GMAT Data Sufficiency Discussions

MBA GMAT
2673
sorry .. misread the question .. as there is no number of terms in the question , we need median as well

so OA is C)

Well now i think your previous reply was . correct. just think of any set of numbers of "say positive" ( thats needed) even integers with range 42 you should have same SD ? isnt it
2674

only A is sufficient

2675
gmatgmat40 Says
Well now i think your previous reply was . correct. just think of any set of numbers of "say positive" ( thats needed) even integers with range 42 you should have same SD ? isnt it


C is the correct answer IMO

For caculating SD we NEED the no of elements in the set, that is why we need both statements.

For eg,
E1= {20,22,24,26} & E2={18,20,22,24,26,28}, both have median as 23.

E1= {0,2,4,.....38,40,42} & E2={2,4,6,...,38,40,42,44}, both have a range of 42.

But when we combine both, we get the Set,
E = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44} or a similar set E' = {0,2,4..........42}
With a median of 23 and Range of 42. And we can calculate the SD.
2676

X and Y are single digit positive interers.

IS X+Y multiple of 6 ?

1. X+4Y is even
2. YX ( y is tenth digit and x is ones digit) is divisible by 3


Solution : second statement is alone insufficient.....

for X+Y=6x i.e 6,12, 18

Using Hit and trial method, if X=9, Y=3
for (1) X+4Y is not even.
(2) YX = 39, multiple of 3
I tried other combinations also , all satifying.

2677

Can i have ans for below question ?
In the fraction x/y, where x and y are positive integers, what is the value of y?
(1) The LCM of x/y and 1/3 is 6. (2) x = 1

2678

Ans: 1 alone is sufficient..

Solution: as LCM of fractions = LCM of numerator/GCF of denominators

given: LCM of x/y & 1/3 is 6
hence, x=6 & y=1


P.S : Anybody please correct , if its wrong.

2679
IS N= 1 ?

1 . N| = 4N-6
2. |N
help

(C) both together are sufficient.

as 1 gives two values of N, and 2) tells that N is positive.
2680
E is a set of consecutive even integers.what is there standard deviations

1. median is 23
2. range is 42.

Help

My answer is "C" . what you guys say ?

My answer : C

as (1) gives whether the number of terms are even or odd....also the mid terms
and (2) range gives the number of terms. hence the whole list.
2681
X and Y are single digit positive interers.

IS X+Y multiple of 6 ?

1. X+4Y is even
2. YX ( y is tenth digit and x is ones digit) is divisible by 3


Solution : second statement is alone insufficient.....

for X+Y=6x i.e 6,12, 18

Using Hit and trial method, if X=9, Y=3
for (1) X+4Y is not even.
(2) YX = 39, multiple of 3
I tried other combinations also , all satifying.



What about 15 , it is divisible by 6 !!:splat:
2682
Is sqrt ( (X-2)^2 ) = 2 - X ....... 1) -XX >0 ......... 2) 2 - X > 0
2683
(C) both together are sufficient.

as 1 gives two values of N, and 2) tells that N is positive.


I guess 1 alone is sufficient . Only one value 2 satisfies this condition.
2684
abhi5m Says
I guess 1 alone is sufficient . Only one value 2 satisfies this condition.

(1) will be required to get the middle term & tells that there are even number of terms (as median is odd)
(2) will give the extreme terms, from which whole sequence can be found.
hence all the parameters to calculate S.D can be get from both statements together.

Correct me, where I am wrong.
2685
gmatgmat40 >0 ......... 2) 2 - X > 0

Both are individually sufficient.
2686
Of 200 members of a certain association each member who speaks German also speaks English and 70 of the
members speak only Spanish. If no members speak all 3 languages; how many members speak 2 of 3
languages?
A. 60 of the members speak only English
B. 20 of the members do not speak any of the three languages.
2687

(C) both statements are required.

2688
Of 200 members of a certain association each member who speaks German also speaks English and 70 of the

members speak only Spanish. If no members speak all 3 languages; how many members speak 2 of 3
languages?
A. 60 of the members speak only English
B. 20 of the members do not speak any of the three languages.


Will go with C

G, E, C, three sets,
G = x (all of these speak E and since no one speaks all three, G is fixed at x)
S = 70 +y(people who speak both, S and E)
E = x+y+z(people who speak only E)

We need to find x+y

Stmt 1 -> z = 60, Not sufficient since we do not know if any of the speakers speak none of the languages

Stmt 2-> 20 = 200 - (G+E+S)
or, G+E+S = 180,
x+70+y+x+y+z = 180
or, 2(x+y)+z = 110
Not sufficient, we do not know z

Combine both, we get the value of x+y. Sufficient.
2689

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

2690
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



This is a very awkwardly worded question, I will give it a shot,
took me almost 3-4 mins to figure it out.

i) 2

GCF of m and p is 2
Plug in nos with GCF 2 and lets see what happens
m p r
4 6 2
6 8 2
6 16 4
8 18 2
4 10 2
Always, r >1, Sufficient

Stmt 2-> LCM of m and p is 30
Again, lets plug in,
m p r
5 6 1
3 10 1
10 15 5
Just in 3 cases, we can say, Not Sufficient

Thus the answer should be A.
2691

Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

2692
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.


This is what we get from the stem: (see image 1)
Veg Non Veg Total
Student
Non Student 15
Total t/2 t/2 t

Stmt 1 -> Veg Students: Veg NonStudents :: 2:3
And NonVeg Students: NonVeg NonStudents :: 4:3 (double the rate)
Put this in the table, we get (See image 2)

Veg Non Veg Total
Student 2x 4y
Non Student 3x 15=3y
Total t/2 t/2 t

or, 7y = t/2
y = t/14

and we know, 3y =15, so t = 70

Thus this statement is sufficient.

Stmt2 -> Veg NonStudents = 3t/10 (30%)

See image 3

Veg Non Veg Total
Student t/5
Non Student 3t/10 15
Total t/2 t/2 t

But we cannot calculate any further, so this Statement is not sufficient.

Thus the answer is A.