Three skills, one big idea: count carefully. Counting tells you how many outcomes there are, probability is just counting the good ones out of all of them, and data analysis is counting, measuring, and reading what a graph is actually telling you. The ISEE loves this section because it's packed with tiny traps: forgetting that order matters, reading the wrong month off a chart, grabbing the middle of a list you never sorted. Let's go find them all.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch on a question to type your answer instead: a whole number like 56, a fraction like 5/36, or a percent like 25. Questions whose answers are words or sentences stay multiple choice.
The most important counting rule is also the simplest: when you make one choice and then another, multiply.
Counting principle: if a task has $a$ ways and another has $b$ ways (independent choices), then there are $a \cdot b$ ways in total.
Here's why. Picture three shirts, and two pairs of pants to go with each one.
Each shirt branches into every pair of pants, so the branches multiply: $3 \cdot 2 = 6$ outfits. Adding them ($3 + 2 = 5$) is the classic first trap.
Add a third choice and the whole tree gets copied again.
Your turn. Every choice in a row multiplies, even when a code lets the same digit come back.
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Now line things up. Seven books on a shelf: 7 choices for the first spot, 6 left for the next, then 5, all the way down. That countdown product has a name.
The factorial of $n$ is $n! = n(n-1)\cdots 2 \cdot 1$, with $0! = 1$.
$0! = 1$. Zero things to arrange, and exactly one way to do it: do nothing. Nailed it.
Factorials get huge fast ($10! = 3{,}628{,}800$), but they cancel beautifully. $\frac{8!}{6!}$ is $8 \cdot 7 \cdot 6!$ over $6!$, so the $6!$ cancels and leaves $8 \cdot 7 = 56$. Never multiply out both factorials.
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Lining up all of a group is $n!$. A rule that pins one person in place just leaves fewer people to arrange.
Five runners line up for a photo, and the tallest one must stand on the left end. How many lineups are possible?
The left end is already decided, so only the other $4$ runners get arranged: $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$. (Counting all $5! = 120$ ignores the rule.)
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Usually, though, the ISEE only asks you to fill some of the spots, and the order you fill them in matters.
A permutation counts ordered selections: changing the order makes a different outcome.
If you're filling $r$ spots from $n$ things, it's the countdown, stopped after $r$ steps. Watch four spots on a shelf fill up from seven books.
The books left in the pile are exactly what the factorial formula divides away:
(ordered selections of $r$ from $n$)
In practice you never need the fraction; just count down $r$ times.
In how many ways can $4$ different books be arranged on a shelf from $7$ distinct books?
Four spots: $7$ choices, then $6$, then $5$, then $4$. So ${}_7P_4 = 7 \cdot 6 \cdot 5 \cdot 4 = 840$.
Whether things can repeat changes the countdown. A lock code shows it best.
No repeats and order matters: that's a permutation. Try a few.
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Sometimes order doesn't matter at all. A committee of Ann, Ben, and Cy is the same committee as Cy, Ann, and Ben.
A combination counts unordered selections: changing the order does not make a new outcome.
Every group of $r$ people shows up $r!$ times in the permutation count, once for each way to order them. Here are all the lineups of Ann, Ben, and Cy collapsing into the one committee they really are.
So divide those copies out:
(unordered selections of $r$ from $n$, read "$n$ choose $r$")
You may see the same count written another way.
Calculators and some books write this as nCr or C(n, r). Same number, different outfit.
Watch it happen: the same tiles, counted with order and then without.
Here's the formula at work on a committee.
How many ways can a committee of $3$ be chosen from $8$ students?
Order doesn't matter (a committee has no first place). ${}_8P_3 = 8 \cdot 7 \cdot 6 = 336$, but every committee got counted $3! = 6$ times. So $\binom{8}{3} = 336 \div 6 = 56$.
The hard part is never the arithmetic. It's deciding which formula the question wants.
