New SAT Maths Detailed Solutions - Sample 2

Detailed solutions to new SAT maths questions are presented.

  1. If point A(-2 , b) lies on the line that passes through the points B(2,4) and C(-1,0), what is the value of b?


    Solution
    For all three points to be on the same line (collinear), the slopes calculated using two different points must be equal. Hence
    Slope of AB = \( \dfrac{4 - b}{2 - (-2)} = \dfrac{4 - b}{4} \)
    Slope of BC = \( \dfrac{0 - 4}{-1 - 2} = \dfrac{4}{3} \)
    Solve the equation \( \dfrac{4 - b}{4} = \dfrac{4}{3} \)
    \( 3(4 - b) = 16 \implies 12 - 3b = 16 \implies b = -\dfrac{4}{3} \)
    Answer: \( b = -\dfrac{4}{3} \)

  2. To which expression is \( \dfrac{\dfrac{x^2+1}{x+2}}{\dfrac{1}{x+2}+4} \) equivalent to for x > 6?


    A) \( \dfrac{x^2+1}{4} \)
    B) \( \dfrac{x^2+1}{4x + 9} \)
    C) \( \dfrac{x^2+1}{x + 2} \)
    D) \( \dfrac{1}{x + 2} \)
    Solution
    Write the main denominator as a single rational expression:
    \( \dfrac{1}{x+2}+4 = \dfrac{1}{x+2} + 4\dfrac{x + 2}{x + 2} = \dfrac{4x + 9}{x+2} \)
    The given expression becomes:
    \( \dfrac{\dfrac{x^2+1}{x+2}}{\dfrac{4x + 9}{x+2}} = \dfrac{x^2+1}{x+2} \cdot \dfrac{x+2}{4x+9} = \dfrac{x^2+1}{4x+9} \)
    Answer: B

  3. Find the real number c such that \( (x - 2)^2 + 4 + c \) has a minimum value equal to 2.


    A) -6
    B) -4
    C) -2
    D) 4
    Solution
    \((x - 2)^2 + 4 + c\) is a quadratic expression with vertex at (2, 4 + c). Since the leading coefficient is positive, the vertex is a minimum point and its y-coordinate \(4 + c\) is the minimum value. Hence:
    \(4 + c = 2 \implies c = -2\)
    Answer: C

  4. t seconds after a ball is thrown vertically from ground, at a velocity of 30 meters per second, its height h changes with time as follows: \( h(t) = -5t^2 + 30t \) where h is in meters. For how long will the height of the ball be more than 25 meters?

    Solution
    Solve \( -5t^2 + 30t = 25 \):
    \( -5t^2 + 30t - 25 = 0 \implies t^2 - 6t + 5 = 0 \implies (t - 1)(t - 5) = 0 \)
    Solutions: t = 1 and t = 5. Since the parabola opens down, the height is above 25 meters between these times.
    Answer: 5 - 1 = 4 seconds

  5. t seconds after a ball is thrown vertically from ground, at a velocity 30 meters per second, its height h changes with time as follows: \( h(t) = -5t^2 + 30t \) where h is in meters. What is the change in the height of the ball between t = 1 and t = 2 seconds?

    Solution
    \( h(1) = -5(1)^2 + 30(1) = 25 \) meters.
    \( h(2) = -5(2)^2 + 30(2) = 40 \) meters.
    Change: \( 40 - 25 = 15 \) meters.
    Answer: The height increases by 15 meters.

  6. The graph of the cost C(x) = m x + b in US dollars as a function of the number of units produced x by a company, is shown below. m is the cost per unit and b is the fixed cost.

    sat question - cost graph

    Use the graph to approximate m, the cost per unit, and the fixed cost b.

