ds - value question

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ds - value question

by cgc » Mon Jan 25, 2010 11:53 am
If a rope is cut into three pieces of unequal length, what is the length of the shortest of these pieces of rope?

(1) The combined length of the longer two pieces of rope is 12 meters.

(2) The combined length of the shorter two pieces of rope is 11 meters.

for rope x + y + z = a

1. x + y = 12

2. y + z = 11

c. can you combine the two equations to create another separate equation? For example...
y = 12 -x
(12-x) + z = 11


Are these not distinct equations to satisfy the equation/variable rule?

Please help explain.

thanks,

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by lalmanistl » Mon Jan 25, 2010 12:54 pm
my answer is E.

This problem has 3 variables but we don't have 3 equations.
Also if you try to solve the equations from stmt1 and stmt2, it doesnt give value
for a variable.

Please correct me, if i'm wrong.

thanks in advance.

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by prinit » Mon Jan 25, 2010 12:59 pm
x+y+z =?

lets assume x is the shortest x<y<z

y+z=12 --eq 1
x+y=11 -- eq 2
------------
z-x=1
=>z=1+x

now put in 1
y=12-z=> 12-1-x

now put in 2
x+12-1-x=11 ...cant solve it for x...

so answer is E ..

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by Stuart@KaplanGMAT » Mon Jan 25, 2010 1:02 pm
cgc wrote:If a rope is cut into three pieces of unequal length, what is the length of the shortest of these pieces of rope?

(1) The combined length of the longer two pieces of rope is 12 meters.

(2) The combined length of the shorter two pieces of rope is 11 meters.

for rope x + y + z = a

1. x + y = 12

2. y + z = 11

c. can you combine the two equations to create another separate equation? For example...
y = 12 -x
(12-x) + z = 11


Are these not distinct equations to satisfy the equation/variable rule?

Please help explain.

thanks,
Hi,

you still only have 3 distinct equations for your 4 variables; you cannot create the 4th equation as you've attempted.

What you've done is a more complicated variation of taking:

y = x + 3

and

4 = 4

and combining them to get:

y + 4 = x + 7

and then using:

y = x + 3
and
y + 4 = x + 7

to solve for x and y; you'll end up with 0=0 if you try to solve those equations.

You'll find if you try to solve the equations you've set up, the same thing will happen; since the third equation is a combination of the first two, it's not distinct.

So, here's the rule:

to solve for a system of n variables, one requires n distinct, linear, equations.

This rule is THE most powerful tool for data sufficiency; the better you know it (and all the exceptions), the fewer calculations you'll have to make on DS questions.
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by cgc » Mon Jan 25, 2010 1:24 pm
great rule!!

thanks for the clarification.

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by VikingWarrior » Mon Jan 25, 2010 4:00 pm
This rule is THE most powerful tool for data sufficiency; the better you know it (and all the exceptions), the fewer calculations you'll have to make on DS questions.
What are the exceptions?

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by Nijeesh » Fri Feb 19, 2010 6:38 am
why can't it be C

(1) The combined length of the longer two pieces of rope is 12 meters.


this statements gives Y+Z=12, so let say 6+6=12 (length of longer ropes 6 and 6)

(2) The combined length of the shorter two pieces of rope is 11 meters.

this gives us X+Y=11 or X+Z=11 , now X can be only 5..so 5+6=11

so shorter rope length is 5

another set of values
1) Y+Z=12, say 7+5=12 (longer ropes 7 and 5)
2) X+z=11, here we can't have X as 6 (to make X+Z as 11 we need X as 6 and Z=5 from (1) )since according to the given data..these are shorter ropes..

so only possible values are Y+Z=12, 6+6=12 (length of longer ropes 6 and 6)
X+Y=11 or X+Z=11 , so 5+6=11


is this not right ?

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by Stuart@KaplanGMAT » Fri Feb 19, 2010 9:52 am
Nijeesh wrote:why can't it be C

(1) The combined length of the longer two pieces of rope is 12 meters.


this statements gives Y+Z=12, so let say 6+6=12 (length of longer ropes 6 and 6)

(2) The combined length of the shorter two pieces of rope is 11 meters.

this gives us X+Y=11 or X+Z=11 , now X can be only 5..so 5+6=11

so shorter rope length is 5

another set of values
1) Y+Z=12, say 7+5=12 (longer ropes 7 and 5)
2) X+z=11, here we can't have X as 6 (to make X+Z as 11 we need X as 6 and Z=5 from (1) )since according to the given data..these are shorter ropes..

so only possible values are Y+Z=12, 6+6=12 (length of longer ropes 6 and 6)
X+Y=11 or X+Z=11 , so 5+6=11


is this not right ?
We still don't know the sum of the pieces, so there are an infinite number of possible lengths.

