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SAT II SAT II -- Math Level 2Math Level 2SAT II SAT II -- Math Level 2Math Level 2SAT II SAT II -- Math Level 2Math Level 2Test #05 SolutionTest #05 SolutionTest #05 SolutionTest #05 SolutionTest #05 SolutionTest #05 Solution

SAT II SAT II -- Math Level 2 Test No. 5Math Level 2 Test No. 5SAT II SAT II -- Math Level 2 Test No. 5Math Level 2 Test No. 5

1. What is the value of ifx ?6411x1. What is the value of ifx ?6411x

(A) (B) (C) (D) (E)623 623 223 322 322 (A) (B) (C) (D) (E)623 623 223 322 322

6411x

6411x

32232264112

x 3223226411 x

But the best way is to use calculator!

Ans. (D)Ans. (D)


2. If 4 and -3 are both zeros of the polynomial p(x),2. If 4 and -3 are both zeros of the polynomial p(x),then a factor of p(x) isthen a factor of p(x) is

(A) x2 – 6 (B) x2 – x – 12 (C) x2 + 6 (D) x2 + x – 12 (E) x2 + x + 12(A) x – 6 (B) x – x – 12 (C) x + 6 (D) x + x – 12 (E) x + x + 12

4 and (-)3 are both zeros of the polynomial p(x).

∴

4 and (-)3 are both zeros of the polynomial p(x).

That means, by Factor Theorem, we get p(4) = 0 and p(-3) = 0.

∴

That means, by Factor Theorem, we get p(4) = 0 and p(-3) = 0.

∴ p(x) must have (x - 4) and (x + 3) factors.

∴

∴ p(x) must have (x - 4) and (x + 3) factors.

∴ p(x) = (x - 4)(x + 3) = x2 – x - 12

∴

∴ p(x) = (x - 4)(x + 3) = x2 – x - 12

∴

∴

Ans. (B)

∴

Ans. (B)


3. What is the volume of a sphere, with center at (1, 2, 3),3. What is the volume of a sphere, with center at (1, 2, 3),that passes through point (1,1,1)?that passes through point (1,1,1)?

(A) 33.2 (B) 46.8 (C) 30 (D) 28.4 (E) 53.8(A) 33.2 (B) 46.8 (C) 30 (D) 28.4 (E) 53.8

Since the equation of the sphere with center at (1, 2, 3) is,Since the equation of the sphere with center at (1, 2, 3) is,

(x - 1)2 + (y - 2)2 + (z - 3)2 = r2, and this equation must satisfy (1, 1, 1),(x - 1)2 + (y - 2)2 + (z - 3)2 = r2, and this equation must satisfy (1, 1, 1),

we get (1 - 1)2 + (1 - 2)2 + (1 - 3)2 = r2we get (1 - 1)2 + (1 - 2)2 + (1 - 3)2 = r2

5 2 r5 2 r

5 r

44But the volume of the sphere is,

5 r

8.4653

4

3

4 33 rVBut the volume of the sphere is,

Ans. (B)

8.4653

3

rV

Ans. (B)


4. Two cities are respectively 3,200 miles and 4,800 miles distant from 4. Two cities are respectively 3,200 miles and 4,800 miles distant from San Francisco, and the distance between the two cities is 5,000 miles. San Francisco, and the distance between the two cities is 5,000 miles. What is the angle between the two cities formedWhat is the angle between the two cities formedfrom the city of San Francisco?from the city of San Francisco?

(A) 65° (B) 38° (C) 74° (D) 89° (E) 122°(A) 65° (B) 38° (C) 74° (D) 89° (E) 122°

Refer to the figure, and also the Law of Cosine,Refer to the figure, and also the Law of Cosine,

we get 5,0002 = 3,2002 + 4,8002 - 2(3,200)(4,800)·cos θ

∴

we get 5,0002 = 3,2002 + 4,8002 - 2(3,200)(4,800)·cos θ

∴∴ θ = cos-1(0.27) = 74°∴

Ans. (C)

∴

Ans. (C)


5. A boy walks diagonally across a square lot. What percent does he 5. A boy walks diagonally across a square lot. What percent does he save by not walking along the edges (approximately)?save by not walking along the edges (approximately)?

(A) 22 (B) 29 (C) 33 (D) 20 (E) 24(A) 22 (B) 29 (C) 33 (D) 20 (E) 24

The distance of walking edges will be (1 + 1) = 2,The distance of walking edges will be (1 + 1) = 2,

while the distance of walking diagonally will be .2while the distance of walking diagonally will be .2

2%3.29%10022

2%3.29%1002

22

Ans. (B)

2

Ans. (B)


6. Find the value of 3 49log6. Find the value of 37 49log

(A) 2/3 (B) -2/9 (C) -4/3 (D) 3/4 (E) -3/2

7

(A) 2/3 (B) -2/9 (C) -4/3 (D) 3/4 (E) -3/2

Refer to the properties of Log function.Refer to the properties of Log function.

We have, log xm = m·log x, and log y x = log x/log y = ln x/ln y.We have, log x = m·log x, and log y x = log x/log y = ln x/ln y.

22221

21

3

21

3

27log

3

27log7log49log49log 7

37

3

12

73

73

7

Ans. (A)

333

77777

Ans. (A)


7. What is the remainder of7. What is the remainder of512

,1512

)( 2345 xxxxxP ,11263

)( xxxxxP

when P(x) is divided by (x – 1)?

1263

when P(x) is divided by (x – 1)?

(A) 3.12 (B) 4.75 (C) 3.58 (D) 4.50 (E) 3.25(A) 3.12 (B) 4.75 (C) 3.58 (D) 4.50 (E) 3.25

Referring to the Remainder Theorem in our previous note,

∴

Referring to the Remainder Theorem in our previous note,

our remainder becomes P(1).

∴

our remainder becomes P(1).

∴ P(1) = 2/3 + 1/6 + 1 + 5/12 + 1 = 39/12 = 3/4∴ P(1) = 2/3 + 1/6 + 1 + 5/12 + 1 = 39/12 = 3/4

Ans. (E)

∴

Ans. (E)

∴

∴

(Please note that P(x) = (x – a)Q(x) + R(x).

∴

(Please note that P(x) = (x – a)Q(x) + R(x).

