GMAT Data Sufficiency Discussions

MBA GMAT
2303
1) If , is

(1) is less than
(2) is less than



Statement 1:
==========
Some ex : x = 2.5 & y = 1 . The eqn is less than 18.
x = 1 & y = - 0.5 .The eqn is less than 18.

Statement 1 is Sufficient .

Statement 2 :
==========

Some ex : x = 2.5 & y = 1. The eqn is less than 18.
x = 6 & y = 1 .The eqn is greater than 18.

Statement 2 is insuff.

I will go with option A
2304
Statement 1 is suff to answer the q.

statement 2 is insuff to ans: x=2 , y=-2 ,z=-5 or x=-4 y = 4 z = -1

So I will go with option A


here is the solution : if you take statement one and square both sides...u get x^4 + y^4 +2(xy)^2 > Z^4.....so from this statement we know that we need to add sumthing positive to X^4 and Y^4 to make it greater....mayb on its own it may b greater ..but we cannot really say that..so insufficient..do the same on the lower question u get sumthing similar..both boil down to the fact that u need to add sumthing to X^4 +y ^4 plus wtw > z^4....whether that sumthing makes a difference or no ..we cannot say...
2305
1) If , is

(1) is less than
(2) is less than

Statement 1:
==========
Some ex : x = 2.5 & y = 1 . The eqn is less than 18.
x = 1 & y = - 0.5 .The eqn is less than 18. (while choosing x,y you have missed the eqn in the question stem is not satisfied => x/y = -2, so there is a problem with this set of values.) Moreover, one thing is for sure that because x/y>2 => x/y is +ve. so both x and y will have the same sign.
you have shown the pair of +ves. Now lets choose 2 values
(x,y) = (-2.5,-1)
=> x/y=2.5>2 (satisfied the question stem)
AND x-y=-1.5 3x+2y = -9.5
Ok now the BIG question is : "IS THERE A SET OF VALUES THAT SATISFIES X/Y>2 AND X-Y18?". Not that I can think of.
so I can say ;
option A is sufficient.

All in ALL we have to always remember that x/2>2
Statement 1 is Sufficient .

Statement 2 :
==========
to prove that statement 2 is not sufficient, lets choose values for which x/y>2 and y-x2.5/1=2.5>2 AND 1-2.5=-1.5-2.5/-1=2.5>2 AND -1-(-2.5)=1.5 -3.5/-1=3.5>2 AND -1-(-3.5)=2.5>2 (does not hold true); it ends here.
Hence statement 2 is not sufficient.

Some ex : x = 2.5 & y = 1. The eqn is less than 18.
x = 6 & y = 1 .The eqn is greater than 18.

Statement 2 is insuff.

I will go with option A

@deepakraam, see my comments in red inline.

Is there a better way of solving this? Like getting a range of values for x,y. So that we can be sure.
2306

The three-digit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?


(1) A + B/10 + C/100 = B + C/10 + A/100
(2) A + B/10 + C/100 = C + A/10 + B/100

2307
@deepakraam, see my comments in red inline.
"(x,y) = (-2.5,-1)
=> x/y=2.5>2 (satisfied the question stem)
AND x-y=-1.5
3x+2y = -9.5
Ok now the BIG question is : "IS THERE A SET OF VALUES THAT SATISFIES X/Y>2 AND X-Y3X+2Y=>18?". Not that I can think of.
so I can say ;
option A is sufficient."
Is there a better way of solving this? Like getting a range of values for x,y. So that we can be sure.

Well ,I have one set of values of x for which all the three equations(In Option A and two in the question) satisfy.
2.1 But the answer still remains the same. Option A.

Edited: You are right. I overlooked the equation 3X+2Y=>18


Thanks,
Anurag...
2308
The three-digit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?


(1) A + B/10 + C/100 = B + C/10 + A/100
(2) A + B/10 + C/100 = C + A/10 + B/100

IMO : D

Reason : From statement 1, A=B, B=C and A=C (on comparing both the sides of the equation, eg. B/10=C/10, therefore B=C, similarly A/100 = C/100, therefore A=C)
Since, A=B=C, the number can be 111,222,333 ... 999 ; 37 is a multiple of 111 (37*3), hence the remainder will be 0 from this statement). Hence A/D are candidates.

From From statement 2, A=B, B=C and A=C (on comparing both the sides of the equation, eg. B/10=A/10, therefore B=C, similarly B/100 = C/100, therefore A=C)
Since, A=B=C, the number can be 111,222,333 ... 999 ; 37 is a multiple of 111 (37*3), hence the remainder will be 0 from this statement). Hence B is also sufficient.

Thus, D
2309
IMO : D

Reason : From statement 1, A=B, B=C and A=C (on comparing both the sides of the equation, eg. B/10=C/10, therefore B=C, similarly A/100 = C/100, therefore A=C)
Since, A=B=C, the number can be 111,222,333 ... 999 ; 37 is a multiple of 111 (37*3), hence the remainder will be 0 from this statement). Hence A/D are candidates.

From From statement 2, A=B, B=C and A=C (on comparing both the sides of the equation, eg. B/10=A/10, therefore B=C, similarly B/100 = C/100, therefore A=C)
Since, A=B=C, the number can be 111,222,333 ... 999 ; 37 is a multiple of 111 (37*3), hence the remainder will be 0 from this statement). Hence B is also sufficient.

