Math Problems with Solutions for Grade 9

Let \( x \) be number of student tickets and \( y \) be number of adult tickets. Use the given information to write the eq equations: \[ x + y = 120 \quad \text{(1)} \quad \text{and} \quad 8x + 12y = 1160 \quad \text{(2)} \] From (1): \[ y = 120 - x \] Substitute into (2): \[ 8x + 12(120 - x) = 1160 \] \[ 8x + 1440 - 12x = 1160 \] \[ -4x = -280 \] Solve for \( x \): \[ x = 70 \] Then \( y = 120 - 70 = 50 \)

\( 70 \) student tickets and \( 50 \) adult tickets were sold.

Let the smallest angle be \( x \), then Second angle = \( 3x \) and the third angle = \( x + 20 \).

Sum of angles in a triangle = \( 180^\circ \): \[ x + 3x + (x + 20) = 180 \] Group like terms: \[ 5x + 20 = 180 \] Solve for \( x \): \[ 5x = 160 \] \[ x = 32 \] Angles: Smallest angle: \( x = 32^\circ \)

Second angle: \( 3x = 96^\circ \)

Third angle: \( x + 20 = 52^\circ \)

To find \( f(2) \), substitute \( x \) by \(2\) in \( f(x) \): \[ f(2) = 2(2)^2 - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3 \] To find \( f(-1) \), substitute \( x \) by \(-1\) in \( f(x) \): \[ f(-1) = 2(-1)^2 - 3(-1) + 1 = 2(1) + 3 + 1 = 6 \]

Volume of the cylinder: \[ V = \pi r^2 h = 3.14 \times 3^2 \times 5 = 3.14 \times 9 \times 5 \] \[ = 3.14 \times 45 = 141.3 \; \text{m^3} \]

There are \( 6 \times 6 = 36 \) possible outcomes with two dice:

(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)

(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)

(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)

(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)

(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)

(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)

The list of all combinations that give a sum of 9 is:

(3,6) , (4,5) , (5,4) , (6,3) \[ \text{Probability} =\dfrac{\text{Total Number combinations that give a sum of 9 }}{\text{Total Number of Combinations}} \] Total Number combinations that give a sum of 9 = 4

Total Number of Combinations = 36

Hence \[ \text{Probability} = \dfrac{4}{36} = \dfrac{1}{9} \]

We write 32 as a power of 2: \[ 32 = 2^5 \] Rewrite the equation as \[ 2^{x+1} = 2^5 \] We deduce the algebraic equation \[ x + 1 = 5 \] Solve for \( x \): \[x = 4 \]

The ladder, the base and the wall form a right triangle with the ladder making the hyporenuse. Let the height be \( h \) and use the Pythagorean theorem: \[ h^2 + 5^2 = 13^2 \] Simplify and rewrite as: \[ h^2 = 144 \] Solve for \( h \): \[ h = \sqrt{144} = 12 \] The ladder reaches 12 meters up the wall.

Let \( x \) be number of $0.25 coins and \( y \) be number of $1 coins, and use the given information to write the equations: \[ x + y = 50 \quad \text{(1)} \quad \text{and} \quad 0.25 x + 1 y = 35 \quad \text{(2)} \] Multiply equation (2) by 100 to eliminate decimals: \[ 25 x + 100 y = 3500 \quad \text(3) \] and from equation (1), deduce: \[ y = 50 - x \] Substitute \( y = 50 - x \) into equation (3): \[ 25x + 100(50 - x) = 3500 \] Expand and group like terms: \[ - 75 x = - 1500 \] Solve for \( x \): \[ x = 20 \] \[ y = 50 - 20 = 30 \] There are 20 coins of $0.25 and 30 coins of $1

The difference between two odd numbers is \( 2 \). If the first number is \( x \), then the secodn is \( x + 2 \) and the third number is \( x + 4 \) and their sum is \( 123 \), hence the equation: \[ x + (x + 2) + (x + 4) = 123 \] Group like tems: \[ 3x + 6 = 123 \] Solve for \( x \): \[ x = 39 \] So the numbers are: \[ x = 39 \; , \; x+ 2 = 39+2 = 41 \; , \; x+4 = 39+4 = 43 \] \[ \{39,41,43\} \]

Let the width be \( x \) meters. Then the length is \( x + 4 \) meters. \[ \text{Area} = \text{ length } \times \text{width} = x(x + 4) = 96 \] \[ x^2 + 4x - 96 = 0 \] Solve using the quadratic formula: \[ x = \dfrac{-4 \pm \sqrt{4^2 + 4 \cdot 96}}{2} = \dfrac{-4 \pm \sqrt{16 + 384}}{2} = \dfrac{-4 \pm \sqrt{400}}{2} \] \[ x = \dfrac{-4 \pm 20}{2} \Rightarrow x = 8 \text{ or } -12 \] We reject the negative value.

