Grade 12 Math Problems with Solutions and Answers
Grade 12 math problems with detailed solutions are presented. More grade 12 math practice tests are included in this website.
Problem 1
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 4 small pumps to fill the swimming pool? (We assume that all large pumps are similar and all small pumps are also similar.)
Let \(R\) and \(r\) be the rate of work of the large and small pumps respectively.
\(4(2R + r) = 1\): 2 large and 1 small work for 4 hours to do 1 job
\(4(R + 3r) = 1\): 1 large and 3 small work for 4 hours to do 1 job
\(T(4R + 4r) = 1\): Find time \(T\) if 4 large and 4 small are to do one job.
Solve for \(R\) and \(r\) the system of first two equations then substitute in the third and solve for \(T\) to find the time: \(T = \dfrac{5}{3}\) hours = 1 hour 40 minutes.
Problem 2
Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.
\(x + y + H = 60\): perimeter, \(x\), \(y\) and \(H\) are the two legs and the hypotenuse of the right triangle
\(\dfrac{1}{2}xy = 150\): area
\(x^2 + y^2 = H^2\): Pythagoras' theorem
3 equations with 3 unknowns.
\((x + y)^2 - 2xy = H^2\): completing the square in the third equation.
\(x + y = 60 - H\): express \(x + y\) using the first equation and use the second equation to find \(xy = 300\) and substitute in equation 5.
\((60 - H)^2 - 600 = H^2\): one equation with one unknown.
Solve for \(H\) to find \(H = 25\) cm. Substitute and solve for \(x\) and \(y\) to find \(x = 15\) cm and \(y = 20\) cm.
Problem 3
A circle of center (-3, -2) passes through the points (0, -6) and (a, 0). Find a.
\(\sqrt{((-6 + 2)^2 + (0 + 3)^2)} = \sqrt{((a + 3)^2 + (0 + 2)^2)}\): distances from center to any point on the circle are equal.
\(25 = (a + 3)^2 + 4\): Simplify and square both sides
\((a + 3)^2 = 21\): Rewrite the above equation
Solve for \(a\):
\(a = -3 + \sqrt{21}\) or \(a = -3 - \sqrt{21}\)
Problem 4
Find the equation of the tangent at (0, 2) to the circle with equation:
\((x + 2)^2 + (y + 1)^2 = 13\)
(-2, -1): center of circle
\(m = \dfrac{2 - (-1)}{0 - (-2)} = \dfrac{3}{2}\): slope of line through the center and the point of tangency (0, 2)
The line through the center and the point of tangency (0, 2) is perpendicular to the tangent.
\(M = -\dfrac{2}{3}\): slope of tangent
\(y = -\dfrac{2}{3}x + 2\): equation of tangent given its slope and point (0, 2).
Problem 5
An examination consists of three parts. In part A, a student must answer 2 of 3 questions. In part B, a student must answer 6 of 8 questions and in part C, a student must answer all questions. How many choices of questions does the student have?
\(_{3}C_{2} \times _{8}C_{6} \times 1 = 84\): Use of fundamental theorem of counting
Problem 6
Solve for x the equation \[ x^2 - 3|x - 2| - 4x = - 6 \]
\( x^2 - 3|x - 2| - 4x = - 6 \) : given
Let \( Y = x - 2 \) which gives \( x = Y + 2 \)
\( (Y + 2)^2 - 3|Y| - 4(Y + 2) = - 6 \) : substitute \( x \) by \( Y + 2 \) in the given equation
\( Y^2 - 3|Y| + 2 = 0 \)
\( Y^2 = |Y|^2 \) : note
\( |Y|^2 - 3|Y| + 2 = 0 \): rewrite equation as
\( (|Y| - 2)(|Y| - 1) = 0 \): factor
\( |Y| = 2 , |Y| = 1 \) : solve for \( |Y| \)
\( Y = 2, -2 , 1 , -1 \) : solve for \( Y \)
\( x = 4 , 0 , 3 , 1 \) : solve for \( x \) using \( x = Y + 2\)
Problem 7
The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the parabola is given by \( y = -x^2 + 4 x + C \), find \( C \) so that the area of the triangle ABC is equal to \( 32 \) square units.
.
\( h = \dfrac{-b}{2a} = 2 \quad \): x-coordinate of the vertex of the parabola
\( k = -(2)^2 + 4(2) + C = 4 + C \quad \) : y-coordinate of the vertex
\( x = \left(2 + \sqrt{4 + C} \right), \quad x = \left(2 - \sqrt{4 + C} \right) \) : the two x-intercepts of the parabola.
