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# Devil's Dozen!!!

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19 Mar 2012, 06:54
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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.

Solution: devil-s-dozen-129312.html#p1063846

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!

Solution: devil-s-dozen-129312.html#p1063847

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

Solution: devil-s-dozen-129312.html#p1063848

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) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia.

Solution: devil-s-dozen-129312.html#p1063863

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.

Solution: devil-s-dozen-129312.html#p1063867

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.

Solution: devil-s-dozen-129312.html#p1063871

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.

Solution: devil-s-dozen-129312-20.html#p1063874

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.

Solution: devil-s-dozen-129312-20.html#p1063884

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.

Solution: devil-s-dozen-129312-20.html#p1063886

10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
(1) There are total 99 snakes in Pandora's box.
(2) From any two snakes from Pandora's box at least one is a viper.

Solution: devil-s-dozen-129312-20.html#p1063888

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. (2)$15 is enough to buy 10 muffins and 8 brownies.

Solution: devil-s-dozen-129312-20.html#p1063892

12. If x>0 and xy=z, what is the value of yz?
(1) $$x^2*y=3$$.
(2) $$\sqrt{x*y^2}=3$$.

Solution: devil-s-dozen-129312-20.html#p1063894

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.
(2) Buster gets to the studio at the same time as Charlie gets to the trailer.

Solution: devil-s-dozen-129312-20.html#p1063897
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26 Mar 2012, 09:43
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Bunuel wrote:
kuttingchai 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.

I didnot understand why 1 is not sufficient, can you elaborate on that ?

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.

Thank you - Bunuel that was easy to understand
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28 Mar 2012, 07:39
I will go with option A as the answer. Its pretty straight forward
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28 Mar 2012, 08:40
amitbehera wrote:
I will go with option A as the answer. Its pretty straight forward

Welcome to GMAT Club.

Which question are you talking about? Anyway, links to the OA's with solutions are given in this post: devil-s-dozen-129312.html#p1060761

Hope it helps.
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04 Apr 2012, 13:43
Bunuel wrote:
SOLUTIONS:

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.

Question: $$x-y=?$$

(1) In one year Jules earned $24 more than Jim from bond M. $$0.12x-0.12y=24$$ --> $$0.12(x-y)=24$$ --> $$x-y=200$$. Sufficient. (2) If the interest were 20% then in one year Jules would have earned$40 more than Jim from bond M. Basically the same type of information as above: $$0.2x-0.2y=40$$ --> $$0.2(x-y)=40$$ --> $$x-y=200$$. Sufficient.

Important note when two quantities are increased (decreased) by the same percent their difference also increase (decrease) by the same percent. For example if you increase 100 and 150 by 20% to 120 and 180 respectively, then their difference will also increase by the same 20% from 50 to 60.

Hey Bunuel,

I am bit confused on how to interpret statement 1.

"In one year Jules earned $24 more than Jim from bond M." Does it mean : $$0.12x-0.12y=24$$ Accounting only for the return on the investment? or $$1.12x-1.12y=24$$ Accounting for the initial principal of the investment? In this question, I was lucky that its didn't matter. We could solve for x-y in both situations. But I can see how reading the question your way could have lead me to the wrong answer. Bunuel, your incredibly gifted dude! Math Expert Joined: 02 Sep 2009 Posts: 34475 Followers: 6287 Kudos [?]: 79762 [1] , given: 10022 Re: Devil's Dozen!!! [#permalink] ### Show Tags 04 Apr 2012, 14:09 1 This post received KUDOS Expert's post alphabeta1234 wrote: Bunuel wrote: SOLUTIONS: 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. Question: $$x-y=?$$ (1) In one year Jules earned$24 more than Jim from bond M. $$0.12x-0.12y=24$$ --> $$0.12(x-y)=24$$ --> $$x-y=200$$. Sufficient.

(2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M. Basically the same type of information as above: $$0.2x-0.2y=40$$ --> $$0.2(x-y)=40$$ --> $$x-y=200$$. Sufficient. Answer: D. Important note when two quantities are increased (decreased) by the same percent their difference also increase (decrease) by the same percent. For example if you increase 100 and 150 by 20% to 120 and 180 respectively, then their difference will also increase by the same 20% from 50 to 60. Hey Bunuel, I am bit confused on how to interpret statement 1. "In one year Jules earned$24 more than Jim from bond M."

Does it mean :

$$0.12x-0.12y=24$$ Accounting only for the return on the investment?

or

$$1.12x-1.12y=24$$ Accounting for the initial principal of the investment?

In this question, I was lucky that its didn't matter. We could solve for x-y in both situations. But I can see how reading the question your way could have lead me to the wrong answer.

(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? Math Expert Joined: 02 Sep 2009 Posts: 34475 Followers: 6287 Kudos [?]: 79762 [1] , given: 10022 Re: Devil's Dozen!!! [#permalink] ### Show Tags 27 Jun 2012, 01:58 1 This post received KUDOS Expert's post 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. 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? 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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03 Jul 2012, 03:39
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.

I don't understand the logic of the following explanation -

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.

I see that you used the above logic to solve the two statements.....I don't get this.
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08 Jul 2012, 18:27
Hi Bunnel,

I didn't really understand the logic that you used for solving the muffin and brownies question, more specifically I have posted my question as above.
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14 Jul 2012, 03:21
for the first question i see either of the options individually giving the shares of the bond...

is that wrong???
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14 Jul 2012, 03:23
mohan514 wrote:
for the first question i see either of the options individually giving the shares of the bond...

is that wrong???

OA for the question #1 is D.

Links to the OA's with solutions are given in this post: devil-s-dozen-129312.html#p1060761
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14 Jul 2012, 21:30
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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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