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If 72.42 = k(24+n/100), where k and n are positive integers and n <100

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If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 01:35
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A
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Difficulty:

  35% (medium)

Question Stats:

44% (00:59) correct 56% (02:29) wrong based on 9 sessions

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Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 03:54
1
Bunuel wrote:
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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html


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Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 03:59
Bunuel wrote:
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



**** we know the range of n. we don't have any idea about k. This question is a perfect one for trail and error method.

So, 0<n<100.

when k=1,

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

\(72.42 = 1*(24+\frac{n}{100})\)

n/100 = 48.42

n = 48.42 *100.

n = 4842, which is far beyond our scope.

So, k can't be 1.

when k = 2,

\(72.42 = 2(24+\frac{n}{100})\)

72.42 = 48 + \(\frac{2n}{100}\)

24.42 =\(\frac{2n}{100}\)

2442/2 = n

still n is out side of the range.

k can't be 2.

when k=3,

72.42 = 72 + 3n/100

0.42 = 3n/100

3n = 42

n= 14

14 could be the value of n .

k can'be 4. if we take k=4, we get negative value for n.

So, k = 3. and n=14.

k + n = 3 + 14 = 17.

A is the correct answer.
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Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 04:11
Bunuel wrote:
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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Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 04:20
chetan2u wrote:
Bunuel wrote:
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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Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 06:51
prabsahi wrote:
chetan2u wrote:
Bunuel wrote:
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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html


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Manager
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Joined: 09 Jun 2014
Posts: 239
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If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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New post 06 Feb 2019, 07:00
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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If 72.42 = k(24+n/100), where k and n are positive integers and n <100   [#permalink] 06 Feb 2019, 07:00
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If 72.42 = k(24+n/100), where k and n are positive integers and n <100

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