∴ P(a) = (a – a)Q(a) + R(a) = 0 + R(a), which is the remainter part.)∴ P(a) = (a – a)Q(a) + R(a) = 0 + R(a), which is the remainter part.)


8. In Figure 2, r cos θ + r sin θ =y

8. In Figure 2, r cos θ + r sin θ =y

( , )P x y

(A) x (B) y (C) r (D) x + y (E) r + y

( , )P x y

(A) x (B) y (C) r (D) x + y (E) r + y

yr

y

x xO x

O x

In polar coordinates, we get x = r cos θ, y = r sin θIn polar coordinates, we get x = r cos θ, y = r sin θ

y

∴

x

yyxr 122 tan and , where

∴

xyxr tan and , where

∴ r cos θ + r sin θ = x + y∴

Ans. (D)

∴

Ans. (D)

SAT II SAT II -- Math Level Math Level 2 2 Test No. Test No. 55SAT II SAT II -- Math Level Math Level 2 2 Test No. Test No. 55

9. Solve the equation .032 xx9. Solve the equation .032 xx

(A) 1 and 1/4 (B) 1/4 (C) 1 (D) 0 (E) 4(A) 1 and 1/4 (B) 1/4 (C) 1 (D) 0 (E) 4

∴

0 ,let now, ,032 uuxxx

Then, u2 + 2u - 3 = 0

∴

∴ ∴

Then, u + 2u - 3 = 0

∴ (u + 3)(u - 1) = 0

∴ ∴

∴ (u + 3)(u - 1) = 0

∴ u = -3 or u = 1, but u > 0 ∴ u = 1 only

∴

∴

∴ u = -3 or u = 1, but u > 0 ∴ u = 1 only

∴

∴

1 xu

∴

∴ ∴

∴

∴

1 xu

∴ ∴

∴

∴ x = 1

∴

∴ x = 1

Ans. (C)

∴

Ans. (C)


10. If f (x) = 5x, then f -1 (x) =10. If f (x) = 5x, then f -1 (x) =

(A) log x for x > 0 (B) log 5 for x > 0 (C) x/5 for x > 0(A) log 5 x for x > 0 (B) log x 5 for x > 0 (C) x/5 for x > 0(D) 5/x for x > 0 (E) x5 for x > 0

5 x

(D) 5/x for x > 0 (E) x5 for x > 0

y = f (x) = 5x, y > 0

Therefore, the inverse function of y = 5x becomes x = 5y by switching x and y.

∴

Therefore, the inverse function of y = 5 becomes x = 5 by switching x and y.

Now, by applying log 5 on both side, we get log 5 x = log 5 5y = y·log 5 5 = y(1) = y

∴

Now, by applying log 5 on both side, we get log 5 x = log 5 5y = y·log 5 5 = y(1) = y

∴ y = log x, x > 0∴ y = log 5 x, x > 0

Ans. (A)

∴

Ans. (A)


11. Find the radius of the circle whose equation is x2 + y2 + 24x – 10y = 011. Find the radius of the circle whose equation is x2 + y2 + 24x – 10y = 0

(A) 11 (B) 12 (C) 13 (D) 14 (E) 15(A) 11 (B) 12 (C) 13 (D) 14 (E) 15

x2 + y2 + 24x - 10y = x2 + 24x + y2 - 10y = 0, or (x + 12)2 + (y - 5)2 = 0 + 122 + 52 = 169

∴

∴

x2 + y2 + 24x - 10y = x2 + 24x + y2 - 10y = 0, or (x + 12)2 + (y - 5)2 = 0 + 122 + 52 = 169

∴ (x + 12)2 + (y - 5)2 = 132

∴

∴ (x + 12)2 + (y - 5)2 = 132

∴ r = 13 with center (-12, 5)

∴

∴ r = 13 with center (-12, 5)

∴

∴

Ans. (C)


12. The converse of ~ p → q is equivalent to12. The converse of ~ p → q is equivalent to

(A) p → ~ q (B) q → p (C) q → ~ p (D) p → q (E) ~ q → p(A) p → ~ q (B) q → p (C) q → ~ p (D) p → q (E) ~ q → p

In our previous note on Logic,In our previous note on Logic,

the converse of ~ p → q is by switching the order of the statement as q → ~ p.the converse of ~ p → q is by switching the order of the statement as q → ~ p.

Ans. (C)Ans. (C)


13. What is the sum of the infinite geometric series13. What is the sum of the infinite geometric series1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + … ?1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + … ?

(A) 1/2 (B) 1 (C) 3/2 (D) 2 (E) 5/2(A) 1/2 (B) 1 (C) 3/2 (D) 2 (E) 5/2

For a given Geometric Sequence; a1, a2, a3, …, an, …,For a given Geometric Sequence; a1, a2, a3, …, an, …,

we get an = a1rn-1 and sn = a1((1 – rn)/(1 – r)).we get an = a1rn-1 and sn = a1((1 – rn)/(1 – r)).

Furthermore, for an infinite series, we have the sum, s = a (1/(1 – r)), |r| < 1.Furthermore, for an infinite series, we have the sum, s = a1(1/(1 – r)), |r| < 1.

Here, since our r = 1/2, and |r| < 1, we get s = (1)(1/(1 - (1/2))) = 2Here, since our r = 1/2, and |r| < 1, we get s = (1)(1/(1 - (1/2))) = 2

Ans. (D)Ans. (D)


14. A pyramid is cut by a plane parallel to its base at a distance from the 14. A pyramid is cut by a plane parallel to its base at a distance from the base equal to one-thirds the length of the altitude. The area of the base equal to one-thirds the length of the altitude. The area of the base is 18. Find the area of the section determined by the pyramidbase is 18. Find the area of the section determined by the pyramidand the cutting plane.and the cutting plane.

(A) 1 (B) 2 (C) 3 (D) 6 (E) 8(A) 1 (B) 2 (C) 3 (D) 6 (E) 8

H:h = 1:(2/3)H:h = 1:(2/3)

Since the ratio of its heights becomes 1:(2/3),Since the ratio of its heights becomes 1:(2/3),

we get the ratio of the base area B:b becomeswe get the ratio of the base area B:b becomes

(1)2:(2/3)2 = 1:(4/9).(1)2:(2/3)2 = 1:(4/9).

