GMAT Data Sufficiency Discussions

Some good DS questions i came across...

1. When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.


2. In a single row of yellow, green and red colored tiles, every red tile is preceded immediately by a yellow tile and every yellow tile is preceded immediately by a green tile. What color is the 24th tile in the row?
(1) The 18th tile in the row is not yellow.
(2) The 19th tile in the row is not green.


3. If a and b are integers, and a > |b|, is a b a - b?
(1) a (2) ab 0




4. What is the standard deviation of Q, a set of consecutive integers?
(1)Q has 21 members.
(2)The median value of set Q is 20

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

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 !!

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

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)

Bringing back a Q discussed som weeks back...

Originally Posted by Govi View Post
Q. The integers m and p are such that 2

1 ?
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30


Is it D?
According to first statement and given data if m=n then p is also even and n+2=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..

:
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 7s 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 lets 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).

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.


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

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

Thanks for clarification....

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.

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.

For all the triangles to be equilateral, all the sides need to be of equal length since the triangles will have common sides. Hence we can answer it using (1) alone.

However, (2) does not ensure that the triangles will be equilateral. The three segments can be any three diameters of a circle and can be drawn in such a way that the sides are not equal.

Hence answer is (1) alone.
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.


Ans. Statement 1 is alone Sufficient to answer dis question...So A

answer is C, thats the way hexagons behave :satisfie:

dumbJoe Says
answer is C, thats the way hexagons behave :satisfie:


Thanks for such a useful information................
For all the triangles to be equilateral, all the sides need to be of equal length since the triangles will have common sides. Hence we can answer it using (1) alone.

However, (2) does not ensure that the triangles will be equilateral. The three segments can be any three diameters of a circle and can be drawn in such a way that the sides are not equal.

Hence answer is (1) alone.


answer is C, thats the way hexagons behave :satisfie:


I disagree. Here is the definition of hexagon - "A polygon of six sides." Nowhere does the question mention a "regular hexagon". Without this consideration, there is no point to the question. Using the explanation I have given in previous post, we can draw a "hexagon" i.e. "a polygon with six sides" in which the triangles are not equilateral.
mealiennothuman Says
I disagree. Here is the definition of hexagon - "A polygon of six sides." Nowhere does the question mention a "regular hexagon". Without this consideration, there is no point to the question. Using the explanation I have given in previous post, we can draw a "hexagon" i.e. "a polygon with six sides" in which the triangles are not equilateral.



point taken, but who do you disagree with ?


i am sorry for not giving a proper explanation. We can draw a non regular hexagon with equal sides by choosing an axis of symmetry passiong through 2 oppostite vertices and then pulling the opposite side closer or farther apart. we would have 3 sets of equal angles, opposite ones. For the second statement imagine three intersecting lines converguing at a single point which happens to be midpoint of them all but sustaining non 60 degree angles among them. Again a hexagon is possible by joining the endpoints of these intersecting lines.
Now coming to a regular hexagon, not only the angles are each 120degree but the figure has absolute symmetry hence both these statements are required and are enough to say that the figure would a regular hexagon and then only we can get equilateral trianlgles.
dumbJoe Says
point taken, but who do you disagree with ?

I disagree that the answer is (C). The answer is (A).

i hope the explanation resolves your contention

point taken, but who do you disagree with ?
i am sorry for not giving a proper explanation.....


I actually started the post saying that I disagree with you but then I re-read the post and have now understood the point. :-P Hmm... This is indeed a good question. Lots of points to think. For those who havent been reading this from start, let me put it all together.

Question:
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.

Analysis:
Definition of a hexagon: A polygon with six sides. The point here to note is that even a polygon with unequal sides is called a hexagon.

Hence (2) alone is definitely not enough to answer the question as we can draw a hexagon with unequal sides which satisfies statement (2). But in this case, the triangles formed will be isoceles and not equilateral.

Now coming to (1). Statement (1) implies that the hexagon has equal sides. However, having equal sides does not necessarily imply that the hexagon is a regular hexagon. We can very well draw a stretched hexagon having equal sides. In this case, the triangles will be scalane. Hence (1) alone is also not enough.

Putting the two together gives us a figure which has to be a regular hexagon, in which case the triangles will be equilateral. Hence answer is (C).