Solving Equations · Lecture 07 · alg-equations
📺 Course materials — Lecture 07:
lecture video · handout PDF (Lecture 7) · board notes (07.jpg) · homework answer key (Chapter 7)
— all in the
amc8/video folder.
·
🖨 Printable PDF (answers & solutions at the back)
🔑 Key facts
Almost every AMC 8 word problem is an equation in disguise. The whole game is translation:
name the unknown ($x$), turn each sentence of the story into one equation, solve, and answer the question
that was asked (not the value of $x$!). Three translations carry the chapter:
ratio $3:4$ → parts $3k$ and $4k$; percent change → multiply by $(1\pm p\%)$;
and average speed $=$ total distance $\div$ total time.
🧰 Instructor's toolbox (from the lecture video & board notes)
1 · One sentence → one equation. Count the conditions: one condition, one unknown;
two conditions, set two unknowns and build a system. The instructor's two exits from a system:
elimination — add or subtract the equations ($A+B=20$, $A-B=2$ → add: $2A=22$) — and
substitution — solve one equation for a variable and plug it into the other ($A=B+2$ into $A+B=20$).
Example 3 (gold coins) is pure substitution; Practice 14 turns a story straight into a quadratic.
2 · Ratios are parts. Heights in ratio $3:4$? Write them as $3k$ and $4k$ —
then "one tree is $16$ feet taller" becomes $4k-3k=16$, so $k=16$ (Practice 3). And keep
part-to-part separate from part-to-whole: if A : B $=30:20$, then A is
$\dfrac{30}{30+20}=\dfrac35$ of the total — the board opens with exactly this part-vs-whole warning.
3 · Chain two ratios with the LCM. To merge $8^\text{th}:6^\text{th}=5:3$ and
$8^\text{th}:7^\text{th}=8:5$ into one continued ratio, rescale the shared member to
$\text{lcm}(5,8)=40$: multiply the first ratio by $8$ and the second by $5$, giving
$6^\text{th}:7^\text{th}:8^\text{th}=24:25:40$. Each part is at least one whole person, so the minimum total is
$24+25+40=89$ (Example 2). The board's motto: find the common quantity and unify it.
4 · Find what doesn't change. When a ratio shifts, anchor the parts to the
invariant. Ages: the difference never changes, so give the difference the same number of parts before and
after. Two people whose ages sum to a constant: rescale both ratios so the sum of parts matches
(the board rescales $3:4$ by $8$ and $5:3$ by $7$ so both totals are $56$). In Practice 4 the green and orange
socks are the invariant — only purple grows.
5 · Percent changes multiply — and mind the base.
Start at $a$: up $x\%$, up $y\%$, down $z\%$ lands on $a(1+x\%)(1+y\%)(1-z\%)$ — never add the percents.
Up then down by the same $y$: $(1+y)(1-y)=1-y^2$, always a net loss (Example 4).
And every percent is measured against its own base, the value right before the change:
$20\to50$ is a $\tfrac{30}{20}=150\%$ increase, but $50\to20$ is only a $\tfrac{30}{50}=60\%$ decrease —
the instructor draws both arrows side by side.
6 · Assume a convenient total. When everything is percents or fractions and no
count is given, invent one: $100$ marbles (Example 1), $200$ homework problems so each half is $100$
(Practice 7), $100$ campers (Challenge C1). Percents become plain counts and the arithmetic does itself.
7 · Percent sentence → fraction equation → cross-multiply. "$60\%$ of the drawer
is purple" becomes $\dfrac{18+x}{36+x}=\dfrac35$, and cross-multiplying gives $5(18+x)=3(36+x)$, so $x=9$
(Practice 4). Put the unknown in the fraction first; clear denominators immediately.
10 · Comparing fractions — four honest ways, one forbidden one.
For $\frac{5}{19},\ \frac{7}{21},\ \frac{9}{23}$ the instructor shows:
reciprocals — flip everything: $\frac{19}{5}=3.8>\frac{21}{7}=3>\frac{23}{9}\approx2.6$, so the original
order reverses to $\frac{5}{19}<\frac{7}{21}<\frac{9}{23}$;
benchmark — compare each to $\frac13$ ($\frac{5}{19}<\frac{5}{15}=\frac13$, $\frac{9}{23}>\frac{9}{27}=\frac13$);
decimal estimate — $0.26,\ 0.33,\ 0.39$.
