1. ## PSAT Help

I have a couple of odd ones I need help with from the practice sheet.

Equal numbers of cards that are marked either r, s, or t are placed in an empty box. If a card is drawn at random fro the box, what is probability that it will be marked either r or s?

A 1/6

B 1/3

C 1/2

D 2/3

E 3/4

I forget how to work with "or". I know its now 3/4, 1/3, or 1/6. They are too high or low.

In the figure below, find the perimeter of square ABCD.

A $\displaystyle 4\sqrt2$

B $\displaystyle 8\sqrt2$

C 8

D 12

E 16

I sloppily answered 8, thinking since the graph only went to two that the length from 0 to c (or any other letter) was 2. I did the Pythagorean theorem and got 2 as one side of the square. Then I added the sides of the perimeter and got 8.

On review of the problem, I saw my mistake and redid it using 3, which gives me 12.

However, that is not the answer.

Let the function f be defined by f(x) = $\displaystyle \sqrt2x$. If f(c) = 4, which is the value of c?

A 4

B 8

C 12

D 32

E 64

I thought this was just a complicated way of saying "plug in x as 4 and see what happens."

Apparently, it's not.

Finally,

*picture what looks like an isosceles triangle (below) that isn't denoted as one. It has angle 'b' as the top angle, angle 'a' as the left angle, and angle 'c' as the right angle.

In the triangle ABC above, if a > b > c, which of the following must be true?

I. 60 < a < 180

II. 45 < b < 90

III 0 < c < 60

A I only

B II only

C I and III only

D II and III only

E I, II, and III

I thought I couldn't be true because an angle at 179.9 would make the triangle something else.
Perhaps it does work in retrospect.
Either II or III is wrong, I'm not sure which one. Two angles with the same properties can't have different properties.

Sigh, I need LaTex help. This is a hard language for me to learn.

Thanks for the help!

2. Originally Posted by Truthbetold
...

In the figure below, find the perimeter of square ABCD.

A $\displaystyle 4/sqrt2$
B $\displaystyle 8/sqrt2$
C 8
D 12
E 16
...
Hello,

use Pythagorean theorem to calculate the length of one side:

$\displaystyle s^2 = 2^2+2^2 = 8~\implies~s=2\sqrt{2}$

That means the perimeter is $\displaystyle 8\sqrt{2}$. So it's answer B

3. Originally Posted by Truthbetold
...
Let the function f be defined by f(x) = $\displaystyle /sqrt/2x$. If f(c) = 4, which is the value of c?
A 4
B 8
C 12
D 32
E 64

...
Hello,

this is your function: $\displaystyle f(x)=\sqrt{2x}$ . If you plug in c instead of x the value of the function is 4:

$\displaystyle 4 = \sqrt{2c}~\implies~16=2c~\implies~c = 8$ So it's answer B again.

4. ## Oye

Hello,

this is your function: . If you plug in c instead of x the value of the function is 4:

OOOh. I see. I forgot to set it $\displaystyle \sqrt2x$ to 4. I set it to 0. Okay, got it.

Hello,

use Pythagorean theorem to calculate the length of one side:

That means the perimeter is . So it's answer B
D'oh!

5. Originally Posted by Truthbetold
Equal numbers of cards that are marked either r, s, or t are placed in an empty box. If a card is drawn at random fro the box, what is probability that it will be marked either r or s?
A 1/6
B 1/3
C 1/2
D 2/3
E 3/4
I forget how to work with "or". I know its now 3/4, 1/3, or 1/6. They are too high or low.

...
Hello,

the general formula is:

$\displaystyle p(r \vee s) = p(r)+p(s)-p(r \wedge s)$

Since it is impossible that one card is marked r and s (that means $\displaystyle p(r \wedge s) = 0$ you only have to add the probabilities p(r) and p(s):

$\displaystyle p(r \vee s) = \frac13+\frac13-0=\frac23$ . So it's answer D