    Solution
    Using points (0, 2000) and (12000, 8000):
    \( m = \dfrac{8000 - 2000}{12000 - 0} = 0.5 \) dollars per unit.
    Fixed cost b is the y-intercept: 2000 dollars.
    Answer: m = 0.5 USD/unit, b = 2000 USD

  7. The graph of the revenue R(x) = a x in US dollars as a function of the number of units x produced by a company, is shown below. a is the revenue per unit sold.

    sat question - revenue graph

    Use the graph to approximate a.

    Solution
    Using points (0, 0) and (8000, 6000):
    \( a = \dfrac{6000 - 0}{8000 - 0} = 0.75 \) dollars per unit.
    Answer: a = 0.75 USD/unit

  8. The graphs of the revenue R(x) = a x and the cost C(x) = m x + b are shown below. R and C are in US dollars and x is the number of units produced. The break even point is the point at which R(x) = C(x).

    sat question - break even graph

    Find the revenue at the break even point in terms of m, a and b.

    Solution
    Set \( a x = m x + b \implies x(a - m) = b \implies x = \dfrac{b}{a - m} \).
    Revenue at break even: \( R = a \cdot \dfrac{b}{a - m} = \dfrac{a b}{a - m} \).
    Answer: \( R = \dfrac{a b}{a - m} \)

  9. The resistances R1 and R2 of two resistors in parallel are equivalent to a resistance Re and are related by \( \dfrac{1}{R_e} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \). Express Re in terms of R1 and R2.

    Solution
    \( \dfrac{1}{R_e} = \dfrac{R_2 + R_1}{R_1 R_2} \implies R_e = \dfrac{R_1 R_2}{R_1 + R_2} \)
    Answer: \( R_e = \dfrac{R_1 R_2}{R_1 + R_2} \)

  10. The magnitude of the universal gravitational force F between two bodies of masses m1 and m2 is given by \( F = \dfrac{G m_1 m_2}{d^2} \). If the force at distance d is 2000 N, what is the force at distance 4d?

    Solution
    \( F_2 = \dfrac{G m_1 m_2}{(4d)^2} = \dfrac{1}{16} \cdot \dfrac{G m_1 m_2}{d^2} = \dfrac{2000}{16} = 125 \) N.
    Answer: 125 Newtons

  11. The graphs of a semi circle of diameter OB and two lines are shown below. Point O is the origin. Line L has equation y = (1/3)x and intersects line M and the semi circle at C. Line M intersects the semi circle and the x-axis at (4, 0). Find an equation for line M.

    sat question - semi circle and lines

    Solution
    OB is a diameter, so \(\angle OCB = 90^\circ\) (Thales' theorem). Thus L ⟂ M.
    Slope of L = \( \frac{1}{3} \), so slope of M = \( -3 \).
    Line M passes through B(4, 0): \( y = -3(x - 4) \implies y = -3x + 12 \).
    Answer: \( y = -3x + 12 \)

  12. A total of 200 students are registered. A student can register in several courses but must register in only one course of geology, geography or chemistry. The table below shows registration. How many boys are registered in chemistry?

    sat question - student registration table

    Solution
    Total students = 200. Sum of known entries: 20 + 30 + 45 + 15 + 35 = 145.
    Boys in chemistry = 200 - 145 = 55.
    Answer: 55

  13. Use the table in the last problem to find the probability p that a student selected at random is registered in geology or geography.

    Solution
    Students in geology or geography: 20 + 30 + 15 + 35 = 100.
    \( p = \dfrac{100}{200} = 0.5 \).
    Answer: 0.5

  14. According to relativity, mass m of a particle at speed v is given by \( m = \dfrac{m_0}{\sqrt{1 - v^2/c^2}} \). What is m if v = 0.9c and \( m_0 = 1 \) gram?