Length of \( BA = k = 4 + C \)
Length of : \( AC = (2 + \sqrt{4 + C}) - 2 = \sqrt{4 + C}\)
\( \text{Area} = \dfrac{1}{2} BA \times AC = \dfrac{1}{2} (4 + C) \times \sqrt{4 + C} \)
\( \dfrac{1}{2} (4 + C) \times \sqrt{4 + C} = 32 \quad \text{: area is equal to 32} \)
\( C = 12 \quad \text{: solve above for } C. \)
Problem 8
The triangle bounded by the lines \( y = 0\), \(y = 2x \) and \( y = -0.5x + k \), with \( k \) positive, is equal to \( 80 \) square units. Find \( k \).
\( A(0,0) \) , \( B(2k/5 , 4k/5) \) , \( C(2k ,0) \) : points of intersection of the 3 lines
Area = \( (1/2) \times (4k/5) \times (2k) = 80 \) : given
\( k = 10 \quad \) : solve the above equation for \( k \) positive. ( given condition).
Problem 9
A parabola has two \( x \) intercepts at \( (-2 , 0) \) and \( (3 , 0) \) and passes through the point \( (5 , 10) \). Find the equation of this parabola.
\(y = a(x + 2)(x - 3) \) : Use the x - intercepts to write the equation of the parabola in factored form
\( 10 = a(5 + 2)(5 - 2) \) since \( (5 , 10) \) is a point on the graph of the parabola and therefore satisfies the equation of the parabola.
\( a = 5/7 \) : solve the above equation for a.
Problem 10
When the polynomial \( P(x) = x^3 + 3 x^2 -2 A x + 3 \), where \( A \) is a constant, is divided by \( x^2 + 1 \) we get a remainder equal to \( 5 x \). Find \( A \).
The division of \( x^3 + 3 x^2 -2 A x + 3 \) by \( x^2 + 1 \) gives the remainder \( -x (1 + 2 A) \)
\( - x (1 + 2A) = 5 x \) : The remainder is given as \( 5 x \)
\( -(1 + 2A) = 5 \) : polynomials are equal if they corresponding coefficient area equal.
\( A = -3 \) : Solve the above for \( A \).
Problem 11
When divided by \( x - 1 \), the polynomial \( P(x) = x^5 + 2 x^3 + A x + B \), where \( A \) and \( B \) are constants, the remainder is equal to \( 2 \). When \( P(x) \) is divided by \( x + 3 \), the remainder is equal \( -314 \). Find \( A \) and \( B \).
\( P(1) = 1^5 + 2(1^3) + A \times (1) + B = 2 \) : remainder theorem
\( P(-3) = (-3)^5 + 2(-3)^3 + A \times (-3) + B = -314 \) : remainder theorem
which gives the system of equations in \( A \) and \( B \). \begin{align*} A + B = -1 \\ -3 A + B = -17 \end{align*}
\( A = 4 \) and \( B = -5 \) : solve the above system of equations.
Problem 12
Find all points of intersections of the 2 circles defined by the equations
(x - 2)^2 + (y - 2)^2 = 4
(x - 1)^2 + (y - 1)^2 = 4
\( x^2 - 4x + 4 + y^2 - 4y + 4 = 4 \quad \text{: expand equation of first circle} \)
\( x^2 - 2x + 1 + y^2 - 2y + 1 = 4 \quad \text{: expand equation of second circle} \)
\( -2x - 2y + 6 = 0 \quad \text{: subtract the left and right terms of the above equations} \)
\( y = 3 - x \quad \text{: solve the above for } y. \)
\( 2x^2 - 6x + 1 = 0 \quad \) substitute \( y \) by \( 3 - x \) in the first equation, expand and group like terms.
\( \left(\dfrac{3}{2} + \dfrac{\sqrt{7}}{2} , \dfrac{3}{2} - \dfrac{\sqrt{7}}{2} \right), \quad \left(\dfrac{3}{2} - \dfrac{\sqrt{7}}{2} , \dfrac{3}{2} + \dfrac{\sqrt{7}}{2} \right) \): solve the above for \(x \) and use \( y = 3 - x \) to find y.
Problem 13
If \( 200 \) is added to a positive integer \( I \), the result is a square number. If \( 276 \) is added to to the same integer \( I \), another square number is obtained. Find \( I \).
