Math Word Problems with Solutions for Grade 11

Grade 11 math word problems with answers and solutions are presented.



  1. An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.

  2. Find the area between two concentric circles defined by

    x2 + y2 -2x + 4y + 1 = 0

    x2 + y2 -2x + 4y - 11 = 0

  3. Find all values of parameter m (a real number) so that the equation 2x2 - mx + m = 0 has no real solutions.

  4. The sum an integer N and its reciprocal is equal to 78/15. What is the value of N?

  5. m and n are integers so that 4m / 125 = 5n / 64. Find values for m and n.

  6. Simplify: 3n + 4001 + 3n + 4001 + 3n + 4001

  7. P is a polynomial such that P(x2 + 1) = - 2 x4 + 5 x2 + 6. Find P(- x2 + 3)

  8. For what values of r would the line x + y = r be tangent to the circle x2 + y2 = 4?

Solutions to the Above Problems

  1. Let x = speed of airplane in still air, y = speed of wind and D the distance between A and B. Find the ratio x / y

    Against the wind: D = 8(x - y), with the wind: D = 7(x + y)

    8x - 8y = 7x + 7y, hence x / y = 15

  2. Rewrite equations of circles in standard form. Hence equation x2 + y2 -2x + 4y + 1 = 0 may be written as

    (x - 1)2 + (y + 2) 2 = 4 = 22
    and equation x2 + y2 -2x + 4y - 11 = 0 as
    (x - 1)2 + (y + 2) 2 = 16 = 42

    Knowing the radii, the area of the ring is Pi (4)2 - Pi (2)2 = 12 Pi

  3. The given equation is a quadratic equation and has no solutions if it discriminant D is less than zero.

    D = (-m)2 - 4(2)(m) = m2 - 8m

    We nos solve the inequality m2 - 8m < 0

    The solution set of the above inequality is: (0 , 8)

    Any value of m in the interval (0 , 8) makes the discriminant D negative and therefore the equation has no real solutions.

  4. Write equation in N as follows

    N + 1/N = 78/15

    Multiply all terms by N, obtain a quadratic equation and solve to obtain N = 5.

  5. 4m / 125 = 5n / 64


    Cross multiply: 64 4m = 125 5n

    Note that 64 = 43 and 125 = 53

    The above equation may be written as: 4m + 3 = 5n + 3

    The only values of the exponents that make the two exponential expressions equal are: m + 3 = 0 and n + 3 = 0, which gives m = - 3 and n = - 3.

  6. 3n + 4001 + 3n + 4001 + 3n + 4001 = 3(3n + 4001) = 3n + 4002

  7. P(x2 + 1) = - 2 x4 + 5 x2 + 6

    Let t = x2 + 1 which also gives x2 = t - 1

    Substitute x2 by t - 1 in P to obtain: P(t) = - 2 (t - 1)2 + 5 (t - 1) + 6 = -2 t 2 + 9t - 1

    Now let t = - x2 + 3 and substitute in P(t) above to obtain

    P(- x2 + 3) = -2 (- x2 + 3) 2 + 9 (- x2 + 3) - 1 = -2 x 4 + 3 x 2 + 8

  8. Solve x + y = r for y: y = r - x

    Substitute in the equation of the circle:

    x2 + (r - x)2 = 4

    Expand: 2 x2 -2 r x + r 2 - 4 = 0

    If we solve the above quadratic equation (in x) we will obtain the x coordinates of the points of intersection of the line and the circle. The 2 points of intersection "become one" and therefore the line and the circle become tangent if the discriminant D of the quadratic equation is zero. Hence

    D = (-2r)2 - 4(2)(r2 - 4) = 4(8 - r2) = 0

    Solve for r to obtain: r = 2sqrt(2) and r = - 2sqrt(2)


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Updated: 2 April 2013

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