Math Word Problems with
Solutions and Explanations - Grade 9

Detailed solutions and full explanations to grade 9 math word problems are presented.

  1. Which number(s) is(are) equal to its (their) square?

    Solution

    If x is the number to find, its square is x2.

    x is equal to its square hence: x = x2

    Solve the above equation by factoring. First write with right side equal to zero.

    x - x2 = 0

    x(1 - x) = 0

    solutions: x = 0 and x = 1

    Check answers.

    1) x = 0 , its square is 02 = 0 . Hence x and its square are equal.

    2) x = 1 , its square is 12 = 1 . Hence x and its square are equal.

  2. Which number(s) is(are) equal to half its (their) square?

    Solution

    Let x be the number to find. Half its square is (1/2) x2.

    x is equal to half its square hence: x = (1/2) x2

    First write with right side equal to zero.

    x - (1/2) x2 = 0

    x(1 - (1/2)x) = 0 , factor

    solutions: x = 0 and x = 2

    Check answers.

    1) x = 0 , half its square is (1/2) 02 = 0 . Hence x and half its square are equal.

    2) x = 2 , half its square is (1/2) 22 = 2 . Hence x and half its square are equal.

  3. Which number(s) is(are) equal to the quarter of its (their) square?

    Solution

    If x is the number to find, then the quarter of its square is (1/4) x2.

    x is equal to the quarter of its square hence: x = (1/4) x2

    First write the above equation with its right side equal to zero.

    x - (1/4) x2 = 0

    x(1 - (1/4)x) = 0 , factor x out

    solutions: x = 0 and x = 4

    Check answers.

    1) x = 0 , quarter of its square is (1/4) 02 = 0 . Hence x and the quarter of its square are equal.

    2) x = 4 , the quarter of its square is (1/4) 42 = 4 . Hence x and the quarter of its square are equal.

  4. A car travels from A to B at a speed of 40 mph then returns, using the same road, from B to A at a speed of 60 mph. What is the average speed for the round trip?

    Solution

    The average speed is given by.

    total distance / total time

    If x is the distance from A to B, then the total distance is equal to 2x (away and return). The total T time is equal to time t1 = x / 40 from A to B (away) plus t2 = x / 60 from B to A (return). Hence

    T = x / 40 + x / 60 = 100 x / 2400 = x / 24

    The average speed is given by

    2x / (x/24) = 48 mph

  5. Tom travels 60 miles per hour going to a neighboring city and 50 miles per hour coming back using the same road. He drove a total of 5 hours away and back. What is the distance from Tom's house to the city he visited?(round your answer to the nearest mile).

    Solution

    Let x be the distance traveled to the city, then the time taken is given by

    distance / speed = x / 60

    and the time to return is given by

    distance / speed = x / 50

    The total time to travel and return is 5 hours. Hence

    x / 60 + x / 50 = 5

    Solve for x

    x = 1500 / 11 = 136 miles (rounded to the nearest mile)

  6. At 11:00 a.m., John started driving along a highway at constant speed of 50 miles per hour. A quarter of an hour later, Jimmy started driving along the same highway in the same direction as John at the constant speed of 65 miles per hour. At what time will Jimmy catch up with John?

    Solution

    Let t be the number of hours, from 11:00 am, when Jimmy catches up with John. The time that Jimmy will have to drive to catch up with John is t - 1/4: he starts a quarter of an hour late. When Jimmy catches up with John, they would have traveled the same distance. Hence

    50 t = 65 (t - 1/4)

    Solve for t

    50t = 65t - 65/4

    t = 65/60 = 1.083 hours = 1hour and 5 minutes

    Jim will catch up with John at

    11:00 am + 1 hour , 5 minutes = 12:05 pm

  7. Find an equation of the line containing (- 4,5) and perpendicular to the line 5x - 3y = 4.

    Solution

    First find the slope of the line 5x - 3y = 4

    -3y = - 5x + 4

    y = (5/3)x - 4/3

    slope = 5/3

    Let m be the slope of the line perpendicular to the line 5x - 3y = 4. Hence

    m * (5/3) = - 1

    m = -(3/5)

    The equation of the line containing (- 4,5) and perpendicular to the line 5x - 3y = 4 is given by

    y - 5 = -(3/5)(x - (-4))

    y = -(3/5) x + 13/5

    5y + 3x = 13

  8. A rectangle field has an area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field?

    Solution

    Let L and W be the length and width of the rectangle. The area and the perimeter of the rectangle may be written as follows

    L * W = 300 , area

    2(L + W) = 80 or L + W = 40 , perimeter

    Solve equation L + W = 40 for W

    W = 40 - L

    Substitute W by 40 - L in the equation L * W = 300

    L(40 - L) = 300

    Write the above equation in standard form

    L2 - 40L + 300 = 0

    (L - 30)(L - 10) = 0

    Solve for L and calculate W

    L = 30 and W = 40 - L = 10

    L = 10 and W = 40 - 10

    Assuming that the length is larger than the width, the rectangle has a length L = 30 meters and a width W = 10 meters

  9. Find the area of a trapezoid whose parallel sides are 12 and 23 centimeters respectively.

    Solution

    The area A of a trapezoid is given by

    A = (1/2) height *(base1 + base2)

    The two bases are given as 12 and 23 cm, however the height is not given and therefore there is not enough information to find the area of the trapezoid.

