Word problems have a bad reputation, and honestly they don't deserve it. The math in this section is stuff you already know: multiplying, dividing, and solving the equations from Section 3. The hard part is the reading. Every word problem is a short story with an equation hiding inside it, and your job is to pull it out. Once you can translate the English into math, these become some of the most predictable points on the whole test.
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 above any question to type your answer instead. Type just the number, no units (a price like 4.50 is fine), and type times like 2 h 24 min or 2:24.
The ISEE uses the same handful of words over and over. Learn what each one means and half the battle is over.
To model word problems, look for these keywords:
Then translate one phrase at a time, left to right, and write it down as you go. Don't try to do it all in your head. Watch how that goes, especially the phrase that has to wait its turn:
Now one where the waiting phrase is "more than." Same move: it lands on the end.
"Three more than twice a number is $19$." What is the number?
"Twice a number" is $2x$. "Three more than" that is $2x+3$. "Is $19$" is $=19$.
Remember the "less than" trap? "$3$ less than twice a number" is $2x-3$, not $3-2x$. "More than" and "less than" both describe what happens to the end of the expression, so the number they mention gets tacked on last.
Some problems hand you two unknowns at once. Call the smaller one $x$ and build the other one from it.
The sum of two numbers is $51$. The larger number is $3$ more than twice the smaller number. What is the smaller number?
Call the smaller number $x$. "Three more than twice" it is $2x+3$, and the two of them add up to $51$:
The smaller number is $16$, and the larger is $2(16)+3=35$. Check: $16+35=51$. ✓
Your turn. Name the smaller one $x$, build the other from it, and add.
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Real-life word problems often come with a flat fee plus a charge for each thing. Peel the fee off first, then divide.
An e-scooter membership costs $\$12$ per month plus $\$0.08$ per minute of ride time. If the total charge for one month is $\$44$, how many minutes were ridden?
So $400$ minutes. If you got $550$, you divided the whole $\$44$ by $0.08$ and forgot that $\$12$ of it was the membership.
Here's that peel as a picture, on a bowling bill:
The fee comes off once; everything left is the per-game charge. Try one:
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When two plans each charge a fee plus a rate, the question is usually "when do they cost the same?" Write both costs and set them equal. The $x$'s land on both sides, just like in Section 3.
Phone plan A costs $\$20$ a month plus $\$0.05$ per minute. Phone plan B costs $\$5$ a month plus $\$0.10$ per minute. For how many minutes do the two plans cost the same?
At $300$ minutes both plans cost $\$35$. Under $300$ minutes plan B is cheaper, and over $300$ plan A is.
Plan B's lower fee is a head start, and its higher rate eats away at it one unit at a time. Add classes and watch:
The head start divided by how fast it shrinks is the break-even point. Find it here:
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Geometry shows up in word problems too. When a problem describes one measurement in terms of another, call the simpler one $w$ (or $x$) and build everything else from it.
The length of a rectangle is $3$ more than twice its width. If the perimeter is $42$ units, what is the width?
Let the width be $w$. Then the length is $2w+3$. A rectangle has two lengths and two widths:
The trap answer is $15$ (that's the length, $2(6)+3$). Always double-check which measurement the question actually asked for.
Notice that one length plus one width is half the perimeter: $15+6=21$. Drag the width and watch the length follow its rule:
On the test you get four answer choices, and one of them has to work. So when the algebra stalls, test the choices:
Halve the perimeter first, then it's a two-numbers problem:
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Age problems have one sneaky rule: time passes for everybody.
Sara is twice Tom's age. In $6$ years, their ages will sum to $51$. How old is Tom now?
Let Tom be $t$, so Sara is $2t$. In $6$ years, Tom is $t+6$ and Sara is $2t+6$. Both of them get $6$ years older!
Slide through the years and watch what changes and what doesn't:
The difference stays fixed, but "times as old" is the one that moves. Watch 4 times as old become twice as old, one birthday at a time:
Every birthday lands on both cakes. Keep that in mind here:
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These problems sound fancy, but they're really just counting.