    Solution
    \( m = \dfrac{1}{\sqrt{1 - 0.9^2}} = \dfrac{1}{\sqrt{0.19}} \approx 2.294 \) grams.
    Answer: ≈ 2.3 grams

  15. The graph of \( x^2 + y^2 - 4x + 2y + 2 = 0 \) is a circle. What is the area enclosed?

    Solution
    Complete squares: \( (x^2 - 4x) + (y^2 + 2y) = -2 \).
    \( (x - 2)^2 - 4 + (y + 1)^2 - 1 = -2 \implies (x - 2)^2 + (y + 1)^2 = 3 \).
    Radius squared \( r^2 = 3 \), area \( = \pi r^2 = 3\pi \).
    Answer: \( 3\pi \)

  16. z is a complex number and \( \overline{z} \) its conjugate. Find z if \( 2z + 3\overline{z} - 3i = 4 + 7i \).

    Solution
    Let \( z = a + bi \), then \( \overline{z} = a - bi \).
    \( 2(a+bi) + 3(a-bi) - 3i = 4 + 7i \implies 5a + (-b - 3)i = 4 + 7i \).
    Equate real/imaginary: \( 5a = 4 \implies a = \frac{4}{5} \); \( -b - 3 = 7 \implies b = -10 \).
    Answer: \( z = \frac{4}{5} - 10i \)

  17. Functions g and f are given by \( g(u) = 3u + 3 \) and \( f(t+2) = g(2t - 3) \). What is \( f(x) \)?

    Solution
    Let \( x = t + 2 \implies t = x - 2 \).
    \( f(x) = g(2(x-2) - 3) = g(2x - 7) = 3(2x - 7) + 3 = 6x - 18 \).
    Answer: \( f(x) = 6x - 18 \)

  18. m is a real number in the equation \( 2x^2 + 3mx = 4 - 2x \). For what value(s) of m is the sum of the solutions greater than 5?

    Solution
    Rewrite: \( 2x^2 + (3m + 2)x - 4 = 0 \).
    Sum of solutions = \( -\dfrac{3m+2}{2} > 5 \implies -(3m+2) > 10 \implies -3m > 12 \implies m < -4 \).
    Answer: \( m < -4 \)

  19. Triangle ABC is equilateral. What are the coordinates of point C?

    sat question - equilateral triangle

    Solution
    Let C = (a, b). AB = 4. For equilateral, AC = 4 and BC = 4.
    \( (a-1)^2 + (b-1)^2 = 16 \) and \( (a-5)^2 + (b-1)^2 = 16 \).
    Subtract: \( (a-1)^2 - (a-5)^2 = 0 \implies (a-1)^2 = (a-5)^2 \implies a = 3 \).
    Substitute: \( (3-1)^2 + (b-1)^2 = 16 \implies 4 + (b-1)^2 = 16 \implies (b-1)^2 = 12 \implies b = 1 \pm 2\sqrt{3} \).
    Using first quadrant, \( b = 1 + 2\sqrt{3} \).
    Answer: \( (3, 1 + 2\sqrt{3}) \)

  20. What is the diameter of the circle whose area and perimeter have the same value?

    Solution
    \( \pi r^2 = 2\pi r \implies r^2 = 2r \implies r = 2 \) (r > 0).
    Diameter \( D = 2r = 4 \).
    Answer: 4 units

  21. John can read 15 pages every 20 minutes. Linda can read 20 pages every 16 minutes. How many minutes are needed for them to read a total of 100 pages together?

    Solution
    John's rate = \( \frac{15}{20} = \frac{3}{4} \) pages/min.
    Linda's rate = \( \frac{20}{16} = \frac{5}{4} \) pages/min.
    Combined rate = \( \frac{3}{4} + \frac{5}{4} = 2 \) pages/min.
    Time for 100 pages = \( \frac{100}{2} = 50 \) minutes.
    Answer: 50 minutes

  22. x and y are integers. Remainder of x/3 is 1, remainder of y/3 is 2. What is the remainder of (2x + 3y)/3?

