Free SAT Math Level 2 Subject Solutions with Explanations
Sample 1

50 Sat Math subject level 2 questions corresponding to those in sample 1, with detailed solutions and explanations are presented.

  1. Two dice are tossed. What is the probability that the sum of the two dice is greater than 3?

    Solution

    When two dice are tossed, there are 6 * 6 = 36 possible outcomes. (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
    (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)
    (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)
    (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)
    (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)
    (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)
    The outcomes that give a sum of 3 or less are (1,1) , (1 , 2) and (2,1)
    There are 36 - 3 = 33 possible outcomes that give a sum that is greater than 3
    The probability that the sum of the two dice is greater than 3 33 / 36 = 11 / 12


  2. If L is a line through the points (2,5) and (4,6), what is the value of k so that the point of coordinates (7,k) is on the line L?

    Solution

    The slope of the line through (2,5) and (4,6) is given by (6 - 5) / (4 - 2) = 1 / 2
    The slope of the line through (4,6) and (7,k) is given by (k - 6) / (7 - 4) = (k - 6) / 3
    The three points are on the same line and therefore the slopes found above must be equal 1 / 2 = (k - 6) / 3
    Solve the above equation for k to find k = 15/2


  3. Find a negative value of k so that the graph of y = x2 - 2 x + 7 and the graph of y = k x + 5 are tangent?

    Solution

    To find points of intersection of the graphs of y = x2 - 2 x + 7 and y = k x + 5, we need to solve the system of the two equations y = x2 - 2 x + 7
    y = k x + 5
    Eliminate y between the two equations to obtain x2 - 2 x + 7 = k x + 5
    The above equation may be written as x2 - x(2 + k) + 2 = 0
    This is a quadratic equation which may have no solutions meaning that the two graphs have no points of intersection, two solutions which means the two graphs have two points of intersection or one solution which means the two graphs are tangent to each other. A quadratic equation have one solution if its discriminant is equal to 0. Hence (2 + k)2 - 4(1)(2) = 0
    Solve for k and select the negative solution k = - 2 - 2 √2


  4. Which of these graphs is the closest to the graph of
    f(x) = |4 - x2| / (x + 2)?


    math sat subject level 2 problem 4 .



    Solution

    Function f(x) = |4 - x2| / (x + 2) is undefined for x = -2 and the only graph that is undefined for x negative is graph B.

  5. The circle of equation (x - 3)2 + (y - 2)2 = 1 has center c. Point M(4,2) is on the circle. N is another point on the circle so that angle McN has a size of 30. Find the coordinates of point N.

    math sat subject level 2 problem 5.



    Solution

    One way to find the coordinates of N is to find the components of vector ON. Vector ON = Vector OC + Vector CN
    The equation of the circle gives the coordinates of the center C which are C (3 , 2)
    The components of vector OC  are  < 3 , 2 >
    Distance CN is equal to the radius of the circle which is 1. Points C and M have the same y coordinates and therefore CM is parallel to the x axis, hence the components of vector CN are given by  <1 . cos(30o) , 1 . sin(30o)> = < √3 / 2 ,  1 / 2>
    Hence the components of vector ON are given by  < 3 , 2 > + < (1 / 2)√3 ,  1 / 2 > = < 3 + √3 / 2 , 5/2 >
    The coordinates of point N are given by ( 3 + √3 / 2 , 5/2 )


  6. Vectors u and v are given by u = < 2 , 0 > and v = < -3 , 1 >. What is the length of vector w given by w = -u - 2v?

    Solution

    We first find w = -u - 2v as follows w = -u - 2v = - < 2 , 0 > -2 < - 3 , 1 >

    = < - 2 , 0> + < 6 , - 2 >

    =  < 4 , - 2 >

    The length of W is given by √(42 + (-2)2) = 2 √5


  7. What is the smallest distance between the point (-2,-2) and a point on the circumference of the circle given by
    (x - 1)2 + (y - 2)2 = 4?