Consecutive integers are integers that follow each other in order, each one $1$ more than the last: $n,\ n+1,\ n+2,\ \dots$
Counting by ones is the plain version. The ISEE also likes counting by twos, which means even and odd numbers:
An even integer is one that $2$ divides exactly, so it can be written as $2k$ for some integer $k$. An odd integer is one that can be written as $2k+1$ for some integer $k$.
Here's the part everyone gets wrong. Consecutive even integers go up by $2$: $10, 12, 14$. And consecutive odd integers also go up by $2$: $29, 31, 33$. Your gut says odd numbers should be $n, n+1, n+3$ or something weird. Your gut is wrong:
Odd or even, you skip one number every time, so it's $n, n+2, n+4$. Here it is with four even numbers:
Find four consecutive even integers that have a sum of $52$.
Call them $n,\ n+2,\ n+4,\ n+6$. Then
The integers are $10, 12, 14, 16$. Quick check: $10+12+14+16=52$. ✓
There's a faster way when the count is odd. Spread out evenly, the numbers balance around the middle one:
Three consecutive odd integers have a sum of $93$. What is the middle integer?
Shortcut: with an odd number of consecutive integers, the middle one is just the average. $93\div 3=31$.
The integers are $29, 31, 33$. And yes, the long way ($n+(n+2)+(n+4)=93$) gives the same thing.
Try the shortcut before you see the list. Where's the middle?
When it's the product that's given, the letter method gets ugly ($n(n+1)(n+2) = 210$ is a cubic). Estimate instead: three numbers in a row multiply to about the cube of the middle one, and $6^3 = 216$ is close to $210$, so try $5 \cdot 6 \cdot 7 = 210$. Dividing $210$ by $3$ is the trap: that's the move for a sum.
The long way always works; the shortcut is a bonus. Use whichever you like:
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A huge chunk of this section is about rates. You use them every day without thinking about it.
A rate compares two quantities measured in different units.
Miles per hour, dollars per pound, pages per minute: each one compares two different kinds of things. The handiest way to write a rate is per one of something.
A unit rate is a rate written "per $1$" of the second unit.
So $60$ miles per $1$ hour is a unit rate, and $120$ miles per $2$ hours is the same speed, just not written per one. "Per" is the magic word. It means "divided by," and it tells you which number goes on the bottom.
A $12$ oz box of cereal costs $\$3.60$. What is the unit price?
The unit price is the price of one unit: here, dollars per ounce, so dollars go on top:
Unit prices are how you figure out which deal is actually better, and the ISEE loves to make two deals look almost identical.
Column A is $\$6\div 12=\$0.50$ per oz. For Column B, notice that $\$4.50$ for $9$ oz would also be exactly $\$0.50$ per oz. But it's $\$4.51$, a penny more, so its unit price is just a hair over $\$0.50$. Column B is greater.
That one penny is the whole question. The test writers put it there hoping you'll round and pick "equal."
Now sort some packages yourself, best deal first. The biggest one isn't always the winner:
Price divided by size, every time. Which package wins?
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Sometimes a rate comes in the wrong units. Miles per hour, but the question wants feet per second. There's a beautiful trick for this.
Unit conversion rewrites a quantity in new units by multiplying by conversion ratios equal to 1.
Why "equal to 1"? Because $60$ minutes and $1$ hour are the same amount of time, so the fraction $\dfrac{60\text{ min}}{1\text{ h}}$ is really just $1$ in disguise. A fraction like that is a conversion ratio. Multiplying by $1$ never changes a quantity. It only changes the units:
Here are the conversion ratios you need to know cold:
A mile is $5{,}280$ feet. Not $5{,}000$. Not $1{,}000$. FIVE THOUSAND TWO HUNDRED EIGHTY. Who approved this?
Each of these fractions is equal to $1$:
The trick is to line the ratios up like pickets in a fence. Each picket can be flipped upside down (it's still equal to $1$), so flip it until the unit you want to get rid of sits on the opposite side from where it already is. Then it cancels, top against bottom. Try it below. Some of the pickets in the tray start out upside down on purpose.