Here B = 18

∴

Here B = 18

∴∴ b = 18 × (4/9) = 8∴ b = 18 × (4/9) = 8

Ans. (E)

∴

Ans. (E)


15. 1/(sin θ·cos θ ) – tan θ =15. 1/(sin θ·cos θ ) – tan θ =

(A) sin θ (B) cos θ (C) tan θ (D) cot θ (E) sec θ(A) sin θ (B) cos θ (C) tan θ (D) cot θ (E) sec θ

22

cot

coscossin1sin1tan

1 22

cotsincossincossincoscossin

tancossin

Ans. (D)

sincossincossincoscossincossin

Ans. (D)


16. If f (x) = |2x – 3|, then f (3) =16. If f (x) = |2x – 3|, then f (3) =

(A) f (-3/2) (B) f (-1) (C) f (0) (D) f (2/3) (E) f (3/2)(A) f (-3/2) (B) f (-1) (C) f (0) (D) f (2/3) (E) f (3/2)

f (3) = |2(3) – 3| = |3| = 3.f (3) = |2(3) – 3| = |3| = 3.

But f (0) = |2(0) – 3| = |-3| = 3But f (0) = |2(0) – 3| = |-3| = 3

Ans. (C)Ans. (C)


17. Given a point P, where rectangular coordinates are (x, y), for which17. Given a point P, where rectangular coordinates are (x, y), for whichx = a/c and y = b/c. What is the ratio of y to x?x = a/c and y = b/c. What is the ratio of y to x?

(A) a/c (B) a/b (C) b/a (D) c/a (E) 1(A) a/c (B) a/b (C) b/a (D) c/a (E) 1

Given x = a/c, y = b/c, we get c = a/x and c = b/y.

∴

Given x = a/c, y = b/c, we get c = a/x and c = b/y.

∴ c = a/x = b/y, or y/x = b/a∴ c = a/x = b/y, or y/x = b/a∴

Ans. (C)


18. Suppose a car is parked at 80 meters north of you, and a patrol car is 18. Suppose a car is parked at 80 meters north of you, and a patrol car is at 100 meter of northeastern corner from you, with an angle of 30°at 100 meter of northeastern corner from you, with an angle of 30°between them. What is the distance, in meters, between them?between them. What is the distance, in meters, between them?

(A) 133 (B) 115 (C) 60 (D) 66 (E) 52(A) 133 (B) 115 (C) 60 (D) 66 (E) 52

Given the information, we get the following figure:Given the information, we get the following figure:

Then, the distance md 60360080100 22 Then, the distance

Ans. (C)

md 60360080100 22

Ans. (C)


19. [(n – 1)!]4/[n!(n – 2)!]2 =19. [(n – 1)!]4/[n!(n – 2)!]2 =

(A) 1/n (B) 1/n2 (C) (n – 1)/n (D) ((n – 1)/n)2 (E) (n – 1)2(A) 1/n (B) 1/n2 (C) (n – 1)/n (D) ((n – 1)/n)2 (E) (n – 1)2

2222241)1(1)!1()!1()!1()!1()!1(

nnnnnnn

2

1

1

)1(1

)!2(

)!1(

!

)!1(

)!2(!

)!1()!1(

)!2(!

)!1(

n

nn

nn

n

n

n

nn

nn

nn

n

2 1)!2(!)!2(!)!2(!

nnnnnnnn

Ans. (D)Ans. (D)

(Note: the best way to solve is, just plug in n = 3)(Note: the best way to solve is, just plug in n = 3)


20. What is the x-intercept of the line containing20. What is the x-intercept of the line containingthe points (9, -5.5) and (-3, 4.5)?the points (9, -5.5) and (-3, 4.5)?

(A) 0.17 (B) 0.83 (C) 1.14 (D) 2.40 (E) 6(A) 0.17 (B) 0.83 (C) 1.14 (D) 2.40 (E) 6

For this problem, we need to review a linear function:For this problem, we need to review a linear function:

That is, for a given linear function, y – y1 = m(x – x1),That is, for a given linear function, y – y1 = m(x – x1),

we have slope m = (y – y )/(x – x ).

∴

we have slope m = (y2 – y1)/(x2 – x1).

∴ With the two points (9, -5.5) and (-3, 4.5),∴ With the two points (9, -5.5) and (-3, 4.5),∴

we get m = (4.5 – (-5.5))/(-3 – 9) = 10/-12 = (-)(5/6)(x – 9).

∴

∴

we get m = (4.5 – (-5.5))/(-3 – 9) = 10/-12 = (-)(5/6)(x – 9).

Since we need to get x-intercept, where y = 0, we get 0 + 5.5 = (-)(5/6).

∴

∴

Since we need to get x-intercept, where y = 0, we get 0 + 5.5 = (-)(5/6).

∴ y – (-5.5) = -(5/6)(x – 9)

∴

∴ y – (-5.5) = -(5/6)(x – 9)

∴ 5.5 = (-5/6)x + 15/2

∴

∴

∴ 5.5 = (-5/6)x + 15/2

∴ (5/6)x = 2

∴

∴

∴

∴ (5/6)x = 2

∴

∴

∴

∴ x = 12/5 = 2.4

∴

∴ x = 12/5 = 2.4

Ans. (D)

∴

Ans. (D)


21. 2/x + y = 4 and x + 2/y = 3, then the ratio of x to y is21. 2/x + y = 4 and x + 2/y = 3, then the ratio of x to y is

(A) 1:2 (B) 2:3 (C) 3:1 (D) 3:2 (E) 3:4(A) 1:2 (B) 2:3 (C) 3:1 (D) 3:2 (E) 3:4

Let 2/x + y = 4 be eq(1), and x + 2/y = 3 be eq(2).Let 2/x + y = 4 be eq(1), and x + 2/y = 3 be eq(2).

Then, multiplying x on both sides of eq(1), we get 2 + xy = 4x, eq(3).Then, multiplying x on both sides of eq(1), we get 2 + xy = 4x, eq(3).

And also multiplying y on both sides of eq(2), we get xy + 2 = 3y, eq(4).And also multiplying y on both sides of eq(2), we get xy + 2 = 3y, eq(4).

