Open ended and multiple choice SAT questions?
Question:
a) 100 b) 120 c) 140 d) 180 e) 200
2) The circumference of circle II is 4 feet longer than the circumference of circle I. How many feet longer is the radius of circle II than the radius of circle I?
a) 1/4pi b) 2/pi c) 1/pi d) 2 e) cannot be determined with info
3) Refer to the link below, C is the center of the circle, what is the value of c?
4) From 1990 until 1998 the value of an intestment increasted by 10% every year. The value of that investment on Jan 1, 1996 was how many timesgreater than the value on Jan 1, 1994?
Answer:
1)
Pythagorean Theorem
a^2 + b^2 = 10^2
a^2 + b^2 = 100
Area:
(1/2)(a)(b) = 20
ab = 40
(a + b)^2
= a^2 + 2ab + b^2
= a^2 + b^2 + 2(ab)
= 100 + 2(40)
= 180
2)
C of II = 2pi(R)
C of I = 2pi(r)
2pi(R) = 2pi(r) + 4
2pi(R) - 2pi(r) = 4
2pi(R - r) = 4
R - r = 4 / 2pi = 2 / pi
3)
The triangle is isosceles (because both legs are radii of the circle). So the other base angle is also 55.
Therefore
c = 180 - 55 - 55 = 70
4)
That's two years difference, so
x = original cost of investement
x(1.10) = cost after one year
x(1.10)(1.10) = cost after two years.
x(1.10)(1.10) = 1.21x
(1.21x - x) = 0.21 times greater, or 21% greater.
1.180
2. 2/pi
3. 70
4. (1.1)^6 -(1.1)^4 /(1.1)^4 = ( 1.1)^2 -- 1 = 0.21
1) d) 180. Use area of triangle formula w/ a, b and Pythagorean th. w/ rt triangle
2) b) 2/pi: pi*(d2 - d1) = 4 = pi*(2r2 - 2r1) = 2pi*(r1 - r2), so
pi*(r2 - r1) = 2, and r2 - r1 = 2/pi
3) 70 deg.: isoceles triangle so other angle is 55, and 180 -2*55 = 70
4) two years of interest, compounded = 1.1^2 = 1.21. So the investment at 1996 was 21% greater than at 1994.
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