Not sure if it's a permutation or a combination? Swap two people and ask: is it a different outcome? President and vice president: yes, swapping them changes things (permutation). A 2-person committee: no (combination). The ISEE always puts the other one in the answer choices.
Try the swap test on a whole stack of situations.
Now a few combinations of your own.
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Sometimes a committee has to come from two groups at once. Choose from each group, then multiply.
A club has $4$ girls and $5$ boys. How many committees of $2$ girls and $2$ boys are possible?
Choose the girls: $\binom{4}{2} = 6$. Choose the boys: $\binom{5}{2} = 10$. Every pair of girls can go with every pair of boys, so it's the counting principle: $6 \cdot 10 = 60$. (Adding, $6 + 10 = 16$, is the trap.)
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Another counting twist: words with repeated letters. Swapping the two L's in BALLOON doesn't make a new word, so the full $7!$ overcounts.
How many distinct arrangements of the letters in BALLOON are there?
$7$ letters, but two L's and two O's. Divide out the swaps that don't change anything: $\dfrac{7!}{2!\,\cdot\,2!} = \dfrac{5{,}040}{4} = 1{,}260$.
See the overcounting for yourself: swap two matching letters and nothing changes.
Every repeated letter needs its own division, which is where students slip.
Divide by $k!$ for every letter that repeats $k$ times. LEVEL has two L's and two E's: $\frac{5!}{2!\,2!} = 30$, not $\frac{5!}{2!} = 60$.
And MISS has only one repeated letter: $\frac{4!}{2!} = 12$.
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Last counting trick: when two people insist on standing together, tape them into one block.
That's the whole method. Here it is written out, along with the opposite question.
Six friends line up for a photo, and Ana and Ben must stand next to each other. How many lineups are possible?
Tape Ana and Ben into one block. Now there are $5$ things to line up: $5! = 120$ ways. Inside the block they can stand Ana-Ben or Ben-Ana, so $2 \cdot 120 = 240$.
If they refuse to stand together, take those away from everything: $6! - 240 = 720 - 240 = 480$.
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Probability starts by listing everything that could happen. Each possible result is called an outcome.
A sample space is the set of all possible outcomes of an experiment.
Then you pick out the outcomes you care about.
An event is any set of outcomes from a sample space.
When every outcome is equally likely, the probability of an event is a fraction:
A probability is just a fraction between $0$ and $1$. Reduce it, flip it to a percent, compare it: everything from Section 1 still works.
All three ideas show up in one roll of a die.
A fair die is rolled once. What is the sample space? What is the event "roll an even number," and what is its probability?
The sample space is every face: $\{1, 2, 3, 4, 5, 6\}$, six equally likely outcomes.
The event is the outcomes you want: $\{2, 4, 6\}$. So $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.
Now build a probability yourself. Paint a spinner until it hits the target.
Or fill a bag. Watch out: one more marble of any color changes every fraction, because the total changes.
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Two dice is the ISEE's favorite sample space. Don't count the sums (2 through 12); count the cells.
Lighting up a sum of $8$ is the classic question.
What is the probability that the sum of two fair dice is $8$?
Two dice make $6 \cdot 6 = 36$ equally likely outcomes. A sum of $8$: $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(6,2)$, so $5$ cells. The probability is $\frac{5}{36}$.
Don't trust the grid? Here are two dice rolled a thousand times.
The ISEE swears every die is fair. I've played board games. I have doubts.
The real rolls pile up exactly where the 36 cells said they would: 7 most, 2 and 12 least.
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When two things happen in a row, the question is whether the first one changes the second.
Independent events are events where the outcome of one does not change the probability of the other.
Most ISEE probability questions use independent events. If they're independent, multiply, exactly like the counting principle:
Two rolls of a die are independent, and so are two draws when the marble goes back in.
A fair die is rolled twice. What is the probability that both rolls are even? And with a jar of $3$ red and $2$ blue marbles, drawing twice with replacement, what's the chance both are red?
