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quants questions

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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quants questions

by nagar.sandeep » Wed Sep 15, 2010 6:59 pm
HI,

I have couple of quants questions, which I could not solve.
Can you please help explaining them:

Question 1. If x, y, and k are positive numbers such that
(x/ (x+y))*10 + (y/(x+y))*20 = k and if x < y,
which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30



Question 2: If xy <> 0, what is the value of 1/x + 1/y?
(1) 1 / (x+y) = -1
(2) xy = 6(x+y)


Thanks,
Regards,
Sandeep
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by Brian@VeritasPrep » Fri Sep 17, 2010 9:42 am
Hey Sandeep,

Good questions - let me tackle that first one, which I think is a pretty good example of GMAT strategy. I did this in less than two minutes using this strategy:

1) Simplify the algebra to get in a position to better solve for k:

x+y is a common denominator on the left, so you really have:

(10x + 20y)/(x+y) = k

Multiply out the denominator to get:

10x + 20y = kx + ky

Now, there are three variables and only one equation, so you can't really solve for k algebraically, so I:

2) Plug in easy-to use numbers to see if you can find a range of values (since they ask "what could be" and not "what is" the value of k, pinning k into a range is probably the best you can do):

Since x < y, let's call them 1 and 2:

10(1) + 20(2) = 1k + 2k
50 = 3k
k = 17.3333

Doing another one, let's try x = 2 and y = 5 to see if different numbers give us something wildly different:

10(2) + 20(5) = 2k + 5k
120 = 7k
k = something just over 17

Since both of our values are in that just-over-17 range, choice D, 18, is looking like a pretty good bet. Maybe try 1-2 more to see if you can get lower than 17 (because 15 looms somewhat close):

Call x 2 and y 3:

10(2) + 20(3) = 2k + 3k
80 = 5k
k = 20

Well, it seems like we can definitely get above the 17 range as high as 20, and now our range of potential values is 17-20, so I'm ready to call it and say that 18 is possible. If we use nonintegers - and we're allowed to...the question doesn't prohibit it - we would be able to tweak our current estimates to get to 18, but with any combinations we've tried already we haven't been able to near 15. The correct answer has to be 18.

A few strategic takeaways here:

1) Cleaning up the algebra is helpful - in most questions that feature variables in a denominator, your goal at some point will be to multiply out that denominator.

2) Because they ask "which COULD BE the value of k", you don't have to solve directly for it if you can use easy-to-plug numbers to establish a range of potential values. You don't need to know exactly what k IS, but rather just what k might look like. Since this question doesn't involve integers specifically, it's not a divisibility problem (so you don't have to worry that 18 might not share necessary factors), so an estimate is probably the most efficient way to tackle this one.
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by debmalya_dutta » Fri Sep 17, 2010 11:52 am
nagar.sandeep wrote:
Question 1. If x, y, and k are positive numbers such that
(x/ (x+y))*10 + (y/(x+y))*20 = k and if x < y,
which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30
k = 10x+10Y+10Y / (x+y) = 10 + 10y / (x+y) = 10 + 10 / (x/y+1)
The value of 10 / (x/y+1) is always < or equal to 10 because 1<x/y+1<2
So the max value 10 + 10 / (x/y+1) can assume is 20 ....
Look at the answer options --- anything above 20 should be the answer ...Hence E
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by debmalya_dutta » Fri Sep 17, 2010 11:56 am
Question 2: If xy <> 0, what is the value of 1/x + 1/y?
(1) 1 / (x+y) = -1
(2) xy = 6(x+y)
The question is asking what is the value of (x+y)/xy

Statement 1 gives me only the value of x+y and hence insufficient
Statement 2 can be reorganised as
(x+y)/xy = 1/6
So statement 2 is sufficient
Hence B
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by Brian@VeritasPrep » Fri Sep 17, 2010 2:07 pm
Hey Deb,

Really nice explanation - I like the way that you worked that out algebraically!