\( I + 200 = A^2 \) : 200 added to \( I \) (unknown integer) gives a square.
\( I + 276 = B^2 \) : 276 added to \( I \) (unknown integer) gives another square.
\( B^2 = A^2 + 76 \) : eliminate \( I \) from the two equations.
Add squares \( A^2 \) (0, 1, 4, 9, 16, 25,...) to 76 until you obtain another square \( B^2 \).
\( 76 + 18^2 = 400 = 20^2 \)
\( A^2 = 18^2 \quad \text{and} \quad B^2 = 20^2 \)
\( I = A^2 - 200 = 124 \)
Problem 14
The sum of the first three terms of a geometric sequence is equal to \( 42 \). The sum of the squares of the same terms is equal to \( 1092 \). Find the first three terms of the sequence.
\(sum_1 = a + a r + a r^2 = 42\) : the sum of the first three terms given, \( r \) is the common ratio.
\( sum_2 = (a)^2 + (a r)^2 + (a r^2)^2 = 1092\) : the sum of the squares of the three terms given.
\( sum_1 = a + ar + a r^2 = \dfrac{ a(r^3 - 1) }{r-1} = 42 \) : apply formula for a finite sum of geometric series.
\( sum_2 = a^2 + a^2 r^2 + a^2 r^4 = \dfrac{a^2 (r^6 - 1) }{(r^2 - 1) } = 1092 \) : the sum of squares is a also a sum of geometric series.
\( \dfrac{sum_2}{(sum_1)^2} = \dfrac{ 1092}{42^2 } = \dfrac{\dfrac{a^2{r^6 - 1}}{r^2 - 1}}{\dfrac{a^2 (r^3 - 1)^2}{(r - 1)^2}}\) : define the ratio.
\( \dfrac{r^2 - r + 1}{r^2 + r + 1} = \dfrac{1092}{42^2} \) : Simplify the above ratio to obtain an equation.
\( r = 4 , r = 1/4 \) : solve for \( r \)
\( a = 2 \) : substitute \( r = 4 \) in \( sum_1\) and solve for \( a \)
\( a = 32 \) : substitute \( r = 1/4 \) in \( sum_1\) and solve for \( a \)
\( a = 2 , a r = 8 , a r^2 = 32 \) : find the three terms for \( r = 4 \)
\( a = 32 , a r = 8 , a r^2 = 2 \) : find the three terms for \( r = 1/4 \)
Problem 15
A rock is dropped into a water well and it travels approximately \( 16 t^2 \) feet in \( t \) seconds. If the splash is heard \( 3.5 \) seconds later and the speed of sound is \( 1087 \) feet/second, what is the height of the well?
\( T_1 + T_2 = 3.5 : \; \) where \( T_1 \) is the time for the rock to reach the bottom of the well, and \( T_2 \) is the time for the sound to reach the top of the well.
\( 16 \times T_1^2 = 1087 \times T_2 : \; \) Since both terms represent the same distance, which is the height of the well.
Solving for \( T_2 \):
\( T_2 = 3.5 - T_1 \)
Substituting \( T_2 \) into the equation:\( 16 \times T_1^2 = 1087 \times (3.5 - T_1) \)
Solving for \( T_1 \):\( T_1 = 3.34 \text{ seconds} \)
Finally, calculating the height of the well:\( \text{Height} = 16 \times (3.34)^2 = 178 \text{ feet} \quad (\text{to the nearest unit}) \)
Problem 16
Two boats on opposite banks of a river start moving towards each other. They first pass each other \( 1400 \) meters from one bank. They each continue to the opposite bank, immediately turn around and start back to the other bank. When they pass each other a second time, they are \( 600 \) meters from the other bank. We assume that each boat travels at a constant speed all along the journey. Find the width of the river?
.
\( S_1 \times t_1 = 1400 : \; \quad S_1 \text{(is the speed of boat 1) } , t_1 \text{ is the time to travel 1400 meters) (boat 1)} \)
\( 1400 + S_2 \times t_1 = X : \; \quad S_2 \text{(is the speed of boat 2)} \)
\( S_1 \times t_2 = X + 600 \quad \text{(} t_2 \text{ is the time to travel } X + 600 \text{ meters for boat 1)} \quad (III) \)
\( S_2 \times t_2 = 2X - 600 \quad (IV) \)
\( S_1 = \dfrac{1400}{t_1} \)
\( S_2 = \dfrac{X - 1400}{t_1} \)
\( \text{Let} \quad T = \dfrac{t_2}{t_1} \)
Substituting \( S_1 \), \( S_2 \), and \( \dfrac{t_2}{t_1} \) into equations (III) and (IV):
\( 1400 \times T = X + 600 \)
\( X \times T - 1400 \times T = 2X - 600 \)
Solving the above system of two equations with two unknowns \( X \) and \( T \) by eliminating \( T \) and solving for \( X \) to obtain width \( X \) of the river :\( X = 3600 \text{ meters} \)
Problem 17
Find the constants \( a \) and \( b \) so that all the 4 lines whose equation are given below pass through the same point.