Each roll is even with probability $\frac{3}{6} = \frac{1}{2}$: $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
With replacement, the marble goes back, so the jar is identical both times: $\frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25}$.
When the marble doesn't go back, the second draw is no longer independent of the first.
Watch for without replacement. Now the first marble stays out, so the second draw has one fewer marble (and maybe one fewer red). With $5$ red out of $10$: $\frac{5}{10} \cdot \frac{4}{9} = \frac{2}{9}$. Using $\frac{5}{10} \cdot \frac{5}{10}$ is the trap.
A probability tree shows both situations. Multiply along each path; add up the paths you want.
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The ISEE's favorite version of "without replacement" tells you how the first draw went and asks only about the second. Then there's nothing to multiply: just count what's left in the bag.
A bag holds $4$ green, $5$ blue, $2$ yellow, and $4$ orange marbles. Kate takes one marble and keeps it; then Joanne takes one. If Kate's marble was yellow, what is the probability that Joanne's is green?
Kate's yellow marble is gone, so Joanne draws from $15 - 1 = 14$ marbles, and all $4$ green ones are still there: $\frac{4}{14} = \frac{2}{7}$.
Multiplying by Kate's $\frac{2}{15}$ is the trap. Her draw already happened, and the question told you how it went.
One fewer in the bag, and one fewer of that color if the first draw matched:
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Last probability tool, and the sneakiest. Sometimes it's way easier to find the chance something doesn't happen.
The complement of an event $A$ is the event "not $A$": every outcome in the sample space that is not in $A$.
Every outcome is either in $A$ or not, so the two probabilities fill up all of $1$:
It saves counting whenever "not" is the smaller pile.
A bag holds $3$ red, $4$ blue, and $5$ green marbles. One marble is drawn at random. What is the probability that it is not green?
There are $3 + 4 + 5 = 12$ marbles, and $5$ are green, so $P(\text{green}) = \frac{5}{12}$.
Not green: $1 - \frac{5}{12} = \frac{7}{12}$. (Counting the others works too: $3 + 4 = 7$ of the $12$.)
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The most common use: "at least one." At least one success covers every outcome except one, the one where you fail every time.
A quiz full of guesses shows how fast "none" shrinks.
A test has four answer choices per question. A student guesses randomly on $5$ questions. What's the probability of getting all $5$ right? All $5$ wrong? At least one right?
All right: $\left(\frac{1}{4}\right)^5 = \frac{1}{1{,}024}$. Ouch.
All wrong: each guess is wrong with probability $\frac{3}{4}$, so $\left(\frac{3}{4}\right)^5 = \frac{243}{1{,}024}$.
At least one right: everything except all wrong, $1 - \frac{243}{1{,}024} = \frac{781}{1{,}024}$.
Here's that complement as a picture: every guess slices off the strip where it's right.
The trap is subtracting the wrong thing.
A biased coin lands heads $\frac{1}{5}$ of the time. At least one head in $5$ flips is $1 - \left(\frac{4}{5}\right)^5$: one minus the chance of no heads. The trap answer $1 - \left(\frac{1}{5}\right)^5$ is one minus all heads, which is a totally different event.
Now test your gut. Is "at least one" more likely than not?
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One last probability idea: what happens on average if you repeat an experiment over and over?
The expected value is the average result over many tries: multiply each possible value by its probability, then add.
A coin tossed three times makes a good first example.
A coin is tossed three times. The table shows the probability of each number of heads. What is the expected number of heads?
| Heads | Probability |
|---|---|
| $3$ | $\frac{1}{8}$ |
| $2$ | $\frac{3}{8}$ |
| $1$ | $\frac{3}{8}$ |
| $0$ | $\frac{1}{8}$ |
$3 \cdot \frac{1}{8} + 2 \cdot \frac{3}{8} + 1 \cdot \frac{3}{8} + 0 \cdot \frac{1}{8}$ $= \frac{3 + 6 + 3 + 0}{8} = \frac{12}{8} = \frac{3}{2}$.
