Data Sufficiency for 780+ Aspirants

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 am
Location: Delhi
Thanked: 84 times
Followed by:9 members

Data Sufficiency for 780+ Aspirants

by sureshbala » Tue Feb 17, 2009 12:07 pm
Folks, in this thread we will get into DS practice questions on a regular basis. I will come up with questions keeping in mind the standard of the questions that one needs to be comfortable with in order to score top in Quant Section of the exam.

Here are the two simple rules that I want all the members to strictly adhere to..

1. Do not post your questions in this thread. Instead start a new thread or post in the relevant thread.

2. Please give a bit of explanation for the answer that you post here.(Don't simply post the answer)

Here is our first question.

X and Y are real numbers such that X/Y > 2. Is 3X + 2Y <17?

I. X - Y < 2

II. Y - X < 2

Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

Re: Data Sufficiency for 780+ Aspirants

by bluementor » Tue Feb 17, 2009 1:43 pm
sureshbala wrote: Here is our first question.

X and Y are real numbers such that X/Y > 2. Is 3X + 2Y <17?

I. X - Y < 2

II. Y - X < 2
x/y > 2 would mean that either both x and y are positive or both x and y are negative. If both x and y are negative, then the answer to the question stem is YES. So only the positive values of x and y need to be considered in trying to prove 3x + 2y<17 to be false.

Statement 1:
x-y < 2

For both x and y being positive, they need to be 'close' enough to zero in order to preserve x/y>2 and x-y<2. Therefore if x and y are small, 3x + 2y will be less than 17.

For eg. x = 3, y=1.1, 3x + 2y = 11.2 < 17. Other possible values of x and y will not be too far from those in this example and will satisfy 3x+2y<17. Therefore, statement 1 is sufficient.

Statement 2:
y-x < 2

If x=10, y=1, then 3x+2y will be larger than 17. Therefore insufficient.

Choose A.

P.S. My solution is a bit crude, especially in solving for statement 1. Would be interested in seeing a more analytical approach.

-BM-

Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

by 4meonly » Tue Feb 17, 2009 2:50 pm

Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

Re: Data Sufficiency for 780+ Aspirants

by x2suresh » Tue Feb 17, 2009 11:27 pm
sureshbala wrote:Folks, in this thread we will get into DS practice questions on a regular basis. I will come up with questions keeping in mind the standard of the questions that one needs to be comfortable with in order to score top in Quant Section of the exam.

Here are the two simple rules that I want all the members to strictly adhere to..

1. Do not post your questions in this thread. Instead start a new thread or post in the relevant thread.

2. Please give a bit of explanation for the answer that you post here.(Don't simply post the answer)

Here is our first question.

X and Y are real numbers such that X/Y > 2. Is 3X + 2Y <17?

I. X - Y < 2

II. Y - X < 2


X/Y >2 -- > both X and Y must be +ve or both -ve

3X + 2Y <17 always true when X and Y are -ve..


1) X - Y < 2 --> eqn(1)

X/Y >2 --> X-2Y>0 --> 2y-x<0 eqn(2)
eqn(1)+eqn(2)
--> y<2 --> X<4

3X + 2Y < 3*4+2*2 <16

Statement 1 is sufficient

2) Y-X <2

WHEN X=2.1 Y=1

3X + 2Y <17 True

WHEN X=30.1 Y=15
3X + 2Y <17 False


not sufficient


A

Senior | Next Rank: 100 Posts
Posts: 44
Joined: Sun Feb 15, 2009 7:37 pm
Thanked: 3 times
Followed by:1 members

Re: Data Sufficiency for 780+ Aspirants

by Bidisha_800 » Tue Feb 17, 2009 11:55 pm

1) X - Y < 2 --> eqn(1)

X/Y >2 --> X-2Y>0 --> 2y-x<0 eqn(2)
eqn(1)+eqn(2)
--> y<2 --> X<4

can we really add inequalities to "solve" like equation ?