Now, eq(3) - eq(4) gives 0 = 4x - 3y

∴

Now, eq(3) - eq(4) gives 0 = 4x - 3y

∴ 4x = 3y

∴

∴ 4x = 3y

∴

∴

∴ x/y = 3/4

∴

∴ x/y = 3/4

Ans. (E)

∴

Ans. (E)


22. A value that satisfies the equation sin2x + 4 sin x = 0 is (in degrees)22. A value that satisfies the equation sin2x + 4 sin x = 0 is (in degrees)

(A) 0 (B) 30 (C) 60 (D) 90 (E) none of these(A) 0 (B) 30 (C) 60 (D) 90 (E) none of these

Using a graphing utility, let y1 = (sin x)2 + 4sin x, we get x = 0, 180° …

∴

Using a graphing utility, let y1 = (sin x)2 + 4sin x, we get x = 0, 180° …

Or, mathematically, sin x(sin x + 4) = 0

∴

Or, mathematically, sin x(sin x + 4) = 0

∴ sin x = 0, sin x ≠ -4

∴

∴ sin x = 0, sin x ≠ -4

∴

∴

∴ x = 0°, 180° …

∴

∴

Ans. (A)

∴

Ans. (A)


23. What is the largest rod that can just fit into 23. What is the largest rod that can just fit into a box 8” × 12” × 9” (in inches)?a box 8” × 12” × 9” (in inches)?

(A) 14 (B) 16 (C) 17 (D) 18 (E) 19(A) 14 (B) 16 (C) 17 (D) 18 (E) 19

The longest diagonal, 179128 222 DThe longest diagonal,

Ans. (C)

179128 D

Ans. (C)


24. What is the smallest positive angle that will make24. What is the smallest positive angle that will make3 + 2 sin (x + π/6) a minimum?3 + 2 sin (x + π/6) a minimum?

(A) 1.05 (B) 2.09 (C) 1.57 (D) 4.19 (E) 5.24(A) 1.05 (B) 2.09 (C) 1.57 (D) 4.19 (E) 5.24

Let y1 = 3 + 2sin (x + π/6) in graphing mode, we see that min occur at x = 4.19 in radian mode.

∴

Let y1 = 3 + 2sin (x + π/6) in graphing mode, we see that min occur at x = 4.19 in radian mode.

Or, mathematically, to get the minimum 2sin (x + π/6) = -2, because -1 ≤ sin (x + π/6) ≤ 1.

∴

Or, mathematically, to get the minimum 2sin (x + π/6) = -2, because -1 ≤ sin (x + π/6) ≤ 1.

∴ sin (x + π/6) = -1, or x + π/6 = 3π/2

∴

∴ sin (x + π/6) = -1, or x + π/6 = 3π/2

∴ x = 3π/2 - π/6 = 9π/6 – π/6 = 4π/3 = 4.19

∴

∴ x = 3π/2 - π/6 = 9π/6 – π/6 = 4π/3 = 4.19

∴

∴

Ans. (D)

∴

Ans. (D)


25. Which of the following is the solution set for25. Which of the following is the solution set for(-)(x + 1)(x – 2)(x – 3) < 0?(-)(x + 1)(x – 2)(x – 3) < 0?

(A) x < -1 (B) -1 < x < 3 (C) -1 < x < 3 or x > 3(A) x < -1 (B) -1 < x < 3 (C) -1 < x < 3 or x > 3(D) x < -1 or 2 < x < 3 (E) -1 < x < 2 or x > 3(D) x < -1 or 2 < x < 3 (E) -1 < x < 2 or x > 3

Using a graph of polynomial function, we get -1 < x < 2, x > 3Using a graph of polynomial function, we get -1 < x < 2, x > 3

Ans. (E)Ans. (E)


26. If f (x) = 3x2 – 5x - 3, then (f (x + a) – f (x))/a =26. If f (x) = 3x2 – 5x - 3, then (f (x + a) – f (x))/a =

(A) 3a2 – 5a – 3 (B) 3x2a2 – 5xa – 3 (C) 6x – 5 + 3a(A) 3a2 – 5a – 3 (B) 3x2a2 – 5xa – 3 (C) 6x – 5 + 3a(D) 6x – 5 (E) none of the above(D) 6x – 5 (E) none of the above

Given f (x) function, we get f (x + a) = 3(x + a)2 - 5(x + a) – 3 =

3x2 + 6ax + 3a2 - 5x - 5a - 3, eq(1).

∴

3x + 6ax + 3a - 5x - 5a - 3, eq(1).

f (x) = 3x2 - 5x - 3, eq(2).

∴

f (x) = 3x2 - 5x - 3, eq(2).

∴ (eq(1) – eq(2))/a = (6ax + 3a2 – 5a)/a = 6x + 3a - 5∴ (eq(1) – eq(2))/a = (6ax + 3a2 – 5a)/a = 6x + 3a - 5

Ans. (C)

∴

Ans. (C)


27. For 0 ≤ x ≤ 2π, if the minimum value of the function27. For 0 ≤ x ≤ 2π, if the minimum value of the functiony = cos (x - π/3) occurs at point P, yy = cos (x - π/3) occurs at point P,then what are the coordinates of P ?then what are the coordinates of P ?

(A) (4π/3, -π ) cos3

y x (A) (4π/3, -π )(B) (4π/3, -1)

cos3

y x

(B) (4π/3, -1)(C) (3π/2, -π )(C) (3π/2, -π )(D) (3π/2, -1)

xO(D) (3π/2, -1)

(E) (3π/2, 0)O

(E) (3π/2, 0)

PP

As seen in the figure, the minimum occurs when cos (x – π/3) = -1,

∴

As seen in the figure, the minimum occurs when cos (x – π/3) = -1,

but cosine function becomes (-)1 at θ = π.

∴

but cosine function becomes (-)1 at θ = π.

∴ x – π/3 = π

∴

∴ x – π/3 = π

∴

∴

∴ x = 4π/3

∴

∴

Ans. (B)

∴

Ans. (B)


28. In how many ways can a committee of 1 woman and28. In how many ways can a committee of 1 woman and3 men be selected from five men and four women3 men be selected from five men and four womenso as to always include a particular man?so as to always include a particular man?

