for the moment)

St1: 7m+11b=15 --(2)
(1)less(2): 4m-4b=0 --> m=b --> substituting m for b in (1), we get 11m+7m=15 --> m=5/6.
We could substitute 5/6 for m and b in the question to verify whether the equation holds good. So St1 is suff.

St2: 10m+8b=15 --(3)
(1)less(3) gives me m= 5/6. Similar to the above, we could substitute 5/6 for m and b in the question to verify it's validity. So St2 is also suff.

So shouldn't the answer be D?

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. _________________ Manager  Joined: 25 Aug 2011 Posts: 135 Location: India GMAT 1: 730 Q49 V40 WE: Operations (Insurance) Re: Devil's Dozen!!! [#permalink] ### Show Tags 1 2 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. I didnt solve this one .. but from statement 1 and II since i know the n and K both I can calculate the probability..IMo C Please correct if wrong Intern  Joined: 21 Feb 2012 Posts: 10 Re: Devil's Dozen!!! [#permalink] ### Show Tags 1 Bunuel wrote: The next set of tough and tricky DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck! 1. Jules and Jim both invested certain amount of money in bond M for one year, which pays for 12% simple interest annually. If no other investment were made, then Jules initial investment in bond M was how many dollars more than Jim's investment in bond M. (1) In one year Jules earned$24 more than Jim from bond M.
(2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M. 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! (2) p is a factor of (n+2)!/n! 3. If x and y are integers, is y an even integer? (1) 4y^2+3x^2=x^4+y^4 (2) y=4-x^2 4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia? (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. (2) 13 patients of Vertigo Hospital have arachnophobia but not acrophobia. 5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach? (1) Of the astronauts who do NOT listen to Bach 56% are male. (2) Of the astronauts who listen to Bach 70% are female. 6. Is the perimeter of triangle with the sides a, b and c greater than 30? (1) a-b=15. (2) The area of the triangle is 50. 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. 8. If p is a positive integer, what is the remainder when p^2 is divided by 12? (1) p is greater than 3. (2) p is a prime. 9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers? (1) The average (arithmetic mean) of the three numbers is 34/3. (2) The largest number of the three distinct numbers is 24. 10. There is at least one viper and at least one cobra in a Pandora's box. How many cobras are there? (1) There are total 99 snakes in Pandora's box. (2) From any two snakes from the a Pandora's box at least one is a viper. 11. 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.
(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 bunuel - i am unable to find the gap in my logic. Can you pls help? Given: 11m+7b=15 -- (1) (let's disregard the inequality for the moment) St1: 7m+11b=15 --(2) (1)less(2): 4m-4b=0 --> m=b --> substituting m for b in (1), we get 11m+7m=15 --> m=5/6. We could substitute 5/6 for m and b in the question to verify whether the equation holds good. So St1 is suff. St2: 10m+8b=15 --(3) (1)less(3) gives me m= 5/6. Similar to the above, we could substitute 5/6 for m and b in the question to verify it's validity. So St2 is also suff. So shouldn't the answer be D? Intern  Joined: 05 May 2012 Posts: 2 Re: Devil's Dozen!!! [#permalink] ### Show Tags 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. 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. Math Expert V Joined: 02 Sep 2009 Posts: 59264 Re: Devil's Dozen!!! [#permalink] ### Show Tags 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. 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. . 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. _________________ Intern  Joined: 11 Mar 2013 Posts: 5 Re: Devil's Dozen!!! [#permalink] ### Show Tags 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. 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 Math Expert V Joined: 02 Sep 2009 Posts: 59264 Re: Devil's Dozen!!! [#permalink] ### Show Tags 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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santorasantu wrote:
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.

Sant

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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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?[/b]

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

Hi,

The distinct number selected can have ONLY one way in which they are in ascending order or for that matter, one way in descending order.
1)Since n <=k and we are selecting n integers, selection can be n! Ways..

2) out of this, only one will be in ascending order.

So PROBABILITY is 1/n!

So we are looking for value of n..
Statement II gives us value of n, hence sufficient

Ans B
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gmat800live wrote:
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 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.

When we consider two statements together then n cannot be 1 because p cannot be a prime which is simultaneous a factor of 5 and 6.
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