Width : \( x = 8 \) m

Length : \( x + 4 = 12 \) m.

If \( x \) is the number to find, its square is \( x^2 \) and since they are equal, we write the equation: \[ x = x^2 \] Write the equation in standard form: \[ x - x^2 = 0 \] Factor \[ x(1 - x) = 0 \] \[ x = 0 \quad \text{or} \quad x = 1 \] Check the answers.

1) If \( x = 0 \), its square is \( 0^2 = 0 \). Hence, \( x \) and its square are equal.

2) If \( x = 1 \), its square is \( 1^2 = 1 \). Hence, \( x \) and its square are equal.

\( x \) is equal to half its square. Hence: \[ x = \dfrac{1}{2}x^2 \] Write in standard form: \[ x - \dfrac{1}{2}x^2 = 0 \] Factor: \[ x\left(1 - \dfrac{1}{2}x\right) = 0 \] Solutions: \[ x = 0 \quad \text{and} \quad x = 2 \] Check answers.

1) For \( x = 0 \): \[ \dfrac{1}{2}(0)^2 = 0 \Rightarrow x = \dfrac{1}{2}x^2 \text{ is satisfied} \] 2) For \( x = 2 \): \[ \dfrac{1}{2}(2)^2 = \dfrac{1}{2}(4) = 2 \Rightarrow x = \dfrac{1}{2}x^2 \text{ is satisfied} \]

The average speed is given by: \[ \text{Average Speed} = \dfrac{\text{Total Distance}}{\text{Total Time}} \] If \( x \) is the distance from A to B, then the total distance is \( 2x \) (away and return).

The total time \( T \) is the sum of time from A to B given by \( \dfrac{x}{40} \) and the time from B to A given by \( \dfrac{x}{60} \), hence: \[ T = \dfrac{x}{40} + \dfrac{x}{60} \] Set to the common denominator \( 120 \) and add : \[ T = \dfrac{3x + 2x}{120} = \dfrac{5x}{120} = \dfrac{x}{24} \] The average speed for the round trip is: \[ \text{Average Speed} = \dfrac{2x}{T} = \dfrac{2x}{\dfrac{x}{24}} = 2x \cdot \dfrac{24}{x} = 48 \text{ mph} \]

Let \( x \) be the distance traveled to the city. Then the time \( T_1 \) taken to go to the city is given by: \[ T_1 = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{x}{60} \] The time \( T_2 \) to return is given by:} \[ T_2 = \dfrac{x}{50} \] The total time to travel and return is 5 hours. Hence: \[ \dfrac{x}{60} + \dfrac{x}{50} = 5 \] Mutliply all terms by the common denominator \( 300 \) \[ 300 \dfrac{x}{60} + 300 \dfrac{x}{50} = 300 \cdot 5 \] Simplify: \[ 50 x + 60 x = 1500 \] Solve for \( x \) \[ x = \dfrac{1500}{11} \approx 136 \text{ miles} \]

Let \( t \) be the time in hours, from 11:00 AM, when Jimmy catches up with John.

Jimmy starts driving a quarter of an hour later, so the time he drives is: \[ t - \dfrac{1}{4} \]. Since they will have traveled the same distance when Jimmy catches up with John, we set up the equation: \[ 50t = 65\left(t - \dfrac{1}{4}\right) \] Put all terms with \( t \) on one side: \[ \dfrac{65}{4} = 15t \] Mutliply all terms of the above equation by \( 4 \) and simplify \[ 65 = 60 t \] Hence \[ t = \dfrac{65}{50} = \dfrac{13}{12} \] Convert to minutes: \[ \dfrac{13}{12} \text{ hours} = \dfrac{13}{12} \times 60 = 65 \text{ minutes} \] Jimmy will catch up with John at: \[ 11:00 \, \text{AM} + 1 \, \text{hour} \, 5 \, \text{minutes} = 12:05 \, \text{PM} \]