I solved this the following way :

3x+2y
=x+2(x+y)
=y{x/y+2(1+x/y)}

x/y >2
x/y = 2+ d (delta) where d >0

3x+2y = y(8+3d)


From (1)

x-y <2
x-y = 2-m (m>0)
y(x/y -1) = 2- m
y(2+d-1) = 2 -m

y = (2-m)/(d+1) <2 ---------(a)

y(d+1) = 2-m < 2
5y < 10 ..............................(b)

now 3x+2y = y(8+3d)
=y {5 + 3(d+1)}
= 5y + 3(2-m)
<10 + 6 -3m
<16 -3m

since m >0 so 3x+2y has to be less than 16
GMATPowerPrep Test1= 740
GMATPowerPrep Test2= 760
Kaplan Diagnostic Test= 700
Kaplan Test1=600
Kalplan Test2=670
Kalplan Test3=570

User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 am
Location: Delhi
Thanked: 84 times
Followed by:9 members

by sureshbala » Wed Feb 18, 2009 10:09 pm
Folks, any take on this....I am looking for more solutions (lucid) before I move away from this....

Master | Next Rank: 500 Posts
Posts: 279
Joined: Wed Sep 24, 2008 8:26 am
Location: Portland, OR
Thanked: 6 times

by pbanavara » Wed Feb 18, 2009 10:58 pm
x/y < 2 From stem

1 x- y < 2

For this to be true both x and y have to be negative. If that's the case then 3x+2y is always < 17 - Sufficient.

2. y-x < 2 ; x<2y : For this to be true x is +ve and y is -ve always .. hence insufficient.

Choose A.

I doubt this is correct - but gave it a shot.

- pradeep
In the land of night, the chariot of the sun is drawn by the grateful dead
https://questor.blocked

GMATPREP1 - 550 (Oct 08)
MGMAT FREE CAT - 600 (Dec 08)
MGMAT CAT1 - 670 (Jan 09)
MGMAT CAT2 - 550 (Jan 09)
MGMAT CAT3 - 640 ( Feb 09)
MGMAT CAT4 - 660 ( Feb 09)

User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 am
Location: Delhi
Thanked: 84 times
Followed by:9 members

by sureshbala » Fri Feb 20, 2009 5:21 am
pbanavara wrote:x/y < 2 From stem

1 x- y < 2

For this to be true both x and y have to be negative. If that's the case then 3x+2y is always < 17 - Sufficient.

2. y-x < 2 ; x<2y : For this to be true x is +ve and y is -ve always .. hence insufficient.

Choose A.

I doubt this is correct - but gave it a shot.

- pradeep
I think you were too quick in concluding the bold statements above..Isn't it?

Master | Next Rank: 500 Posts
Posts: 279
Joined: Wed Sep 24, 2008 8:26 am
Location: Portland, OR
Thanked: 6 times

by pbanavara » Fri Feb 20, 2009 8:52 am
sureshbala wrote:
pbanavara wrote:x/y < 2 From stem

1 x- y < 2

For this to be true both x and y have to be negative. If that's the case then 3x+2y is always < 17 - Sufficient.

2. y-x < 2 ; x<2y : For this to be true x is +ve and y is -ve always .. hence insufficient.

Choose A.

I doubt this is correct - but gave it a shot.

- pradeep
I think you were too quick in concluding the bold statements above..Isn't it?

So x-y < 2 and x > 2y or x< y+2 and x > 2y

Try x +ve and y +ve - doesn't hold true
Try x -Ve and y +ve - doesn't hold true
x +ve and y -ve - doesn't not true

x -ve and y -ve - Holds true. Unless am missing something here.

Hence the conclusion.

I had to edit the post multiple times .. darn this is what happens when work and GMAT prep clash.
In the land of night, the chariot of the sun is drawn by the grateful dead
https://questor.blocked

GMATPREP1 - 550 (Oct 08)
MGMAT FREE CAT - 600 (Dec 08)
MGMAT CAT1 - 670 (Jan 09)
MGMAT CAT2 - 550 (Jan 09)
MGMAT CAT3 - 640 ( Feb 09)
MGMAT CAT4 - 660 ( Feb 09)

Junior | Next Rank: 30 Posts
Posts: 29
Joined: Sun Feb 08, 2009 11:29 am
Thanked: 3 times

by ven4gmat » Sat Feb 21, 2009 10:06 am
I somehow managed to get the answer as A. But not confident enough... sureshbala, i hope you are coming up with the explanation

User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 am
Location: Delhi
Thanked: 84 times
Followed by:9 members

by sureshbala » Mon Feb 23, 2009 5:47 am
Given x/y>2…..implies both x and y can be positive or both of them are negative
If both of them are negative, it is clear that 3x+2Y<0.
If both of them are positive we have x > 2y…......(1)

Statement1: x-y < 2 ...(2)
From (1) we have x-y > y .....(3)
From (2) and (3), y<x-y<2, so y<2.
From (2), we have x<y+2, so x<4.
Since y<2 and x<4, 3x+2y<16.
Hence statement (1) alone is sufficient

Coming to statement 2, we can think along the same lines and easily find two contradicting examples.
Let x=100 and y=1
Clearly x/y > 2, and y-x < 2 and the value of 3x+2y=302 (>18)
Let x=3 and y=1
Clearly x/y>2 and y-x<2 and the value of 3x+2y=11(<18)
So statement(2) is not sufficient
Hence the answer is Choice A.