You can never actually get $1.5$ heads, and that's fine: it's the long-run average. Do the three tosses $100$ times and you'd expect about $100 \cdot 1.5 = 150$ heads.
The word "average" tempts people into the wrong average.
Don't just average the values. A spinner that pays $\$1$ half the time and $\$5$ or $\$10$ a quarter of the time each has an expected value of $\$4.25$, not $\frac{1 + 5 + 10}{3}$. The likely outcomes get more weight.
Here is that spinner, spun a thousand times.
The average lands on $\$4.25$, just as the expected value promised. That's how you decide whether a game is worth playing.
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Four ways to summarize a pile of numbers with just one number. Start with the one everybody knows.
The mean (average) is the sum of the values divided by how many values there are.
The middle child:
The median is the middle value when the data are in order (or the average of the two middle values).
The popular one:
The mode is the most frequent value.
And one that measures spread instead of center:
The range of a data set is the maximum minus the minimum.
Here are the two that need the most care, the median and the mean, side by side.
For the data $9, 4, 7, 2, 6$, what is the median? And what's the mean of $12, 18, 10, 20$ cups sold over four days?
Sort first: $2, 4, 6, 7, 9$. The middle value is $6$. (Grabbing the middle of the unsorted list gives $7$, which is wrong.)
Mean: $\frac{12 + 18 + 10 + 20}{4} = \frac{60}{4} = 15$ cups per day.
Sorting is the whole trick for the median, so watch it happen.
Now build your own data set and watch the median, mode, and range react.
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The mode and the range need no arithmetic beyond one subtraction, which is exactly why careless reading costs points.
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The mean is the balance point of the data, and that's why one wild value can drag it around.
The median, on the other hand, barely cares about how far out the wild value sits.
Mean vs. median on the ISEE: one big value pulls the mean toward it, but the median barely moves. In $2, 2, 3, 9$, the mean is $4$ and the median is $2.5$. If the data are lopsided to the right, bet on mean $>$ median.
One more disguise for the mean. Sometimes a graph's horizontal axis is the value (books read) and each bar's height is how many students had that value. Then the bar heights are counts, not data.
In a class, $3$ students read $1$ book over the summer, $5$ read $2$ books, and $2$ read $4$ books. What is the mean number of books read?
Each bar is a stack of students: the total is $1 \cdot 3 + 2 \cdot 5 + 4 \cdot 2 = 3 + 10 + 8 = 21$ books, shared by $3 + 5 + 2 = 10$ students. Mean: $\frac{21}{10} = 2.1$.
Averaging the bar heights, $\frac{3 + 5 + 2}{3}$, averages the counts, which is not a number of books at all.
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The mean also works backward. If you know the mean you want, you know the total you need.
Theo's first four quiz scores are $7$, $9$, $6$, and $8$. What does Theo need on the fifth quiz for a mean of exactly $8$?
A mean of $8$ over $5$ quizzes means a total of $5 \times 8 = 40$. Theo has $7 + 9 + 6 + 8 = 30$, so the fifth quiz needs $40 - 30 = 10$.
(Answering $8$, the mean Theo wants, forgets that the low scores have to be made up.)
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What happens to these measures when you change every score at once? Here's the whole story:
Adding slides the whole data set; multiplying stretches it.
Try both moves yourself, in any order you like.
Then test yourself: call which measures move before you see the dots go.
On the test it comes as a story.
A teacher adds $6$ points to every test score. Which measure changes the least: mean, median, mode, or range? And on a different quiz with range $5$, every score is doubled. What's the new range?
Adding $6$ slides everything up by $6$, including the highest and lowest scores, so the gap between them, the range, doesn't change at all.
Doubling stretches every gap, so the range doubles: $5 \times 2 = 10$.