Here's that first fence written out as math:
Convert $60$ mph to feet per second.
Miles cancel miles, hours cancel hours, minutes cancel minutes. Only feet and seconds survive. That's how you know you set it up right.
One more, this time in metric:
Convert $72\ \text{km/h}$ to meters per second.
(One hour is $60\times 60=3600$ seconds, so that last picket does two jobs at once.)
Sometimes the test skips the arithmetic entirely and asks which expression equals the converted speed. Then the answer choices are four fences, and only one has every unit canceling. Check each ratio: the unit you're getting rid of must sit on the opposite side of the fraction bar from where it started.
Your turn to build a fence:
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A rate that runs for a while works the same way: get the time into the rate's unit first, then multiply.
A printer prints $25$ pages per minute. How many pages does it print in $2$ hours?
The rate is per minute, so turn the hours into minutes: $2\times 60=120$ minutes.
If you got $50$, you multiplied by the hours without changing them to minutes.
Line up the units before you multiply:
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One habit makes every conversion safer: keep the units in your work.
If your answer comes out with units like $\frac{\text{mi}^{2}}{\text{ft}}$, a picket is upside down. The units are a built-in answer checker: if they're wrong, the number is wrong too.
Speed is the most famous rate of all: miles per hour. It gets its own formula.
Distance, Speed, Time: $\text{Distance}=\text{Speed}\times\text{Time}$
Divide both sides by the time and you get the speed. That's all a speed question is:
A car travels $210$ miles in $3.5$ hours. What is its average speed?
Whichever of the three is missing, the formula hands it to you:
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Now the trap. Time on the ISEE often comes in hours and minutes, and there are $60$ minutes in an hour, not $100$.
If Devin is traveling at $54$ mph for $3$ hours $20$ minutes, how far does Devin travel?
$20$ minutes is $\frac{20}{60}=\frac{1}{3}$ of an hour, so the time is $3\frac{1}{3}=\frac{10}{3}$ hours. Not $3.2$ hours!
If you used $3.2$ hours, you'd get $172.8$ miles, and the test writers know it: that's exactly the wrong answer they put in the choices. Notice that nobody wrote "3.2" in the problem. You'd be inventing that mistake all by yourself.
The trap runs the other way too: $2.4$ hours is not $2$ hours $40$ minutes. Slide it and see:
Each tenth of an hour is $6$ minutes. Practice the switch on its own first:
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Then put it to work in a trip:
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Comparing two trips works the same way: find each time, then compare.
Time $=$ Distance $\div$ Speed. Column A: $90\div 60=1.5$ hours. Column B: $100\div 50=2$ hours. Column B is greater.
"Average speed" has its own formula, and it is not the average of the speeds.
Average Speed:
So add up all the miles, add up all the hours, and divide.
Kai drives for $2$ hours at $30$ mph, then for $1$ hour at $60$ mph. What is Kai's average speed for the whole drive?
Distance: $2\times 30+1\times 60=120$ miles. Time: $2+1=3$ hours.
Not $45$, the average of $30$ and $60$: Kai spent twice as long at $30$ mph.
The same thing happens on a round trip, and it fools almost everybody.
Drive $60$ miles there at $30$ mph and back at $60$ mph and your average is $\frac{120}{2+1}=40$ mph, not $45$. You spend more time at the slow speed, so it counts for more. Total distance over total time, every single time.
If "average of the speeds" really were the average speed, these two cars would tie:
Set your own speeds and watch where the real average lands:
Total distance over total time:
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Every so often the ISEE sneaks in a problem that needs a tool you haven't officially learned yet.
The next problem uses the Pythagorean theorem: in a right triangle with legs $a$ and $b$ and long side $c$, $a^{2}+b^{2}=c^{2}$. You'll get the full story in Section 6. For now, notice that $24, 32, 40$ is just the famous $3, 4, 5$ set scaled up by $8$.
Picture two riders heading off at a right angle:
Here are the same numbers written out.