Forbidden: adding tops and bottoms — $\frac{1+2}{3+5}$ is not how fractions add (the board crosses it out in red).
📖 Worked examples
2019 AMC 8 #8 · successive percents
Example 1. Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
(A) 20(B) $33\frac{1}{3}$(C) 38(D) 45(E) 54
Key idea: start with $100$ marbles and just follow the story.
After Pedro: $100(1-20\%)=80$. After Ebony: $80(1-10\%)=72$. After Jimmy: $72(1-25\%)=54$.
That's $54$ of the original $100$ — each percent acted on the
current amount, not the original.
Answer: E
2013 AMC 8 #16 · continued ratio
Example 2. A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$-graders to $6^\text{th}$-graders is $5:3$, and the ratio of $8^\text{th}$-graders to $7^\text{th}$-graders is $8:5$. What is the smallest number of students that could be participating in the project?
(A) 16(B) 40(C) 55(D) 79(E) 89
Key idea: the $8^\text{th}$ grade appears in both ratios with different numbers ($5$ and $8$) — unify it at $\text{lcm}(5,8)=40$.
$\;8^\text{th}:6^\text{th}=5:3=40:24$ and $8^\text{th}:7^\text{th}=8:5=40:25$, so
$6^\text{th}:7^\text{th}:8^\text{th}=24:25:40$.
Each part is at least one student, so the minimum is $24+25+40=89$.
Answer: E
2017 AMC 8 #17 · system of equations
Example 3. Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
(A) 9(B) 27(C) 45(D) 63(E) 81
Key idea: two sentences, two equations. Let $x$ = coins, $y$ = chests.
"Nine per chest, two chests empty": $x=9(y-2)$. "Six per chest, three left over": $x=6y+3$.
Substitute: $9(y-2)=6y+3 \Rightarrow 9y-18=6y+3 \Rightarrow 3y=21 \Rightarrow y=7$, so $x=6\cdot7+3=45$.
Answer: C
2019 AMC 8 #22 · symmetric percent change
Example 4. A store increased the original price of a shirt by a certain percent and then decreased the new price by the same percent. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased?
(A) 16(B) 20(C) 28(D) 36(E) 40
Key idea: up $p$ then down $p$ multiplies the price by $(1+p)(1-p)=1-p^2$.
So $1-p^2=0.84 \Rightarrow p^2=0.16 \Rightarrow p=0.4=40\%$.
(Sanity check from the board: $100 \to 140 \to 84$. ✓ And note the answer is
not $16\%$ — that trap is choice A.)
Answer: E
2010 AMC 8 #8 · relative speed
Example 5. As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $\frac12$ mile in front of her. After she passes him, she can see him in her rear-view mirror until he is $\frac12$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?
(A) 6(B) 8(C) 12(D) 15(E) 16
Key idea: freeze Emerson. Relative to him, Emily moves at $12-8=4$ mph.
She must travel from $\tfrac12$ mile behind him to $\tfrac12$ mile ahead — a relative distance of $1$ mile.
Time $=\dfrac{1}{4}$ h $=15$ minutes.
Answer: D
✏️ Practice
Try each one before opening the solution. The set walks from ratios through percents to motion — same order as the lecture.
2020 AMC 8 #1
1. Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
(A) 6(B) 8(C) 12(D) 18(E) 24
Solution
Chain the two comparisons: water : sugar : lemon juice $=8:2:1$.
Lemon juice is $3$ cups, so water $=3\cdot8=24$.
Answer: E
2022 AMC 8 #6
2. Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
(A) 4(B) 5(C) 6(D) 7(E) 8
Solution
Equally spaced means the middle is the mean: smallest $+$ largest $=2\cdot15=30$.
With smallest $x$ and largest $4x$: $x+4x=30 \Rightarrow x=6$.
Answer: C
2010 AMC 8 #11
3. The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$. In feet, how tall is the taller tree?
(A) 48(B) 64(C) 80(D) 96(E) 112
Solution
Heights $3k$ and $4k$: the difference is one part, $4k-3k=k=16$.
Taller tree $=4k=64$ feet.
