[dumbJoe]

1. E
the satement itself says no is of form 4y+3
option A says its an odd multiple of 3 and Op B says multiple of 5, combining all 15z+60 . no unique no

2.C
statement gives the sequence GYR (1,2,3...............)
op A says that either R or G at 18 so no solution but B says no R at 18. So gives a unique answer when put together.

but since the statement says about all the tiles in row, and only G has no precedence requirement. The pattern GYR should follow from the first tile and hence no need of any other statement to deduce answer, but it has no corresponding answer choice.

3.C
rearranging says if b<a(1- mod b)
option A and B put together say a,b are both non positive. which gives the answer.

4.C
i guess you need deviation from mean of all elements to get SD, op A and B combined gives the elements as well as deviations.

[padfoot]

Well here is where i bounce in !!! Will try to contribute as much as I can !! Well this was a spam to subscribe.


""" I DID'NT MEANT TO """ --- MODS PLEASE FORGIVE !!

[dumbJoe]

Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?
(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

[quizophobic]

My take (A) Stmt 1 is sufficient...

STarting with the fact that mode=7,
This implies that 7 occurs max times:: surely more than once..
Now if 7 appeared 4 times, the sequence in increasing order will be:
7,7,7,7,a,b OR a,7,7,7,7,b OR a,b,7,7,7,7 In none of the cases, we can hav median=6 as mentioned

So 7 occurs twice or thrice, and since 7> median the two/three occurances must be on right hand side of mid-point
Now if 7 were the greatest number, then all numbers have to be equal to 7 ( 'coz mean=7)
which is not the case...(median=6)

So 7 occurs twice and on 4th and 5th order in increasing sequence...
The series as of now stands at: a,b,c,7,7,d

Since median=6, => c=5
So the seuence becomes: a,b,5,7,7,d
Mean=7 => sum =42
a+b+5+7+7+d=42
or a+b+d=23 (3)

Now taking condition (1) (a+b) = (7+d)/5

Putting in(3) (7+d)/5 +d =23 This gives d=18...

So statement (1) is sufficient

(2) tells us what we already know.... So not sufficient

Hence answer = (A)

[quizophobic]

Bringing back a Q discussed som weeks back...

Originally Posted by Govi (GMAT Data Sufficiency Discussions)
Q. The integers m and p are such that 2<m<p and m is not a factor of p. If r is the remainder when p is divided by m, is r > 1 ?
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30

Originally Posted by nitiman

Is it D?
According to first statement and given data if m=n then p is also even and n+2=< p =<2n-2. In this case the remainder can be anything between 2 to 2n-2 which is always greater than 1.
According to sec statement LCM of m and p is 30. Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. m>2 So possible values of m and p are 3 and 10 or 5 and 6 and in both the cases remainder is 1.
So both statements are individually sufficient.

It must be (A) actually..
For (2), consider the pair 10 and 15...
It satisfies (2), but here r=5


Tell me if i'm getting this wrong..

[dumbJoe]

choice (A). We can learn a number of things about the set even before examining the statements. The fact that the mode of the set equals 7 tells us that we have at least two 7’s in the set. Since the median is 6, and there is an even number of numbers in the set, we know that numbers four and five must be 7. (If numbers five and six were 7, then three and four would be six, and 7 would no longer be the mode.) So since the fourth number is 7, the third number must be 5 in order for the median to be 6. Now the set looks like the following: {x, y, 5, 7, 7, z}. The mean of the set is 7 and the set has six numbers, therefore, we know the sum of the numbers is 42. This means that the remaining variables must equal 23, so x + y + z = 23. We also know that z – x = 16 (the range is the difference between the largest and smallest number). When we plug in the numbers to statement (1), we find that . Now we have three distinct equations and three variables, so statement (1) is sufficient. Because statement (1) is sufficient, we can eliminate choices (B), (C), and (E). We now have to choose between (A) and (D) so let’s look at statement (2). Statement (2) tells us that numbers three and four are 5 and 7 respectively, but we can already calculate those values given the information in the question. Therefore, we know that statement (2) is not sufficient, which rules out choice (D) and leaves us with the credited answer, (A).

[Bhola]

Hi,

This question is from OG10. My answer is different from OG10.Plz verify.

A. If r and s are positive integers, r is what percent of s ?
1. r= (3/4)s
2. r+s = 75/100


My answer is A while OG10 answer is D.

[dumbJoe]

coincidently, i did the same problem yesterday but in the second statement instead of a '+' sign there was a division sign, and the answer was 'd' , so i think it is a typo that has created the confusion

[Bhola]

IF it is / instead of +, then there is no point in discussing this question. :-)

Thanks for clarification....

[dumbJoe]

Three segments are drawn from opposite corners of a hexagon to form six triangles.These segments all bisect each other at point A. Are all of the triangles equilateral?

1. all six sides of hexagon are the same length
2. the three segments drawn between the opposite corners are equal length.