    Solution

    Let us find the center and radius of the circle The equation of the circle is in standard form and its center has coordinates (1 , 2) and its radius is equal to 2.
    The distance between the center of the circle and point (-2, -2) is given by √( (-2 -1)2 + (-2 -2)2 ) = 5
    Point (-2,-2) is outside the circle. The shortest distance between point (-2,-2) and a point on the circle is given by 5 - 2 = 3


  8. What is the equation of the horizontal asymptote of function
    f(x) = 2 / (x + 2) - (x + 3) / (x + 4)?


    Solution

    If x increases indefinitely, 2 / (x + 2) will approach 0 and - (x + 3) / (x + 4) will approach - 1. If x decreases indefinitely, 2 / (x + 2) will approach 0 and - (x + 3) / (x + 4) will approach - 1. Hence the equation of the horizontal asymptote is y = - 1.

  9. The lines with equations x + 3 y = 2 and -2 x + k y = 5 are perpendicular for k =

    Solution

    Let us find the slope of the line with equation x + 3y = 2 3y = - x + 2

    y = -(1 / 3)x + 2/3

    slope = -1 / 3

    The slope of the line with equation -2x + ky = 5 is calculated as follows k y = 2x + 5

    y = (2 / k) x + 5 / k

    slope = 2 / k

    For two lines to be perpendicular, the product of their slopes must be equal to - 1. Hence (-1 / 3)*(2 / k) = -1
    Solve the above equation for k to find k = 2 / 3


  10. If f(x) = (x - 1)2 and g(x) = √x, then (g o f)(x) =

    Solution

    (g o f) (x)  is the composition of functions defined as follows  (g o f) (x) = g ( f (x) )
    Substitute f (x) by (x - 1) 2 (g o f) (x) = g ( f (x) )

    = g( (x - 1) 2 )

    = √(x - 1) 2 

    = | x - 1 |

    NOTE THAT √ x 2 = | x |  not x  



  11. The domain of f(x) = √(4 - x2) / √(x2 - 1) is given by the interval

    Solution

    Remember that the domain of a function is the set of all values of x that make f(x) a real number. The given function has square roots which must be positive or zero because the square root of a negative number is not a real number. The expression in the denominator cannot be zero because we cannot divide by zero. Hence the two conditions to satisfy

    1) (x2 - 1) > 0

    2) (4 - x2) ≥ 0

    Factor (x2 - 1) and solve the inequality 1)

    (x - 1)(x + 1) > 0

    The sign chart of (x - 1)(x + 1) is shown below

    math sat subject level 2, solution problem 11.

    and the solution set of (x - 1)(x + 1) > 0 is given by the interval

    [-1 , 0) U (0 , +1]

    We now factor (4 - x2) and solve the inequality 2)

    (2 - x)(2 + x) ≥ 0

    The sign chart of (2 - x)(2 + x) is shown below

    math sat subject level 2, solution problem 11.

    and the solution set of (2 - x)(2 + x) ≥ 0 is given by the interval

    [-2 , 2]

    For a given value of x to be in the domain of f it must satisfy both inequalities above which means the domain is the set of x values belonging to the intersection of the intervals [-1 , 0) U (0 , +1] and [-2 , 2] as shown below

    math sat subject level 2, solution problem 11 .

    The domain of f is given by
    [-2, -1) U (1, 2]

  12. The area of the circle x2 + y2 - 8 y - 48 = 0 is

    Solution

    We first need to write the given equation in standard form by completing square x 2 + y 2 - 8 y - 48 = 0
    Write the term (y 2 - 8 y)  as a square plus a constant : add and subtract 16 y 2 - 8 y  = y 2 - 8 y + 16 - 16

    = (y - 4) 2 - 16

     

    Substitute by in the given equation and write the equation of the circle in standard form x 2 + (y - 4) 2 - 16 - 48 = 0

    or

    x 2 + (y - 4) 2  = 64

    Hence the radius R of the circle is equal to 8. The area is given Pi R 2  = 64 Pi