Two cyclists leave the same point, one going east at $12$ mph and the other north at $16$ mph. After $2$ hours, how many miles apart are they?
After $2$ hours, one is $24$ miles east and the other is $32$ miles north. East and north make a right angle, so the distance between them is the long side of a right triangle:
The trap answer is $56$ ($24+32$), which is how far apart they'd be if they went in opposite directions.
One more twist: what if one traveler is chasing the other?
John has jogged $500$ meters when Erin starts jogging on the same path. Each jogs at a constant speed, and Erin is faster. Which one piece of information (in meters per minute) would you need to find how long it takes Erin to catch up?
The one number you need is the difference in their speeds. Erin closes the gap by that much every minute, so the time to catch up is
If Erin jogs $100$ m/min and John jogs $80$ m/min, she gains $20$ meters a minute: $500 \div 20 = 25$ minutes.
Here's that chase, minute by minute:
Chasing uses the difference. Heading toward each other is the other half of the story:
Heading toward each other (or apart)? The gap changes by the sum of the speeds. Chasing in the same direction? It changes by the difference. Mixing these up is the whole trap.
Pick a direction, set the speeds, and drag the clock to watch the gap change:
Toward each other or apart, add the speeds:
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One ISEE question gives no numbers to crunch: "Which one piece of information would be needed to find how long until she catches up?" Think about the formula. Catch-up time is the head start divided by the difference in speeds, so if the head start is given, the one missing ingredient is the difference in speeds. Not one speed alone, and not the sum: that's for heading toward each other.
Same direction, subtract them:
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Here's a classic: two kinds of things with two different prices, and you know the total count and the total cost. That's two facts, so you need two equations.
Total Value: $(\text{number of items})\times(\text{value per item})=\text{total value}$
Each kind of item gets its own copy of that formula, and the two copies add up to the total. Then a second equation counts the items:
A theater sells $16$ tickets for a total of $\$104$. Adult tickets are $\$9$ and student tickets are $\$5$. How many of each were sold?
Let $a$ = adult tickets and $s$ = student tickets. One equation counts tickets, the other counts dollars:
From the first equation, $s=16-a$. Swap that into the second: $9a+5(16-a)=104$, so $4a+80=104$ and $a=6$.
That means $6$ adult tickets and $s=16-6=10$ student tickets. Check: $9(6)+5(10)=54+50=104$. ✓
Same setup with fruit:
A grocery store sells apples for $\$0.60$ each and oranges for $\$0.40$ each. If you buy a total of $13$ pieces of fruit for $\$6.80$, how many apples did you buy?
$a+o=13$ and $0.60a+0.40o=6.80$. Swap in $o=13-a$: $0.60a+5.20-0.40a=6.80$, so $0.20a=1.60$ and $a=8$.
And coins, the ISEE's favorite:
A jar has quarters and dimes. There are $17$ coins totaling $\$3.05$. How many quarters are there?
Work in cents to dodge the decimals: $q+d=17$ and $25q+10d=305$. Swap in $d=17-q$: $25q+170-10q=305$, so $15q=135$ and $q=9$.
The trap answer is $8$ (the number of dimes). Read the last line of the question again before you bubble.
Every one of these solutions ends by dividing by the difference between the two values ($25-10=15$, $0.60-0.40=0.20$). Here's why. Start with every coin a dime and trade them in one at a time:
Each swap adds the difference in value, so the number of swaps is the extra money divided by that difference. And when the clock is ticking, there's an even quicker way in:
Out of time? Test the answer choices. Pick one, figure out the other count, and check the total. On a multiple-choice test, one of the four has to work.
Two equations, one for the count and one for the value:
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The total-value problems hid a tool that deserves its own name. Two unknowns need two facts, and two facts written as equations are a system.
A system of equations is two (or more) equations about the same unknowns. A solution is a set of values that makes every equation true at once.
There are two ways in, and the shape of the equations tells you which to use:
Solve: $y = 2x + 1$ and $3x + 2y = 16$.
