Math Problems with Solutions and Explanations for 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 *(π * 52) = 250 π , where π = 3.14
    The volume of water and stone is given by
    V2 = 13.2 *(π * 52) = 330 π
    The volume of the stone is given by
    V2 - v1 = 330 π - 250 π = 80 π
    = 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 = π (D/2)2 = π D2 / 4 = 200 π cm3
    = 628 cm3

  14. The numbers 2 , 3 , 5 and x have an average equal to 4. What is x?
    Solution
    The average of 2 , 3 , 5 and x is given by
    (2 + 3 + 5 + x) / 4
    and it is known and equal to 4. Hence
    (2 + 3 + 5 + x) / 4 = 4
    Solve the above equation for x to obtain
    x = 6

  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.

    Solution
    The average of x , y , z and w is equal to 25. Hence
    (x + y + z + w) / 4 = 25
    The average of x , y and z is equal to 27. Hence
    (x + y + z ) / 3 = 27
    The above gives
    x + y + z = 3 27 = 81
    substitute x + y + z by 81 in the equation (x + y + z + w) / 4 = 25 to obtain the equation
    (81 + w) / 4 = 25
    Solve the above for w
    w = 19

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

    Solution
    The arithmetic mean of 41 , 46 , x , y , z is equal to 50. Hence
    (41 + 46 + x + y + z) / 5 = 50
    Rearrange the equation to obtain
    x + y + z = 163
    The mode is the repeated value. We cannot have x = y = z = 45 because their sum will not be equal to 163. The only possibility is that x = y = 45. Hence the above equation may be written as
    45 + 45 + z = 163
    Solve the above for z
    z = 73

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

    Solution
    Substitute x by 2 in the given equation
    2(2)+ 1 = 2A + 3(2 + A)
    Solve for A
    A = - 1 / 5

  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?

    Solution
    dm is the symbol of a decimeter and is equal to 10 cm. We convert all given units in dm.
    radius: r = 50 cm = 5 dm
    height : h = 1 m = 10 dm
    The volume of water in the cylindrical container is given by:
    πr2 h = 3.14 (5 md) 2 10 dm = 785 dm3
    1 dm3 weights 1 kg. Hence the total weight of water in the container is equal to
    785 dm3 1 kg / dm3 = 785 Kg

  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.


    Solution
    ABC is isosceles; hence
    AB = AC
    ABM and ACM have equal sides and one common side and equal angles are therefore congruent triangles. Hence the area of triangle AMC is half the area of ABC.
    In triangle MAC angles MAC and MCA are both equal to 45 hence triangle MAC is an isosceles right triangle and therefore trianles MAN and MCN are congruent (similar proof to above). Therefore area of triangle MNC is half the area of triangle AMC. Hence the ratio of the area of triangle MNC to the area of triangle ABC is equal to 1 / 4.

  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).
    Solution
    Pump A can fill 1 tank in 4 hours; hence the rate of pump A is
    1 / 4 (tank / hour)
    Pump B can fill 1 tank in 6 hours; hence the rate of pump B is
    1 / 6 (tank / hour)
    Let t be the number of hours that takes pump A to fill the tank. Hence during the t hours pump A fills
    t (1 / 4) of the tank
    Pump B was stopped for one hour and hence will pump water into the tank for t -1 hours; hence during t - 1 pump B fills
    (t - 1) (1 / 6) of the tank
    The two pumps work together to fill 1 tank; hence
    t (1 / 4) + (t - 1) (1 / 6) = 1
    Multiply all terms in the above equation by 24 and simplify
    24 t (1 / 4) + 24 (t - 1) (1 / 6) = 24 1
    6 t + 4(t - 1) = 24
    Solve for t
    10 t = 28
    t = 2.8 hours = 2 hours + 0.8 60 minutes = 2 hours 48 minutes
    The tank is full at
    8 a.m + 2 hours 48 minutes = 10:48 a.m.

