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Three photographers, Lisa, Mike, and Norm, take photos of a wedding. The total of Lisa and Mikes's photos is 50 less tha

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by Vincen » Sat Dec 05, 2020 4:59 am

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Three photographers, Lisa, Mike, and Norm, take photos of a wedding. The total of Lisa and Mikes's photos is 50 less than the sum of Mike's and Norms. If Norm's 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

Answer: E

Source: EMPOWERgmat
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Re: Three photographers, Lisa, Mike, and Norm, take photos of a wedding. The total of Lisa and Mikes's photos is 50 less

by Brent@GMATPrepNow » Sat Dec 05, 2020 6:00 am
Vincen wrote: ↑
Sat Dec 05, 2020 4:59 am
 Three photographers, Lisa, Mike, and Norm, take photos of a wedding. The total of Lisa and Mikes's photos is 50 less than the sum of Mike's and Norms. If Norm's 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

Answer: E

Source: EMPOWERgmat
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
Brent Hanneson - Creator of GMATPrepNow.com
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