SAT math problem on the area of a rectangle

[m_0]

Hello,

I'm not sure if this is the correct forum, but I'm confused about an SAT math problem...

The length and width of a certain rectangle are both decreased by 50%. If the L and W of the new rectangle are then increased by 40%, the area of the resulting rectangle is what percent less than the area of the original rectangle?

The answer is 51%, but the explanation is rather confusing. Could someone explain why this is the case (and why the answer isn't 10%)?

 

[Deleted member 4993]

Hello,

I'm not sure if this is the correct forum, but I'm confused about an SAT math problem...

The length and width of a certain rectangle are both decreased by 50%. If the L and W of the new rectangle are then increased by 40%, the area of the resulting rectangle is what percent less than the area of the original rectangle?

The answer is 51%, but the explanation is rather confusing. Could someone explain why this is the case (and why the answer isn't 10%)?

Are you just guessing it to be 10%?

Can you show us what would be the percent loss in area when the length and the width were decreased by 50%?

 

[m_0]

Are you just guessing it to be 10%?

Can you show us what would be the percent loss in area when the length and the width were decreased by 50%?

Well, I originally subtracted 50 from 100 to get 50% remaining. Add in 40%, that's 90%. 100 - 90 -> 10%. That clearly isn't right, though...

 

[pka]

If $\displaystyle L=0.5l$ then $\displaystyle L+0.4L=0.7l$ where $\displaystyle l$ is the original length. What is the new area?

 

[m_0]

If $\displaystyle L=0.5l$ then $\displaystyle L+0.4L=0.7l$ where $\displaystyle l$ is the original length. What is the new area?

$\displaystyle L = .14l$

Correct?

 

[pka]

$\displaystyle L = .14l$
Correct?

NO!
If $\displaystyle l$ is the original length then $\displaystyle L=0.5l$.
The increase that by $\displaystyle 0.4$ or $\displaystyle .7l$.

 

[Mrspi]

Hello,

I'm not sure if this is the correct forum, but I'm confused about an SAT math problem...

The length and width of a certain rectangle are both decreased by 50%. If the L and W of the new rectangle are then increased by 40%, the area of the resulting rectangle is what percent less than the area of the original rectangle?

The answer is 51%, but the explanation is rather confusing. Could someone explain why this is the case (and why the answer isn't 10%)?

Maybe a concrete example might help you understand (since you appear to be a little shaky with general expressions involving percent).

Let's start with a rectangle with a length of 20 units and a width of 10 units. What is the area of this rectangle? 200 units², right?

Now, DECREASE both the length and the width by 50%. The new length will be 20 units - 50% of 20 units, or 20 - 10, or 10 units. The new width will be 10 units - 50% of 10 units, or 10 - 5, or 5 units. And we've got a new rectangle with length 10 units and width 5 units. What is the area of the NEW rectangle? Well, it is 50 units². And if we look at the percent of area LOST, we see that there has been a decrease of 150 units² (200 units² - 50 units²) which is a decrease of (150/200)*100%, or 75%.

Now....take that 10 by 5 rectangle, and INCREASE both the length and width by 40%:

10 + .40(10) = 14 units for the new length
5 + .40(5) = 7 units for the new width

And the new area is 14(7), or 98 units²

How does the area of this new rectangle compare to the area of the rectangle we started with? It is smaller by 200 - 98, or 102 units². The percent of decrease is (102/200)*100%.....which is 51%.

I hope this helps you understand what is happening here.

 

[m_0]

Mrspi,

That makes perfect sense. Thank you very much (thanks to everyone else who helped, too!)

Maybe a concrete example might help you understand (since you appear to be a little shaky with general expressions involving percent).

Let's start with a rectangle with a length of 20 units and a width of 10 units. What is the area of this rectangle? 200 units², right?

Now, DECREASE both the length and the width by 50%. The new length will be 20 units - 50% of 20 units, or 20 - 10, or 10 units. The new width will be 10 units - 50% of 10 units, or 10 - 5, or 5 units. And we've got a new rectangle with length 10 units and width 5 units. What is the area of the NEW rectangle? Well, it is 50 units². And if we look at the percent of area LOST, we see that there has been a decrease of 150 units² (200 units² - 50 units²) which is a decrease of (150/200)*100%, or 75%.

Now....take that 10 by 5 rectangle, and INCREASE both the length and width by 40%:

10 + .40(10) = 14 units for the new length
5 + .40(5) = 7 units for the new width

And the new area is 14(7), or 98 units²

How does the area of this new rectangle compare to the area of the rectangle we started with? It is smaller by 200 - 98, or 102 units². The percent of decrease is (102/200)*100%.....which is 51%.

I hope this helps you understand what is happening here.