The first equation hands you $y$, so substitute: $3x + 2(2x + 1) = 16$, which is $7x + 2 = 16$, so $x = 2$. Then $y = 2(2) + 1 = 5$.
Check both: $5 = 2(2) + 1$ ✓ and $3(2) + 2(5) = 16$ ✓.
When nothing is solved for you, line up the matching terms and subtract:
Solve: $2x + 3y = 12$ and $2x - y = 4$.
Both equations have $2x$, so subtract the second from the first: $(2x + 3y) - (2x - y) = 12 - 4$, which is $4y = 8$, so $y = 2$. Then $2x - 2 = 4$ gives $x = 3$.
If the coefficients don't match, multiply one equation first: for $3x + 2y = 19$ and $2x + 5y = 20$, multiply the first by $2$ and the second by $3$ so both have $6x$, then subtract.
Read the last line of the question twice. After all that work, the ISEE might ask for $x + y$, not $x$. And the subtraction step is where signs get lost: $3y - (-y)$ is $4y$, not $2y$.
Substitute when a letter is handed to you, eliminate when the terms line up:
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Two people working together finish faster than either one alone. But how much faster? Not "average the times," and definitely not "add the times."
Combined Work Rate:
where $A$ and $B$ are the times each person takes working alone.
Where does that come from? Think in rates. If Kristy takes $4$ hours, she does $\frac{1}{4}$ of the job every hour. If David takes $6$ hours, he does $\frac{1}{6}$ every hour. Together they knock out $\frac{1}{4}+\frac{1}{6}$ per hour, and the formula is just what you get when you add those fractions and flip. Watch it happen:
Here's the Kristy and David problem written out.
Kristy can do a job alone in $4$ hours, and David can do it alone in $6$ hours. Working together, how long will it take them?
Now convert carefully: $0.4$ hours is $0.4\times 60=24$ minutes. So the answer is $2$ h $24$ min.
The trap answer is $2$ h $40$ min, which is what you get if you read $2.4$ hours as "2 hours 40 minutes." Same mistake as Devin's road trip: an hour has $60$ minutes, not $100$.
Now run it backward: you know the time together, and you want one person's time alone.
Alex and Blake are painting equal-sized fences. Blake takes twice as long to paint a fence as Alex, but the two of them working together can paint a fence in $40$ minutes. How long does it take Alex to paint a fence by herself?
Let Alex take $t$ minutes, so Blake takes $2t$. Use the formula:
Alex takes $60$ minutes alone (and Blake takes $120$). Check: $\frac{60\times 120}{180}=40$. ✓
Forward or backward, the rates add:
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The rates idea also handles a twist the formula can't: someone working against you.
A hose can fill a pool in $4$ hours, and a drain can empty the full pool in $6$ hours. If the drain is accidentally left open, how long does the hose take to fill the empty pool?
The drain undoes some of the hose's work every hour, so subtract its rate instead of adding it:
One twelfth of the pool every hour means $12$ hours. The trap answer is $2.4$ hours, which is what you'd get if the drain were a second hose helping out.
Change the hose, change the drain, and open or close it:
When one side works against the other, subtract:
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Mixture problems look like chemistry, but they're really just bookkeeping. The secret is to stop thinking about the liquid and track only the stuff dissolved in it (the salt, the acid, whatever).
In a mixture problem, a solution is a liquid with something dissolved in it.
Salt water is the classic one. A solution's percent tells you what share of it is the dissolved stuff, so the amount of salt is just a percent of the whole:
How many liters of salt are in $20$ L of a $10\%$ salt solution?
$10\%$ of $20$ L is $0.10\times 20=2$ L of salt. The other $18$ L is water.
Every mixture problem is built out of that one multiplication. Count the salt in each solution, add, and you get the formula:
Mixture Formula:
Each side of that formula is counting the same thing: how much salt there is. Salt from beaker one plus salt from beaker two equals salt in the mix. Before you pour, guess where the mix will land:
Count the salt, then divide by the whole mix:
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Now with an unknown amount. The formula doesn't care which piece is missing:
A chemist has $10$ liters of a solution that is $10\%$ salt. How many liters of a $40\%$ salt solution must be added to create a final mixture that is $25\%$ salt?