  21. Are the lines with equations 2x + y = 2 and x - 2y = 0 parallel, perpendicular or neither?
    Solution
    Write each of the given equations in the form y = m x + b and identify the slope m.
    2x + y = 2 gives y = - 2 x + 2 , slope m1 = - 2
    x - 2y = 0 gives y = x / 2 , slope m 2= 1 / 2
    m1 and m2 are not equal and therefore the lines are not parallel. The product of the two slopes is
    m1 m2 = - 2 (1 / 2) = -1
    The two lines are perpendicular

  22. What are the dimensions of the square that has the perimeter and the area equal in value?
    Solution
    Let x be the side of the square. The perimeter 4x and the area x 2 are equal; hence
    4 x = x 2
    Solve the above equation
    4 x - x 2 = 0
    x (4 - x) = 0
    Two solution
    x = 0 for which the square does not exist
    x = 4 which is a square of side 4 units.

  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?
    Solution
    Let L be the length and W be the width. "a length 3 meters more that its width" is written as
    L = W + 3
    The perimeter of the rectangle is given by
    2 L + 2 W
    The are of the rectangle is given by
    L W
    "perimeter equal in value to its area" is written as
    2 L + 2 W = L W
    Substitute L by 3 + W in the above equation to obtain
    2 (W + 3)+ 2 W = (3+ W) W
    Expand and write the equation in standard form
    W 2 - W - 6 = 0
    Solve the above quadratic equation to obtain
    W = 3 and W = - 2
    The width of a rectangle is positive hence W = 3 and
    L = W + 3 = 6

  24. Find the circumference of a circular disk whose area is 100π square centimeters.
    Solution
    Area A = 100π; hence
    A = π R 2 = 100π , where R is the radius of the disk.
    Hence the equation
    π R 2 = 100π
    Solve for R.
    R 2 = 100 give R = 10 cm
    The circumference C is given by .
    C= 2 π R = 20 π cm

  25. The semicircle of area 50 π centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.
    Solution
    If R is the radius of a semicircle; its area is known and is given by the formula
    (1/2)πR 2 = 50 π
    Solve the above equation for R
    R 2 = 100 gives R = 10
    The length L of the rectangle is equal to the diameter; hence
    L = 2 R = 20 cm
    The semicircle is inscribed inside the rectangle and therefore it radius is equal to width W of the rectangle; hence
    W = R = 10 cm
    The area of the rectangle is equal to
    L W = 20 10 = 200 cm 2

  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.
    Solution
    The area A of a triangle given its two sides a and b making an angle α is given by
    A = (1/2) a b sin(α)
    Use the cosine rule to find cos(α) as
    cos (α) = (a 2 + b 2 - c 2) / 2 a b , c is the third side of the triangle.
    Use the formula sin(α) = √(1- cos 2 (α)) to rewrite the formula for the area as
    A = (1/2) a b sin(α) = (1/2) a b √(1- ((a 2 + b 2 - c 2) / 2 a b) 2)
    Simplify to obtain
    A = (1 / 4) √(4 a 2 b 2 - (a 2 + b 2 - c 2) 2)
    Solve for c to obtain two solutions
    c = √ [ a 2 + b 2 ~+mn~ √(4 a 2 b 2 - 16 A 2) ]
    In this problem a = 26, b = 40 and A = 200. Substitute and solve to find two solutions
    c1 = √ (356) = 2 √ (89) ; c2 = √ (4196) = 2 √ (1049)

Answers to the Above Questions

  1. 0 , 1
  2. 0 , 2
  3. 0 , 4
  4. 48 miles per hour
  5. 136 miles
  6. 12:05 a.m.
  7. 5y + 3x = 13
  8. length = 30 meters , width = 10 meters
  9. not enough information to solve the problem.
  10. 50 meters
  11. 72 √3 square centimeters.
  12. 80 π cubic centimeters
  13. 200 π square centimeters
  14. x = 6
  15. w = 19
  16. x = 45 , y = 45 and z = 73
  17. A = - 1/5
  18. 785 kg
  19. 1:4
  20. 10:48 a.m.
  21. perpendicular
  22. square with side length = 4 units
  23. length = 6 units and width = 3 units
  24. 20π cm
  25. 200 cm 2
  26. two solutions: 2 √ (89) and 2 √ (1049)

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