For the 1st problem I am getting definite values of -3 and 7/5... havent got a solution set

Havent understood the second one

third one (A)

HSPA wrote:For the 1st problem I am getting definite values of -3 and 7/5... havent got a solution set

Havent understood the second one

third one (A)

Can you please explain how you get the 7/5... I only get the -3!

1) 3x-2 <= 2x-5 b) 2-3x <= 2x-5 c) 3x-2 <=5-2x d) 2-3x <= 5-2x

I solved these 4 equations.....

3) 10x < 3y -6
multiply with 1/10
x < 3y/10 - (+ve)
multiply with 1/12 on both sides
x/12 < y/40 - (+ve)
SO conclusion is Y/40 is greater than X/12 by some +ve value..... So A is sufficient

B) Now multiply with 1/144 on both sides
x/12 - (+ve) > y/36 ....How to make Y/36 as Y/40 ... I havent got this So I kept B insufficent

For QUESTION ONE: Is that a < or a > between the absolute values? Regardless, you'll need to split it into FOUR inequalities:

Just like

| 3x - 2 | > 10

would become

3x - 2 > 10

AND

3x - 2 < -10

If you double the absolute values, you're going to have four inequalities instead of two.

For QUESTION TWO:

s - 1 / s < 1 / t - t

Is this correct? t - t = 0, which will be undefined since it's in the denominator. Just want to make sure there's no error here in the way you've copied it....

For QUESTION THREE:

x /12 > y / 40? Yes/no DS.

To determine sufficiency, we need to know the value of x AND the value of y.

(1) Gives us the range for the relationship between x and y.

Let's say x = 3, then the inequality would become 30 < 3y - 6, or 12 < y. Since y can be anything as long as it is greater than 12, we could say y = 20. That would make our inequality 3/12 > 20/40, or 1/4 > 1/2 which is FALSE.

Let's say x = -3, then the inequality would become -30 < 3y - 6, or -24 < 3y, or -8 < y. Again, no matter what we choose, it will make our original inequality FALSE.

Let's say x = 0, then the inequality would becomes 0 < 3y - 6, or 6 < 3y, or 2 < y. Let's say y = 40. Then we'd get 0/12 > 40/40 for our original inequality. That becomes 0 > 1, another FALSE statement.

We can see that (1) is sufficient, since when we choose very different sets of numbers, we answer our original question in the same way.

(2) Gives another range for the relationship between x and y. Let's try some numbers again to test the results.

If x = 0, the inequality becomes 12(0) - 7 > 4y, then 0 - 7 > 4y, then -7/4 > y. Let's say y = -2. Plugging those back in to the original inequality: 0 > -1/20. This is a TRUE statement.

If x = -1, the inequality becomes 12(-1) - 7 > 4y, then -12 - 7 > 4y, then -19 > 4y, then -19/4 > y. Let's say y = -5. The original inequality becomes -1/12 > -5/40, or -1/12 > -1/8. This is another TRUE statement.

If x = 1/2, then 12(1/2) - 7 > 4y, then 6 - 7 > 4y, then -1/4 > y. Let's say y = -1/2. Plugging that into the original inequality: (1/2)/12 > (-1/2)/40, becomes 1/2 x 1/12 > -1/2 x 1/40, becomes 1/24 > -1/80. Again, a TRUE statement.

IMO, the answer is [spoiler](C)[/spoiler].

hi guys
for Q1 my answer
-3<x<2/3, and 2/3<x<7/5

for Q2 i did not understand the figures also

for Q3 my answer is A, tend to disagree with Vivian

two approaches to tackle this problem - solution areas { nightmare here, with six areas --> (3x-2)>0, (3x-2)<0, (3x-2)=0, (2x-5)>0, (2x-5)<0, (2x-5)=0 } OR squaring both sides. I chose the second -less dreary way

|3x - 2| ≤ |2x - 5|
(3x-2)^2 ≤ (2x-5)^2
9x^2 -12x +4 ≤ 4x^2 -20x +25
5x^2 +8x -21 ≤ 0,
...

5x^2+8x-21=0, |Discriminant=Sqrt(64+20*21)=22, x(1,2)=(-8+-22)/10= {-3; 1.4}

(x+3)(x-1.4) ≤ 0, x ≤ -3 and x ≤ 1.5. The solution set for x is (-3; 1.4).

garuhape wrote: 1. Find the solution set for the following inequality |3x - 2| ≤ |2x - 5|?

hi night reader
are you dead sure in your solution? if i got your answer right it is x<-43
if x=-50 then
|3(-50)-2|=|-152|=152
|2(-50)-2|=|-102|=102, but 152>102
and if x=0
then
|-2|=2
|-5|=5. 2<5

please check solution above

clock60 wrote:hi night reader
are you dead sure in your solution? if i got your answer right it is x<-43
if x=-50 then
|3(-50)-2|=|-152|=152
|2(-50)-2|=|-102|=102, but 152>102
and if x=0
then
|-2|=2
|-5|=5. 2<5

ok your final solution x<-4.2
x=-5
|-15-2|=|-17|=17
|-10-5|=|-15|=15. but again 17>15, but left part must be less than right part, according to the inequality
|3x - 2| < |2x - 5|

for the second I got B
given: s-1/s<1/t-t, Is s < t?
st(1) s>1, s-1/s>0. Since 1/t-t>s-1/s, 1/t-t>0 and t<1 -hence Sufficient to answer No, s>t
st(2) t>0, It can be 1/t-t>0 or 1/t-t<0. If 1/t-t>0 then 0<t<1 then it must be s>1 -So s>t. And when 1/t-t<0 then t>1 and s<1 -So t>s Insufficient, as two cases are possible s>t and s<t

IOM B

garuhape wrote: 2. If s-1/s<1/t-t is s < t?
s > 1
t > 0

please check solution above

clock60 wrote:ok your final solution x<-4.2
x=-5
|-15-2|=|-17|=17
|-10-5|=|-15|=15. but again 17>15, but left part must be less than right part, according to the inequality
|3x - 2| < |2x - 5|

i also think that working with solution areas works better here.
squaring better works with equations
but honestly i don`t think we must bother about this problem, i never saw similar problems in og or gprep
with given answers it is easier to find right one

what about 3 problem it seems pretty more gmat stylish

please do not look after 6 areas/intervals above - square both sides

clock60 wrote:i also think that working with solution areas works better here.
squaring better works with equations
but honestly i don`t think we must bother about this problem, i never saw similar problems in og or gprep
with given answers it is easier to find right one

what about 3 problem it seems pretty more gmat stylish

@ night reader
may be it is better to put aside this
it closer to the truth, but probably you commit math mistake, you answer -3<x<1.5. as i see
try x=1.45
3*1.45-2=2,35
|2*1.45-5|=|-2.1|=2.1
2.35>2,1 not true