Find the slope of the line \[ 5x - 3y = 4 \] Rewriting in slope-intercept form: \[ -3y = -5x + 4 \] \[ y = \dfrac{5}{3}x - \dfrac{4}{3} \] So, the slope of the line is: \[ \text{slope} = \dfrac{5}{3} \] The product of the slopes of two perpendicular lines is equal to \( -1 \). Let \( m \) be the slope of the line perpendicular to the line \( 5x - 3y = 4 \). Then: \[ m \cdot \dfrac{5}{3} = -1 \] Solve for \( m \). \[ m = -\dfrac{3}{5} \] The equation of the line passing through the point \( (-4, 5) \) and perpendicular to \( 5x - 3y = 4 \) is: \[ y - 5 = -\dfrac{3}{5}(x - (-4)) = -\dfrac{3}{5}(x + 4) \] \[ y = -\dfrac{3}{5}x - \dfrac{12}{5} + 5 \] \[ y = -\dfrac{3}{5}x + \dfrac{13}{5} \] Multiplying both sides by 5 to eliminate fractions: \[ 5y = -3x + 13 \] Final equation: \[ 3x + 5y = 13 \]

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: \[ L \cdot W = 300 \quad \text{(area)} \] \[ 2(L + W) = 80 \quad \text{or} \quad L + W = 40 \quad \text{(perimeter)} \] Solve the equation \( L + W = 40 \) for \( W \): \[ W = 40 - L \] Substitute \( W \) by \( 40 - L \) in the equation \( L \cdot W = 300 \): \[ L(40 - L) = 300 \] Write the above equation in standard form: \[ L^2 - 40L + 300 = 0 \] Factor the left side: \[ (L - 30)(L - 10) = 0 \] Solve for \( L \) to obtain two solutions: \[ L = 30 \quad , \quad L = 10 \] Solve for \( L \) and calculate \( W \):

For \( L = 30 \), then \( W = 40 - L = 10 \)

For \( L = 10 \), then \( W = 40 - 10 = 30 \)

Assuming that the length is larger than the width, the rectangle has a length \[ L = 30 \, \text{meters} \quad \text{and} \quad W = 10 \, \text{meters} \]

Let \( x \) be the side of the swimming pool. If the area not covered by the swimming pool is half of the rectangular garden, then the other half is covered by the swimming pool, whose area is \( x^2 \). Hence, \[ x^2 = \text{Half the area of the rectangle} \] \[ = \dfrac{1}{2} ( 100 \times 50 ) = 2500 \] Solve for \( x \): \[ x = \sqrt{2500} = 50 \text{ meters} \] Side of the swimming pool = 50 meters

Note that since \( \triangle ABC \) is equilateral and \( BH \perp AC \), we have: \[ HC = \dfrac{50}{2} = 25 \, \text{cm} \] Since \( MN \parallel HC \), triangles \( \triangle BMN \) and \( \triangle BHC \) are similar, and the ratios of corresponding sides are equal: \[ \dfrac{BN}{BC} = \dfrac{BM}{BH} = \dfrac{MN}{HC} = \dfrac{12}{25} \] The area \( A \) of triangle \( \triangle BMN \) is given by: \[ A = \dfrac{1}{2} \cdot BM \cdot MN \] The area \( B \) of triangle \( \triangle BHC \) is: \[ B = \dfrac{1}{2} \cdot BH \cdot HC \] Now consider the ratio of the areas: \[ \dfrac{A}{B} = \dfrac{\dfrac{1}{2} \cdot BM \cdot MN}{\dfrac{1}{2} \cdot BH \cdot HC} = \dfrac{BM}{BH} \cdot \dfrac{MN}{HC} = \left( \dfrac{12}{25} \right)^2 \] Which gives: \[ A = \left( \dfrac{12}{25} \right)^2 B \] Let us now find \( BH \) using the Pythagorean theorem: \[ BH = \sqrt{50^2 - 25^2} = \sqrt{2500 - 625} = \sqrt{1875} = 25\sqrt{3} \] Then the area \( B \) of triangle \( \triangle BHC \) is: \[ B = \dfrac{1}{2} \cdot BH \cdot HC = \dfrac{1}{2} \cdot 25\sqrt{3} \cdot 25 = \dfrac{1}{2} \cdot 25^2 \cdot \sqrt{3} \] Now use \( A = \left( \dfrac{12}{25} \right)^2 B \) found above to find area \( A \) of triangle \( \triangle BMN \): \[ A = \left( \dfrac{12}{25} \right)^2 \cdot B = \left( \dfrac{12}{25} \right)^2 \cdot \dfrac{1}{2} \cdot 25^2 \cdot \sqrt{3} \] \[ A = \dfrac{144}{625} \cdot \dfrac{625}{2} \cdot \sqrt{3} = 72\sqrt{3} \, \text{cm}^2 \]

