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tough math problem - measurement

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by gupta.amit3 » Sun Nov 09, 2008 8:41 am
Two measure standards R and S. 24 and 30 measured with R are 42 and 60 when they are measured with S, respectively. If 100 is acquired with S, what would its value be measured with R?

Can somebody help with explanation.
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Re: tough math problem - measurement

by logitech » Sun Nov 09, 2008 9:31 am
gupta.amit3 wrote:Two measure standards R and S. 24 and 30 measured with R are 42 and 60 when they are measured with S, respectively. If 100 is acquired with S, what would its value be measured with R?

Can somebody help with explanation.
Gupta, can you pleae post the answer choices and official answer. I also want to know the source of your problems.

Thanks!
LGTCH
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by sgarnepudi » Sun Nov 09, 2008 9:38 am
Yeah .. the ratios are so distinct .. can you please re-verify the numbers too
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by stop@800 » Sun Nov 09, 2008 10:20 am
The equations will be
Let relation be
S= aR + b

42 = 24a + b
60 = 30a + b

solve for A and B
we get a=3 and b = -30

S = 3R - 30

now we have
100 = 3R - 30

so R = 130/3

OA please??


I also think there is some gap in numbers. :)
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by scoobydooby » Mon Nov 10, 2008 3:08 am
50 as per R?

Could be that numbers below 30 in R, the digits gets reversed in S and the numbers 30 and above get doubled
so a value of 100 in S, greater than 30 would read 50 in R
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what I think

by technofreak » Thu Nov 13, 2008 12:23 pm
Hi,

I am a newbie on this site and just getting started with the preperations.

I came across this problem and here is what I think the solution is:

Case 1: when R weighs 24, S weighs 42
Case 2: when R weighs 30, S weighs 60

Thus,

S = (R*3)-30 (How I got to this is by looking at the numbers and differences and believing that GMAT questions don't require complicated calculations :))

for Case 1;

42=(24*3)-30

for case 2:
60=(30*3)-30

Now,

R = (S+30)/3

Thus, when S=100

R=(100+30)/3 = 130/3 =43.33

Thus, the answer is R = 43.33

I hope this helps, I couldn't really figure out any other explanation
:?
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by navdeepbajwa » Tue Oct 20, 2009 2:26 am
Case 1 R=24 & S=42
Case 2 R=30 & S=60

Diff in R=6 , diff in S=18

so diff in S=3* diff in R
I assumed this relation to hold

Let new R=X and S=100

3(X-30)=100-60

X=43.33
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by sanjana » Tue Oct 20, 2009 3:43 am
IMO : 43.33

Given :
R measures : 24 and 30
S measures : 42 and 60

Change in R : 30-24 = 6
Change in S : 60-42 = 18

A change in 6 on the R scale corresponds to a change in 18 on the S scale.

Hence,change in r/change in s = 6/18 = 1/3

Now S goes from 60 - 100(given) which implies a change of 40
hence
change in R = 40*1/3 = 13.33

R started at 30,hence when S is 100,R is 30+13.33 = 43.33
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by spidermanadda » Wed May 09, 2012 9:02 am
I think the Answer is 54..
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by neelgandham » Wed May 09, 2012 1:17 pm
spidermanadda, some archaeologist you are( Just kidding!). Now that you have extracted an old piece let me try to answer this question.

Let the scales R and S be linearly related. let S = aR+b, where a and b are real numbers.
From the question stem
42 = 24a + b
60 = 30a + b
Solving for a and b, we get the value of a = 3 and b = -30
If S = 100, 100 = R*3 - 30
So the value of R = 130/3

Let the scales R and S are not linearly related. and let the relation be in the form S = (a/R) + b
From the question stem
42 = (a/24) + b => 1008 = a + 24b
60 = (a/30) + b => 1800 = a + 30b
Subtract the first equation from the second.
1800 - 1008 = a + 30b - (a + 24b)
792 = 6b
a = -2160 and b = 132
So if S = 100 R = -2160/(100-132) = 67.5

Two different answers. This is definitely not a GOOD GMAT question!
Anil Gandham
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