GMAT Data Sufficiency Discussions

MBA GMAT
2343
dude, integers are numbers without any fractonal/decimal part .. Includes the counting numbers {1,2,3,}, zero {0}, and the negative of the counting numbers {-1, -2, -3, }
so while taking i=j=-9 , the sum becomes -18 which is an ineteger .. since u have already got the concept why statement 2 was correct for postive nos. i believe your doubts are cleared now ..


Thanks for the explanation, actually one of my teachers said that -ve numbers are not even or odd.....thats y i got ocnfused.
2344

Is xy = x^2

(1) x + y (2) xy

2346
ankitgarg20 Says
Is the ans D?


OA is B.

Hard to comprehend 😞
2347
Is xy = x^2

(1) x + y (2) xy

Ans: B
Need to prove if x=y .
1st statement: x+y -5,-6 or -5,1
2nd statement:xy Both are of opposite signs . -3,6 OR 1,-2
" So from statement2 you can say that x!=y" hence answer can be drawn from statement2

*was confused with options before. Editing done
2349
Is xy = x^2

(1) x + y (2) xy
ankitgarg20 Says
Is the ans D?

OA is B.

Hard to comprehend :(


I think I got the method here.

xy = x^2
> x = 0 or x=y

1) x+yx can be 0 and y -ve, but nothing concrete can be said

2) xy> x and y have opposite signs
so x=y cannot be possible
x=0 cannot be possible

Hence B.

Silly me!
2350
ankitgarg20 Says
Well statement 2 is definitely sufficient to answer the question. I found statement 1 to be equally sufficient to answer the question and thus picked D. Is there no Explanation provided with the question?


Can you explain how you solved this?
2351

Is the value of a prime number?

(1) 3a b = 1
(2) b is a non-zero integer.

2352
Is the value of a prime number?

(1) 3a b = 1
(2) b is a non-zero integer.


Simplying the eqn we get ,

3 ^( 3a-b) / 2^b = ?

Statement 1: 3a-b =1. The eqn will become a prime no only when b= 0. For other values of b,the eqn cannot become a prime number.So this is insuff

statement 2: b is a non-zero integer. When we look at the modified eqn, numerator is always odd and denominator is always even except b=0.So when b is non-zero, eqn cannot turn out be a prime number.So this statement is suff.


I will go with option B


-Deepak.
2355
Is the value of a prime number?

(1) 3a b = 1
(2) b is a non-zero integer.


nothing can be said, oA pls
2356
Is the value of a prime number?

(1) 3a b = 1
(2) b is a non-zero integer.

Simplying the eqn we get ,

3 ^( 3a-b) / 2^b = ?

Statement 1: 3a-b =1. The eqn will become a prime no only when b= 0. For other values of b,the eqn cannot become a prime number.So this is insuff

statement 2: b is a non-zero integer. When we look at the modified eqn, numerator is always odd and denominator is always even except b=0.So when b is non-zero, eqn cannot turn out be a prime number.So this statement is suff.


I will go with option B


-Deepak.

I agree to ans as B, but do not agree with the underlined explanation above.

When b is non zero, the expression can never turn out to be a Prime number. A fraction with an odd numerator and an even denominator can never solve to give you a prime factor. This is because an odd number is not completely divisible by any even number.

asbhat Says
nothing can be said, oA pls



OA is C

Lets consider the modified expression:
3^(3a-b)/2^b

1) 3a-b = 1
> 3/2^b
Can be prime if b=0. Else not prime. Insuff

2) b is non-zero integer
> 3^(3a-b)/2^b
If b = +ve > not prime
If b = -ve > not prime
However, if b = -1 > can be prime iff 3a-b=0
Hence insuff.

Take both. A and B.
> Not prime.
Hence suff.
2357

If a, b, and c are real numbers, is ?

(1) a > c
(2) ac > b2

2358
If a, b, and c are real numbers, is ?

(1) a > c
(2) ac > b2



A =is a/b > b /c?


Statement 1:
===========
b = 3 ; a= 4 , c =1 . A is False.
b = 2 ; a = 4, c = 2 . A is true.

So statement 1 is insuff.