We first find the volume of water without the stone: \[ V_1 = 10 \times (\pi \times 5^2) = 250\pi \; \text{cm}^3 \] The volume of water and stone is given by: \[ V_2 = 13.2 \times (\pi \times 5^2) = 330\pi \; \text{cm}^3 \] The volume of the stone is given by the difference: \[ V_2 - V_1 = 330\pi - 250\pi = 80\pi \] \[ \approx 251.1 \, \text{cm}^3 \] Using \( \pi \approx 3.14 \)

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 \[ x^2 = 400, \quad \text{hence} \quad x = 20 \, \text{cm} \] The diagonal \( D \) of the square is found using Pythagoras' theorem: \[ D^2 = x^2 + x^2 = 800 \] The area \( A \) of the circle is given by \[ A = \pi \left( \dfrac{D}{2} \right)^2 = \dfrac{\pi D^2}{4} = 200\pi \, \text{cm}^2 \] \[ A \approx 628 \, \text{cm}^2 \]

The average of 2, 3, 5, and \( x \) is given by \[ \dfrac{2 + 3 + 5 + x}{4} \] and it is known to be equal to 4. Hence, \[ \dfrac{2 + 3 + 5 + x}{4} = 4 \] Solve the above equation for \( x \) to obtain \[ x = 6 \]

The average of \( x \), \( y \), \( z \), and \( w \) is equal to 25. Hence, \[ \dfrac{x + y + z + w}{4} = 25 \] The average of \( x \), \( y \), and \( z \) is equal to 27. Hence, \[ \dfrac{x + y + z}{3} = 27 \] The above gives \[ x + y + z = 3 \times 27 = 81 \] Substitute \( x + y + z \) by 81 in the equation \( \dfrac{x + y + z + w}{4} = 25 \) to obtain the equation: \[ \dfrac{81 + w}{4} = 25 \] Solve the above for \( w \): \[ 81 + w = 25 \times 4 = 100 \] \[ w = 100 - 81 = 19 \] Hence, \( w = 19 \).

The arithmetic mean of \(41\), \(46\), \(x\), \(y\), and \(z\) is equal to 50. Hence, \[ \dfrac{41 + 46 + x + y + z}{5} = 50 \] 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 \(z\): \[ z = 73 \]

Substitute \( x \) by 2 in the given equation: \[ 2(2) + 1 = 2A + 3(2 + A) \] Solve for \( A \): \[ A = \dfrac{-1}{5} \]

The symbol \( \text{dm} \) represents a decimeter, which is equal to 10 cm. We convert all given units to \( \text{dm} \).

Radius: \( r = 50 \, \text{cm} = 5 \, \text{dm} \)

Height: \( h = 1 \, \text{m} = 10 \, \text{dm} \)

The volume of water in the cylindrical container is given by: \[ V = \pi r^2 h = 3.14 \times (5 \, \text{dm})^2 \times 10 \, \text{dm} = 785 \, \text{dm}^3 \] Since 1 dm\(^3\) weighs 1 kg, the total weight of water in the container is: \[ 785 \, \text{dm}^3 \times \dfrac{1 \, \text{kg}}{\text{dm}^3} = 785 \, \text{kg} \]

ABC is isosceles; hence \[ AB = AC \] Triangles \( ABM \) and \( ACM \) have equal sides, one common side, and equal angles. Therefore, they are congruent triangles. Hence, the area of triangle \( AMC \) is half the area of triangle \( ABC \). In triangle \( MAC \), angles \( \angle MAC \) and \( \angle MCA \) are both equal to \( 45^\circ \). Hence, triangle \( MAC \) is an isosceles right triangle. Therefore, triangles \( MAN \) and \( MCN \) are congruent (similar proof to above). Thus, the 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 \[ \dfrac{1}{4}. \]