  13. The y coordinates of all the points of intersection of the parabola y2 = x + 2 and the circle x2 + y2 = 4 are given by

    Solution

    The points of intersections of the parabola and the circle are solutions to both equations simultaneously and therefore to find these points we need to solve the system of equations defined by y2 = x + 2

    x2 + y2 = 4

    Substitute y2 by x + 2 in the second equation and rewrite the equation obtained in standard form x2 + x + 2 = 4

    x2 + x - 2 = 0

    Solve the quadratic equation obtained x = 1 and x = -2
    We now substitute the values of x obtained in the equation y2 = x + 2 to find y For x = 1 , and y2 = 1 + 2 = 3

    y = + or - √3

    for x = -2 , and y2 = -2 + 2 = 0

    hence y = 0

    The y coordinates of the points of intersection are - √3 , 0 and √3


  14. What is the smallest positive zero of function f(x) = 1/2 - sin(3x + Pi/3)?

    Solution

    The zeros of function f are found by solving the equation 1/2 - sin(3x + Pi/3) = 0
    Rearrange the equation to obtain sin(3x + Pi/3) = 1 / 2
    In order to solve the trigonometric equation above, we need to solve the two equations below. Note that because sine functions are periodic, there is an infinite number of solutions. 3x + Pi/3 = Pi / 6 + 2 k Pi and 3x + Pi/3 = 5 Pi / 6 + 2 k Pi for k any integer
    We now solve the above equations. First equation: 3x + Pi/3 = Pi / 6 + 2 k Pi

    3x = - Pi / 6 + 2 k Pi

    x = - Pi / 18 + 2 k Pi / 3

    Second equation: 3x + Pi/3 = 5 Pi / 6 + 2 k Pi

    3x = Pi / 2 + 2 k Pi

    x = Pi / 6 + 2 K Pi / 3

    We now need to find the smallest positive solution. For k = 0 we have two solutions: x = - Pi / 6 and Pi / 6 and hence the smallest zero is equal to Pi / 6


  15. If x - 1, x - 3 and x + 1 are all factors of a polynomial P(x) of degree 3, which of the following must also be a factor of P(x)?

    I) x2 + 1
    II) x2 - 1
    III) x2 - 4x + 3

    Solution

    We first use the factors to write polynomial P as follows P(x) = k (x - 1)(x - 3)(x + 1) , k is a nonzero real number
    Other possible factors of P(x) are (x - 1)(x - 3) = x2 - 4 x + 3

    (x - 1)(x + 1) = x2 - 1

    and (x - 3)(x + 1) = x2 - 2x -3

    Of those listed in I, II and III, II and III are also factors of P(x).


  16. A cylinder of radius 5 cm is inserted within a cylinder of radius 10 cm. The two cylinders have the same height of 20 cm. What is the volume of the region between the two cylinders?

    Solution

    Let V1 be the volume of the small cylinder and V2 the volume of the large cylinder where V1 = (52 Pi) * 20

    V2 = (102 Pi) * 20

    The volume V of the region bewteen the two cylinders in the difference between the two volumes V = V2 - V1 = 1500 Pi cm3



  17. A data set has a standard deviation equal to 1. If each data value in the data set is multiplied by 4, then the value of the standard deviation of the new data set is equal to

    Solution

    The standard deviation involves the square root of the sum of(xi - m)2 where xi are the data values and m is the mean. If each data value is multiplied by k then the mean will be k times. Hence the new standard deviation will involve the square root of the sum of (k xi - k m)2 and therefore the new standard deviation will be k times the old standard deviation. In this case k = 4 and the old standard deviation is one. The standard deviation after multiplication will be equal to 4.


  18. A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector?

    Solution

    When cut open along the slant height H and flattened, the cone becomes a sector of a circle as shown in the figure below right and the perimeter of the base of the cone is equal to arc length of the sector shown on the right. The perimeter P of the base of the cone is given by:

    math sat subject level 2 problem 18 - Cone .



