Having Problems with % and Rate Problems.

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Having Problems with % and Rate Problems.

by Makushr1 » Mon Aug 16, 2010 6:23 am
I have always been very good at math and algebra, but for some reason these types of questions keep giving me problems. Any suggestions on what to do for these types? After seeing how to do the problem, I understand it, but keep makin the same mistakes. What messes me up are when there are multiple variables in the question. I usually sub in some numbers, but it gets confusing at the end when there are variables in the answers.

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?
y/200
2y
50y
50/y
5000/y

90 students represent x percent of the boys at Jones Elementary School. If the boys at Jones Elementary make up 40% of the total school population of x students, what is x?
125
150
225
250
500

It takes machine A x hours to manufacture a deck of cards that machine B can manufacture in y hours. If machine A operates alone for z hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?
y(100x - z) /(x + y)
Source: — Quantitative Reasoning |

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by selango » Mon Aug 16, 2010 7:50 am
m>0,y>0

x=m/100*2y

m=a/100*x?[we need to find a]

sub x value in above equation

=a/100 * m/100 * 2y

m=am*y/5000

a=5000/y
--Anand--

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by selango » Mon Aug 16, 2010 8:00 am
Let the total number of Boys be B.

90=x/100*B

B=40/100*x

Sub B value in first equation.

90=x/100*40/100*x

x^2=22500-->x=150
--Anand--

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by selango » Mon Aug 16, 2010 8:15 am
A take x hrs to manufacture deck.

So for 1 hr it do 1/x of work.

For z hr,z/x of work.

Remaining Work done need to be done by both A and B=(100-Z/x)

Together they take xy/x+y hours to manufacture 1 deck.

To manufacture (100-Z/x) decks they take (100-Z/x)*xy/(x+y) hrs

y*(100x - z)/(x + y
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by Gurpinder » Mon Aug 16, 2010 9:42 am
selango wrote:Let the total number of Boys be B.

90=x/100*B

B=40/100*x

Sub B value in first equation.

90=x/100*40/100*x

x^2=22500-->x=150
hey dude,

i get 135, what am i doing wrong?

im using proportions. 40/60=90/x -------- 4x=540 >> x = 135

thanks
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by Makushr1 » Mon Aug 16, 2010 9:57 am
Selango, thanks for the help with those problems.

However, I was more wondering if anyone had suggestions on how to practice for these problems? For some reason, they are the only ones I continually have difficulty with.

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by claymauldin » Tue Sep 14, 2010 2:00 pm
Multiply x by (x/100) first, then multiply each side by (100/40). this will equal 225 = (x^2)/100. Next, multiply by 100 on each side and take the square root of 22,500.
Gurpinder wrote:
selango wrote:Let the total number of Boys be B.

90=x/100*B

B=40/100*x

Sub B value in first equation.

90=x/100*40/100*x

x^2=22500-->x=150
hey dude,

i get 135, what am i doing wrong?

im using proportions. 40/60=90/x -------- 4x=540 >> x = 135

thanks

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by Tani » Fri Sep 17, 2010 6:04 am
you are assuming that the 90 students are all boys - in fact, there are only 60 boys.
Tani Wolff

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by fatalityish » Tue Sep 21, 2010 2:35 pm
Makushr the best way to tackle these problems is to get the basic concept of percent and work related problems, clear. Since the questions which you have given in this thread are the simplest ones, i would recommend that go through OG (math review section) for the explanation of these things. OG has given easy to understand methods and formulae for this. Also there are a couple pf videos on You tube for the explantion of the same.
Links: -
percentage
https://www.youtube.com/watch?v=7JYaE0ogtMI

work related problem
https://www.youtube.com/watch?v=64MKDGF_D08
Ishaan

Makushr1 wrote:Selango, thanks for the help with those problems.

However, I was more wondering if anyone had suggestions on how to practice for these problems? For some reason, they are the only ones I continually have difficulty with.