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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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Math Expert
Joined: 02 Sep 2009
Posts: 58340
If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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06 Feb 2019, 02:35
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35% (medium)

Question Stats:

61% (01:39) correct 39% (02:02) wrong based on 28 sessions

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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

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Math Expert
Joined: 02 Aug 2009
Posts: 7960
Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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06 Feb 2019, 04:54
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
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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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06 Feb 2019, 04:59
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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06 Feb 2019, 05:11
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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Joined: 09 Jun 2014
Posts: 352
Location: India
Concentration: General Management, Operations
Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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06 Feb 2019, 05:20
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.

Math Expert
Joined: 02 Aug 2009
Posts: 7960
Re: If 72.42 = k(24+n/100), where k and n are positive integers and n <100  [#permalink]

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06 Feb 2019, 07:51
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.

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

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06 Feb 2019, 08:00
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.

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.
If 72.42 = k(24+n/100), where k and n are positive integers and n <100   [#permalink] 06 Feb 2019, 08:00
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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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