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

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