Let $x$ be the liters of $40\%$ solution. Count the salt:
Makes sense, too: $25\%$ is exactly halfway between $10\%$ and $40\%$, so you need equal amounts of each.
Sometimes the total is fixed instead, and you split it between the two solutions:
A lab tech needs $20$ L of a $20\%$ salt solution, made by mixing a $10\%$ solution with a $50\%$ solution. How many liters of each go in?
This time the total is fixed. Let $x$ be the liters of $50\%$ solution, so the other $20-x$ liters are $10\%$. Count the salt:
So $5$ L of the $50\%$ and $15$ L of the $10\%$. Check: $2.5+1.5=4$ L of salt in $20$ L is $20\%$. ✓
Same bookkeeping, one unknown:
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Plain water counts as a mixture too. It's just a solution with $0\%$ of the stuff in it:
You have $200$ mL of a $30\%$ acid solution. How much water must be added to make a $20\%$ solution?
Water is just a $0\%$ solution! The acid doesn't change: $200\times 0.30=60$ mL of acid. For that to be $20\%$ of the new total:
Pour in your own water and watch the percent drop while the acid stays put:
Boiling water off is adding water in reverse. The salt stays put and only water leaves:
Here it is with the numbers:
A tank holds $40$ L of water that is $5\%$ salt. How many liters of water must evaporate to make it $8\%$ salt?
The salt is $40\times 0.05=2$ L, and it doesn't go anywhere. For $2$ L to be $8\%$ of what's left:
So $40-25=15$ L must evaporate. The trap answer is $25$, which is what's left in the tank, not what boiled off.
Guess first, then boil and see where your guess lands:
One last habit for every mixture answer:
Sanity check every mixture answer: the final percent always lands between the two percents you mixed. If your mix of $10\%$ and $40\%$ somehow comes out at $45\%$, something went wrong.
Find the salt, then the total it has to be a percent of:
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One kind of word problem lives somewhere else: a price that changes by two percents in a row (a markup, then a sale) is Section 1's.
Remember Two Percent Changes in a Row? Percent changes multiply, they don't add: up $30\%$ and then down $20\%$ is $1.30\times 0.80=1.04$, a $4\%$ increase, not $10\%$.
One more word problem with its own formula. Money in a savings account earns a percent of itself every year, and with simple interest that percent is always taken of the original deposit.
Simple interest is interest earned on the original amount only: the same amount every year, a fixed percent of the principal.
$P$ is the principal (the amount deposited), $r$ is the yearly rate as a decimal, and $t$ is the time in years. The total in the account is $P + I$.
Leo deposits $\$2{,}000$ in an account that pays $6\%$ simple interest per year. How much interest does it earn in $3$ years, and how much is in the account then?
$I = Prt = 2{,}000 \cdot 0.06 \cdot 3 = 360$ dollars: $\$120$ a year, three times.
The account holds the principal plus the interest: $2{,}000 + 360 = \$2{,}360$. (Answering $\$360$ when the question asks for the total, or $\$2{,}360$ when it asks for the interest, is the most common slip.)
The formula runs backward too: given the interest, you can find the rate, the time, or the deposit by dividing.
Two conversions before you multiply: the percent becomes a decimal ($6\%$ is $0.06$, never $6$), and the time is in years ($18$ months is $1.5$ years). Forgetting the first makes the interest $100$ times too big; forgetting the second makes it $12$ times too big.
Percent to a decimal, months to years, then $Prt$:
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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.
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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 a number answer; answers in hours and minutes stay multiple choice down here.
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Alright, time for something with a little heat. A boat going upstream fights the current, so its speed is (boat $-$ current). Coming back downstream, the current gives it a push: (boat $+$ current). Find both of those speeds from the distances and times, and the current is half the gap between them. Planes with a headwind and a tailwind work exactly the same way.
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