Solve the problem.
-The perimeter of a rectangle is 50 m. If the width were doubled and the length were increased by 5 m, the perimeter would be 74 m. What are the length and width of the rectangle?
A)width 7 m, length 12 m
B)width 12 m, length 12 m
C)width 7 m, length 18 m
D)width 18 m, length 7 m

Question

Quizplus · Accepted Answer

Let's say the width is "w" and the length is "l". 
According to the problem, 
2l + 2w = 50 
l + w = 25 
Also, 
2(l+5) + 2(2w) = 74 (width doubled and length increased by 5) 
2l + 4w + 10 = 74 
2l + 4w = 64 
l + 2w = 32

Now we have two equations: 
l + w = 25 
l + 2w = 32

Solving for "l" and "w", we get: 
w = 7 and l = 18

Therefore, the width is 7 m and the length is 18 m. Let's represent the width with "w" and the length with "l".
From the problem, we have two equations:
2w + 2l = 50  --> equation 1 
2(w x 2) + (l + 5) x 2 = 74  --> equation 2

To solve for w and l, we need to solve for either variable in terms of the other. From equation 1, we can solve for l:
2w + 2l = 50
2l = 50 - 2w
l = 25 - w

Now we can substitute this into equation 2:
2(w x 2) + (l + 5) x 2 = 74
4w + 2(l + 5) = 74
4w + 2(25 - w + 5) = 74
2w + 60 = 74
2w = 14
w = 7

Now we can use equation 1 to solve for l:
2w + 2l = 50
2(7) + 2l = 50
14 + 2l = 50
2l = 36
l = 18

Therefore, the width is 7 m and the length is 18 m. So, the answer is C.

Solve the Problem. -The Perimeter of a Rectangle Is 50 M. If the M