Free SAT Maths Level 1 Subject Solutions with Explanations
Sample 1

Detailed solutions and explanations to the 31 Sat Maths subject level 1 questions corresponding to those in sample 1, are presented.

  1. In the figure below, AB and GE are parallel. Triangle ACD is isosceles with the lengths of CA and CD equal. The measures of angles FDE and GDH are 60° and 65° respectively. What is the measure of angle CAB?

    math sat subject level 1 problem 1.


    Solution
    Since AB and GE are parallel, angles BAD and EDF are corresponding angles and therefore equal in size. Hence the size of angle BAD is 60°
    Angles CDE and GDH are vertical angles and therefore equal in size. Hence the size of angle CDE is 65°. The size of angle CDA is given by
    180° - (60° + 65°) = 55°
    The size of angle CAB is given by
    size of angle BAD - size of angle CAD = 60° - 55° = 5°

  2. If a, b and y are positive real numbers such that none of them is equal to 1 and b 2a + 6 = y 2, which of these must be true?
    Solution
    Since a and b are positive
    √(b 2a + 6) = b (1/2) (2a + 6) = b a + 3
    Since y is positive
    √(y 2) = y
    Since b 2a + 6 = y 2, taking the square root of both sides of the equation, we obtain
    y = b a + 3

  3. What is true about the graph of function f defined by
    f(x) = 4 + | | x | - 3 |

    (I) Symmetric with respect to y-axis
    (II) Has no x intercepts
    (III) Has a y intercept at (0 , 7)
    Solution
    (I): Determine f(-x) and simplify it
    f(-x) = 4 + | | - x | - 3 | = 4 + | | x | - 3 | since |-x| = |x|
    Hence f(x) is even and its graph is symmetric with to the y axis.
    (II): Find any x intercepts by solving f(x) = 0
    4 + | | x | - 3 | = 0
    Add -4 to both sides of the above equation to obtain
    | | x | - 3 | = - 4
    The absolute value of an expression is never negative, hence the above equation has no solutions and therefore the graph of the given function has no x intercepts.
    (III): Find the y intercept by finding f(0)
    f(0) = 4 + | | 0 | - 3 | = 7
    The graph of f has a y intercept at (0 , 7).
    All three statements in (I), (II) and (III) are true.

  4. A number x is divided by 0.71 and the result decreased by 0.2. If the fifth root of the final result equals -0.15, what is the value of x rounded to the nearest thousandth?
    Solution
    x divided by 0.71 and the result decreased by 0.2
    x / 0.71 - 0.2
    The fifth root of the final (above) result equals -0.15
    (x / 0.71 - 0.2)1/5 = -0.15
    We need to solve the above equation in order to find x. Raise both sides of the equation to the power 5.
    (x / 0.71 - 0.2) = (-0.15) 5
    Solve for x.
    x = 0.71( (-0.15) 5 + 0.2 ) (appro.)= 0.14194608
    Round to the nearest thousanth
    x (appro.)= 0.142

  5. Find the constant k so that the perpendicular bisector of the line segment with end points (k , 0) and (4 , 6) has a slope of -3.
    Solution
    We first the slope m1 of the line through the given points
    m1 = (6 - 0) / (4 - k) = 6 / (4 - k)
    The slope of the perpendicular to the line segment is equal to -3. For two lines to be perpendicular, the product of their slopes must be equal to - 1. Hence
    ( - 3 )*(6 / (4 - k) ) = - 1
    Solve the above for k.
    -18 / (4 - k) = - 1
    -18 = - 4 + k
    k = - 14

  6. Forty students are each registered to study one or more of three courses. 23 students are registered in one course only and 14 students are registered in two courses only. How many are registered in all 3 courses?
    Solution
    We first use Venn diagram to represent the students registered in one course only which is represented in yellow and the students who study two courses only which is represented in blue.

    math sat subject level 1 solution problem 6.


    If the total number of students studying one, two or three courses is 40, then the number of students studying three course is given by
    40 - 23 - 14 = 3

  7. Which of the statements below are true about the graphs of the equations x 2 + y 2 = 9 and (x + 6) 2 + y 2 = 9?
    (I) Both graphs are circles.
    (II) The two graphs touch at one point.
    (I) The two graphs have two distinct points of intersection.
    Solution
    The two equations are of the form.
    (x - h) 2 + (y - k) 2 = r 2
    which are equations of circles.
    Let us now find the points of intersection of the two circles by solving the system of equations.
    (x + 6) 2 + y 2 = 9
    x 2 + y 2 = 9
    Subtract the left hand sides and right hand sides of the two equations as follows
    (x + 6) 2 + y 2 - [ x 2 + y 2 ] = 9 - 9
    Which gives an equation with one variable only
    12x + 36 = 0
    x = - 3
    Use equation x 2 + y 2 = 9 to find y
    3 2 + y 2 = 9
    y = 0
    The two circles have one point in common and therefore touch at one point.
    Only statement (I) and (II) are true.