Thus, D


answer is D. try this
Is

(1) x is not equal to 3

(2) - x x > 0
2310
answer is D. try this
Is

(1) x is not equal to 3

(2) - x x > 0

Phew .. this is a toughie .. what's the answer to this, is it C ??
2311
answer is D. try this
Is

(1) x is not equal to 3

(2) - x x > 0

My take E:
sqrt((x-3)^2) = +-(x-3)

if x-3 = 3-x => x=3.
if -(x-3) = (3-x) => 0=0 (no solution)

(1) alone - There is no solution => so the eqn is not satisfied.
(2) alone - No value that satisfies => so the eqn is not satisfied.

combining (1) and (2) : No values that satisfies => so the eqn is not satisfied.

so the solution is E
2312
My take E:
sqrt((x-3)^2) = +-(x-3)
if x-3 = 3-x => x=3.
if -(x-3) = (3-x) => 0=0 (no solution)
(1) alone - There is no solution => so the eqn is not satisfied.
(2) alone - No value that satisfies => so the eqn is not satisfied.
combining (1) and (2) : No values that satisfies => so the eqn is not satisfied.
so the solution is E


well, the official answer is B. explanation is not given but i am trying my way: let me know if i am going wrong!!
sqrt((x-3)^2) = x-3|(as Gail.Wynand explained) which is equal to (3-x), this is true only when x
so x-3 = 3-x only when x
1) alone suggest x is not equal to 3. can be greater or smaller than 3.

2) alone suggest that x has to be less than zero for -x|x to be greater than zero. So S2 sufficient.

answer is B. Is this approach right? quant gurus plz help...
2313
well, the official answer is B. explanation is not given but i am trying my way: let me know if i am going wrong!!
sqrt((x-3)^2) = x-3|(as Gail.Wynand explained) which is equal to (3-x), this is true only when x
so x-3 = 3-x only when x
1) alone suggest x is not equal to 3. can be greater or smaller than 3.

2) alone suggest that x has to be less than zero for -xx to be greater than zero. So S2 sufficient.

answer is B. Is this approach right? quant gurus plz help...


Maybe I missed something. from where do you get x-3| = 3-x ? Is it like sqrt((x-3)^2) = x-3? Pardon my understanding
2314

Hi

I had a simple question.Say we pick a no like 5 .Is 0 considered a multiple of 5

as I was doing the following problem and I marked the answer as B but actually it was E

If x is a prime no what is the value of x

(1) x (2) (x-2) is a multiple of 5

The problem is from the Kaplan premier book.

Also wanted to ask you is there any good book / any best practices for permutations/combinations/probablity as I am weak in this area.

2315
Gail.Wynand? Pardon my understanding


yeah right. sqrt((x-3)^2) = x-3. is it not right?
2316

try this now: similar type problem
Is ?

1) -x x > 0

2) 5 - x > 0

2317
try this now: similar type problem
Is ?

1) -x x > 0

2) 5 - x > 0


I will start the problem with statement 2

5-x > 0. Assume x = 4 & x = -6. On both the occasions,I have the eqn satisfied.So Statement 2 is suff.

Statement 1. X has to be negative. assume x = -3. Even with this value the prblm statement is satisfied.
So statement 1 is also suff.

I will opt for option D.
2318
MissionPGPX. is it not right?

Gotch Ya buddy. My approach was wrong. Corrected it though. Thanks.
2319
try this now: similar type problem
Is ?

1) -x x > 0

2) 5 - x > 0

the same approach that @deepakraam suggested:

(1) alone this is only true for x(I have assumed a rule here that the sqrt of a square can not be -ve on(in) GMAT; Is this correct ?)

(2) alone : x indeed.
either 0
My take D.
2320
try this now: similar type problem
Is ?

1) -x x| > 0

2) 5 - x > 0

IMO - A

Using your method :
sqrt((x-5)^2) = x-5 which is equal to (5-x), this is true only when x
so x-5 = 5-x only when x
1) alone suggest that x has to be less than zero for -x|x to be greater than zero. So S1 sufficient.

2) alone suggest x is lesser than 5, so it can be positive as well as negative .. Hence S2 insufficient.
2321
IMO - A

Using your method :
sqrt((x-5)^2) = x-5| which is equal to (5-x), this is true only when x
so x-5 = 5-x only when x
1) alone suggest that x has to be less than zero for -x|x to be greater than zero. So S1 sufficient.

2) alone suggest x is lesser than 5, so it can be positive as well as negative .. Hence S2 insufficient.

Dude the equation is satisfied for +ve values equal to or lesser than 5 and 0 ofcourse. I think you missed that
2322
IMO - A

Using your method :
sqrt((x-5)^2) = x-5| which is equal to (5-x), this is true only when x
so x-5 = 5-x only when x

1) alone suggest that x has to be less than zero for -xx to be greater than zero. So S1 sufficient.

2) alone suggest x is lesser than 5, so it can be positive as well as negative .. Hence S2 insufficient.


ohh boy... i posted my approach to get it checked. Thanks for helping me to find the catch in my approach. refer to the blue bold part. I think i made a mistake there. Modulus of any value yield negative result only when the entire value has a negative sign before it, which in our case is (x-5). x-5 = 5-x only when x
|x-5 = 5-x only when x-5Any value less than 5 should satisfy the equation. this is true for previous equation as well.

@deepakram, can you comment on this approach? is this ok?