    P = 2 Pi r = 2 Pi 6 = 12 Pi

    The slanted height H is calculated using Pythagora theorem

    H = √(r2 + h2) = √(62 + 82) = 10

    The arc length of S the sector is given by

    S = H * t = 10 t , where t is the sector angle to find.

    S = P gives 12 Pi = 10 t

    t = 1.2 Pi = 216 degrees.


  19. If sin(x) = -1/3 and Pi ≤ x ≤ 3Pi/2, then cot(2x) =

    Solution

    We first express cot(2x) as follows

    cot(2x) = 1 / tan(2x) = ( 1 - tan2 ) / ( 2 tan(x) )

    We now need to find tan(x) using the identity

    tan(x) = sin(x) / cos(x)

    We know the value of sin(x), we need to find the value of cos(x) using the identity sin2(x) + cos2(x) = 1 as follows

    cos2(x) = 1 - sin2(x)

    Solving the above for cos(x), we obtain two solutions

    cos(x) = √(1 - sin2(x)) and cos(x) = - √(1 - sin2(x))

    Angle x satisfies the condition Pi ≤ x ≤ 3Pi/2 for which cos(x) is negative. Hence

    cos(x) = - √(1 - sin2(x))

    We now substitute and determine cot(2 x) as follows

    cos(x) = - √(1 - (-1/3)2)

    = - √(8/9) = - 2 √2 / 3

    tan(x) = sin(x) / cos(x) = (-1/3) / ( - 2 √2 / 3 ) = 1 / 2 √2

    cot(2 x) = ( 1 - tan2 ) / ( 2 tan(x) ) = ( 1 - 1/8 ) / ( 1 / √2 ) = 7 √ 2 / 8 = 7 / ( 4 √ 2 )

  20. Which of the following functions satisfy the condition f(x) = f -1(x)?

    I) f(x) = - x
    II) f(x) = √x
    III) f(x) = - 1/x

    Solution

    Let us find the inverse of f(x) = - x

    y = - x , write given function as equation

    x = - y , change x to y and y to x

    y = - x = f -1(x) , solve for y to find inverse

    The given function satisfies f(x) = f -1(x).

    Let us find the inverse of f(x) = √x

    y = √x

    x = √y

    y = x2 and is different from f(x) = √x

    The given function does not satisfy the condition f(x) = f -1(x)

    Let us find the inverse of f(x) = - 1/x

    y = - 1/x

    x = - 1/y

    x y = - 1

    y = - 1/x = f -1(x)

    The given function satisfies the condition f(x) = f -1(x)

  21. If f(x) = 1 / (x - 2), which of the following graphs is closest to the graph of | f(x) |?

    math sat subject level 2 problem 21 .



    Solution

    A function of the form y = | f(x) | take positive values or zero. The graphs in A) and B) have negative values and therefore neither can be the graph of y = | f(x) |. In fact B) is not the graph of a function. Also f(x) is undefined at x = 2 and also has a vertical asymptote at x = 2. The only graph that seems to have a vertical asymptote at x = 2 is graph C) and is therefore the closet to the graph of | f(x) |.

  22. If in a triangle ABC, sin(A) = 1/5, cos(B) = 2/7, then cos(C) =

    Solution

    The sum of all three angles in a triangle is equal to 1800. Hence

    A + B + C = 1800

    or A + B = 1800 - C

    The above gives

    cos( 1800 - C ) = cos ( A + B )

    Use trigonometry formula to rewrite above as

    cos( 1800 ) cos( C ) + sin( 1800 ) sin( C ) = cos (A) cos (B ) - sin( A ) sin( B )

    Simplify and rewrite the above as

    cos ( C ) = sin( A ) sin( B ) - cos( A ) cos( B )