  8. If the perimeter of a regular hexagon is equal to 6 a, then the area of this hexagon if given by:
    Solution
    A regular hexagon has 6 sides of equal size. Hence if the perimeter is 6 a, each side has a length equal to a. Also this regular hexagon may be considered as made up of 6 equilateral triangles of side a.

    math sat subject level 1 solution problem 8.


    We first find the area A of an equilateral triangle of side a. The height h of the triangle is found using Pythagora's theorem

    math sat subject level 1 solution problem 8.


    h = √( a2 - (a/2)2 ) = a √3 / 2
    The area A of the equilateral triangle of side a is given by
    A = (1/2)* a √3 / 2 * a = a2 √3 / 4
    The given hexagon is made up of 6 equilateral triangle each of side a, hence the are of the hexagon is given by
    6 * ( a2 √3 / 4 ) = 1.5 √3 a2

  9. AB is a diameter to the circle in the figure below and point C is on the circle. The measure of the diameter is 10 units and side AC has a length of 5 units. Find the measure of angle CBA

    math sat subject level 1 problem 9.


    Solution
    Since AB is the diamter of the circle, then triangle ABC is a right triangle and
    sin(angle CBA) = 5 / 10 = 1/2
    Hence angle CBA has a measure of arcsin(1/2) = 30 degrees

  10. For what value of positive k does the equation have one solution only?
    (x + k) x = - 4

    Solution
    We first expand the left side and write the equation in standard form
    x 2 + k x + 4 = 0
    The equation is quadratic and in order to have one solution only, its discriminant D must be equal to 0. Let us first find the discriminant D.
    D = k 2 - 4 (1)(4) = k 2 - 16
    Solve D = k 2 - 16 = 0 and select the positive solution
    k = 4

  11. The average of the roots of a quadratic equation is equal to 3 and the difference of the roots is equal to 2. Which of these could be the equation?
    Solution
    Let A and B be the roots of the equation. First their average is 3. Hence
    (A + B) / 2 = 3 or A + B = 6
    The difference of the two roots is 2. Hence
    A - B = 2
    Solve the system of equations A + B = 6 and A - B = 2 to find
    A = 4 and B = 2
    A quadratic equation with roots 4 and 2 and leading coefficient 1 has the form
    (x - 4)(x - 2) = 0
    Expand the left side of the equation.
    x 2 - 6 x + 8 = 0

  12. If points M, B(2 , 6) and C(4 , 8) are such that MBC is a right triangle with hypotenuse BC, then the 3 points M, B and C are on a circle of radius

    Solution
    Use the converse of Thales' theorem that states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle to draw triangle MBC as shown below.

    math sat subject level 1 solution problem 12.


    If the diameter is BC then the radius R is half the distnace BC. Hence
    R = (1/2) √( (4 - 2) 2 + (8 - 6) 2 ) = √2

  13. The volume of a rectangular solid is equal to 1000 m 3. If the length, width and height are increased by 50%, the volume of the new rectangular solid is equal to
    Solution
    If L, W and H are the three dimensions of a rectangular solid then its volume V is given by
    V = L * W * H
    If L is increased by 50%, the new value of L will be
    L + 50%L = 1.5 L
    Therefore increasing each dimension of the rectangle by 50% is the same as multiplying each of the 3 dimensions by 1.5. Hence the volume of the new rectangular solid is
    1.5L * 1.5W * 1.5H = (1.5) 3 L * W * H
    = 3.375 * 1000 = 3,375 m 3

  14. In the figure below angle ADE is right and BC and DE are parallel. The length of AC is 5 and the length of BC is 4. Find tan(angle CED).

    math sat subject level 1 problem 14.


    Solution
    Since BC is parallel to DE, triangles ABC and ADE are similar and therefore triangle ABC is also a right triangle. The corresponding angles of the two triangle are congruent and therefore equal in size. Hence
    tan(CED) = tan(ACB)
    Use Pythagora's theorem to find the length of side AB and find tan(ACB)
    AB = sqrt(52 - 452) = 3
    tan(ACB) = 3/4 = tan(CED)

  15. What are the coordinates of the center of the circle that is entirely in quadrant II and is tangent to the lines y = 8, y = 2 and x = -2?
    Solution
    Since the circle is tangent to y = 8 and y = 2, the y coordinate of the center is equal to
    (8 + 2) / 2 = 5
    and its radius is equal to
    (8 - 2) / 2 = 3
    The x coordinate of the center is
    (-2 - 3) = -5
    The coordinates of the center are
    (-5 , 5)

  16. In the figure below BC is parallel to DE. The length of side AC is x and the length of side CE is 4x. What is the area of triangle ADE if the area of ABC is 100 cm 2?

    math sat subject level 1 problem 16.


