GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Nov 2019, 17:37 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Devil's Dozen!!!

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

60
225
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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: https://gmatclub.com/forum/devil-s-doze ... l#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: https://gmatclub.com/forum/devil-s-doze ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#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: http://gmatclub.com/forum/devil-s-dozen ... l#p1063897
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

27
56
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.

_________________
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

5
1
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
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

28
37
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?

Also very tricky.

(1) Of the astronauts who do NOT listen to Bach 56% are male --> if # of astronauts who do NOT listen to Bach is $$x$$ then $$0.56x$$ is # of males who do NOT listen to Bach. Notice that $$0.56x=\frac{14}{25}x$$ must be an integer. Hence x must be a multiple of 25: 25, 50, 75, ... But $$x$$ (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So $$x$$ can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient.

(2) Of the astronauts who listen to Bach 70% are female. Now, if we apply the same logic here we get that, if # of astronauts who listen to Bach is $$y$$ then $$0.7y$$ is # of females who listen to Bach: $$0.7y=\frac{7}{10}y$$ must be an integer. Hence it must be a multiple of 10, but in this case it can take more than 1 value: 10, 20, 30. So, this statement is not sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

17
33
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ 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. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: $$Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50$$. Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

13
18
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

13
12
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

12
25
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

11
20
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?

700 question.

Let the three integers be $$a$$, $$b$$, and $$c$$, where $$0<a<b<c$$. Given: $$abc=c^2$$ --> $$ab=c$$. Question: $$ab=c=?$$

(1) The average (arithmetic mean) of the three numbers is 34/3 --> $$a+b+c=34$$ --> $$a+b+ab=34$$ --> $$(a+1)(b+1)=35$$. Now, since $$a$$ and $$b$$ are integers, then $$a+1=5$$ and $$b+1=7$$. $$a=4$$ and $$b=6$$ --> $$ab=24$$. Sufficient. (Notice that $$a+1=1$$ and $$b+1=35$$ is not possible since in this case $$a=0$$ and we are told that all integers are positive).

(2) The largest number of the three distinct numbers is 24. Directly give the value of c. Sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

9
23
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

7
8
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

6
10
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 [ 7.34 KiB | Viewed 85393 times ]
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

5
13
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.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

### Show Tags

4
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.

(1) p is greater than 3. Clearly insufficient: different values of $$p$$ will give different values of the remainder.
(2) p is a prime. Also insufficient: if $$p=2$$ then the remainder is 4 but if $$p=3$$ then the remainder is 9.

(1)+(2) You can proceed with number plugging and try several prime numbers greater than 3 to see that the remainder will always be 1 (for example try $$p=5$$, $$p=7$$, $$p=11$$).

If you want to double-check this with algebra you should apply the following property of the prime number: any prime number greater than 3 can be expressed either as $$p=6n+1$$ or $$p=6n-1$$.

If $$p=6n+1$$ then $$p^2=36n^2+12n+1$$ which gives remainder 1 when divided by 12;

If $$p=6n-1$$ then $$p^2=36n^2-12n+1$$ which also gives remainder 1 when divided by 12.

_________________
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

3
1
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
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

2
1
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
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

2
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
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

2
1
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
Manager  Joined: 25 Aug 2011
Posts: 135
Location: India
GMAT 1: 730 Q49 V40 WE: Operations (Insurance)

### Show Tags

2
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.

IMO D.

each statement gives a linear eqn in 2 variables. since i just need the diffrenec between both and not the exact value for Each . I and II are individually sufficient Hence D
Intern  Joined: 25 Oct 2010
Posts: 1

### Show Tags

2
Thanks for welcoming, Bunuel..awaiting ur OAs n explanations for some of those tricky qns in the list.

@dyaffe55 - Yeah, thts rgt, i overlooked the given data for qn #5. Thanks for the explanation, it makes more sense.

There are 2 more wrong answers frm me other than this in the qn list, to my knwledge till thn..sry if thr is any more..
9. the answer is B (and not D, i mistyped the answer choice in my prev post)
10. guess its B (and not E..came across a similar qn post frm Bunuel..guess i should be more alert on the 'atleast' keyword usage in the qn stems, hereafter.) Re: Devil's Dozen!!!   [#permalink] 23 Mar 2012, 01:35

Go to page    1   2   3   4   5    Next  [ 91 posts ]

Display posts from previous: Sort by

# Devil's Dozen!!!   