Answer: B
2020 AMC 8 #13
4. Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
(A) 6(B) 9(C) 12(D) 18(E) 24
Solution
The $6+12=18$ non-purple socks never change. With $x$ new purple socks,
$\dfrac{18+x}{36+x}=\dfrac35$. Cross-multiply: $5(18+x)=3(36+x) \Rightarrow 90+5x=108+3x \Rightarrow x=9$.
Answer: B
2020 AMC 8 #15
5. Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
(A) 5(B) 35(C) 75(D) $133\frac13$(E) 300
Solution
$0.15x=0.2y \Rightarrow y=\dfrac{0.15}{0.2}\,x=0.75x$ — so $y$ is $75\%$ of $x$.
(The board's shortcut: divide both sides by $20\%$, leaving $x\cdot\frac{15}{20}=y$.)
Answer: C
2012 AMC 8 #20
6. What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?
(A) $\frac{9}{23}<\frac{7}{21}<\frac{5}{19}$(B) $\frac{5}{19}<\frac{7}{21}<\frac{9}{23}$(C) $\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$(D) $\frac{5}{19}<\frac{9}{23}<\frac{7}{21}$(E) $\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
Solution
Flip to reciprocals: $\frac{19}{5}=3.8$, $\frac{21}{7}=3$, $\frac{23}{9}\approx2.6$.
Bigger reciprocal = smaller fraction, so $\frac{5}{19}<\frac{7}{21}<\frac{9}{23}$.
(Benchmark check: $\frac{7}{21}=\frac13$ exactly; $\frac{5}{19}<\frac13<\frac{9}{23}$.)
Answer: B
2017 AMC 8 #14
7. Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
(A) 89(B) 92(C) 93(D) 96(E) 98
Solution
Say the assignment has $200$ problems — $100$ alone, $100$ together.
Chloe: $80$ right alone, and $\dfrac{80+x}{200}=88\%$ gives $x=96$ correct in the joint half (shared with Zoe).
Zoe: $\dfrac{90+96}{200}=\dfrac{186}{200}=93\%$.
Answer: C
2022 AMC 8 #16
8. Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
(A) 24(B) 25(C) 26(D) 27(E) 28
Solution
$a+b=42$, $b+c=52$, $c+d=60$. Add the first and last, subtract the middle:
$a+d=(a+b)+(c+d)-(b+c)=42+60-52=50$, so the average is $25$.
No need to find any individual number!
Answer: B
2015 AMC 8 #16
9. In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
(A) $\frac{2}{15}$(B) $\frac{4}{11}$(C) $\frac{11}{30}$(D) $\frac{3}{8}$(E) $\frac{11}{15}$
Solution
Pairs are one-to-one, so $\tfrac13 n=\tfrac25 s$, giving $\dfrac{n}{s}=\dfrac65$: say $n=6k$ ninth graders and $s=5k$ sixth graders.
Buddied students: $\tfrac13\cdot6k+\tfrac25\cdot5k=2k+2k=4k$ out of $11k$ total $=\dfrac{4}{11}$.
Answer: B
2018 AMC 8 #6
10. On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
(A) 50(B) 70(C) 80(D) 90(E) 100
Solution
Coastal: $10$ miles in $30$ min $=\tfrac13$ mile per minute. Highway: $3\times$ faster $=1$ mile per minute,
so $50$ miles takes $50$ minutes. Total $30+50=80$ minutes.
Answer: C
2019 AMC 8 #16
11. Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
(A) 45(B) 62(C) 90(D) 110(E) 135
Solution
Average speed is total distance over total time — set up exactly that:
$\dfrac{15+x}{0.5+\frac{x}{55}}=50$ (the first leg takes $\tfrac{15}{30}=0.5$ h).
Then $15+x=25+\dfrac{10x}{11} \Rightarrow \dfrac{x}{11}=10 \Rightarrow x=110$.
Answer: D
2018 AMC 8 #12
12. The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping, he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?
(A) 5:50(B) 6:00(C) 6:30(D) 6:55(E) 8:10
Solution
Treat the two clocks as runners at constant speeds: real : car $=30:35=6:7$.
When the car clock has "run" $7$ hours (noon → 7:00), real time has run $6$ hours — it's 6:00 PM.
Answer: B
2019 AMC 8 #20
13. How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
(A) 0(B) 1(C) 2(D) 4(E) 8
Solution
A square equal to $16$ can be $+4$ or $-4$: $x^2-5=\pm4$, so $x^2=9$ or $x^2=1$,
giving $x=\pm3,\ \pm1$ — four real solutions. (Forgetting the negative branch is the whole trap.)