    Solution
    Since BC is parallel to DE, the triangles ADE and ABC are similar and the lengths of their corresponding sides related as follows
    AB / AD = BC / DE = h / H = AC / AE = x / 5x = 1/5, where h is the altitude of triangle ABC and H is the altitude of triangle ADE.
    The areas A1 and A2 of triangles ABC and ADE are given by
    A1 = (1/2)BC * h = 100 , A2 = (1/2)DE * H
    Since BC / DE = h / H = 1/5
    (BC * h) / (DE * H) = 1 / 25
    and hence
    DE * H = 25 (BC * h)
    the area A2 is given by
    A2 = (1/2)DE * H = 25 (1/2)(BC * h) = 25 * 100 = 2500 cm 2

  17. In the figure below is shown a right triangle ABC. What is the volume of the cone obtained by rotating the triangle about the side AB?

    math sat subject level 1 problem 17.

    Solution
    The cone will have radius AC determined using Pythagora's theorem
    AC = sqrt(102 - 62) = 8
    The volume V of the cone of radius r = 8 and height h = 6 is given by
    V = (1/3) ? r2 h = (1/3) ? 64 * 6 = 128?

  18. If 3(x + y) = 27 and x and y are positive integers, which of these cannot be the value of x / y?
    A) 1/2
    B) 8
    C) 2
    D) 1/8
    E) 1/4
    Solution
    Let us try to solve the system of equation for each case. We start with case A)
    A) 3(x + y) = 27 and x / y = 1 / 2
    Rewrite system of equation as
    x + y = 9 and x / y = 1 / 2 or 2x = y
    Solve to find x and y
    x = 3 and y = 6
    We solve case B)
    B) 3(x + y) = 27 and x / y = 8
    solutions: x = 8 and y = 1
    We solve case C)
    C) 3(x + y) = 27 and x / y = 2
    solutions: x = 6 and y = 3
    We solve case D)
    D) 3(x + y) = 27 and x / y = 1 / 8
    solutions: x = 1 and y = 8
    We solve case E)
    E) 3(x + y) = 27 and x / y = 1 / 4
    solutions: x = 9/5 and y = 36/5
    In case E) x and y are not positive integers and is therefore in contradiction. Hence the answer to the question is E)

  19. [ cos(x) sin(2x) - 2 sinx ] / [sin2x cos(x)] =
    Solution
    Use trigonometric identity sin(2x) = 2 sin(x)cos(x) to rewrite numerator as follows
    cos(x) sin(2x) - 2 sinx = cos(x) 2 sin(x)cos(x) - 2 sin(x)
    = 2sin(x) (cos(x)2 - 1) = - 2 sin3(x)
    We now substitute the expression in the numerator by - 2 sin3(x) and simplify
    [ cos(x) sin(2x) - 2 sinx ] / [sin2x cos(x)] = - 2 sin3(x) / [sin2x cos(x)]
    = -2 sin(x) / cos(x) = -2 tan(x)

  20. Find the smallest value of a positive integer x such that x 2 + 4 is divisible by 25.
    Solution
    For x 2 + 4 to be divisible by 25, it must of the form
    x 2 + 4 = 25 k , where k is a positive integer greater than 1.
    Substitute different values of k and solve for x
    k = 1; x 2 + 4 = 25 k = 25 , x 2 = 21 , x is not a positive integer.
    k = 2; x 2 + 4 = 25 k = 50 , x 2 = 46 , x is not a positive integer.
    k = 3; x 2 + 4 = 25 k = 75 , x 2 = 71 , x is not a positive integer.
    k = 4; x 2 + 4 = 25 k = 100 , x 2 = 96 , x is not a positive integer.
    k = 5; x 2 + 4 = 25 k = 125 , x 2 = 121 , x = 11 and is a positive integer.
    The smallest, positive integer, value of x such that x 2 + 4 is divisible by 25 is 11.

  21. Find the value of constant a such that x = 1 is a solution to the equation
    a√(x + 3) - 4a | 2x - 1 | = 4

    Solution
    Substitute x by 1 in the given equation and simplify
    a√(1 + 3) - 4a | 2(1) - 1 | = 4
    2a - 4a = 4
    Solve for a
    a = -2

  22. m and n are positive integers such that m > n and m 2n = 46656. Find m.
    Solution
    We first write m 2n as (mn)2
    m 2n = (mn)2 = 46656
    which gives
    mn = √ 46656 = 216
    We now write 216 as the product of powers of prime numbers
    216 = 2*2*2*3*3*3 = 23 33 = 63
    Hence
    mn = 63 , where m = 6 and n = 3 and is smaller than m.