(A) 84 (B) 70 (C) 48 (D) 24 (E) 126(A) 84 (B) 70 (C) 48 (D) 24 (E) 126

Out of 5 men, a particular man is already fixed!

∴

Out of 5 men, a particular man is already fixed!

∴∴ We only have 4 men and 4 women.

∴

∴ We only have 4 men and 4 women.

Now, selecting 1 woman from 4 women becomes 4C1, and also 3 men from 4 men becomes 4C3.

∴

∴

Now, selecting 1 woman from 4 women becomes 4C1, and also 3 men from 4 men becomes 4C3.

∴ 4C2 × 4C1 = 24∴ 4C2 × 4C1 = 24

Ans. (D)

∴

Ans. (D)


29. If and then x =3x ,23 xx 29. If and then x4 =31 x ,231 nn xx 29. If and then x4 =1 ,231 nn xx

(A) 1.732 (B) 2.542 (C) 2.843 (D) 2.947 (E) 2.982(A) 1.732 (B) 2.542 (C) 2.843 (D) 2.947 (E) 2.982

54.232323 xx 54.232323 12 xx

84.2)54.2(2323 23 xx

947.2)84.2(2323

84.2)54.2(2323 23

xx

xx

947.2)84.2(2323 34 xx

Ans. (D)Ans. (D)


30. Suppose the graph of f (x) = x2 – 4x + 7 is translated 2 units left30. Suppose the graph of f (x) = x2 – 4x + 7 is translated 2 units leftand 3 unit down. If the resulting graph represents g(x),and 3 unit down. If the resulting graph represents g(x),what is the value of g(-0.5)?what is the value of g(-0.5)?

(A) 26.25 (B) 20.25 (C) 0.25 (D) -0.25 (E) -20.25(A) 26.25 (B) 20.25 (C) 0.25 (D) -0.25 (E) -20.25

Since x is translated 2 units left, we get, x → x - (-2) = x + 2.Since x is translated 2 units left, we get, x → x - (-2) = x + 2.

∴

Also, y is 3 down, y → y - (-3) = y + 3

∴

Also, y is 3 down, y → y - (-3) = y + 3

∴ g(x) + 3 = f (x + 2), or g(x) = f (x + 2) – 3 = (x + 2)2 - 4(x + 2) + 7 - 3∴ g(x) + 3 = f (x + 2), or g(x) = f (x + 2) – 3 = (x + 2) - 4(x + 2) + 7 - 3

Now, g(-1/2) = (3/2)2 – 4(3/2) + 4 = 9/4 - 6 + 4 = 1/4

∴

Now, g(-1/2) = (3/2)2 – 4(3/2) + 4 = 9/4 - 6 + 4 = 1/4

Ans. (C)

∴

Ans. (C)


31. 2 tan x – 2 sin2x·tan x equals which one of the following?31. 2 tan x – 2 sin2x·tan x equals which one of the following?

(A) cos 2x (B) 2 sin x (C) 2 cos x (D) cos2x (E) sin 2x(A) cos 2x (B) 2 sin x (C) 2 cos x (D) cos2x (E) sin 2x

The best way to solve quickly is to use a graphing utility by letting, y1 = 2 tan x - (2 sin2x)(tan x).The best way to solve quickly is to use a graphing utility by letting, y1 = 2 tan x - (2 sin x)(tan x).

And then, try answer choice (A) = y2, (B) = y3, … so on.And then, try answer choice (A) = y2, (B) = y3, … so on.

We will see answer choice (E) becomes the overlapping one.We will see answer choice (E) becomes the overlapping one.

Now, if we solve mathematically, 2 tan x – 2 sin2x·tan xNow, if we solve mathematically, 2 tan x – 2 sin2x·tan x

= 2(sin x/cos x) – 2(sin2x)(sin x/cos x)= 2(sin x/cos x) – 2(sin2x)(sin x/cos x)

= 2[(sin x – sin3x)/cos x] = 2[(sin x(1 – sin2x))/cos x]= 2[(sin x – sin x)/cos x] = 2[(sin x(1 – sin x))/cos x]

= 2[(sin x·cos2x)/cos x] = 2 sin xcos x = sin 2x.= 2[(sin x·cos x)/cos x] = 2 sin xcos x = sin 2x.

Ans. (E)Ans. (E)


32. The Fibonacci sequence can be defined recursively as32. The Fibonacci sequence can be defined recursively asa1 = 1a1 = 1a2 = 1a2 = 1a = a + a for n ≥ 3.an = an-1 + an-2 for n ≥ 3.What is the 12th term of this sequence?What is the 12th term of this sequence?

(A) 12 (B) 36 (C) 72 (D) 108 (E) 144(A) 12 (B) 36 (C) 72 (D) 108 (E) 144

a3 = a2 + a1 = 1 + 1 = 2a3 = a2 + a1 = 1 + 1 = 2

a = a + a = 1 + 2 = 3a4 = a3 + a2 = 1 + 2 = 3

a = a + a = 3 + 2 = 5a5 = a4 + a3 = 3 + 2 = 5

a = a + a = 5 + 3 = 8a6 = a5 + a4 = 5 + 3 = 86 5 4

……

But from this sequential pattern, we may extend the sequence as,

∴

But from this sequential pattern, we may extend the sequence as,

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

∴

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

∴ The 12th term is 144∴ The 12th term is 144

Ans. (E)

∴

Ans. (E)


33. How many odd numbers greater than 30,000 may be formed using 33. How many odd numbers greater than 30,000 may be formed using the digits 1, 2, 3, 4, and 5 if each digit must be used exactly once in the digits 1, 2, 3, 4, and 5 if each digit must be used exactly once in each number?each number?

(A) 42 (B) 48 (C) 64 (D) 96 (E) 112(A) 42 (B) 48 (C) 64 (D) 96 (E) 112

1st, 2nd, 3rd, 4th, 5th1st, 2nd, 3rd, 4th, 5th

Now, to be greater than 30,000, we could only have either 3, 4 or 5 on the 1st spot.Now, to be greater than 30,000, we could only have either 3, 4 or 5 on the 1st spot.

That is, for example, 3, , , , .That is, for example, 3, , , , .

But we also need ODD number, thus the last spot should be 1 or 5.But we also need ODD number, thus the last spot should be 1 or 5.

