Three photographers, Lisa, Mike and Norm, take photos of a w

[BTGmoderatorDC]

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

OA E

Source: EMPOWERgmat

[GMATGuruNY]

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

The sum for Lisa and Mike is 50 less than the sum for Norm and Mike.
Since Mike is common to both sums, the 50-photo difference must be yielded solely by Lisa and Norm.
Implication:
Lisa took 50 fewer photos than Norm.

We can PLUG IN THE ANSWERS, which represent the number of photos taken by Norm.
When the correct answer is plugged in, Norm will take 50 more photos than Lisa.
Let N = Norm and L = Lisa.

D: N=80
Since Norm's total is 10 more than twice Lisa's, we get:
2L = 70.
L = 35.
In this case, N-L = 80-35 = 45.
Since the difference is too small, a greater answer choice is needed.

The correct answer is E.

[Jay@ManhattanReview]

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

OA E

Source: EMPOWERgmat

Say the number of photos Lisa, Mike and Norm, take are L, M, and N, respectively.

Given the total of Lisa and Mikes photos is 50 less than the sum of Mike's and Norms, we have

L + M = M + N - 50
L = N - 50 ---(1)

Given Norms photos number 10 more than twice the number of Lisa's photos, we have

N = 2L + 10 ---(2)

Plugging-in the value of L from eqn (1) in eqn (2), we have

N = 2(N - 50) + 10

N = 90

The correct answer: E

Hope this helps!

-Jay
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[Brent@GMATPrepNow]

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

Answer: E

Cheers,
Brent

[Scott@TargetTestPrep]

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

OA E

Source: EMPOWERgmat

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

Answer: E