Statement 2:
===========
b = 2 ; a = 4 , c =2 . A is true .
b = 3 ; a = -2 ; c =-5 .A is false.

So statement 2 is insuff.

Combing also we cannot arrive at the answer.

So I will go with option E

-Deepak.
2359
If a, b, and c are real numbers, is ?

(1) a > c
(2) ac > b2


OA is E.

But I'm not able to understand where did I go wrong.

Here is my approach:

1) This is clearly insuff.

2) ac > b^2
=> ac/b^2 > 1
=> a/b * c/b > 1
=> (a/b)/(b/c) > 1
=> LHS > 1

If a/b = b/c
LHS of above expression becomes 1, hence the exp doesn't hold true.
so a/b != b/c

If a/b > b/c, then LHS becomes greater than one. Hence the statement holds true.
Hence B.

a/b
I initially thought E as the answer, but changed it to B after this *silly* thing clicked :(

Any comments?
2360
OA is E.

But I'm not able to understand where did I go wrong.

Here is my approach:

1) This is clearly insuff.

2) ac > b^2
=> ac/b^2 > 1
=> a/b * c/b > 1
=> (a/b)/(b/c) > 1
=> LHS > 1

If a/b = b/c
LHS of above expression becomes 1, hence the exp doesn't hold true.
so a/b != b/c

If a/b > b/c, then LHS becomes greater than one. Hence the statement holds true.
Hence B.

a/b
I initially thought E as the answer, but changed it to B after this *silly* thing clicked :(

Any comments?

(a/b) * (c/b) is not equal to (a/b)/(b/c).
Actually, (a/b)/(b/c) = (a/c).
THis is were it goes wrong.

Thanks,
Anurag...
2361
(a/b) * (c/b) is not equal to (a/b)/(b/c).
Actually, (a/b)/(b/c) = (a/c).
THis is were it goes wrong.

Thanks,
Anurag...



(a/b)/(b/c) = ac/b^2 😐
2362
PearlCrystal


Oh yes... You are right....What a mess. That's the mistake you make when you solve quant without pen and paper...:-(
Ok let's examine your approach..
hmm.. you are right that a/b is not equal to b/c.
But the value of (a/b) and (b/c) both could be either negative or positive and still keep the equation (a/b)/(b/c) > 1.
Suppose, (a/b) =4 and (b/c) = 2, then (a/b) > (b/c). But if we take (a/b) = -4 and (b/c) = -2 then (b/c) > (a/b).

Thanks,
Anurag...
2363
Oh yes... You are right....What a mess. That's the mistake you make when you solve quant without pen and paper...:-(
Ok let's examine your approach..
hmm.. you are right that a/b is not equal to b/c.
But the value of (a/b) and (b/c) both could be either negative or positive and still keep the equation (a/b)/(b/c) > 1.
Suppose, (a/b) =4 and (b/c) = 2, then (a/b) > (b/c). But if we take (a/b) = -4 and (b/c) = -2 then (b/c) > (a/b).

Thanks,
Anurag...


Haan yarr .. Correct.

I really want to :banghead:
2364

a grams of a mixture contains X percentage of sugar by weight,b grams of a mixture contains Y percentage of sugar by weight,c grams of a mixture contains Z percentage of sugar by weight.
What percentage of the sugar the mixture contains by weight?

a) a=b=c

b) X=Y=Z

Can someone please workout the solution for this?

2365
Haan yarr .. Correct.

I really want to :banghead:

When we compare two fractions a/b and c/d cross multiplication and checking will tell which fraction is larger. i.e. if ad>bc then a/b > c/d.

Thus when we have a/b and b/c and given that ac>b^2, doesn't it give us answer that a/b > b/c????

What is wrong here 😞
2366
a grams of a mixture contains X percentage of sugar by weight,b grams of a mixture contains Y percentage of sugar by weight,c grams of a mixture contains Z percentage of sugar by weight.
What percentage of the sugar the mixture contains by weight?

a) a=b=c

b) X=Y=Z

Can someone please workout the solution for this?


statement 1:
===========
No information about X.So insuff

Statement 2:
==========
No information about a.So insuff.

combining both it is still insuff as there is no concrete values of either a or x.So insuff.

I will go with option E

-Deepak.