    Given that sin(A) = 1/5 and cos(B) = 2/7, We now need to find cos (A) and sin (B) using identities as follows:

    cos (A) = √(1 - (1/5)2) = √24 / 5

    sin (B) = √(1 - (2/7)2) = √45 / 7

    Substitute the values of sin(A), sin(B), cos(A) and cos(B) into cos ( C ) = sin( A ) sin( B ) - cos( A ) cos( B ) we obtain

    cos(C) = (√45 - 2√24) / 35

  23. Find the sum
    100
    ∑ (3 + k)
    k=1


    Solution

    Rewriting the sum explicitly, we have

    100
    ∑ (3 + k) = 4 + 5 + 6 +...+ 103
    k=1


    The terms in the sum are those of an arithmetic sequence with common difference 1. Hence the sum is equal to the average the first term and the last term multiplied by the number of terms which is 100. Hence

    100
    ∑ (3 + k) = 100*(4 + 103)/2 = 5350
    k=1

  24. What value of x makes the three terms x, x / (x + 1) and 3x / [(x + 1)(x + 2)] those of a geometric sequence?

    Solution

    Let r be the common ratio of the geometric sequence. Hence the relationship between the first and second term and the second and third term are as follows

    x / (x + 1) = r x and 3x / [(x + 1)(x + 2)] = r x / (x + 1)

    Solve both equations above for x to obtain

    r = 1 / (x + 1) and r = 3 / (x + 2)

    r is constant, hence

    1 / (x + 1) = 3 / (x + 2)

    Solve the above for x to obtain

    x = -1 / 2

  25. As x increases from Pi/4 to 3 Pi/4, |sin(2x)|

    A) always increases
    B) always decreases
    C) increases then decreases
    D) decreases then increases
    E) stay constant

    Solution

    As x changes g(x) = |sin(2x)| takes either positive values or zero.

    For x = Pi / 4, g(Pi/4) = | sin(2 Pi/4) | = 1

    For x = Pi / 2, g(Pi/2) = | sin(2 Pi/2) | = 0

    For x = 3 Pi / 4, g(3 Pi/4) = | sin(3 Pi/2) | = 1

    As x increases from Pi/4 to 3Pi/4, |sin(2x)| decreases then increases.

  26. If ax3 + bx2 + c x + d is divided by x - 2, then the remainder is equal

    Solution

    Let P(x) = a x3 + b x2 + c x + d. According to the remainder theorem, the remainder r when dividing P(x) by x - 2 is equal to P(2). Hence

    r = P(2) = a 23 + b 22 + 2 c + d

    = 8a + 4b + 2c + d

  27. A committee of 6 teachers is to be formed from 5 male teachers and 8 female teachers. If the committee is selected at random, what is the probability that it has an equal number of male and female teachers?

    Solution

    Let C(n,r) be the number of samples of r elements to be selected from a set of n elements (combinations). Then

    C(n,r) = n! / r! (n - r)!

    The committee has a total of 6 members. If the number of males and females in this committee are equal then the committee has 3 males and 3 females. There are C(5,3) ways to select 3 males from 5, C(8,3) ways to select 3 females from 8 and C(13,6) ways to select 6 persons from 13. The probability is equal to

    P = C(5,3)*C(8,3) / C(13,6) = (5! / 3!2!)*(8! / 3!5!) / (13! / 6!7!) = 140 / 429

  28. The range of the function f(x) = -|x - 2| - 3 is

    Solution

    The absolute value is either positive or zero. Hence

    |x - 2| ≥ 0

    Multiply both sides of the inequality by -1 and reverse the symbol of inequality to obtain

    - |x - 2| ≤ 0

    Subtract 3 from both sides of the inequality to obtain

    - |x - 2| - 3 ≤ - 3

    The right side of the above inequality is equal to f(x). Hence the inequality gives

    f(x) ≤ - 3

    The above inequality states that the range of y = f(x) is given by

    y ≤ -3

  29. What is the period of the function f(x) = 3 sin 2(2x + Pi/4)?

    Solution

    The period P is defined for sine and cosine functions of the form A sin (Bx + C) + D or A cos (BX + C) + D as follows

    P = 2 Pi / | B |

    Hence we need to reduce the power of the given function using the identity sin2(x) = 1/2 - 1/2 cos(2x) as follows

    f(x) = 3 sin 2(2x + Pi/4) = 3 [ 1/2 - 1/2 cos 2((2x + Pi/4)) ]

    = -3/2 cos (4x + Pi/2) + 3/2

    Hence the period P of f(x) is given by

    P = 2 Pi / | 4 | = Pi / 2

  30. It is known that 3 out of 10 television sets are defective. If 2 television sets are selected at random from the 10, what is the probability that 1 of them is defective?