User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 am
Location: Delhi
Thanked: 84 times
Followed by:9 members

by sureshbala » Sun Mar 01, 2009 10:28 pm
Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800
sureshbala wrote:Here is our first question.

X and Y are real numbers such that X/Y > 2. Is 3X + 2Y <17?

I. X - Y < 2

II. Y - X < 2
A bit of thought and number picking is often the quickest way to solve high end DS questions. Remember, no bonus points on test day for elegant solutions!

Original: x and y have the same sign and x/y > 2. As noted, double negative values will give us a "yes" answer all the time, so we just need to try to get a "no" answer using positive values.

(1) x - y < 2

OK, now we need numbers that are less than 2 apart but for which x is more than double y. Well, if x=4 and y=2, then they're exactly 2 apart and x is exactly double y. If we pick numbers bigger than 4 and 2, then for x to be more than double y, it will have to be more than 2 greater than y. Therefore, x < 4 and y < 2. If x=4 and y=2 then we'd have:

3*4 + 2*2 = 16 which is less than 17, so x<4 and y<2 are definitely going to give us a "no" answer: sufficient.

(As an aside, if the question had been "Is 3x + 2y < 16" it would have been more interesting!)

(2) y - x < 2

Here we can quickly pick y = 1 and x = 100 to see that we can satisfy both rules and get a "yes" answer; since we already know that negatives are possible and that all negatives give us a "no" answer, insufficient.

(1) is suff, (2) isn't: choose (A).

This might seem time consuming, but that's only because I explained every step out loud. If you train yourself to think this way, this question is finished in under a minute.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Master | Next Rank: 500 Posts
Posts: 137
Joined: Thu Jan 08, 2009 1:27 am
Thanked: 7 times

by welcome » Mon Mar 02, 2009 12:55 am
Dear Suesh,
Thanks for a good Qs and a nice answer.

Dear Stuart,
I made this question right by making 1st condition by inequalities calculations, but in condition 2 I did not had an option but for to choose nos.

Can you or Suresh, please explain us wha all we need to consider before choosing smart nos. for cheking the answers, if we are choosing not to go by algebric methods.

Thanks.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590

Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Mon Mar 02, 2009 3:13 am
sureshbala wrote:Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.
Basic rules:
a>0, this simply means the curve ax^2+bx+c=0 is 'U' shaped, having a single minimum point.
The sign of -b/2a can be used to identify if this single minimum point occurs in the positive x region (-b/2a > 0) or in the negative x region (-b/2a < 0).
The sign of c determines whether the curve intersects the y-axis in the positive y region (c>0) or the negative y region(c<0).

Statement 1: Sum of the roots is positive
This could mean either:
(i) both roots are positive (with c>0), or
(ii) one positive root and one negative root (with c<0, and the positive root having the larger absolute value)

For both these cases, the minimum point is in the positive x region. So we have:

-b/2a > 0
Since we know, a>0, therefore b must be less than 0.

a + b + c>0
We know that a>0 and b<0, but c could be either positive or negative. Therefore, we are unable to determine which of these two cases (i and ii) is true. Insufficient.

Statement 2: Product of the roots is positive
This could mean either:
(iii) both roots are positive (with -b/2a >0), or
(iv) both roots are negative (with -b/2a <0).

For both these cases, the y-intercept must be positive (c>0).

a + b + c>0
We know that a>0 and c>0, but b could be either positive or negative. Therefore, we are unable to determine which of these two cases (iii and iv) is true. Insufficient.

Both statements together:
From statement 1, we have established that b<0. This produces -b/2a > 0 (minimum point occuring in the positive x region)
From statement 2, we have established that c>0. The y-intercept in positive.

We can conclude now that both roots are positive.

Choose C.

-BM-