Devil's Dozen!!!

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.

Bunnel can you please elaborate Q 5....Why x cannot take a value of 50


14x/25-14*2=28 less than 35...

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.

Answer: B.

we could also conclude that there is also one viper snake??

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. .

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: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

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.

Hi Bunuel,
I have a question.From 1),cant we have : as p is a factor of (n+2)!-n! so, n![(n+2)(n+1)-1].

.so p is a factor of n! as it is some value * n! ??

Please clarify.

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.

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 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
add both equations:
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

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.

Answer: B.

One doubt dear
Lets assume k=1 2 3 4 5 6 7 8 9 10
Now if first select the number 4 then the total number i.e. k is definitely going to affect the total probability
Please clear.

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 Bunnel,

Thanks for updating such elaborate answers ,I have one confusion regarding the answer posted above,which I have marked in red.

If we can substitute one muffin with one brownie why can't we be sure to replace 2 brw with 2 muffins.


Thanks in advance.

Regards,
Countdown

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.