That means, 3, , , , 1 or 3, , , , 5 with

∴

That means, 3, , , , 1 or 3, , , , 5 with

the possible choices of 3 × 2 × 1 = 6 combination each!!

∴

the possible choices of 3 × 2 × 1 = 6 combination each!!

∴ 6 + 6 = 12 combination for this case!!∴ 6 + 6 = 12 combination for this case!!

We also have two other cases such as: 4, , , , 1, 3 or 5, or 3 × 2 × 1 × (3) = 18 combinations!!

∴

We also have two other cases such as: 4, , , , 1, 3 or 5, or 3 × 2 × 1 × (3) = 18 combinations!!

∴

Or, 5, , , , 1 or 3 with 3 × 2 × 1 × (2) = 12 combinations.Or, 5, , , , 1 or 3 with 3 × 2 × 1 × (2) = 12 combinations.

Therefore, the total = 12 + 8 + 12 = 42.Therefore, the total = 12 + 8 + 12 = 42.

Ans. (A)Ans. (A)


34. If matrix A has dimensions m × n and matrix B has34. If matrix A has dimensions m × n and matrix B hasdimensions n × m, where m and n are distinct positive integers, which dimensions n × m, where m and n are distinct positive integers, which of the following statements must be true?of the following statements must be true?

I. The product BA does not exist.I. The product BA does not exist.II. The product AB exists and has dimensions m × m.II. The product AB exists and has dimensions m × m.II. The product AB exists and has dimensions m × m.III. The product AB exists and has dimensions n × n.III. The product AB exists and has dimensions n × n.

(A) I only (B) II only (C) III only (D) I and II (E) I and III(A) I only (B) II only (C) III only (D) I and II (E) I and III

The product of the matrix must match the dimension of columns of the first matrixThe product of the matrix must match the dimension of columns of the first matrix

equal to the rows of the second matrix, such as [p × q][q × r] = [p × r].equal to the rows of the second matrix, such as [p × q][q × r] = [p × r].

∴ ∴

Here, (I) BA = [n × m][m × n] = [n × n]

∴ ∴∴ BA exist ∴ (I) is not true!

∴

∴ BA exist ∴ (I) is not true!

(II) AB = [m × n][n × m] = [m × m]

∴

∴ ∴

(II) AB = [m × n][n × m] = [m × m]

∴ AB exist with [m × m]

∴ ∴

∴ AB exist with [m × m]

(III) Not true!

∴

(III) Not true!

∴

Ans. (B)


35. When drawn on the same set of axes, the graphs of35. When drawn on the same set of axes, the graphs ofx2 – y2 = 4 and 4(x – 1)2 + y2 = 25 have in common exactlyx – y = 4 and 4(x – 1) + y = 25 have in common exactly

(A) 0 points (B) 1 point (C) 2 points (D) 3 points (E) 4 points(A) 0 points (B) 1 point (C) 2 points (D) 3 points (E) 4 points

2 2( 1)x y

2 2

2 2

( 1)1

x y

2 2

155

552

Ans. (C)Ans. (C)


36. If sin θ·cos θ < 0 and sin θ·tan θ > 0,y

36. If sin θ·cos θ < 0 and sin θ·tan θ > 0,y

then θ must be in which quadrantthen θ must be in which quadrantin the figure above?in the figure above?

(A) I (B) II (C) III (D) IV(A) I (B) II (C) III (D) IV(E) There is no quadrant in which both conditions

xO(E) There is no quadrant in which both conditions

are true.O

are true.

sin θ·cos θ < 0

∴

sin θ·cos θ < 0

∴

∴

∴ sin θ and cos θ must have opposite signs!

∴

∴ sin θ and cos θ must have opposite signs!

∴ II or IV

∴

∴

∴ II or IV

sin θ·tan θ > 0

∴

∴

sin θ·tan θ > 0

∴ sin θ and tan θ must have the same signs.

∴

∴

∴ sin θ and tan θ must have the same signs.

∴ I or IV

∴

∴ I or IV

∴

∴

Now, IV is the only quadrant that satisfy both!!

∴

Now, IV is the only quadrant that satisfy both!!

Ans. (D)Ans. (D)


37. If f (-x) = - f (x) for all real numbers x and if (a, b) is a point37. If f (-x) = - f (x) for all real numbers x and if (a, b) is a pointon the graph of f, which of the following points must also beon the graph of f, which of the following points must also beon the graph of f?on the graph of f?

(A) (-b, -a) (B) (-a, -b) (C) (-a, b) (D) (a, -b) (E) (b, a)(A) (-b, -a) (B) (-a, -b) (C) (-a, b) (D) (a, -b) (E) (b, a)

The property f (-x) = -f (x) is for ODD function, which is symmetric to origin (0, 0),The property f (-x) = -f (x) is for ODD function, which is symmetric to origin (0, 0),

we must have (a, b) in quadrant I and also (-a, -b) in quadrant III on the curve of f (x).we must have (a, b) in quadrant I and also (-a, -b) in quadrant III on the curve of f (x).

Ans. (B)Ans. (B)


38. Lines A͞͞͞B and A͞C are tangents to a circle at points B and C 38. Lines A͞͞͞B and A͞C are tangents to a circle at points B and C respectively. Minor arc BC is 2.5 π inches, and the diameter ofrespectively. Minor arc BC is 2.5 π inches, and the diameter ofthe circle is 12 inches. What is the number of degrees in angle BAC?the circle is 12 inches. What is the number of degrees in angle BAC?

