GmatFoc

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GmatFoc

by gbb » Mon Mar 23, 2009 4:19 pm
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

answer D
Source: — Problem Solving |

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by cramya » Mon Mar 23, 2009 4:44 pm
https://www.beatthegmat.com/tough-algebr ... 13568.html

Awesome explanation from Ian.



My approach if we dint recongnize this. For a word prob if the answer choices are not in terms of variables but actual numbers then reverse engineering would be a good approach.

Given : x<y

Start from C

k=15

k=10x+20y/x+y

10x+20y = 15x+15y

5y=5x

y=x

Cant be true since we are given x<y so k must be greater than 15

k=18


10x+20y = 18x+18y

2y = 8x

y=4x

Possible

k=30

10x+20y = 30x+30y

-10y = 20x
y = -2x

Not possible since x and y are both positive and x<y

Choose D)


P.S: Ian's solution would be the easiest and the best but instead of randomly guessing(if we dint recognize this as a weighted average problem) the reverse engineering could serve as a back up method for solving the problem. In GMAT we dont have to go with a text book approach if somethign else strikes us first i.e as long as we can get to the answer within 2 minutes.


Just my 2 cents.... Good luck!

Regards,
CR

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by cramya » Mon Mar 23, 2009 4:50 pm
Sometimes TacoBell's "Think outside the box" helps in tests like GMAT. Lets use it with caution though.... :D

If u had the time lets check D

y=4x

x=1 y=4

10(1) + 20(4) / 1+4 = 18

10+80 / 5 = 18
90 = 90

Bingo... Move on.....

Regards,
CR

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Re: GmatFoc

by logitech » Mon Mar 23, 2009 8:16 pm
This equation can be condensed to:

10(x+y)/(x+y) + 10(y/(x+y))

We know the first part of the equation is 10 - so actually the problem asks what y/(x+y) can be...and we have a tip x<y

if x=y , our equation is 0.5

so we know that the answer needs to be more than 15 :idea:

If x=0 , our equation gives us = 1 so the answer needs to be less than 20

15 < ANSWER < 20

And only 18 fits the bill B-)
LGTCH
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