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Question

Rectangle A has a length of 300 inches and a width of 200 inches. Rectangle A has the same area as Rectangle B, but has different dimensions. The width of Rectangle B is 10% shorter than the width of Rectangle A as shown.    Which percent represents how much longer the length of Rectangle B must be to maintain the same area as Rectangle A?  

 

It must be 10% longer than Rectangle A.  

It must be 1119% longer than Rectangle A.  

It must be 200% longer than Rectangle A.  

It must be 400% longer than Rectangle A.

Answer 1

This problem is a great example of T.M.I. (too much information).  It gives percents and asks for percents, so we really don't need to know any dimensions.

 

The areas are the same:    L_a*W_a = L_b*W_b

How much longer (in percent) does L_b need to be than L_a?   That is, what is 100%*(L_b - L_a)/L_a

            L_a*W_a = L_b*W_b

            L_a*W_a = L_b*(0.9*W_a)

            (1/0.9)*W_a/W_a  = L_b/L_a

            111.1%  = L_b/L_a

             L_b = 111.1% * L_a         or    L_b is 11.1% longer than L_a     (near 10%)

Answer 2

First, we write assign variables to represent length and width.

 

Let x = length

Let y = width

Area = xy

 

For Rectangle A:

 

x = 300 in

y = 200 in

Area = 60000 in²

 

For Rectangle B:

 

y = 200 - 200(0.10) = 180 in

Area = 60000 in²

 

 

Now that we written down our known parameters, we must find the length of Rectangle B.  Then compare it to the length of Rectangle A to get the percent.

 

x = Area / y

 

x = 60000 in² / 180 in

x = 333.33 in

 

Since the length of rectangle B is longer than the length of rectangle A, we are on the right track being that the answer choices has the key term "longer".

 

Now we set up an equation to find the percent.

 

300 + 300r = 333.33

 

r is the percent in decimal form.

 

300r = 33.33

     r = 0.1111

 

Multiply the value of r by 100. 

 

0.1111 * 100 = 11%

 

This is close to 10%.

 

Therefore, the length of rectangle B must be 10% longer than rectangle A.