    Solution

    Out of the 10 tv sets, 3 are defective and therefore 7 are non defective. If you select 2 out of 10 tv and 1 is defective the second one is non defective. There are C(3,1) ways of selecting 1 defective out of 3 , C(7,1) ways of selecting 1 non defective out of 7 and C(10,2) ways of selecting 2 tv sets out of 10. Hence the probability P is given by

    P = C(3,1)*C(7,1) / C(10,2) where C(n,r) is the combination of n terms taken r at the time.

    = 7/15

  31. In a triangle ABC, angle B has a size of 50o, angle A has a size of 32o and the length of side BC is 150 units. The length of side AB is

    Solution

    We first find the size of the third angle C

    C = 180o - (50o + 32o) = 98o

    We now use the sine law to find the length of side AB

    AB / sin(C) = BC / sin(A)

    Solve the above for AB, substitute and evaluate

    AB = sin(C) * BC / sin(A) = sin(98) * 150 / sin(32) = 280 (rounded to the nearest unit)

  32. For the remainder of the division of x3 - 2x2 + 3kx + 18 by x - 6 to be equal to zero, k must be equal to

    Solution

    Let P(x) = x3 - 2x2 + 3kx + 18. Using the remainder theorem, the remainder R of the division of P(x) by x - 6 is equal to P(6). Hence

    R = P(6) = 63 - 2(6)2 + 3(6) k + 18

    = 216 - 72 + 18 k + 18 = 162 + 18 k

    Set R = 0 and solve for k

    162 + 18 k = 0 , gives k = - 9

  33. It takes pump (A) 4 hours to empty a swimming pool. It takes pump (B) 6 hours to empty the same swimming pool. If the two pumps are started together, at what time will the two pumps have emptied 50% of the water in the swimming pool?

    Solution

    The rates RA and RA of pumps A and B are given by

    RA = 1/4 and RB = 1/6

    Let t be the time that the two pumps will take to empty 50% of the pool. Hence

    t(1/4 + 1/6) = 0.5

    Multiply all terms in the above equation by the lcm of 4 and 6 which 12, simplify and solve

    t(3 + 2) = 6 t = 6/5 = 1.2 hours = 1 hour 12 minutes

  34. The graph of r = 10 cos(Θ) , where r and Θ are the polar coordinates, is

    A) a circle
    B) an ellipse
    C) a horizontal line
    D) a hyperbola
    E) a vertical line

    Solution

    If r and Θ are the polar coordinates then the x and y coordinates are given by

    x = r cos(Θ) and y = r sin(Θ) and x2 + y2 = r2cos2(Θ) + r2sin2(Θ) = r2

    Multiply both sides of the given equation r = 10 cos(Θ) by r

    r2 = 10 r cos(Θ)

    Substitute r2 by x2 + y2 and r cos(Θ) by x in the above equation to obtain

    x2 + y2 = 10 x

    The above equation may be written as

    x2 - 10 x + y2 = 0

    (x - 5)2 + y2 = 25

    The above equation is that of a circle of center (5,0) and radius 5. Hence the given equation r = 10 cos(Θ) is that of a circle.

  35. If (2 - i)*(a - bi) = 2 + 9i, where i is the imaginary unit and a and b are real numbers, then a equals

    Solution

    We first expand the left side of the given equation.

    2 a - 2b i - i a - b = 2 + 9 i

    which may be written as

    2 a - b + i(-2 b - a) = 2 + 9 i

    Two complex numbers are equal if their real parts and imaginary are equal. Hence the above equation gives two other equations

    2 a - b = 2 and - 2 b - a = 9

    Solve the above system of equation to find a = - 1.

