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 variableNorms photos number 10 more than twice the number of Lisa's photosLet 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 NormsWe 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 =
40How 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