Thread: PSAT questions: Probablility & Area of Circles

1. PSAT questions: Probablility & Area of Circles

I am surprised to see these problems on the PSATs. I can't imagine what the SATS will be like.

I did not understand problems 15 and 18. They qualify as 'Hard' questions. Do you think so?

I ransacked my brain, trying to find a solution but my efforts were to no avail. (Pardon the messy work ) The diagram on number 18 included everything except the line of congruencies and the area marked 'x' and 'y'.

2. The circle one.

Each unshaded reason equals x. You know the sum of the two unshaded reasons, 2x, equals the arithmetic mean, or average, of the shaded reasons. The two shaded regions are equal to eachother, as they have the same central angle since they are vertical angles. Thus, the average area of the shaded reagions is the value of each shaded region. So, each shaded region is equal to 2 unshaded regions, or 2x. You now have two regions with values of x degrees, and two more regions with values of 2x degrees, altogether for a total of 6x.

6x = 360
x = 60

3. For the probability one, I think it is like this.

If you spin a 1 on the first spinner, you have a 0 chance of getting an 8.
If you spin a 2, you need a 6, so that is a 1/4 x 1/6 chance, or 1/24
If you spin a 3, you need a 5 OR 6, so that is a 1/4 x 1/3 chance, or 1/12.
If you spin a 4, you need a 4, 5, OR 6, so that is a 1/4 x 1/2 chance, or 1/8.

1/24 + 1/12 + 1/8
1/24 + 2/24 + 3/24 = 6/24 = 1/4

4. Originally Posted by Masterthief1324
I am surprised to see these problems on the PSATs. I can't imagine what the SATS will be like.

I did not understand problems 15 and 18. They qualify as 'Hard' questions. Do you think so?

I ransacked my brain, trying to find a solution but my efforts were to no avail. (Pardon the messy work ) The diagram on number 18 included everything except the line of congruencies and the area marked 'x' and 'y'.
For Problem 18.

One diameter divides the circle in two equal regions. There are two diameters and they divide four equal regions, two of them shaded and other two unshaded.

We known circle has 360 degree at its center.

Therefore, 4x = 360 => x=90.

EDIT: Unshaded regions = Average area of shaded regions, I misinterpreted this.

mattio is right, actually. Apology.

5. Originally Posted by geton
For Problem 18.

One diameter divides the circle in two equal regions. There are two diameters and they divide four equal regions, two of them shaded and other two unshaded.

We known circle has 360 degree at its center.

Therefore, 4x = 360 => x=90.
The shaded regions are 2x and the unshaded regions are x, they are not equal in area.

6. Originally Posted by mattio
The shaded regions are 2x and the unshaded regions are x, they are not equal in area.
Look, diameter always divide a circle equally. If two diameter intersect, there will be four equal regions in the circle.

In the question they said, unshaded regions is equal to the area of the shaded regions.

7. Originally Posted by geton
Look, diameter always divide a circle equally. If two diameter intersect, there will be four equal regions in the circle.

In the question they said, unshaded regions is equal to the area of the shaded regions.
Two diameters intersecting do NOT divide a circle into four equal regions unless they are perpendicular, and it does not state that. Moreover, it states that the SUM of the unshaded regions is equal to the AVERAGE area of the shaded regions, and since the shaded regions are equal area because of their equal central angle (vertical angles), the average area of the two shaded regions is the actual area of each shaded region. Thus, two unshaded regions = one shaded region. If an unshaded region is x and a shaded region is two of those, the shaded regions are 2x. Each half of the circle is 3x, two halves make 6x. 360 = 6x, x = 60.

8. Originally Posted by mattio
The circle one.

Each unshaded reason equals x. You know the sum of the two unshaded reasons, 2x, equals the arithmetic mean, or average, of the shaded reasons. The two shaded regions are equal to eachother, as they have the same central angle since they are vertical angles. Thus, the average area of the shaded reagions is the value of each shaded region. So, each shaded region is equal to 2 unshaded regions, or 2x. You now have two regions with values of x degrees, and two more regions with values of 2x degrees, altogether for a total of 6x.

6x = 360
x = 60
I think I see where you're getting at. The bolded statement is stated in the problem as a conditional-fact: "If the sum of the areas of the unshaded region is equal to the average of the areas of the shaded region".

1. 2*(measure angle of x) = (measure angle y); Stated by the problem

2. Written in terms of Area (respectively):

3. Written in terms of Degrees (Respectively):
2x = (1)y

4. 4x = (2)y

5. 2x + 2y = 360 or ...
2x + (4x) = 360

6. Simplify: 6x = 360; x = 60

Is this proof correct?

9. Originally Posted by geton
For Problem 18.

One diameter divides the circle in two equal regions. There are two diameters and they divide four equal regions, two of them shaded and other two unshaded.

We known circle has 360 degree at its center.

Therefore, 4x = 360 => x=90.
If x were 90, then x = y which cannot be. (It cannot be because the sum of the area of two angle x's is equal to ONE area of angle 'y', therefore the angle y has the value of 2 angle x's)

x = y is not a statement of equality.
x = x is a statement of equality.