(A) 90° (B) 95° (C) 70° (D) 100° (E) 105°(A) 90° (B) 95° (C) 70° (D) 100° (E) 105°

B͡C = 2.5π = r·θ, where r = 6

∴

B͡C = 2.5π = r·θ, where r = 6

∴ B͡

∴

∴ B͡C = 2.5π = 6·θ

∴

∴ ∠

∴ B͡C = 2.5π = 6·θ

∴ θ = 2.5π/6 = 2.5(180°)/6 = 75°

∴ ∠

∴

∴ θ = 2.5π/6 = 2.5(180°)/6 = 75°

∴ ∠ BAC = 180° - 75° = 105°

∴

∴ ∠ BAC = 180° - 75° = 105°

Ans. (E)

∴

∴ ∠

Ans. (E)


39. The points in the rectangular coordinate plane are transformed in 39. The points in the rectangular coordinate plane are transformed in such a way that each point P(x, y) is moved to the point P'(3x, 3y).such a way that each point P(x, y) is moved to the point P'(3x, 3y).If the distance between a point P and the origin is d,If the distance between a point P and the origin is d,then the distance between the points P and P' isthen the distance between the points P and P' is

(A) 1/d (B) d/2 (C) d (D) 2d (E) 3d(A) 1/d (B) d/2 (C) d (D) 2d (E) 3d

dyxyxyyxxPP 2244)3(3' 222222 dyxyxyyxxPP 2244)3(3' 222222

Ans. (D)Ans. (D)


40. The portion of the plane, whose equation is 12x + 15y + 10z = 120, 40. The portion of the plane, whose equation is 12x + 15y + 10z = 120, that lies in the first octant forms a pyramid with the coordinate that lies in the first octant forms a pyramid with the coordinate planes. Find its volume.planes. Find its volume.

(A) 90 (B) 120 (C) 140 (D) 150 (E) 160(A) 90 (B) 120 (C) 140 (D) 150 (E) 160

For the x-intercept, let y = 0, z = 0, then x = 120/12 = 10.For the x-intercept, let y = 0, z = 0, then x = 120/12 = 10.

Similarly, we put x = 0 and z = 0, to get 15y = 120

∴

Similarly, we put x = 0 and z = 0, to get 15y = 120

∴ y = 120/15 = 8.∴ y = 120/15 = 8.

∴

∴

Also, x = 0 and y = 0 gives 10z = 120

∴

∴

Also, x = 0 and y = 0 gives 10z = 120

∴ z = 12∴ z = 12

The volume of pyramid becomes,

∴

The volume of pyramid becomes,

V = (1/3)(Base area) × height

∴

V = (1/3)(Base area) × height

= (1/3) × (1/2 × 10 × 8) × 12 = 160= (1/3) × (1/2 × 10 × 8) × 12 = 160

Ans. (E)Ans. (E)


41. Which of the following shifts of the graph of y = 2x241. Which of the following shifts of the graph of y = 2x2

would result in the graph of y = 2x2 – 4kx + 2k2 + k,would result in the graph of y = 2x – 4kx + 2k + k,where k is a constant greater than 2?where k is a constant greater than 2?

(A) Left 2 units and up k units(A) Left 2 units and up k units(B) Left k unit and up k + 1 units(B) Left k unit and up k + 1 units(C) Right k unit and down k + 1 units(C) Right k unit and down k + 1 units(D) Left k unit and up k units(D) Left k unit and up k units(E) Right k unit and up k units(E) Right k unit and up k units

y = 2x2 - 4kx + 2k2 + k = 2(x2 - 2kx + k2) + k, ory = 2x2 - 4kx + 2k2 + k = 2(x2 - 2kx + k2) + k, or

y = 2(x - k)2 + k → this shows that y = 2x2 has moved to right k units and up k units.y = 2(x - k)2 + k → this shows that y = 2x2 has moved to right k units and up k units.

Ans. (E)


42. For what positive value of a will the line y = ax - 5 be42. For what positive value of a will the line y = ax - 5 betangent to the circle x2 + y2 = 4?tangent to the circle x + y = 4?

21 21 31(A) 1 (B) (C) (D) (E) 2

2

21

3

21

4

31(A) 1 (B) (C) (D) (E) 2

2 3 4

Referring to the distance formula from a point (x , y ) toReferring to the distance formula from a point (x1, y1) to1 1

CByAx a line Ax + By + C = 0 is ,11 CByAx

d

a line Ax + By + C = 0 is

we have (x , y ) = (0, 0) and

,22 BA

d

we have (x1, y1) = (0, 0) and

d = r = 2, and A = a, B = -1, C = -5

BA

d = r = 2, and A = a, B = -1, C = -5

)5()0)(1()0( a )5()0)(1()0(2

ard

12

2

ard

52

1

a

1

52

2

a

215

12 a

21or ,

51 22 aa

4or ,

21 aa

21a

2

21a

Ans. (B)

2

Ans. (B)


43. (x – 1)3 – 2(x – 1)2 – (x – 1) + 2 =43. (x – 1)3 – 2(x – 1)2 – (x – 1) + 2 =

(A) x(x – 2)(x – 3) (B) (x – 1)(x – 2)(x – 3) (C) x(x – 1)(x – 2)(A) x(x – 2)(x – 3) (B) (x – 1)(x – 2)(x – 3) (C) x(x – 1)(x – 2)(D) (x + 1)(x + 2)(x + 3) (E) x(x – 1)(x – 3)(D) (x + 1)(x + 2)(x + 3) (E) x(x – 1)(x – 3)

Let A = x 1, then A3 - 2A2 - A + 2

= A2(A - 2) - (A - 2)= A (A - 2) - (A - 2)

= (A - 2)(A2 - 1)= (A - 2)(A2 - 1)

= (A - 2)(A + 1)(A - 1)= (A - 2)(A + 1)(A - 1)

Now, replacing A = x - 1, we get (x - 3)(x)(x - 2).Now, replacing A = x - 1, we get (x - 3)(x)(x - 2).

Ans. (A)


44. When a certain radioactive element decays, the amount that exists44. When a certain radioactive element decays, the amount that existsat any time t can be calculated by the function A(t) = A0e

-t /1000,at any time t can be calculated by the function A(t) = A0e ,where A0 is the initial amount and t is the elapsed time in years.where A0 is the initial amount and t is the elapsed time in years.How many years would it take for an initial amount of 800 How many years would it take for an initial amount of 800 milligrams of this element to decay to 200 milligrams?milligrams of this element to decay to 200 milligrams?

(A) 0.5 (B) 500 (C) 1,386 (D) 1,443 (E) 5,704(A) 0.5 (B) 500 (C) 1,386 (D) 1,443 (E) 5,704

1000800200t

e

1000800200 e

10001200

t

e

1000

4

1

800

200e

Now, applying “ln” on both sides, we get

4800

).1(lnln1

ln 1000t

et

et

Now, applying “ln” on both sides, we get ).1(1000

ln1000

ln4

1ln 1000

te

te

100010004

386,11

ln1000 t 386,14

1ln1000 t

Ans. (C)

4

Ans. (C)


45. If the values of the function g(x) represent the slope of the line 45. If the values of the function g(x) represent the slope of the line tangent to the graph of the function f (x), shown on the right, at each tangent to the graph of the function f (x), shown on the right, at each point (x, y), which of the following could be the graph of g(x)?point (x, y), which of the following could be the graph of g(x)?