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Sometimes a score counts more than once.
In a weighted mean, each value is multiplied by its weight (how much it counts), the products are added, and the sum is divided by the total weight.
The everyday version is a final exam that counts twice: it's simply written down twice before you average.
Lisa scored $93$, $89$, and $95$ on three tests. The final exam counts twice in the mean. What is the lowest final score that gives a mean of at least $93$?
Counting the final twice means $5$ scores: $93, 89, 95, f, f$. A mean of $93$ needs a total of $5 \times 93 = 465$.
So $277 + 2f \ge 465$, which means $2f \ge 188$ and $f \ge 94$. (Counting the final once would wrongly give $95$.)
Drag the final's bar and change how much it counts.
Combining two groups is a weighted mean too: the bigger group counts more.
Class A has $10$ students with a mean score of $70$. Class B has $30$ students with a mean of $90$. What is the mean of all $40$ students?
Totals first: $10 \times 70 = 700$ and $30 \times 90 = 2{,}700$. All together: $\dfrac{700 + 2{,}700}{40} = \dfrac{3{,}400}{40} = 85$.
Not $80$! Averaging $70$ and $90$ pretends the classes are the same size, but Class B has three times as many students.
Put both classes on a balance beam and you can see why $80$ fails.
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All of these numbers are only as good as the data behind them.
A random sample gives every member of the whole group the same chance of being chosen.
A sample that only asks one special part of the group leans one way: it's biased.
To learn about a whole school, survey a random sample of the whole school. Your friends, the football crowd, or the kids in the library before school are all special groups, so their answers are biased. Even "a random sample of the library" is still only the library.
Run the survey both ways and compare.
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Half of reading a graph is reading the labels: the axes, the legend, the key. The other half is not being fooled by the picture.
A bar chart compares counts across categories.
Put two months side by side and the legend becomes the most important thing on the chart.
Here is that chart, read three ways.
The chart shows library checkouts by genre for January and February. (i) Which genre had the most checkouts in February? (ii) What is the total of Nonfiction across both months? (iii) By how many did Mystery increase from January to February?
(i) February is the red bars: Fiction, with $30$.
(ii) Nonfiction: $15 + 18 = 33$.
(iii) Mystery went from $20$ to $28$: up $8$.
The ISEE rarely asks the easy version of question (i).
Read the legend before you read a bar. If the question says January, cover the February bars with your finger. In January the tallest bar is Mystery ($20$), but the tallest bar on the whole chart is February's Fiction ($30$). The ISEE loves to make the tallest bar on the whole chart the wrong answer.
Bars aren't the only way to draw counts. Sometimes they're pictures:
A pictograph shows counts with repeated pictures; a key tells you what one picture is worth.
The key is where the points are won or lost.
The pictograph shows how many books four students read. How many books did Dee read? How many more did Ben read than Cy?
| Ana | |
| Ben | |
| Cy | |
| Dee |
Key: = $4$ books, and = $2$ books
Dee has $5$ circles: $5 \times 4 = 20$ books.
Ben has $4$ circles and a half: $16 + 2 = 18$. Cy has $2$ and a half: $8 + 2 = 10$. So Ben read $18 - 10 = 8$ more. (Counting circles, $4\tfrac{1}{2} - 2\tfrac{1}{2} = 2$, forgets the key.)
And when the numbers are parts of one whole, slice a pie:
A circle graph (pie chart) shows how a whole is split into parts; the slices add up to 100%.
To turn a slice into a number, take that percent of the whole.
Food is $25\%$ of $\$2{,}000$, which is $\$500$. Here's one data set wearing all five outfits.
Now the dirty trick. The same four numbers can look very different, depending on where the y-axis starts.
Nothing about the data changed. Only the bottom of the axis moved.
Always check where the y-axis starts. When it starts at $15$, Thursday's bar looks $2.5$ times as tall as Wednesday's, but it's $20$ vs. $17$. A chart that starts at $15$ turns a tiny difference into a skyscraper.