\( x + y = - 1 \)
\( - x + 3 y = - 11 \)
\( a x + b y = 4 \)
\( 2 a x - b y = 2 \)
Solve the system of the first two equations to obtain the solution \( (2, -3) \).
The solution \( (2, -3) \) must also be the solution to the last two equations since the 4 lines passes through the same point \( (2, -3) \).
\( a(2) + b(-3) = 4 : \;\) substitute x by 2 and y by -3 in the third equation
\( 2a(2) - b(-3) = 2 : \; \) substitute x by 2 and y by -3 in the fourth equation
Solving for \( a \) and \( b \) the above system of equations to obtain:\( a = 1, \quad b = -\dfrac{2}{3} \)
Problem 18
Find the area of the right triangle shown below.
Similarly, the coordinates of point \( A \) are found by solving the system: \[ y = -3x + 20 \] \[ y = 2x \]
Solving for \( (x,y) \), we get \( A(4,8) \).
Now, we calculate the distance \( \overline{OB} \): \[ \overline{OB} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \] And the distance \( \overline{AB} \): \[ \overline{AB} = \sqrt{(6-4)^2 + (2-8)^2} = \sqrt{2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} \] The area of triangle \( OAB \) is: \[ \text{Area} = \dfrac{1}{2} \times \overline{OB} \times \overline{AB} \] \[ = \dfrac{1}{2} \times \sqrt{40} \times \sqrt{40} \] \[ = \dfrac{1}{2} \times 40 \] \[ = 20 \]
Problem 19
It takes pump A \( 2 \) hours less time than pump B to empty a swimming pool. Pump A is started at 8:00 a.m. and pump B is started at 10:00 a.m. At 12:00 p.m. \( 60\% \) of the pool is empty when pump B broke down. How much time after 12:00 p.m. would it take pump A to empty the pool?
Let \( t \) be the time (in hours) it takes pump B to empty the pool alone.
Pump A takes \( t - 2 \) hours to empty the pool alone.
Pump A works for 4 hours (from 8:00 a.m. to 12:00 p.m.) at the rate \( \dfrac{1}{t - 2} \)
Pump B works for 2 hours (from 10:00 a.m. to 12:00 p.m.) at the rate \( \dfrac{1}{t} \)
Total work done by both pumps by 12:00 p.m. is 60% of the pool: \[ \dfrac{4}{t - 2} + \dfrac{2}{t} = 0.6 \]
Multiply all terms of the above equation by \( t(t - 2) \) and simplify:
\[ 4t + 2(t - 2) = 0.6t(t - 2) \]The above equation may be written as :
\[ 0.6t^2 - 7.2t + 4 = 0 \]Using the quadratic formula, we find: \[ t = \dfrac{36 \pm 4\sqrt{66}}{6} = 6 \pm \dfrac{2\sqrt{66}}{3} \]
The valid solution is \( t = 6 + \dfrac{2\sqrt{66}}{3} \).
Pump A's time to empty the pool alone is \( t - 2 \): \[ t - 2 = 4 + \dfrac{2\sqrt{66}}{3} \]
Pump A's rate is \( r = \dfrac{1}{4 + \dfrac{2\sqrt{66}}{3}} \).
Time required to empty 40% of the pool: \[ \text{Time} = \dfrac{0.4}{r} = 0.4 \times \left(4 + \dfrac{2\sqrt{66}}{3}\right) \]
The time after 12:00 p.m. that pump A takes to empty the pool is: \[ \boxed{ 0.4 \times \left(4 + \dfrac{2\sqrt{66}}{3}\right) \approx 3.766 \; \text{hours}} \]
Problem 20
The number of pupils in school A is equal to half the number of pupils in school B. The ratio of the boys in school A and the boys in school B is \( 1 : 3 \) and the ratio of the girls in school A and the girls in school B is \( 3 : 5 \). The number of boys in school B is 200 higher than the number of boys in school A. Find the number of boys and girls in each school.