(A) (B) (C) (D) (E)(A) (B) (C) (D) (E)g(x) g(x) g(x) g(x)

x x x x

By examining the movement of the slopes:By examining the movement of the slopes:

m → m → m → m → m → m ,m1 → m2 → m3 → m4 → m5 → m6,

we get the graph of g(x) is approximately similar towe get the graph of g(x) is approximately similar to

answer choice (D).answer choice (D).

Ans. (D)Ans. (D)


46. If f (x) = (x – 1)/x and f 2(x) = f (f (x)), f 3(x) = f (f 2(x)), ..., f n(x)46. If f (x) = (x – 1)/x and f 2(x) = f (f (x)), f 3(x) = f (f 2(x)), ..., f n(x)= f (f n-1(x)), where n is a positive integer greater than 1,= f (f (x)), where n is a positive integer greater than 1,what is the smallest value of n such that f n(x) = f (x)?what is the smallest value of n such that f n(x) = f (x)?

(A) 2 (B) 3 (C) 4 (D) 6 (E) No value of n works.(A) 2 (B) 3 (C) 4 (D) 6 (E) No value of n works.

1x 1)(

x

xxf

11 x

x

1

11

1

1)(

xx

x

xf 1

11

11)(

))(()(2

x

x

x

xxfxffxf

111)())(()(

xxxxf

xffxf

xx

11

1

xx

1

11

1

1

1))(()(3 2

x

xxfxffxf

1

1

1

1

1

1))(()(3 2

x

xx

xfxffxf

1

1

1

11

xx

x

)())(()(4

113

xfxffxf

xx

)())(()(4 3 xfxffxf

4n

Ans. (C)

4n

Ans. (C)


47. Figure 8 shows a triangle inscribed in47. Figure 8 shows a triangle inscribed ina semicircle. If the area of the trianglea semicircle. If the area of the triangleis 8·sin θ·cos θ, what is the length ofis 8·sin θ·cos θ, what is the length ofthe radius of the semicircle?the radius of the semicircle?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

△ ∠

∴

Since △ABC is inscribed in a semicircle, we have ∠B = 90°.

∴

Since △ABC is inscribed in a semicircle, we have ∠B = 90°.

∴ sin θ = h/2r, or h = (2r) sin θ.

∴ △

△ ∠

∴ sin θ = h/2r, or h = (2r) sin θ.

Also, cos θ = b/2r, or b = (2r) cos θ.

∴ △

∴

Also, cos θ = b/2r, or b = (2r) cos θ.

∴Area of △ABC = (1/2)bh = (1/2)(2r cos θ)(2r sin θ) =

∴

∴Area of △ABC = (1/2)bh = (1/2)(2r cos θ)(2r sin θ) =

2r2 sin θ·cos θ = 8 sin θ·cos θ.

∴

∴ △

2r2 sin θ·cos θ = 8 sin θ·cos θ.

∴

∴

∴ △

∴ 2r2 = 8 or r2 = 4

∴

∴ 2r = 8 or r = 4

∴ r = 2

∴

∴ r = 2

Ans. (B)

∴

Ans. (B)


48. What is the set of all points in space that is equidistance48. What is the set of all points in space that is equidistancefrom any two fixed points?from any two fixed points?

(A) a triangle (B) a perpendicular bisector plane(A) a triangle (B) a perpendicular bisector plane(C) a circle (D) a line (E) a sphere(C) a circle (D) a line (E) a sphere

The definition of a perpendicular bisector line in plane is the set of all points that is equidistance The definition of a perpendicular bisector line in plane is the set of all points that is equidistance

from any two fixed points. Therefore, in space, we have P.B. plane.from any two fixed points. Therefore, in space, we have P.B. plane.

Ans. (B)Ans. (B)


49. The graph of xy – x – y + 2 = 0 can be expressed as a set of parametric 49. The graph of xy – x – y + 2 = 0 can be expressed as a set of parametric equations. If y = t/(t + 1) and x = f (t), then f (t) =equations. If y = t/(t + 1) and x = f (t), then f (t) =

(A) t + 1 (B) t – 1 (C) t + 2 (D) t – 2 (E) (t – 3)/2(A) t + 1 (B) t – 1 (C) t + 2 (D) t – 2 (E) (t – 3)/2

02)()( , and )( Replacing t

tft

tft

ytfx 021

)(1

)( ,1

and )( Replacing

t

ttf

t

ttf

t

tytfx

222)1(21

111

ttttttt

ttt

1

2

1

22

1

)1(2

1

1)(or 2

11

1)(

t

t

t

tt

t

tt

ttf

t

t

t

ttf

1111)(or 2

11

1)(

tttt

tftt

tf

21)2(

)(

t

tttf 2

11)(

t

ttf

Ans. (C)

11 t

Ans. (C)


50. Given the following parametric equations:50. Given the following parametric equations:x = 2 cos θ + 1, y = 3 sin θ, which one of the following curve x = 2 cos θ + 1, y = 3 sin θ, which one of the following curve represents the above equations in rectangular form?represents the above equations in rectangular form?

(A) a circle (B) a parabola (C) a hyperbola (D) an ellipse (E) a line(A) a circle (B) a parabola (C) a hyperbola (D) an ellipse (E) a line

1 yxSince squaring both side of the above equations,,sin

3 and cos

2

1

yxSince squaring both side of the above equations,,sin

3 and cos

2

22we get .sin and cos

)1( 22

22

yx

we get .sin3

and cos2 22

32

)1( 2222

yx

which is an ellipse with center (1, 0),1sin cos32

)1( 22

22

yx

and major axis on y-axis with its length of 6.32 22

and major axis on y-axis with its length of 6.

Ans. (D)Ans. (D)

sat ii -math level 2 test #05 solutionberkeleyus.com/documents/mathiic_test_05_powerpoint.pdf ·...

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