For a long list of numbers, like 15 quiz scores, there are three more graphs worth knowing. The first keeps every number.
A stem-and-leaf plot lists data by splitting each number into a stem (the leading digits) and a leaf (the last digit).
Group the data into equal intervals and draw bars, and you get this:
A histogram is a bar graph of how many data values fall into each interval; the bars touch because the intervals do.
Or squash the data down to five numbers. Two of them need a name first.
The quartiles of ordered data are $Q_1$, the median of the lower half, and $Q_3$, the median of the upper half (with an odd number of values, the median itself is left out of both halves).
With those, the five-number picture is easy to describe.
A box plot shows the minimum, the lower quartile $Q_1$, the median, the upper quartile $Q_3$, and the maximum of a data set.
Here are the same 15 scores as all three pictures.
The stem-and-leaf keeps every number, the histogram shows where the scores pile up (more land in the 70s than anywhere else), and the box plot shows the middle half lives between $64$ and $84$. Build a histogram yourself from a fresh set of scores.
Back to the box. The five numbers cut the data into four quarters, each holding about $25\%$ of the values. The middle two quarters live in the box, and the width of the box gets its own name.
The interquartile range (IQR) is $Q_3 - Q_1$.
It's the width of the box, so it measures how spread out the middle half of the data is. Finding the quartiles takes one careful split.
Find $Q_1$, $Q_3$, and the IQR of $4, 7, 8, 10, 11, 13, 15$.
Seven values, already in order, so the median is the 4th: $10$. Leave it out and split the rest: lower half $4, 7, 8$ and upper half $11, 13, 15$.
$Q_1$ is the middle of the lower half, $7$, and $Q_3$ is the middle of the upper half, $13$. So the IQR is $13 - 7 = 6$.
Watch a box plot get built from scratch, one step at a time.
Then build one with your own hands: drag the five handles to their numbers.
Back to the 15 quiz scores in the stem-and-leaf plot, for every kind of question at once.
Use the $15$ quiz scores in the stem-and-leaf plot above. (i) How many students scored in the $80$s? (ii) What is the median? (iii) What is the range? (iv) How many scores are below $Q_1$? (v) What is the IQR?
(i) The stem $8$ has leaves $1, 4, 8$: three students ($81$, $84$, $88$).
(ii) The leaves are already in order, so count to the middle. With $15$ scores, that's the $8$th: $75$.
(iii) Largest minus smallest: $97 - 52 = 45$.
(iv) $Q_1 = 64$, and the scores below it are $52$, $58$, $61$: three of them.
(v) $Q_3 - Q_1 = 84 - 64 = 20$. (The range, $45$, uses the whiskers; the IQR uses only the box.)
A histogram's bars are intervals, so a question about "above $80$" is a question about which bars to add.
A histogram of exam scores has bars $51$–$60$ ($1$ student), $61$–$70$ ($2$), $71$–$80$ ($4$), $81$–$90$ ($3$), and $91$–$100$ ($1$). How many students scored above $80$? And which is bigger, the median score or the range of the scores?
Above $80$ means the last two bars: $3 + 1 = 4$ students. (The $71$–$80$ bar ends at $80$, so it doesn't count.)
There are $11$ scores, so the median is the $6$th from the bottom. The first two bars hold $3$ scores and the third bar holds the $4$th through $7$th, so the median is somewhere in $71$–$80$: at least $71$. The range can't be more than $100 - 51 = 49$. So the median is bigger, even though you never learn the exact scores.
That second question is pure ISEE: you can't find either number exactly, but you can still tell which is bigger.
If a question says the data are symmetric about the mean, every bar to the right of the mean is a mirror image of the bar the same distance to the left. So an unfinished bar graph can be finished in your head: copy each bar across. (Symmetric data also have mean $=$ median, which is why the question mentions both.)