Pump A can fill 1 tank in 4 hours; hence the rate of pump A is \[ \dfrac{1}{4} \quad \text{(tank/hour)}. \] Pump B can fill 1 tank in 6 hours; hence the rate of pump B is \[ \dfrac{1}{6} \quad \text{(tank/hour)}. \] Let \( t \) be the time in hours that it takes pump A to fill the tank. Hence, during the \( t \) hours, pump A fills \[ t \times \dfrac{1}{4} \quad \text{of the tank}. \] Pump B was stopped for one hour and hence will pump water into the tank for \( t - 1 \) hours; hence, during \( t - 1 \) hours, pump B fills \[ (t - 1) \times \dfrac{1}{6} \quad \text{of the tank}. \] The two pumps work together to fill 1 tank; hence, \[ t \times \dfrac{1}{4} + (t - 1) \times \dfrac{1}{6} = 1. \] Multiply all terms in the above equation by 24 and simplify: \[ 24 \times t \times \dfrac{1}{4} + 24 \times (t - 1) \times \dfrac{1}{6} = 24 \times 1. \] \[ 6t + 4(t - 1) = 24. \] Simplify: \[ 10t = 28 \quad \] Solve for \( t \): \[ t = \dfrac{28}{10} = 2.8 \, \text{hours} \] \[ = 2 \, \text{hours} + 0.8 \times 60 \, \text{minutes} \] \[ = 2 \, \text{hours} \, 48 \, \text{minutes}. \] The tank is full at \[ 8:00 \, \text{a.m.} + 2 \, \text{hours} \, 48 \, \text{minutes} = 10:48 \, \text{a.m.}. \]

Write each of the given equations in the form \( y = mx + b \) and identify the slope \( m \). \[ 2x + y = 2 \quad \text{gives} \quad y = -2x + 2, \quad \text{slope} \, m_1 = -2 \] \[ x - 2y = 0 \quad \text{gives} \quad y = \dfrac{x}{2}, \quad \text{slope} \, m_2 = \dfrac{1}{2} \] \( m_1 \) and \( m_2 \) are not equal and therefore the lines are not parallel. However, The product of the two slopes is: \[ m_1 \times m_2 = -2 \times \dfrac{1}{2} = -1 \] The two lines are perpendicular.

If \( R \) is the radius of a semicircle, its area is given by the formula: \[ \dfrac{1}{2} \pi R^2 = 50 \pi \] Solve the above equation for \( R \): \[ R^2 = 100 \quad \Rightarrow \quad R = 10 \] The length \( L \) of the rectangle is equal to the diameter, hence: \[ L = 2R = 20 \, \text{cm} \] The semicircle is inscribed inside the rectangle, and therefore its radius is equal to the width \( W \) of the rectangle. Hence: \[ W = R = 10 \, \text{cm} \] The area of the rectangle is: \[ L \times W = 20 \times 10 = 200 \, \text{cm}^2 \]

The area \( A \) of a triangle given its two sides \( a \) and \( b \), making an angle \( \alpha \), is given by \[ A = \dfrac{1}{2} ab \sin(\alpha) \] Use the cosine rule to find \( \cos(\alpha) \) as \[ \cos(\alpha) = \dfrac{a^2 + b^2 - c^2}{2ab} \] where \( c \) is the third side of the triangle. Use the formula \[ \sin(\alpha) = \sqrt{1 - \cos^2(\alpha)} \] to rewrite the formula for the area as \[ A = \dfrac{1}{2} ab \sin(\alpha) = \dfrac{1}{2} ab \sqrt{1 - \left( \dfrac{a^2 + b^2 - c^2}{2ab} \right)^2} \] Simplify to obtain \[ A = \dfrac{1}{4} \sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2} \] Solve for \( c \) to obtain two solutions: \[ c = \sqrt{a^2 + b^2 \pm \sqrt{4a^2b^2 - 16A^2}} \] In this problem, \( a = 26 \), \( b = 40 \), and \( A = 200 \). Substitute and solve to find two solutions: \[ c_1 = \sqrt{356} = 2\sqrt{89}, \quad c_2 = \sqrt{4196} = 2\sqrt{1049} \]