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Three photographers, Lisa, Mike and Norm, take photos of a wedding.

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Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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Updated on: 21 Jul 2015, 07:46
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Difficulty:

15% (low)

Question Stats:

84% (02:01) correct 16% (02:13) wrong based on 71 sessions

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Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take?

A. 40
B. 50
C. 60
D. 80
E. 90

Originally posted by kk07734 on 21 Jul 2015, 07:16.
Last edited by kk07734 on 21 Jul 2015, 07:46, edited 1 time in total.
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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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21 Jul 2015, 07:32
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kk07734 wrote:
Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take?

A. 40
B. 50
C. 60
D. 80
E. 90

L+M = M+N-50
i.e. L = N-50

Also N = 10+2L

i.e. L = (10+2L)-50
i.e. L = 40
i.e. N = 90

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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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21 Jul 2015, 09:03
L+M = (M+N) - 50
L = N-50

N = 2L + 10
N = 2(N-50)+10
N = 2N - 100 + 10
N = 90. Ans (E).
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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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22 Jul 2015, 08:35
Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take?

L+M=M+N-50 / N=2L+10

50=M+N-L-M
50=N-L
50=2L+10-L
40=L
2(40)+10=90

A. 40
B. 50
C. 60
D. 80
E. 90
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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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23 Oct 2018, 19:27
kk07734 wrote:
Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take?

A. 40
B. 50
C. 60
D. 80
E. 90

We can create the equations:

L + M = M + N - 50

L = N - 50

and

N = 10 + 2L

Substituting, we have:

N = 10 + 2(N - 50)

N = 10 + 2N - 100

N = 90

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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.  [#permalink]

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09 Dec 2018, 10:31
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kk07734 wrote:
Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take?

A. 40
B. 50
C. 60
D. 80
E. 90

We can also solve the question using 1 variable

Norms photos number 10 more than twice the number of Lisa's photos
Let x = the number of photographs that Lisa took.
So, 2x + 10 = the number of photographs that Norm took.

The total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms
We can write: (# of Lisa's photos) + (# of Mike's photos) = (# of Mike's photos) + (# of Norm's photos) - 50
Subtract (# of Mike's photos) from both sides to get: (# of Lisa's photos) = (# of Norm's photos) - 50
Plug in pre-defined values to get: (x) = (2x + 10) - 50
Simplify: x = 2x - 40
Solve: x = 40

How many photos did Norm Take?
2x + 10 = the number of photographs that Norm took.
So, 2x + 10 = 2(40) + 10 = 90

Cheers,
Brent
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Re: Three photographers, Lisa, Mike and Norm, take photos of a wedding.   [#permalink] 09 Dec 2018, 10:31
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