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Devil's Dozen!!!

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Devil's Dozen!!! [#permalink] New post   19 Mar 2012, 06:54
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Welcome to the "Devil's Dozen", a compilation of challenging data sufficiency questions designed to push your problem-solving skills to the limit. Ready to take on the challenge?

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) 32 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 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.


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.


12. If x>0 and xy=z, what is the value of yz?

(1) \(x^2*y=3\).
(2) \(\sqrt{x*y^2}=3\).


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.

The solutions can be found below on this page.
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:03
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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 \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient.

(2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\).

We simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\), then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient.

(1)+(2) Simplifying the equation from (1) gives \((n+2)!-n!=n!((n+1)(n+2)-1)\). 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, so the only common factor they share is 1. As from (2), \(p\) is a factor of \((n+1)(n+2)\), it cannot be a factor of \((n+1)(n+2)-1\). Thus, for \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), it should be a factor of the first multiple of this expression: \(n!\). Sufficient.


Answer: C
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:34
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5. If at least one astronaut does NOT listen to Bach at the Solaris space station, then how many of the 35 astronauts at the station listen to Bach?

This is a tricky question.

(1) Of the astronauts who do NOT listen to Bach, 56% are male.

If we denote the number of astronauts who do NOT listen to Bach as \(x\), then the number of male astronauts who do NOT listen to Bach can be represented as \(0.56x\). It is important to note that \(0.56x\) can be expressed as \(\frac{14}{25}x\), and it needs to be an integer. Consequently, \(x\) must be a multiple of 25, such as 25, 50, 75, and so on. Additionally, since \(x\) cannot exceed the total number of astronauts, which is 35, the only feasible value for \(x\) is 25. Therefore, the number of astronauts who do listen to Bach is given by subtracting \(x\) from the total number of astronauts: \(35 - 25 = 10\). Sufficient..

(2) Of the astronauts who listen to Bach, 70% are female.

Applying the same logic as before, let's denote the number of astronauts who listen to Bach as \(y\). In this case, the number of female astronauts who listen to Bach can be expressed as \(0.7y\), which is equivalent to \(\frac{7}{10}y\). To ensure this value is an integer, \(y\) must be a multiple of 10. However, unlike the previous scenario, there are multiple possible values that satisfy this condition: 10, 20, and 30. Therefore, this statement alone is not sufficient to determine the exact number of astronauts who listen to Bach.


Answer: A
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:38
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6. Is the perimeter of a triangle with sides \(a\), \(b\), and \(c\) greater than 30 centimeters ?

750+ Level Question.

(1) \(a-b=15\).

Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

Hence, we have \(a+b > c\) (since the length of any side of a triangle must be less than the sum of the other two sides) and \(c > 15\) (since the length of any side of a triangle must be greater than the positive difference of the other two sides). These imply \(a+b > c > 15\). As both \(a + b\) and \(c\) exceed 15, it follows that the perimeter, \(a+b+c\), will be greater than \(15+15=30\). Sufficient.

(2) The area of the triangle is 50 centimeters .

For a given perimeter, an equilateral triangle has the largest area. Now, if the perimeter were equal to 30 centimeters, then it would have the largest area if it were equilateral.

Let's find out what this area would be: \(Area=s^2*\frac{\sqrt{3} }{4}=(\frac{30}{3})^2*\frac{\sqrt{3} }{4}=25*\sqrt{3} < 50\). Since even an equilateral triangle with a perimeter of 30 cannot have an area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:41
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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?

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

It is important to note two things:

1. The probability of selecting any subset of \(n\) numbers from the set is the same. No particular subset of \(n\) numbers is favored over any other subset of \(n\) numbers, so they all have an equal probability of being selected.

2. Any subset of \(n\) distinct numbers can be selected in \(n!\) ways, and only one out of these \(n!\) possible ways will be in ascending order. For example, subset of {1, 2, 3} can be selected in 6 ways: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) or (3, 2, 1). As you can see, out of these \(3!\) possible ways, only one is in ascending order, (1, 2, 3). Therefore, the probability of selecting a subset in ascending order is \(\frac{1}{n!}\).

Thus, to answer the given question, we only need to know the size of the subset (\(n\)) that we are selecting from the set \(A\). Therefore, the firs statement is not sufficient, while the second one is.


Answer: B
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:05
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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, \(y^2(y^2-4)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\). 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.
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:52
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9. The product of three distinct positive integers equals the square of the largest of these three numbers. What is the product of the two smaller numbers?

This is a 750 level question.

Let's denote the three integers as \(a\), \(b\), and \(c\), with \(0 < a < b < c\). Given: \(abc=c^2\), thus, \(ab=c\). The question is: what is the value of \(ab=c\)?

(1) The average (arithmetic mean) of the three numbers is \(\frac{34}{3}\).

This essentially states that \(a+b+c=34\). Replace \(c\) with \(ab\) to yield \(a+b+ab=34\). From this, we can infer that \((a+1)(b+1)=35\). Given \(a\) and \(b\) are integers, this implies \(a+1=5\) and \(b+1=7\), leading to \(a=4\) and \(b=6\). Therefore, \(ab=24\). This is sufficient. (Note that the scenario where \(a+1=1\) and \(b+1=35\) isn't valid, as it results in \(a=0\), contradicting the condition that all integers are positive).