Let \( x \) be the number of pupils (boys and girls) in school A
Let \( y \) be the number of pupils (boys and girls) in school B.
We are given that the number of pupils in school A is half of those in school B:
\[ x = \dfrac{y}{2} \]Let \( b_A \) and \( b_B \) be the number of boys in schools A and B, respectively. The ratio of boys in school A to school B is: \[ b_A : b_B = 1 : 3 \] This means: \[ b_B = 3 b_A \]
We are also given that the number of boys in school B is 200 more than in school A:
\[ b_B = b_A + 200 \] Using \( b_B = 3b_A \), we substitute: \[ 3b_A = b_A + 200 \]Solving for \( b_A \):
\[ 3b_A - b_A = 200 \] \[ 2b_A = 200 \] \[ b_A = 100 \] Thus, \[ b_B = 3(100) = 300 \]Let \( g_A \) and \( g_B \) be the number of girls in schools A and B, respectively. The ratio of girls in school A to school B is: \[ g_A : g_B = 3 : 5 \]
This means:
\[ g_B = \dfrac{5}{3} g_A \]We know:
\[ x = b_A + g_A \]and
\[ y = b_B + g_B \]Substituting \( x = \dfrac{y}{2} \):
\[ b_A + g_A = \dfrac{b_B + g_B}{2} \]Substituting known values:
\[ 100 + g_A = \dfrac{300 + g_B}{2} \]Using \( g_B = \dfrac{5}{3} g_A \):
\[ 100 + g_A = \dfrac{300 + \dfrac{5}{3} g_A}{2} \]Multiply both sides by 2:
\[ 200 + 2g_A = 300 + \dfrac{5}{3} g_A \]Multiply all terms by 3 to eliminate fractions:
\[ 600 + 6g_A = 900 + 5g_A \] \[ 6g_A - 5g_A = 900 - 600 \] \[ g_A = 300 \] So, \[ g_B = \dfrac{5}{3} \times 300 = 500 \]Final Answer:
There ara 100 boys, 300 girls, and a total of 400 pupils in school A.
There ara 300 boys, 500 girls, and a total of 800 pupils in school B.
Problem 21
Four large and 2 small pumps can fill a swimming pool in 2 hours. Two large and 6 small pumps can also fill the same swimming pool in 2 hours. How long does it take 8 large and 8 small pumps to fill 50% of the swimming pool. (NOTE: all the large pump have same power and all the small pumps have the same power).
Let \( L \) be the rate of one large pump (pool per hour).
Let \( S \) be the rate of one small pump (pool per hour).
Four large and two small pumps fill the pool in 2 hours, hence:
\[ 2(4L + 2S) = 1 \]Two large and six small pumps fill the pool in 2 hours, hence:
\[ 2(2L + 6S) = 1 \]Rewrite above system of the two equations for \( L \) and \( S \) as: \[ 8L + 4 S = 1 \] \[ 4L + 12 S = 1 \]
Multiply the second equation by 2:
\[ 8L + 24S = 2 \]Subtract the first equation from the one above: \[ (8L + 4 S) - (8L + 24S) = 1 - 2 \] \[ -20 S = -1 \] \[ S = \dfrac{1}{20} \]
Substitute \( S = \dfrac{1}{20} \) into equation \( 8L + 4 S = 1 \): \[ 8L + 4 (\dfrac{1}{20}) = 1 \]
Solve the above equation for \( L \) \[ L = \dfrac{1}{10} \]
Find the Rate of 8 Large and 8 Small Pumps
\[ \text{Total rate} = 8L + 8S \] \[ = 8 \times \dfrac{1}{10} + 8 \times \dfrac{1}{20} \] \[ = \dfrac{8}{10} + \dfrac{8}{20} \] \[ = \dfrac{4}{5} + \dfrac{2}{5} \] \[ = \dfrac{6}{5} \]Since the pumps together fill \( \dfrac{6}{5} \) of the pool per hour, the time required to fill \( 50\%\) of the pool is:
\[ {\dfrac{6}{5}} t = 50\% \]Hence \[ t = \dfrac{50\%}{\dfrac{6}{5}} = \dfrac{1}{2} \times \dfrac{5}{6} \] \[ = \dfrac{5}{12} \text{ hours} \] \[ = 25 \text{ minutes} \]
It will take 25 minutes to fill \( 50\% \) of the pool with 8 large and 8 small pumps.