Just one quick edit to your calculation last line - you say that the MAX value is less than 20. and we can also prove that the MINIMUM value is greater than 15:

1 < x/y + 1 < 2

So:

10/(x/y + 1) will be greater than 5 (which is 10/2 and less than 10 (which is 10/1), so the entire value:

10 + 10/(x/y + 1) will be greater than 15 and less than 20. D is therefore correct.

Great explanation!
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by nagar.sandeep » Sat Sep 18, 2010 7:26 am
Thanks alot Brian.

/Sandeep

Brian@VeritasPrep wrote:Hey Sandeep,

Good questions - let me tackle that first one, which I think is a pretty good example of GMAT strategy. I did this in less than two minutes using this strategy:

1) Simplify the algebra to get in a position to better solve for k:

x+y is a common denominator on the left, so you really have:

(10x + 20y)/(x+y) = k

Multiply out the denominator to get:

10x + 20y = kx + ky

Now, there are three variables and only one equation, so you can't really solve for k algebraically, so I:

2) Plug in easy-to use numbers to see if you can find a range of values (since they ask "what could be" and not "what is" the value of k, pinning k into a range is probably the best you can do):

Since x < y, let's call them 1 and 2:

10(1) + 20(2) = 1k + 2k
50 = 3k
k = 17.3333

Doing another one, let's try x = 2 and y = 5 to see if different numbers give us something wildly different:

10(2) + 20(5) = 2k + 5k
120 = 7k
k = something just over 17

Since both of our values are in that just-over-17 range, choice D, 18, is looking like a pretty good bet. Maybe try 1-2 more to see if you can get lower than 17 (because 15 looms somewhat close):

Call x 2 and y 3:

10(2) + 20(3) = 2k + 3k
80 = 5k
k = 20

Well, it seems like we can definitely get above the 17 range as high as 20, and now our range of potential values is 17-20, so I'm ready to call it and say that 18 is possible. If we use nonintegers - and we're allowed to...the question doesn't prohibit it - we would be able to tweak our current estimates to get to 18, but with any combinations we've tried already we haven't been able to near 15. The correct answer has to be 18.

A few strategic takeaways here:

1) Cleaning up the algebra is helpful - in most questions that feature variables in a denominator, your goal at some point will be to multiply out that denominator.

2) Because they ask "which COULD BE the value of k", you don't have to solve directly for it if you can use easy-to-plug numbers to establish a range of potential values. You don't need to know exactly what k IS, but rather just what k might look like. Since this question doesn't involve integers specifically, it's not a divisibility problem (so you don't have to worry that 18 might not share necessary factors), so an estimate is probably the most efficient way to tackle this one.
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by fatalityish » Tue Sep 21, 2010 1:52 pm
Hey Brian and Deb good ways to solve the first problem. Second one is pretty much simple and is to be done the way Deb did it.
For the first question, we can solve it in a slightly different way, which i felt was faster.

1. Rearrange the equation as follows:-

(10x+20y)/(x+y) = k
or
(x+2y)/(x+y) = k/10 ..............eqn (I)

2. Since we have all numerical values for possible values of k, just put the values from the answers in the equation (I)

Case (A)
(x+2y)/(x+y) = 1
=>
x+2y = x+y (we can cancel x and y since they are greater than 0)
=>
2=1 (we all know tat is not possible :P)

Case (B)
(x+2y)/(x+y) = 1.2
=>
0.8y = 0.2x
=>
4y = x
not possible as x<y

Case (C)
Similarly as above we will get y = x which is again not possible.

Case (D)
Same as Case B when we solve this we get y = 4x
which is possible.

We can stop solving at this point but if required to be sure we can solve for Case (E) as well and check. For Case (E) equation comes as x = -0.5y . This is not possible either as both x and y are positive numbers.


Solving the sum by this method took me around 30-35 secs.
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