  36. Lines L1 and L2 are perpendicular that intersect at the point (2 , 3). If L1 passes through the point (0 , 2), then line L2 must pass through the point

    A) (0 , 3)
    B) (1 , 1)
    C) (3 , 1)
    D) (5 , 0)
    E) (6 , 7)

    Solution

    We first find the slope m1 of line L1 since we know two points on this line.

    m1 = (2 - 3) / (0 - 2) = 1 / 2

    Since L1 and L2 are perpendicular, the slope m2 of line L2 is related to m1 as follows

    m1*m2 = -1 or (1/2) * m2 = -1 which gives m2 = -2

    We now know a point and a slope of line L2, hence its equation is given by

    y - 3 = -2 (x - 2)

    We now substitute the coordinates in the equation above and determine which point is on the line L2. Point (3 , 1) in C) is on L2.

  37. A square pyramid is inscribed in a cube of total surface area of 24 square cm such that the base of the pyramid is the same as the base of the cube. What is the volume of the pyramid?

    math sat subject level 2 problem 37.

    Solution

    The cube has 6 square faces. The total surface is 24 square cm and therefore the area of one square face of the cube is

    24 / 6 = 4 square cm and the side of one square face is 2 cm

    The volume of the pyramid is equal to the one third the area of the base which is 4 square cm times the height. Since the pyramid is inscribed in the cube, the height is equal to a side of the square.

    Volume = (1/3) * 4 * 2 = 8/3 cubic cm

  38. The graph defined by the parametric equations

    x = cos2t

    y = 3 sint -1

    is

    Solution

    Let us solve the given equation y = 3 sint -1 for sin t

    sin t = (y + 1) / 3

    Square both sides of the above equation

    sin2t = (y + 1)2 / 9

    The above equation may be written

    1 - cos2t = (y + 1)2 / 9

    We now substitute cos2t by x in the above equation to obtain

    1 - x = (y + 1)2 / 9

    Solve the above for x

    x = - (y + 1)2 / 9 + 1

    The above equation is that of a parabola with axis parallel to x axis. However there are restrictions on both x and y. Since x = cos2t and y = 3 sint -1, x cannot be greater than 1 for example. So the graph defined by the parabolic equations is not a whole parabola but a part of a parabola.


  39. math sat subject level 2 problem 39.


    Solution

    Calculating the limit as is gives an undetermined form 0 / 0 . Rewrite the numerator in factored form and simplify

    (x4 - 16) / (x - 2) = (x2 - 4)(x2 + 4) / (x - 2)

    = (x - 2)(x + 2)(x2 + 4) / (x - 2)

    = (x + 2)(x2 + 4) , after simplification

    The limit as x approaches 2 is given by

    (2 + 2)(22 + 4) = 32

  40. For x > 0 and x not equal to 1, log16(x) =

    Solution

    Use the change of base formula and log with base to rewrite log16(x) as follows

    log16(x) = log2(x) / log2(16)

    = log2(x) / log2(24)

    = log2(x) / 4

    = 0.25 log2(x)

  41. The value of k that makes function f, defined below, continous is

    math sat subject level 2 problem 41.



    Solution

    We first find the limit of f(x) as x approaches 0.

    f(x) = (2x 2 + 5x) / x = 2x + 5

    The limit of the above as x approaches 0 is.

    2(0) + 5 = 5

    For a function to be continuous at x = 0, the following condition must be satisfied

    lim f(x) as x approaches = f(0)

    5 = 3k - 1

    Solve the above for k.

    k = 2

  42. If logb(a) = x and logb(c) = y, and 4x + 6y = 8, then logb(a2. c3)

    Solution

    We first divide all terms in the equation by 2 to obtain

    2x + 3y = 4

    We next substitute x by logb(a) and y by logb(c) in the above equation.