  10. A rectangular garden in Mrs Dorothy's house has a length of 100 meters and a width of 50 meters. A square swimming pool is to be constructed inside the garden. Find the length of one side of the swimming pool if the remaining area (not occupied by the pool) is equal to one half the area of the rectangular garden.

    Solution

    let x be the side of the swimming pool. If the area not covered by the swimming pool is half the rectangular garden, then the other half is covered by the swimming pool whose area is x2. Hence

    x2 = (1/2) 100*50 = 2500

    Solve for x

    x = 50 meters , side of the swimming pool

  11. ABC is an equilateral triangle with side length equal to 50 cm. BH is perpendicular to AC. MN is parallel to AC. Find the area of triangle BMN if the length of MN is equal to 12 cm.

    problem 11 .



    Solution

    Note that since ABC is an equilateral triangle and BH is perpendicular to AC, HC = 50/2 = 25 cm. Also since MN and HC are parallel, the two triangles BMN and BHC are similar and the sizes of their sides are proportional. Hence

    BN / BC = MN / HC = BM / BH = 12 / 25

    The area of triangle BMN is given by

    A = (1/2) BM * MN

    The area of triangle BHC is given by

    B = (1/2) BH * HC

    Note the ratio of the two areas

    A / B = [ (1/2) BM * MN ] / [ (1/2) BH * HC ] = (BM/BH) * (MN/HC) = (12/25)*(12/25) = (12/25)2

    let us now find BH using Pythagora's theorem.

    BH = sqrt(502 - 252) = 25 sqrt(3)

    The area B of triangle BHC is given by.

    B = (1/2) BH * HC = (1/2) 25 sqrt(3) 25 = (1/2) 252 sqrt(3)

    We now use the ratio A / B = (12/25)2 to find A the area of triangle BMN.

    A = (12/25)2 * B = (12/25)2 * (1/2) 252 sqrt(3) = 72 sqrt(3) cm2

  12. The height h of water in a cylindrical container with radius r = 5 cm is equal to 10 cm. Peter needs to measure the volume of a stone with a complicated shape and so he puts the stone inside the container with water. The height of the water inside the container rises to 13.2 cm. What is the volume of the stone in cubic cm?

    problem 12 .



    Solution

    We first find the volume of water without the stone.

    V1 = 10 *(Pi * 52) = 250 Pi , where Pi = 3.14

    The volume of water and stone is given by

    V2 = 13.2 *(Pi * 52) = 330 Pi

    The volume of the stone is given by

    V2 - v1 = 330 Pi - 250 Pi = 80 Pi

    = 251.1 cm3

  13. In the figure below the square has all its vertices on the circle. The area of the square is equal to 400 square cm. What is the area of the circular shape?

    problem 13.



    Solution

    Note the size of the diagonal of the square is equal to the size of the diameter of the circle. The side x of the square is such that

    x2 = 400 , hence x = 20 cm

    The diagonal D of the square is found using Pythagora's theorem

    D2 = x2 + x2 = 800

    The area A of the circle is given by

    A = Pi (D/2)2 = Pi D2 / 4 = 200 Pi cm3

    = 628 cm3

  14. The numbers 2 , 3 , 5 and x have an average equal to 4. What is x?

  15. The numbers x , y , z and w have an average equal to 25. The average of x , y and z is equal to 27. Find w.

  16. Find x , y , z so that the numbers 41 , 46 , x , y , z have a mean of 50 and a mode of 45.

  17. A is a constant. Find A such that the equation 2x + 1 = 2A + 3(x + A) has a solution at x = 2.

  18. 1 liter is equal to 1 cubic decimeter and 1 liter of water weighs 1 kilogram. What is the weight of water contained in a cylindrical container with radius equal to 50 centimeters and height equal to 1 meter?

  19. In the figure below triangle ABC is an isosceles right triangle. AM is perpendicular to BC and MN is perpendicular to AC. Find the ratio of the area of triangle MNC to the area of triangle ABC.

    problem 18.



  20. Pump A can fill a tank of water in 4 hours. Pump B can fill the same tank in 6 hours. Both pumps are started at 8:00 a.m. to fill the same empty tank. An hour later, pump B breaks down and took one hour to repair and was restarted again. When will the tank be full? (round your answer to the nearest minute).

  21. Are the lines with equations 2x + y = 2 and x - 2y = 0 parallel, perpendicular or neither?

  22. What are the dimensions of the square that has the perimeter and the area equal in value?

  23. Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?

  24. Find the circumference of a circular disk whose area is 100pi square centimeters.

  25. The semicircle of area 50 pi centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.

  26. A triangle has an area of 200 cm2. Two sides of this triangle measure 26 and 40 cm respectively. Find the exact value of the third side.


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Updated: 21 May 2009 (A Dendane)