Two questions on reading charts, then two on box plots.
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Last graph type: what if each data point has two numbers, like hours studied and test score?
A scatter plot shows pairs of numbers as points, to see whether two quantities are related.
The cloud of points tells you the relationship: uphill, downhill, or nowhere at all.
Sort a few clouds yourself. Some are loose, and you still have to call a direction.
When the cloud does have a direction, you can draw a line through it.
A trend line is a straight line drawn through the middle of a scatter plot's points.
It shows the relationship, and it makes predictions. Its slope tells you how much one thing changes when the other goes up by one.
Here's how a trend line gets read on the test.
A trend line for test scores passes through $(0, 50)$ and $(8, 98)$, where $x$ is hours studied. What does its slope mean? What score does it predict for $5$ hours of studying?
Slope: $\dfrac{98 - 50}{8 - 0} = \dfrac{48}{8} = 6$. Each extra hour of studying goes with about $6$ more points.
At $5$ hours: start at $50$ and add $6$ points per hour, $50 + 6 \cdot 5 = 80$. That's a prediction, not a promise: real scores scatter around the line.
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This is the big one. A line graph doesn't just show numbers, it shows how fast they're changing.
A line graph shows how a quantity changes over time, with its points joined by straight segments.
Remember slope, rise over run? On a graph of miles vs. hours, the rise is miles and the run is hours. So the slope is miles per hour: a speed!
Every slope comes with units: the $y$-units per one $x$-unit. That's what makes it a relationship between the two variables, not just a number.
Uphill: $y$ is growing. Flat: $y$ isn't changing at all. Downhill: $y$ is shrinking. On a distance-from-home graph, that's riding away, stopped, and heading home.
Watch two cyclists ride out a whole day on one graph.
Every piece of that picture answers a question.
Two cyclists, A and B, ride for 8 hours. (i) Who is traveling faster between $t = 0$ and $t = 4$? (ii) At $t = 4$, who is farther from home? (iii) During which interval does cyclist B stop moving? (iv) At $t = 8$, who is farther from home?
(i) A climbs $12$ miles in $4$ hours ($3$ mph). B only climbs $6$ by $t = 3$ and then stalls. A's line is steeper, so A is faster.
(ii) Read the heights at $t = 4$: A is at $12$, B is at $6$. A is farther.
(iii) Stopped means flat (slope $0$): B is flat from $t = 3$ to $t = 5$.
(iv) At $t = 8$: A has come back down to $12$, while B has climbed to $18$. B is farther, even though A led most of the day.
This is the same $d = rt$ from Section 4. Solve it for $r$ and you get $r = \frac{d}{t}$: change in distance over change in time. The slope.
Now drive it yourself: the riders on the road follow the graph.
A flat line on a distance graph means the cyclist stopped. Snack break. Respect.
Two questions, one graph, two different readings.
Steepness is speed, but height is distance. "Who is faster?" means compare slopes. "Who is farther?" means compare heights. The ISEE asks both about the same graph, hoping you'll mix them up.
Can you read a whole ride from its graph alone?
It isn't just bikes. Any line graph over time is a rate in disguise.
The test also runs this backward: a story in words, four graphs to choose from. Turn the story into a shape first. A cold potato in a hot oven heats up fast at first, then levels off near the oven's temperature, so its graph climbs steeply and flattens out; it can't keep climbing forever, and it can't start hot.
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What if the graph is curvy? Then the speed changes from moment to moment, and finding the slope at a single instant is exactly where calculus begins.
That's the whole section: count carefully, compare the right numbers, and read the labels before the picture.
Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game. Watch for "cannot be determined": a bag with an unknown number of blue marbles, or a data list with one unknown value, can go either way.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the ones that need their choices stay multiple choice.
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Alright, if you've made it this far, it's time for something with a little heat. Draws without replacement and random committees: the second pick depends on the first, so draw the tree or count the groups.
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