    2 logb(a) + 3 logb(c) = 4

    We next use the power formula to rewrite the above as .

    logb(a2) + logb(c3) = 4

    Using the product formula, the above is rewritten as

    logb(a2. c3) = 4

  43. The point (0 , -2 , 5) lies on the

    Solution

    The x coordinate of the given point is the only coordinate equal to 0 and therefore the given point is on the yz plane

  44. Curve C is defined by the equation y = √(9 - x2) with x≥ 0. The area bounded by curve C, the x axis and the y axis is

    Solution

    Let us square both sides of the given equation to obtain

    y 2 = 9 - x 2

    Rewrite as the equation of a circle in standard form.

    x 2 + y 2 = 9

    The above equation is that of a circle with radius 3 and center at (0,0). The region bounded by curve C, the x axis and the y axis and such that x≥ 0 is a quarter of this circle and therefore its area is the quarter of the area of the circle and is given by

    (1/4) Pi (3 2) = 9 Pi / 4

  45. In a plane there are 6 points such that no three points are collinear. How many triangles do these points determine?



    Solution

    A triangle is defined by any 3 non collinear points. Hence the number of triangles that 6 points can determine is equal to the number of combinations of 6 points taken 3 at the time and is given by

    C(6,3) = 6! / (3!(6-3)!) = 6*5*4*3! / (3!3!) = 20


  46. math sat subject level 2 problem 46.

    Find a, b and c.



    Solution

    What is given is the multiplication of two matrices A.B = c. The product of the first row of A by the the first column of B gives

    3(0) + 2(b) = - 4

    Solve for b

    b = - 2

    The product of the first row of A by the the second column of B gives

    3 a + 2 c = 9

    The product of the second row of A by the the second column of B gives

    - a - 2 c = - 7

    Solve the last two equations for a and c simultaneously to obtain

    a = 1 and c = 3

  47. If the sum of the repeating decimals 0.353535... + 0.252525... is written as a fraction in lowest terms, the product of the numerator and denominator is

    Solution

    We first write 0.353535...as a sum as follows

    0.353535... = 0.35 + 0.0035 + 0.000035 ....

    The above is an infinite geometric series with common ration 0.001. Hence the sum

    = 0.35 / (1 - 0.01) = 0.35 / 0.99 = 35 / 99

    We proceed in the same way for 0.252525...

    0.252525... = 0.25 + 0.0025 + 0.000025 +....

    = 0.25 / (1 - 0.01) = 25 / 99

    The sum is given by

    0.353535... + 0.252525... = 35/99 + 25/99 = 60/99 = 20/33

    The product of numerator and denominator

    20 * 33 = 660

  48. sin(tan-1√2) =

    Solution

    Define arc t as follows

    t = tan-1√2

    tan t = opposite side / adjacent side = √2 = √2 / 1

    Hypotenuse = √(1 + 2) = √3

    sin(tan-1√2) = sin t

    = opposite side / hypotenuse = √2 / √3 = 0.82 (approximated at two decimal places)

  49. If 8x = 2 and 3x+y = 81, then y =



    Solution

    Equation 8x = 2 may be written as follows

    (23)x = 2

    Hence

    3x = 1 or x = 1/3

    Equation 3x+y = 81 may be written as follows

    3x+y = 81 = 34

    Hence

    x + y = 4 , y = 4 - x = 4 - 1/3 = 11/3

  50. Let f(x) = -x2 / 2. If the graph of f(x) is translated 2 units up and 3 units left and the resulting graph is that of g(x), then g(1/2) =



    Solution

    Let us first find the equation for g(x).

    When - x2 / 2 is translated 2 units up it becomes - x2 / 2 + 2.

    g(x) = - (x + 3)2 / 2 + 2 , translated 3 units left

    g(1/2) = - (1/2 + 3)2 / 2 + 2 = - 33/8



More Math Practice Tests

Free Practice for SAT, ACT Math tests

Free Practice for GAMT Math tests

Free Compass Math tests Practice

Free GRE Quantitative for Practice

Free AP Calculus Questions (AB and BC) with Answers

privacy policy