[Video] Q27: A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. if these alterations decreased the area of the rectangle by 12 percent, what is the value of p ?

Explanation for Question 27 From the Math (Calc) Section on the Official Sat Practice Test 3

So looking at number 27, we're talking about a rectangle, 2 right? And it's all its dimensions are altered. 3 So it started out with the length times within an area. 4 And then we see that first off, the length was increased by 10%. 5 So it was changed to 1.1 times L this additional 0.1 6 gives us that additional 10% and then the width was, 7 um, decreased by P percent, right? So we'll do one minus P, 8 right? Because that gets rid of that P percent that we lost times the 9 original width. And then our areas decreased by 12%, 10 right? And one minus 0.12 is 0.8, 11 eight. So we now have only 88% of the original area. 12 And so those are a new dimensions. And we're going to use this information 13 over here to write ourselves an equation, right? 14 Where we have 0.8, eight times area is equal to 1.1 L 15 our length times one minus P times w our 16 whip right now, I'll get rid of the parentheses here. 17 Those aren't necessary. And so what you might notice right, 18 is that I have three variables happening, but only one equation. 19 But if I look closely at them, I know that length times width actually 20 gives me area. So I'm going to rewrite this equation so that my length 21 and my width are next to each other, 22 rather than separate. This is the same thing. 23 I've simply moved the width, um, 24 earlier up in the equation that doesn't change anything. 25 And I know that length times width is actually area. 26 So I'm going to rewrite this part. Here's an a instead, 27 well,...