<p>hey all just came across this question on a practice test.The question fairly simple-thing is when i solve it using algebra and when i plug in i get get 2 different answers. would really appreciate help on it...</p>
<p>What happens to the area of a rectangle with breadth h^2 and length 3p when h is doubled and p is halved?</p>
<p>(A)The area is squared.
(B)The area is multiplied by 4.
(C)The area is doubled.
(D)The area is halved.
(E)The area remains the same.</p>
<p>its c i think</p>
<p>i have no idea lol</p>
<p>the area without changes would be 3ph^2. with the change, it would be (4h^2)(1.5p), so it will be 6ph^2, which is doubled the original.</p>
<p>If you have trouble seeing the answer quickly I recommend that for those kinds of problems sometimes its just easier to assign a value to h and p, and quickly work it out and see which answer fits.</p>
<p>mkevb1 how is h^2 squared 4h^2??</p>
<p>skywalker: With these type of problems, if you're comfortable with your algebra, then just keep your head straight, write everything out, and you'll be fine.</p>
<p>Originally the width was h^2 and the length was 3p, so that the area is 3(h^2)(p); I'm moving the 3 to the beginning, as multiplication is commutative and it will make the problem easier. They say that h is doubled and that p is halved. So that your h becomes 2h and your p becomes p/2:</p>
<p>3[(2h)^2][p/2]</p>
<p>It is important not to write:</p>
<p>3[2(h)^2][p/2]</p>
<p>as you would be doubling h^2, not h.</p>
<p>If you work with the first problem, and square 2h, it becomes 4h^2:</p>
<p>3(4h^2)(p/2)</p>
<p>At this point, you can combine the constants. 4 and 3 is 12, divided by 2 is 6:</p>
<p>6(h^2)(p)</p>
<p>which, if we compare to the original 3(h^2)(p) is double the original area. If you want to compare them more mathematically, you would ask what "the new area" is to "the old area," or:</p>
<p>[6(h^2)(p)]/[3(h^2)(p)]</p>
<p>and, by canceling like terms, and simple division, you get the answer 2, or verbally, the area has been doubled.</p>
<p>What is breadth?</p>
<p>h^2 squared is not 4h^2. however, when you double the "h" in "h^2", it becomes "(2h)^2", which in result is 4h^2.</p>
<p>(C) The area is doubled</p>
<p>U've got: h^2 and 3p
Hypothetically, lets make h=5 and p=2
plug it into the equation
thus the area would equal 150 <------
Now, lets double h, but halve p
then, h=10 and p=1
(10)^2 times 3(1) =300 <------</p>
<p>It is c</p>
<p>Original Area = h^2(3p) = 3(h^2)p</p>
<p>new area = (2h)^2(1.5p) = 4h^2(1.5p) = 4 * 1.5 * h^2 *p = 6(h^2)p</p>
<p>remember h^2 is h times h. h^2 when doubled works to 4h. now half the original p from 3p to 1.5p or 3/2. the orig area is 6hp, and the new area is 4h times 3/2p. this works to 12/2hp or 6hp. same deal *****es</p>
<p>just kidding, but it was cool wasnt it</p>
<p>syn thanks a lot... i was making that very mistake(doubling h^2)...</p>
<p>definitely c!</p>