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Bunuel
If \(72.42 = k(24+\frac{n}{100})\), where k and n are positive integers and n < 100, then k + n =

A 17
B 16
C 15
D 14
E 13

\(72.42 = k(24+\frac{n}{100})\)

equate both sides
72.42 = 3 *( 24+ 14/100)
k=3 , n= 14
K+N= 17
IMO A
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Bunuel
If \(72.42 = k(24+\frac{n}{100})\), where k and n are positive integers and n < 100, then k + n =

A 17
B 16
C 15
D 14
E 13


Let us get both sides in similar terms..

\(72.42 = k(24+\frac{n}{100})\).....\(72+0.42 =72+\frac{42}{100}=3(24+\frac{14}{100})= k(24+\frac{n}{100})\)
Therefore equating both sides, we get k as 3 and n as 14, so k+n=3+14=17..

A



Thanks for the nice solution..

However I wanted to know whats wrong with my approach..

If I adjust the main equation ..I get K= 7242 % (2400+N)

Now I was looking for value of N that will make K as a positive integer..so the value of N was
N=42 (since 2400+ 42 can divide 7242 giving K=3 as positive integer)
However this means N+K= 3+42=45 (Out of range of the options)

With my approach I will and up taking 3 mins in the test and no solution.


Is there any alternative approach to try those problems.

Most important question:How do you identify what approach needs to be taken in these problems.

Please help !!
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prabsahi
chetan2u
Bunuel
If \(72.42 = k(24+\frac{n}{100})\), where k and n are positive integers and n < 100, then k + n =

A 17
B 16
C 15
D 14
E 13


Let us get both sides in similar terms..

\(72.42 = k(24+\frac{n}{100})\).....\(72+0.42 =72+\frac{42}{100}=3(24+\frac{14}{100})= k(24+\frac{n}{100})\)
Therefore equating both sides, we get k as 3 and n as 14, so k+n=3+14=17..

A



Thanks for the nice solution..

However I wanted to know whats wrong with my approach..

If I adjust the main equation ..I get K= 7242 % (2400+N)

Now I was looking for value of N that will make K as a positive integer..so the value of N was
N=42 (since 2400+ 42 can divide 7242 giving K=3 as positive integer)
However this means N+K= 3+42=45 (Out of range of the options)

With my approach I will and up taking 3 mins in the test and no solution.


Is there any alternative approach to try those problems.

Most important question:How do you identify what approach needs to be taken in these problems.

Please help !!


You are perfectly fine with the solution, but you have gone wrong in the highlighted part..
\(k=\frac{7242}{2400+n}\).. Looking at 7242 and 2400 doen below, you should realize 2400*3=7200..
7242 divided by 3 is 2414=2400+14 so our fraction becomes \(3=\frac{7242}{2400+14}\)
k+n=3+14=17
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Let us get both sides in similar terms..

\(72.42 = k(24+\frac{n}{100})\).....\(72+0.42 =72+\frac{42}{100}=3(24+\frac{14}{100})= k(24+\frac{n}{100})\)
Therefore equating both sides, we get k as 3 and n as 14, so k+n=3+14=17..

A[/quote]



Thanks for the nice solution..

However I wanted to know whats wrong with my approach..

If I adjust the main equation ..I get K= 7242 % (2400+N)

Now I was looking for value of N that will make K as a positive integer..so the value of N was
N=42 (since 2400+ 42 can divide 7242 giving K=3 as positive integer)
However this means N+K= 3+42=45 (Out of range of the options)

With my approach I will and up taking 3 mins in the test and no solution.


Is there any alternative approach to try those problems.

Most important question:How do you identify what approach needs to be taken in these problems.

Please help !![/quote]


You are perfectly fine with the solution, but you have gone wrong in the highlighted part..
\(k=\frac{7242}{2400+n}\).. Looking at 7242 and 2400 doen below, you should realize 2400*3=7200..
7242 divided by 3 is 2414=2400+14 so our fraction becomes \(3=\frac{7242}{2400+14}\)
k+n=3+14=17[/quote]



Just got my mistake.Thank you so much for seeing it through and pointing it out :)

I guess the mistake was my calculation 2442*3 will give 7326 and not 7242.
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