(2) The largest of the three numbers is 24.

This statement directly provides the value of \(c\), which is sufficient.


Answer: D
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:55
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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?

This is a 700+ level question.

Given: \(11m+7b \le 15\), where \(m\) and \(b\) are the prices of one muffin and one brownie, respectively.

Question: Is \(27m+27b \le 45\)? Reduce by 3: \(9m+9b \le 15\). The question essentially asks whether we can substitute 2 muffins with 2 brownies.

If \(m \gt b\), we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m \lt b\), we won't know this for sure.

However, consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m \gt b\) or \(m \lt b\)), it would mean that we can substitute 2 (fewer than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies.

\(7m+11b \le 15\): We can substitute 4 muffins with 4 brownies, so according to the above, we can certainly substitute 2 muffins with 2 brownies. Sufficient.

(1) $15 is enough to buy 10 muffins and 8 brownies.

\(10m+8b \le 15\): We can substitute 1 muffin with 1 brownie, so according to the above, this does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.


Answer: A
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:53
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10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

This question is challenging and requires logical reasoning. Once you understand the logic behind it, the question becomes simple. However, if you have never encountered this type of problem before, it can be difficult to solve. If you made an error, it's important to review the solution and understand the reasoning, as it can be a valuable tool for similar problems on actual test.

(1) From any two snakes from Pandora's box at least one is a viper

Statement (1) states that from ANY two snakes in the box, at least one is a viper. This means that there cannot be two or more cobras in the box, because if there were, then we could choose two cobras and that would contradict the statement. Therefore, there must be at most one cobra in the box. And since we also know that there is at least one cobra in the box, there must be exactly one cobra. So, statement (1) is sufficient to answer the question.

(2) The total number of snakes Pandora's box is 99. Clearly insufficient.


Answer: A
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:33
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4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

This is one of the trickiest questions in our question pool, and it is crucial to approach it with great care and attention to detail when reading the solution.

(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:



• "Both" and "Neither" are both equal to \(x\);

• Green box \(= 45 - x\);

• Blue box \(= (45 - x) + x = 45\);

• Yellow box \(= 58 - 45 = 13\).

As you can see, the number of patients who have acrophobia is 13. Sufficient.

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


Answer: A
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:57
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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.

Since Charlie was able to cover the same distance as Buster in 20 minutes less time, it can be inferred that Charlie walks at a faster rate than 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). Therefore, when they pass each other, Buster will be closer to the studio (destination point) than to the trailer. Sufficient.


Answer: B
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Re: Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:02
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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.
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Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:50
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8. If \(p\) is a positive integer, what is the remainder when \(p^2\) is divided by 12?

(1) \(p\) is greater than 3.

This is clearly insufficient as different values of \(p\) yield different remainders.

(2) \(p\) is a prime.

This too is insufficient. For instance, if \(p=2\), the remainder is 4, whereas if \(p=3\), the remainder is 9.

(1)+(2) By plugging in several prime numbers greater than 3, one can observe that the remainder is consistently 1 (for example, using \(p=5\), \(p=7\), or \(p=11\)).

To further verify this using algebra, consider the property of prime numbers: any prime number greater than 3 can be represented as either \(p=6n+1\) or \(p=6n-1\).

For \(p=6n+1\), \(p^2\) becomes \(36n^2+12n+1\), which gives a remainder of 1 when divided by 12.

For \(p=6n-1\), \(p^2\) is \(36n^2-12n+1\), which also results in a remainder of 1 upon division by 12.

Answer: C.
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Devil's Dozen!!! [#permalink] New post   23 Mar 2012, 03:56
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12. If \(x \gt 0\) and \(xy = z\), what is the value of \(yz\)?

(1) \(x^2*y=3\).

If \(x=1\), then \(y=z=3\), yielding \(yz=9\). However, if \(x=3\), then \(y=\frac{1}{3}\), \(z=1\), and \(yz=\frac{1}{3}\). Not sufficient.

(2) \(\sqrt{x*y^2}=3\).

Squaring this equation, we get: \(x*y^2=9\), which can be rewritten as: \((xy)*y=9\). Since \(xy=z\), this implies: \(z*y=9\). Sufficient.

Answer: B.
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:03
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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.

let the 3 numbers be a<b<c. abc=c^2. therefore,c(c-ab)=0.. since a,b,c are positive int. c !=0.. therfore c-ab=0. Hence c=ab

from statement2 : ab =24
statement 1 NS

IMO B.
Pleas correct if wrong
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:05
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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.

from statement 1, we get diffrent remainders for different values of p NS
statement 2 : all primes greater than 3 give 1 as remainder hence sufficient but less than 3 has diff remainders

IMO C
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:08
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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?
(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
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:15
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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.

statement I since a-b = 15. the third side c should be greater than 15. for convenience I set this as 15.1 and b= 0.1 therefore a = 15.1 and a+b+c = 30.2 >30 hence Suff
Statement II :NS

IMo A
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:18
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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.

IMO E statements together are not suff
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Re: Devil's Dozen!!! [#permalink] New post   20 Mar 2012, 02:23
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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. Clearly insufficient.

On solving statement I : 13 ppl have acrophobia hence, suff.

IMO A
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