    Solution
    \( x = 3k + 1,\ y = 3j + 2 \).
    \( 2x + 3y = 2(3k+1) + 3(3j+2) = 6k + 2 + 9j + 6 = 6k + 9j + 8 \).
    \( 6k + 9j + 8 = 3(2k + 3j + 2) + 2 \). Remainder = 2.
    Answer: 2

  23. Find the smallest positive integer x so that when divided by 5 remainder is 4 and when divided by 6 remainder is 5.

    Solution
    \( x = 5k + 4 = 6j + 5 \implies 5k = 6j + 1 \).
    Try j = 4: \( 5k = 25 \implies k = 5 \).
    \( x = 5(5) + 4 = 29 \).
    Answer: 29

  24. 1 liter of water is poured into a cylindrical container of radius 10 cm. How high in cm will the water rise?

    Solution
    1 liter = 1000 cm³. Volume = \( \pi r^2 h = 1000 \).
    \( h = \dfrac{1000}{\pi (10)^2} = \dfrac{1000}{100\pi} = \dfrac{10}{\pi} \) cm.
    Answer: \( \frac{10}{\pi} \) cm

  25. If AB is 1/3 of BO, what is the length of chord MN of the circle with center O and radius 5 cm?

    sat question - right similar triangle

    Solution
    Let \( BO = x \), then \( AB = x/3 \). \( x + x/3 = 5 \implies \frac{4x}{3} = 5 \implies x = 15/4 \).
    In right \( \triangle MBO \): \( MB^2 + OB^2 = 5^2 \implies MB^2 = 25 - \frac{225}{16} = \frac{175}{16} \implies MB = \frac{5\sqrt{7}}{4} \).
    \( MN = 2 \cdot MB = \frac{5\sqrt{7}}{2} \).
    Answer: \( \frac{5\sqrt{7}}{2} \) cm

  26. In the figure below, AB is tangent to a circle with diameter BC. Find the length of segment AD.

    sat question - chord of circles

    Solution
    \( \triangle ABC \) is right at B, \( \triangle BDC \) is right at D (Thales).
    \( \triangle ABC \sim \triangle BDA \) (common angle A).
    \( AC = \sqrt{6^2 + 4^2} = 2\sqrt{13} \).
    \( \dfrac{AC}{AB} = \dfrac{BC}{BD} \implies BD = \dfrac{AB \cdot BC}{AC} = \dfrac{6 \cdot 4}{2\sqrt{13}} = \dfrac{12}{\sqrt{13}} \).
    \( AD = \sqrt{AB^2 - BD^2} = \sqrt{36 - \frac{144}{13}} = \sqrt{\frac{324}{13}} = \frac{18}{\sqrt{13}} \).
    Answer: \( \frac{18}{\sqrt{13}} \)

  27. Lines m and n are parallel and lines k, n and J all intersect at point O. Find angle x.

    sat question - parallel lines and intersects

    Solution
    Angle adjacent to 116°: \( \alpha = 180 - 116 = 64° \).
    Alternate interior angles: \( \alpha = \beta = 64° \).
    In triangle: \( x + \beta + 48 = 180 \implies x + 64 + 48 = 180 \implies x = 68° \).
    Answer: 68°

  28. Points A, B, C on a circle radius 5 cm, O is center, \( \angle CAB = 30° \). Find shaded area to one decimal place.

    sat question - circle and right triangle

    Solution
    \( \angle COB = 2 \times 30° = 60° \), \( \angle AOC = 120° \).
    Area sector AOC? Shaded = half circle area minus areas of triangles AOC and COB.
    Area triangle AOC = \( \frac{1}{2} R^2 \sin 120° = \frac{25}{2} \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \).
    Area triangle COB = \( \frac{1}{2} R^2 \sin 60° = \frac{25}{2} \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \).
    Shaded area = \( \frac{1}{2} \pi (25) - \frac{25\sqrt{3}}{4} - \frac{25\sqrt{3}}{4} = \frac{25\pi}{2} - \frac{25\sqrt{3}}{2} \approx 22.2 \) cm².
    Answer: ≈ 22.2 cm²