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Source: — Data Sufficiency |
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by force5 » Mon Mar 21, 2011 7:47 am
IMO A

question says 3x+2y>20
stmnt 1 (sufficient)

x>y+1
( multiplying by 2 and adding to stem)

x>4.4 and y>3.2
which gives us 4x+y>20 ( for all values)

stmnt 2 (insufficient)

y<5 (can have various values)
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by HSPA » Mon Mar 21, 2011 9:06 am
I got Y<5 is sufficient here...

using A

if x > Y+1 then x~= y+2
3(y+2)+2y>20 => 5y+6>20
y ~= 3 x=5

is 4x+y>20 yes

IMO D here
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by srcc25anu » Mon Mar 21, 2011 11:18 am
i tried to plot the above inequalities on a graph and feel its E (cannot be solved even by both statements)

any more explanations on this one???
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by force5 » Mon Mar 21, 2011 11:35 am
Hi Anu i really feel Stmnt A will give you the answer.

BTW Kevin what is the OA????
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by clock60 » Mon Mar 21, 2011 12:51 pm
hi guys
got A here, my reasoning
if 3x+2y>20 does 4x+y>20
(1) x-y>1, here summing both 3x+2y>20 and x-y>1. gives us 4x+y>21, thus if 4x+y>21, 4x+y must be greater that 20
suff
(2)y<5 insuff to prove this, again sum 3x+2y>20, and y<5, this lead us to 3x+y>15
x=5, y=1 then 3*5+1=16>15, and 4x+y=4*5+1=21>20 the answer is yes
x=1,y=15, then 3*1+15=18>15, but 4*1+15=19<20 the answer is no
insuff
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by Anurag@Gurome » Mon Mar 21, 2011 7:41 pm
srcc25anu wrote:i tried to plot the above inequalities on a graph and feel its E (cannot be solved even by both statements)

any more explanations on this one???
If 3x + 2y > 20, is 4x + y > 20 ?

(1) x > y + 1
(2) y < 5

Solution:
Consider first (1) alone.
It says x > y + 1.
Or x - y > 1.------------(a)
The main question says that 3x + 2y > 20.------------(b)
Adding (a) and (b) we get that 4x + y > 21.
This means that 4x+2y > 20.
Or (1) alone is sufficient.
Next consider (2) alone.
It says that y < 5.
Or -5y > -25.----------(c)
Now, 3x + 2y > 20 implies that 12x + 8y > 80.---------(d)
Adding (c) and (d), we get that 12x + 3y > 55.
Or 4x + y > 18 1/3.
This is not sufficient to tell us whether 4x + y > 20 or not.
Or (2) alone is not sufficient.

The correct answer is (A)
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by Night reader » Mon Mar 21, 2011 8:55 pm
Statement (1) x>y+1, x-y>1
3x + 2y > 20 Plus x-y>1 ---> 4x+y>21 Sufficient;

Statement (2) y<5 OR -y>-5 Plus 3x + 2y > 20 ---> 3x+y>15. Verifying x ---> -y>-5 Plus 3x+y>15 x>10/3
 Not Sufficient

IOM A
kevincanspain wrote:If 3x + 2y > 20, is 4x + y > 20 ?

(1) x > y + 1
(2) y < 5
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by HSPA » Mon Mar 21, 2011 10:11 pm
Let me use Y<5 one more time with Y= 4, Y= -2, (Y= 6 and Y = 20)

at y = 4 ; 3x+2y>20 will give x>4 => 4x+y > 20 satisfies
at y = -2; 3x+2y>20 will give x>8 => 4x+y > 20 satisfies
at y = 6; 3x+2y>20 will give x>3 => 4x+y >20 is not satisfied
at y =20; x is -ve so 4x is more negitive... so not satisfied

So Y< 5 is doing good.. what am I missing Night reader/Anurag
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by Night reader » Mon Mar 21, 2011 10:56 pm
nobody said that y can't be 4 and x can't be >4

conditions given y<5 and x>10/3 and in fact y could be -2,000,000 and/or x could be 2,000,000
the discomfort (trouble/problem/issue/question/doubt) with your ascertainment about y=4 and x>4 is fixing areas which are fixed already that is 5<y and x>10/3. We need 100% validation, BUT here it's missing; the ranges are not fixed the way we like OR you like. x could be less OR greater than (20-15) ... the missing quantity RHS
HSPA wrote:Let me use Y<5 one more time with Y= 4, Y= -2, (Y= 6 and Y = 20)

at y = 4 ; 3x+2y>20 will give x>4 => 4x+y > 20 satisfies
at y = -2; 3x+2y>20 will give x>8 => 4x+y > 20 satisfies
at y = 6; 3x+2y>20 will give x>3 => 4x+y >20 is not satisfied
at y =20; x is -ve so 4x is more negitive... so not satisfied

So Y< 5 is doing good.. what am I missing Night reader/Anurag
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by Geva@EconomistGMAT » Mon Mar 21, 2011 11:26 pm
HSPA wrote:Let me use Y<5 one more time with Y= 4, Y= -2, (Y= 6 and Y = 20)

at y = 4 ; 3x+2y>20 will give x>4 => 4x+y > 20 satisfies
at y = -2; 3x+2y>20 will give x>8 => 4x+y > 20 satisfies
at y = 6; 3x+2y>20 will give x>3 => 4x+y >20 is not satisfied
at y =20; x is -ve so 4x is more negitive... so not satisfied

So Y< 5 is doing good.. what am I missing Night reader/Anurag
Your calculations show that the further y grow from 5, the more certain you are that 4x+y is indeed greater than 20. At y=-2, 4x alone is already at 32. Do the opposite: try using a 4<y<5 (get y closer to 5)

at y = 4.5 ; 3x+2y>20 will give x>11/3 => 4x+y will not necessarily satisfy >20
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by kevincanspain » Tue Mar 22, 2011 2:55 am
kevincanspain wrote:If 3x + 2y > 20, is 4x + y > 20 ?

(1) x > y + 1
(2) y < 5
4x + y > 20 if 4x + y >= 3x + 2y i.e x >= y

If x < y, 4x + y < 3x + 2y, and 4x + y may be less than, greater than or equal to 20

(1) IF x > y + 1, x > y SUFF
(2) If y < 5 , since x > (20 - 2y)/3, x > 10/3.


If y is slightly less than 5 and x slightly greater than 10/3, 4x + y will be close to 40/3 + 5 = 55/3 < 20

However, if y < 0, x> 0 and the answer would be yes

NOT SUFF

Arurag's comments on (2) are very nice!
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