For example, we could have:

y=6.4, z=5.6, x=5.4
y=6.3, z=5.7, x=5.3

and we could go on with different numbers forever (remember, we're not limited to integers, since nowhere does it say that the ropes must be cut into integer lengths).
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by Nijeesh » Fri Feb 19, 2010 10:48 am
thanks Stuart..
Last edited by Nijeesh on Fri Feb 19, 2010 2:31 pm, edited 1 time in total.

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by Ian Stewart » Fri Feb 19, 2010 12:31 pm
VikingWarrior wrote:
This rule is THE most powerful tool for data sufficiency; the better you know it (and all the exceptions), the fewer calculations you'll have to make on DS questions.
What are the exceptions?
I discussed the many exceptions (and there are many exceptions) in this post:

www.beatthegmat.com/n-variables-n-disti ... 20728.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by Stuart@KaplanGMAT » Fri Feb 19, 2010 3:56 pm
Ian's post is very informative. We can summarize the exceptions as follows:

if you have n distinct linear equations and n variables, you can always solve the entire system.

Most of the "exceptions" that appear on the GMAT aren't actually exceptions to the rule; in the cases in which it turns out that it looks like we have enough information to solve but we don't, it's not because the rule doesn't apply: it's because our equations either aren't linear or aren't distinct. Ian does a great job of explaining those terms, so I won't bother!

The more common situation we find ourselves in, especially in DS, is when we don't actually need n distinct linear equations to solve the problem.

We only need those n distinct linear equations for sure if we want to solve the entire system. When we're solving for part of the system, or for a relationship among the variables rather than the variables themselves, we can often solve with as few as 1 equation.

Here are a couple of examples:

Q: what's the value of 3x - 4y?

(1) 9x - 12y = 21

(2) 9x + 12y = 33

Our first thought may be: hey, 2 equations for 2 unknowns, the answer is "C".

However, if we examine (1) more closely, we see that we can divide both sides by 3 to get:

3x - 4y = 7

which certainly answers the question "what's the value of 3x - 4y?".

So, since (1) gives us the exact relationship for which we're looking, it's good enough all by itself.

Q: what's the value of c?

(1) a - 2b - 2c = 14

(2) 12 - c = 2b - a

At first glance, it looks as though we have 2 equations but 3 variables, so there's no way to solve. However, if we rearrange the second equation and line it up under the first, we have:

a - 2b - 2c = 14
a - 2b - c = -12

and now when we subtract the second equation from the first, we get:

-c = 26

which certainly allows us to solve for c.

So, even though we only have 2 equations, combined they allow us to solve for the 1 variable about which the question asks: together sufficient.

Here's a list of questions to ask yourself in these situations:

1) do I need to solve the entire system?

if yes, then I need 1 distinct linear equation for each variable; if no, then I should look for special cases.

2) are the equations distinct and linear?

3) am I being asked to solve for the values of the variables or for a relationship among variables?

And here's the big thing to watch out for:

1) has the testmaker made it difficult to compare the statements? (e.g. are the variables on different sides of the equal sign?)

If so, that's because the testmaker doesn't want you to compare them; when in doubt, rearrange them to make them easier to compare.
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by shashank.ism » Sun Feb 21, 2010 4:57 am
cgc wrote:If a rope is cut into three pieces of unequal length, what is the length of the shortest of these pieces of rope?
(1) The combined length of the longer two pieces of rope is 12 meters.
(2) The combined length of the shorter two pieces of rope is 11 meters.
for rope x + y + z = a
1. x + y = 12
2. y + z = 11
c. can you combine the two equations to create another separate equation? For example...
y = 12 -x
(12-x) + z = 11
Are these not distinct equations to satisfy the equation/variable rule?
Please help explain.
thanks,
let the length of three unequal parts be x,y,z so x+y+z=total length = L
St.1) let z be smallest so x+y =12 it doesn't solve the problem...====not suff.
St.2) The combined length of the shorter two pieces of rope is 11 meters. so y+z =11 ====not suff.
combined)
x+y =12
y+z =11both also insuff.
Ans E
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by Spidy001 » Sun Apr 03, 2011 6:01 pm
let x,y,z be lengths of the rope and x<y<z.

x=?

1. Not sufficient.

y+z = 12
we dont the total rope length to find x.

2. Not sufficient.

x+y = 11

we dont know y to find x.

Together z-x = 1=> x= z-1. we still dont know z to find x.

Answer is E.

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by sushantgupta » Tue May 03, 2011 9:13 pm
Answer should be E.

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by jcc0523 » Sat Jun 11, 2011 8:04 pm
Answer E ,, if we knew how long the rope was totally, then it would seem to be solvable.
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