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MCR3U1 Practice Problems - 6PM Drill Session

Problem 1: Finding zeros

Solve: $3x^2-7x-6=0$

3 or -2/3

Problem 2: Optimization word problem

A farmer has 100m of fencing to make a rectangular pen against a barn wall. What dimensions give maximum area?

Area = W*H
100 - 2w = length

$$\begin{array}{rl}& A(w)=w(100-2w)=100-2w^2\end{array}$$

Problem 3: Vertex form

Convert $y=2x^2-12x+10$ to vertex form and find the vertex.

$$\begin{array}{rl}& 2(x^2-6x+9-9)+10\\ & 2(x-3{)}^2+10-18\\ & 2(x-3{)}^2-8\\ & V(3,-8)\end{array}$$

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TRIGONOMETRY (Sine/Cosine Law)

Problem 4: Which law to use?

Triangle ABC: angle A = 60°, side a = 8, side b = 10. Find angle B.

Decision: Two sides + one non-included angle → Sine Law $\frac{\sin A}{a}=\frac{\sin B}{b}$ $\frac{\sin 60°}{8}=\frac{\sin B}{10}$ $\sin B=\frac{10\sin 60°}{8}=\frac{10\times 0.866}{8}=1.083$

Problem: sin B > 1, so no solution exists!

Problem 5: Cosine Law application

Triangle with sides a = 7, b = 9, c = 12. Find angle C.

Use: $c^2=a^2+b^2-2ab\cos C$ $144=49+81-126\cos C$ $144=130-126\cos C$ $14=-126\cos C$ $\cos C=-\frac{14}{126}=-\frac{1}{9}$ $C={\cos }^{-1}(-\frac{1}{9})=96.4°$

Problem 6: Real-world application

You’re 100m from the base of a tower. The angle of elevation to the top is 35°. How tall is the tower?
t = OA
tan 35 = x/100
tan 35 * 100 = x = 70.0m
Setup: $\tan 35°=\frac{h}{100}$ Solution: $h=100\tan 35°=100\times 0.700=70.0$m

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EXPONENTIAL FUNCTIONS

Problem 7: Compound interest

You invest $5000 at 4% annually, compounded quarterly. How much after 6 years?

Formula: $A=P(1+\frac{r}{n}{)}^{nt}$ Where P=5000, r=0.04, n=4, t=6

Solution: $A=5000(1+\frac{0.04}{4}{)}^{4\times 6}=5000(1.01{)}^{24}=5000\times 1.270$ = $6350$

Problem 8: Exponential decay

A substance has half-life of 8 years. If you start with 200g, how much remains after 20 years?

Formula: $A_t=A_0(\frac{1}{2}{)}^{\frac{t}{h}}$ Where $A_0=200$, $h=8$, $t=20$

Solution: $A_{20}=200(\frac{1}{2}{)}^{\frac{20}{8}}=200(\frac{1}{2}{)}^{2.5}=200\times 0.177=35.4$g

Problem 9: Growth equation

A population grows from 1000 to 1500 in 3 years. What’s the annual growth rate?

Setup: $1500=1000(1+r{)}^3$ $1.5=(1+r{)}^3$ $(1+r)=\sqrt[3]{1.5}=1.145$ Growth rate: $r=0.145=14.5\%$ per year

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SEQUENCES & SERIES

Problem 10: Arithmetic sequence

First term is 7, common difference is 4. Find the 15th term and sum of first 15 terms.

15th term: $t_{15}=a+(n-1)d=7+(15-1)(4)=7+56=63$

Sum: $S_{15}=\frac{n}{2}[2a+(n-1)d]=\frac{15}{2}[2(7)+(15-1)(4)]=\frac{15}{2}[14+56]=\frac{15\times 70}{2}=525$

Problem 11: Geometric sequence

First term is 3, common ratio is 2. Find the 8th term and sum of first 8 terms.

8th term: $t_8=a{r}^{n-1}=3(2{)}^{8-1}=3(2{)}^7=3\times 128=384$

Sum: $S_8=\frac{a(r^n-1)}{r-1}=\frac{3(2^8-1)}{2-1}=\frac{3(256-1)}{1}=3\times 255=765$

Problem 12: Sequence identification

What type of sequence: 5, 8, 11, 14, 17…?

Check differences: 8-5=3, 11-8=3, 14-11=3, 17-14=3 Answer: Arithmetic with $a=5$, $d=3$

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ANNUITIES (Financial applications)

Problem 13: Ordinary annuity

You save $200/month for 4 years at 6% annual interest, compounded monthly. How much do you have?

Given: R=200, i=0.06/12=0.005, n=4×12=48

Formula: $A=\frac{R[(1+i{)}^n-1]}{i}=\frac{200[(1.005{)}^{48}-1]}{0.005}$

Calculate: $(1.005{)}^{48}=1.270$ $A=\frac{200[1.270-1]}{0.005}=\frac{200\times 0.270}{0.005}=\frac{54}{0.005}=$10,800$

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QUICK DRILL: Formula Recognition (2 minutes)

For each scenario, identify which formula you’d use:

1. Triangle with three sides given → Cosine Law
2. Investment with regular monthly payments → Annuity Formula
3. Finding zeros of $a{x}^2+bx+c=0$ → Quadratic Formula
4. Population doubles every 5 years → Exponential Growth
5. Triangle with two angles and one side → Sine Law
6. Substance loses half its mass every decade → Exponential Decay

Time Check: You should finish these in 90 minutes

Next step: Pick the topic you struggled with most and do 2-3 more similar problems. Then move to trigonometric identities and function transformations.