  23. The root of an equation of the form f(x) = 2 is x = 5. The solution of the equation defined by f(- 2 x + 1) = 2 is equal to
    Solution
    If s be the solution of f(-2x + 1) = 2, then f(- 2 s + 1) = 2
    But we also know that f(5) = 2 since x = 5 is a solution to f(x) = 2; hence
    f(- 2 s + 1) = f(5)
    which gives
    - 2 s + 1 = 5
    which gives
    s = -2

  24. For what value of b will the system of equations 2x + 5y = 3 and -5x + by = 14 have no solution?
    Solution
    The lines corresponding to the two equations given above have different x-intercepts and therefore the given system of equations will have no solution if the slopes of the two lines are equal. We now write the two equations in slope intercept form
    2x + 5y = 3 may be written as y = -(2/5) x + 3/5
    -5x + by = 14 may be written as y = (5/b) x + 14/b
    The slopes of the two lines are equal if
    -2/5 = 5/b
    Solve for b
    b = -25 / 2
    If b = -25/2, the given system of equation has no solutions.

  25. Find the real numbers a and b such that 3 a - b i = (2 - i)(4 + i) where i = √ -1.
    Solution
    Expand the right hand side of the given equation
    3 a - b i = 9 - 2 i
    Two complex numbers are equal if their real parts are and their imaginary parts are equal. Hence
    3a = 9 and -b = -2
    Solve to find
    a = 3 and b = 2

  26. A dealer increased the price of an item by 20%, then increased the price of the same item by 30%. If x is the original price, what is the price after the two increases?
    Solution
    Price after the first increase
    x + 20%x = x + 0.2 x
    Price after the second increase
    x + 0.2 x + 30% (x + 0.2x) = x + 0.2x + 0.3(x + 0.2x)
    = 1.2x + 0.3x + 0.06x = 1.56x

  27. Two dice are thrown. What is the probability that the sum of the two numbers obtained is greater than 10?
    Solution
    When two dice are thrown, there are 36 possible outcomes:(1,1),(1,2),... Three of these outcomes have a sum greater than 10 and these are: (5,6), (6,5) and (6,6). Hence the probability that the sum of the two numbers obtained is greater than 10 is given by
    3 / 36 = 1 / 12

  28. If x and y are two real numbers such that 3x + 2y = 5 and 5x + 4y = 9, then 4x + 3y =
    Solution
    Add the right and the left hand sides of the two equations to obtain a new equation.
    (3x + 2y) + (5x + 4y) = 5 + 9
    Group like terms.
    8x + 6y = 14
    Divide all terms by to obtain a new equation.
    4x + 3y = 7

  29. The solution set of the inequality is given by the interval
    A) (5 , + infinity)
    B) (-infinity , -1)
    C) (-infinity , 1) U (5 , +infinity)
    D) (-infinity , -1) U (5 , +infinity)
    E) (-infinity , 0) U (5 , +infinity)
    Solution
    The graphs of |2x - 4| (brown) and that of x + 1 (in blue) intersect at x = 1 and x = 5. From the graph, it can be deducted that |2x - 4| is greater than x + 1 over the intervals
    (-infinity , 1) and (5 , + infinity)

    math sat subject level 1 problem 29.

  30. If m and n are positive integers are such that m / n = 2 / 3, which of these values cannot be values of m and n?
    A) m = 12 and n = 18
    B) m = 60 and n = 90
    C) m = 34 and n = 51
    D) m = 7 and n = 21
    E) m = 102 and n = 153
    Solution
    Set and reduce the fraction m / n for the given values
    A) m = 12 and n = 18 , m / n = 12/18 = 2/3
    B) m = 60 and n = 90 , m / n = 60 / 90 = 6 / 9 = 2/3
    C) m = 34 and n = 51 , m / n = 34 / 51 = 2*17 / 3*17 = 3/2
    D) m = 7 and n = 21 , m / n = 7 / 21 = 1/3
    m = 7 and n = 21 cannot be the values of m and n such that m / n = 2/3

  31. If 35 is the median of the data set including 21, 7, 45, 33, 62 and x, then x =
    Solution
    We first order the known values if the given set
    7 , 21 , 33 , 45 , 62
    If x = 3 or 14, the mean will be equal to
    (21 + 33) / 2 = 27
    If x = 33, the mean will be equal to
    (33 + 33) / 2 = 33
    If x = 37, the mean will be equal to
    (33 + 37) / 2 = 35
    So for x = 37, the data set is
    7 , 21 , 33 , 37 , 45 , 62 and the mean is 35


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