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Spring has passed, summer is gone and winter is here. And the song that I meant to sing remains unsung. For I have spent my days stringing and un-stringing my instrument.

Last edited by walker on 16 Jul 2013, 19:18, edited 3 times in total.

From Statement 1: a + b + c + d = 240 c + d = 190 a+ b = 50

We can have a situation where c < 50 say c = 40 and d = 150.

Here the no of integers less than 50 = 3.

Or we can have c > 50 c = 90

Here the no of integers less than 50 = 2.

Not Sufficient

IMO the answer should be B
_________________

Spring has passed, summer is gone and winter is here. And the song that I meant to sing remains unsung. For I have spent my days stringing and un-stringing my instrument.

Last edited by AbhiJ on 08 Sep 2012, 22:09, edited 1 time in total.

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190.

(2) The median of the four integers is 50.

The answer should be C. Explanation: From the main statement, it is given that a+b+c+d = 240, now if you look at the statement (1) , it says that c+d = 190 ( assuming that c and d are two largest integers out of 4 positive integers. which also means that a+b = 240-190 = 50 from statment (2), it says that median of the four integers is 50. which means that (b+c)/2 = 50 ( assuming that a<b<c<d or assume that b and c are the middle two integers. so, it means b+c = 100 Now if you combine both statements, you will find that a+b = 50 and b+c = 100 now, if a+b = 50 so the largest value of b = 49 and if so then then c > 50 because b+c = 100. and d of course would be greater than 50 from the statement c+d = 190. So, we can deduce that both statements together are sufficient to answer ( ans : c & d) this question. So, the answer will be C.

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If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190. Means that two of the numbers are greator than 60. avg be 95. so we need to have two nos. whose avg should be 25. so rest two no. should always be less than 50. therefore it give answer to above question. (2) The median of the four integers is 50.

Median is 50 means middle two no. avg is 50. so one of them should be greator than 50 other less than 50. so from rest two no. one no. should be less than 50 and other more than 50. therefore this also gives the answer.

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190.

(2) The median of the four integers is 50.

THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW:

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive> integers is 4*60=240.

Say four integers are a, b, c and d so that 0<a<b<c<d. So, we have that a+b+c+d=240.

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of \{b,c,d\} is 51 means that c=51. Now, if b=50, then only a, will be less than 50, but if b<50, then both a and b, will be less than 50. But we are also given that c+d=190. Substitute this value in the above equation: a+b+190=240, which boils down to a+b=50. Now, since given that all integers are positive then both a and b must be less than 50. Sufficient.

(2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so median=\frac{b+c}{2}=50. Since given that b<c then b<50<c, so both a and b are less than 50. Sufficient.

Re: If the average of four distinct positive integers is 60 [#permalink]
09 Sep 2012, 10:23

Hi Bunuel,

Isn’t the first data in Statement (1) of the problem (median 51) redundant? Because, even if we just know that the sum of the two largest integers is 190, we can answer the question – that there are 2 integers less than 50.

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190.

(2) The median of the four integers is 50.

THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW:

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive> integers is 4*60=240.

Say four integers are a, b, c and d so that 0<a<b<c<d. So, we have that a+b+c+d=240.

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of \{b,c,d\} is 51 means that c=51. Now, if b=50, then only a, will be less than 50, but if b<50, then both a and b, will be less than 50. But we are also given that c+d=190. Substitute this value in the above equation: a+b+190=240, which boils down to a+b=50. Now, since given that all integers are positive then both a and b must be less than 50. Sufficient.

(2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so median=\frac{b+c}{2}=50. Since given that b<c then b<50<c, so both a and b are less than 50. Sufficient.

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190.

(2) The median of the four integers is 50.

THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW:

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive> integers is 4*60=240.

Say four integers are a, b, c and d so that 0<a<b<c<d. So, we have that a+b+c+d=240.

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of \{b,c,d\} is 51 means that c=51. Now, if b=50, then only a, will be less than 50, but if b<50, then both a and b, will be less than 50. But we are also given that c+d=190. Substitute this value in the above equation: a+b+190=240, which boils down to a+b=50. Now, since given that all integers are positive then both a and b must be less than 50. Sufficient.

(2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so median=\frac{b+c}{2}=50. Since given that b<c then b<50<c, so both a and b are less than 50. Sufficient.

Answer: D.

(1) The median of the three largest integers is 51 - where did this come from? I don't see it in the question...
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If the average of four distinct positive integers is 60 [#permalink]
09 Sep 2012, 21:24

Expert's post

I think there are two versions of this question:

Version 1:

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The sum of two largest integers is 190.

(2) The median of the four integers is 50.

Here the answer is (B)

Here statement 1 is not sufficient. Let me show by taking 2 cases: Case 1: Numbers are 24, 26, 30, 160 Case 2: Numbers are 20, 30, 90, 100 Number of numbers which are less than 50 are different in these two cases. So statement 1 alone is not sufficient.

Version 2: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.

(2) The median of the four integers is 50.

Answer (D) Here statement 1 is sufficient too. Since median of the three largest integers is 51, the middle of the three largest integers must be 51. So the numbers are <51, <51, 51, >51 Since the sum of the two largest numbers is 190, the largest number must be 190 - 51 = 139 i.e. 79 more than the average of 60. Since 51 is 9 less than 60, we need to balance out another 70 in the two smallest numbers to get an average of 60. The numbers must be all distinct. Can one of the two smallest numbers be equal to 50? No! If it is equal to 50, it will balance out 10 of the 70 and we will need to balance out 60 from the smallest number. That will make the smallest number = 0 but all numbers must be positive. So, the two smallest numbers must be less than 50.