How do you solve a work rate problem with different pump sizes?

Define a rate for each pump type (e.g., $R$ for large, $r$ for small). Set up a system of equations based on the combined work over time (Rate $\times$ Time = Work). Solve the system to find individual rates and then apply them to the required conditions.

How do you calculate the area of a triangle inscribed in a parabola?

Identify the coordinates of the vertices. Use the vertex formula ($h = -b/(2a)$) to find the maximum point and solve for x-intercepts to find the base. The height of the triangle is typically the y-coordinate of the parabola's vertex.

Master Grade 12 Math: 21 Comprehensive Problems & Solutions

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.)

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Solution:

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.

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Solution:

$x + y + H = 60$: 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.

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Solution:

$\sqrt{((-6 + 2)^2 + (0 + 3)^2)} = \sqrt{((a + 3)^2 + (0 + 2)^2)}$: 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:

$(x + 2)^2 + (y + 1)^2 = 13$

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Solution:

(-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?

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Solution:

$_{3}C_{2} \times _{8}C_{6} \times 1 = 84$: Use of fundamental theorem of counting

Problem 6

Solve for x the equation \[ x^2 - 3|x - 2| - 4x = - 6 \]

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Solution:

$ 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.

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Solution:

$ h = \dfrac{-b}{2a} = 2 $: x-coordinate of the vertex of the parabola

$ k = -(2)^2 + 4(2) + C = 4 + C $: y-coordinate of the vertex

$ x = (2 + \sqrt{4 + C}), x = (2 - \sqrt{4 + C}) $: 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 $: area is equal to 32

$ C = 12 $: 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 $.

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Solution:

$ 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 $: solve the above equation for $ k $ positive.

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.

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Solution:

$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 - 3) $ 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 $.

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Solution:

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 $.

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Solution:

$ 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 $:

\[ A + B = -1 \] \[ -3 A + B = -17 \]

$ 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 $

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Solution:

$ x^2 - 4x + 4 + y^2 - 4y + 4 = 4 $: expand equation of first circle

$ x^2 - 2x + 1 + y^2 - 2y + 1 = 4 $: expand equation of second circle

$ -2x - 2y + 6 = 0 $: subtract the left and right terms of the above equations

$ y = 3 - x $: solve the above for $ y $.

$ 2x^2 - 6x + 1 = 0 $: substitute $ y $ by $ 3 - x $ in the first equation, expand and group like terms.

$\left(\dfrac{3}{2} + \dfrac{\sqrt{7}}{2} , \dfrac{3}{2} - \dfrac{\sqrt{7}}{2} \right), \left(\dfrac{3}{2} - \dfrac{\sqrt{7}}{2} , \dfrac{3}{2} + \dfrac{\sqrt{7}}{2} \right)$:

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 $.

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Solution:

$ I + 200 = A^2 $: 200 added to $ I $ gives a square.

$ I + 276 = B^2 $: 276 added to $ I $ 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 $ and $ 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.

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Solution:

$sum_1 = a + a r + a r^2 = 42$: 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?

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Solution:

$ 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} $ (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?

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Solution:

$ S_1 \times t_1 = 1400 $: $ S_1 $ is the speed of boat 1, $ t_1 $ is the time to travel 1400 meters.

$ 1400 + S_2 \times t_1 = X $: $ S_2 $ is the speed of boat 2, $ X $ is the width.

$ S_1 \times t_2 = X + 600 $: $ t_2 $ is the time to travel $ X + 600 $ meters for boat 1.

$ S_2 \times t_2 = 2X - 600 $

$ S_1 = \dfrac{1400}{t_1} $

$ S_2 = \dfrac{X - 1400}{t_1} $

Let $ T = \dfrac{t_2}{t_1} $

Substituting into equations: $ 1400 \times T = X + 600 $ and $ X \times T - 1400 \times T = 2X - 600 $

Solving the system for $ X $: $ 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 $

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Solution:

Solve the system of the first two equations to obtain the solution $ (2, -3) $.

The point $ (2, -3) $ must satisfy the last two equations:

$ a(2) + b(-3) = 4 $

$ 2a(2) - b(-3) = 2 $

Solving for $ a $ and $ b $: $ a = 1, b = -\dfrac{2}{3} $.

Problem 18

Find the area of the right triangle shown below.

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Solution:

The slope of $ AB $ is $ -3 $ since it is perpendicular to $ OB $ (slope $ 1/3 $).

Equation of $ AB $: $ y - 5 = -3(x - 5) \implies y = -3x + 20 $.

Point $ B $: solve $ y = -3x + 20 $ and $ y = x/3 \implies B(6,2) $.

Point $ A $: solve $ y = -3x + 20 $ and $ y = 2x \implies A(4,8) $.

$ \overline{OB} = \sqrt{6^2 + 2^2} = \sqrt{40} $

$ \overline{AB} = \sqrt{(6-4)^2 + (2-8)^2} = \sqrt{40} $

$ \text{Area} = \dfrac{1}{2} \times \sqrt{40} \times \sqrt{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?

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Solution:

Let $ t $ be B's time; A takes $ t - 2 $. A works 4h, B works 2h.

$ \dfrac{4}{t - 2} + \dfrac{2}{t} = 0.6 \implies 4t + 2(t - 2) = 0.6t(t - 2) $

$ 0.6t^2 - 7.2t + 4 = 0 $. Valid solution $ t = 6 + \dfrac{2\sqrt{66}}{3} $.

Pump A's time $ t - 2 = 4 + \dfrac{2\sqrt{66}}{3} $.

Time for remaining 40%: $ \text{Time} = 0.4 \times (4 + \dfrac{2\sqrt{66}}{3}) \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.

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Solution:

$ x = y/2 $. $ b_B = 3b_A $. Given $ b_B = b_A + 200 \implies 2b_A = 200 \implies b_A = 100, b_B = 300 $.

$ g_B = \dfrac{5}{3} g_A $. Since $ 2(100 + g_A) = 300 + g_B $:

$ 200 + 2g_A = 300 + \dfrac{5}{3}g_A \implies \dfrac{1}{3}g_A = 100 \implies g_A = 300, g_B = 500 $.

School A: 100 boys, 300 girls. School B: 300 boys, 500 girls.

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?

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Solution:

$ 2(4L + 2S) = 1 $ and $ 2(2L + 6S) = 1 $. Solving: $ L = 1/10, S = 1/20 $.

Total rate for 8L and 8S: $ 8(1/10) + 8(1/20) = 6/5 $ of a pool per hour.

Time for 50%: $ \dfrac{6}{5}t = 0.5 \implies t = 5/12 \text{ hours} = 25 \text{ minutes} $.

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