Answer: D
2009 AMC 8 #23
14. On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had $6$ jelly beans left. There were two more boys than girls in her class. How many students were in her class?
(A) 26(B) 28(C) 30(D) 32(E) 34
Solution
Girls $x$, boys $x+2$; the handout used $400-6=394$ jelly beans:
$x^2+(x+2)^2=394 \Rightarrow 2x^2+4x-390=0 \Rightarrow x^2+2x-195=0 \Rightarrow (x-13)(x+15)=0$.
So $x=13$: $13$ girls $+$ $15$ boys $=28$ students.
Answer: B
🆕 Fresh from the latest exams
The same ideas, exactly as they appeared on the most recent tests.
2026 AMC 8 #4
15. Brynn's savings decreased by $20\%$ in July, then increased by $50\%$ in August. Brynn's savings are now what percent of the original amount?
(A) 80(B) 90(C) 100(D) 110(E) 120
Solution
Multiply the factors: $(1-20\%)(1+50\%)=0.8\times1.5=1.2$ — that's $120\%$ of the original.
(Adding $-20\%+50\%=+30\%$ would be wrong: each percent has its own base.)
Answer: E
2026 AMC 8 #5
16. Casey went on a road trip that covered $100$ miles, stopping for a lunch break along the way. The trip took $3$ hours in total and her average speed while driving was $40$ miles per hour (mph). In minutes, how long was the lunch break?
(A) 15(B) 30(C) 40(D) 45(E) 60
Solution
Driving time $=\dfrac{100}{40}=2.5$ h. The trip took $3$ h, so lunch was $0.5$ h $=30$ minutes.
Answer: B
2024 AMC 8 #21
17. A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was $3:1$. Then $3$ green frogs moved to the sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4:1$. What is the difference between the number of green frogs and the number of yellow frogs now?
(A) 10(B) 12(C) 16(D) 20(E) 24
Solution
Start with $3x$ green, $x$ yellow. After the swap: green $3x-3+5=3x+2$, yellow $x+3-5=x-2$.
New ratio: $3x+2=4(x-2) \Rightarrow x=10$. Now green $=32$, yellow $=8$, difference $=24$.
Answer: E
⭐ Challenge (AMC 10 level)
2009 AMC 10A #18 · percents + Venn
C1. At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
(A) 30%(B) 40%(C) 49%(D) 51%(E) 70%
Solution
Take $100$ campers: $60$ play soccer, $30$ swim, and $60\cdot40\%=24$ do both.
Non-swimmers: $100-30=70$; of these, soccer players are $60-24=36$.
$\dfrac{36}{70}\approx51.4\% \to 51\%$.
Answer: D
2002 AMC 10A #12 · system + motion
C2. Mr. Earl E. Bird leaves home every day at 8:00 AM to go to work. If he drives at an average speed of $40$ miles per hour, he will be late by $3$ minutes. If he drives at an average speed of $60$ miles per hour, he will be early by $3$ minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
(A) 45(B) 48(C) 50(D) 55(E) 58
Solution
Let $s$ = distance in miles, $t$ = minutes available. Convert: $40$ mph $=\tfrac23$ mi/min and $60$ mph $=1$ mi/min.
The two sentences become $s=\tfrac23(t+3)$ and $s=1\cdot(t-3)$.
Substitute: $\tfrac23(t+3)=t-3 \Rightarrow 2t+6=3t-9 \Rightarrow t=15$, $s=12$.
On-time speed: $\dfrac{12}{15}$ mi/min $=0.8\cdot60=48$ mph.
Answer: B
🗝 Answer key
| Example | 1 | 2 | 3 | 4 | 5 |
|---|
| E | E | C | E | D |
|---|
| Practice | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
|---|
| E | C | B | B | C | B | C | B | B | C | D | B | D | B | E | B | E |
|---|
| Challenge | C1 | C2 |
|---|
| D | B |
|---|
Sources: prep-course Lecture 07 (video + board notes + 36 worked examples, translated to English and curated to the most instructive) and official AMC 8 problems from the local bank
(kb/content/banks/). Answers cross-checked between the course solutions and the bank's official answer keys.