QuanticaPrealgebraOpen in the courseA word problem is one fact written twice, once in plain English and once in algebra, and the two versions say the identical thing. A sentence like a number tripled and then reduced by four gives eleven is just \(3x - 4 = 11\) put into words. The same five steps handle every one of them. Read the whole problem, name the unknown with a letter, translate each phrase by its meaning, solve, then answer the exact question asked and check it against the original story.
Problem
An arcade prize machine doubles the bronze tokens you feed it and drops in \(7\) bonus tokens, and after Priya's turn the screen reads \(31\). In other words, seven more than twice her tokens is thirty-one. How many bronze tokens did Priya feed the machine?
Show a hint
- Give the unknown a name first. Let \(x\) be the number of tokens Priya fed the machine. Now translate the sentence one phrase at a time. What does "twice her tokens" become, and what does "seven more than" that become? The words "is thirty-one" hand you the equals sign.
- Twice her tokens is \(2x\), and seven more than that is \(2x + 7\). The phrase "is thirty-one" sets it equal to \(31\), so your equation is \(2x + 7 = 31\). Subtract \(7\) from both sides, then divide by \(2\).
Show the full solution
Let \(x\) be the tokens Priya fed in. Twice her tokens plus seven is thirty-one, so \(2x + 7 = 31\). Subtract \(7\) to get \(2x = 24\), then divide by \(2\), so she fed in \(\boxed{12}\) tokens. Translate one phrase at a time. Twice gives \(2x\), seven more than adds \(7\), and is hands you the equals sign.
Problem
A beekeeper scoops out exactly a third of whatever a jar holds, pours it into a bottle, then stirs in \(9\) more ounces. The bottle then weighs \(17\) ounces, since here the word of means multiply. How many ounces of honey did the jar hold at the start?
Show a hint
- Let \(x\) stand for the number of ounces the jar held at the start. Now read the words one piece at a time and turn each piece into math. "A third of the amount" becomes \(\tfrac{1}{3}x\), "increased by nine" adds \(9\), and "is seventeen" gives you the equals sign and the \(17\).
- Your equation is \(\tfrac{1}{3}x + 9 = 17\). Subtract \(9\) from both sides to get \(\tfrac{1}{3}x = 8\), then multiply both sides by \(3\) to undo the one third.
Show the full solution
Let \(x\) be the ounces the jar held at the start. A third of that plus nine ounces is seventeen, so \(\tfrac{1}{3}x + 9 = 17\). Subtract \(9\) to get \(\tfrac{1}{3}x = 8\), then multiply both sides by \(3\), so the jar held \(\boxed{24}\) ounces. Check it. A third of \(24\) is \(8\), and nine more is \(17\).
Problem
Maya tapes streamers to a wall, and the number on the wall is nineteen fewer than the number in her box, which counts out to \(40\). Watch the order. Fewer than subtracts from the number, so this is \(x - 19\), not \(19 - x\). How many streamers were in Maya's box to start with?
Show a hint
- Let \(x\) be the number of streamers in the box. The phrase "nineteen fewer than the number" means you start with the number and take 19 away, so it becomes \(x - 19\), not \(19 - x\). Set that equal to 40.
- Solve \(x - 19 = 40\) by adding 19 to both sides. Then read your answer back into the words to make sure nineteen fewer really lands on forty.
Show the full solution
Let \(x\) be the streamers in Maya's box. Nineteen fewer than that number is \(x - 19\), and it equals forty, so \(x - 19 = 40\). Add \(19\) to both sides, so the box held \(\boxed{59}\) streamers. Fewer than reverses the order you hear it in. Reading it as \(19 - x\) would give \(-40\) streamers.
Problem
A workshop has two number machines. The first triples a secret number and subtracts \(5\), and its readout shows \(40\). The second doubles that same number and adds \(2\). First solve for the secret number, then feed it into the second machine. What does the second machine's readout show?
Show a hint
- Give the secret number a name, say \(n\). The first machine triples it and subtracts five to read forty, so that sentence becomes an equation in \(n\). Solve that equation first, then look again at what the question is actually asking for.
- Solve \(3n - 5 = 40\) to find \(n = 15\), but do not stop there. The question asks for the second machine's reading, which is \(2n + 2\). Put \(n = 15\) into \(2n + 2\) and compute that number.
Show the full solution
Let \(n\) be the secret number. The first machine triples it and subtracts five to show forty, so \(3n - 5 = 40\), giving \(3n = 45\) and \(n = 15\). The question wants the second machine, which reads \(2n + 2 = 2 \cdot 15 + 2 = \boxed{32}\). Solving for the variable is not always the last step. Check what the question actually asked for before you answer.
Problem
A rooftop garden has \(96\) plants, all marigolds and lavender, with three times as many lavender as marigolds. Let \(m\) be the number of marigolds, so the lavender count is \(3m\) and the two add to \(96\). The question asks for lavender, not \(m\). How many lavender plants are there?
Show a hint
- Pick the smaller, simpler quantity to name with a variable. Let \(m\) be the number of marigolds. Then write the lavender count in terms of \(m\), and remember the two kinds together make \(96\) plants.
- There are three times as many lavender as marigolds, so the lavender count is \(3m\). The total ties them together, so \(m + 3m = 96\). Combine to get \(4m = 96\), solve for \(m\), then read the question again. It asks for lavender, which is \(3m\), not \(m\).
Show the full solution
Let \(m\) be the marigolds, so the lavender count is \(3m\). The two kinds are the whole garden, so \(m + 3m = 96\), which gives \(4m = 96\) and \(m = 24\) marigolds. The question asks for lavender, which is \(3 \cdot 24 = \boxed{72}\) plants. Check it. \(24 + 72 = 96\), and \(72\) is three times \(24\).
Problem
A \(150\) cm ribbon is cut into two pieces, with the longer piece \(30\) cm more than the shorter. Let \(x\) be the shorter piece, so the longer is \(x + 30\) and the two add to \(150\). The question asks for the longer piece, not \(x\). How long is the longer piece, in centimeters?
Show a hint
- You have two unknown lengths, but you do not need two letters. Let \(x\) stand for the shorter piece. Since the longer piece is \(30\) more than the shorter one, you can write it as \(x + 30\). Now both pieces are written with the same single letter.
- The two pieces together make the whole \(150\) centimeter ribbon, so \(x + (x + 30) = 150\). Combine to get \(2x + 30 = 150\), then \(2x = 120\), so \(x = 60\). That is the shorter piece. The question asks for the longer piece, which is \(x + 30\).
Show the full solution
Let \(x\) be the shorter piece in centimeters, so the longer piece is \(x + 30\). The two came from one ribbon, so \(x + (x + 30) = 150\), which is \(2x + 30 = 150\), then \(2x = 120\) and \(x = 60\). The question asks for the longer piece, \(x + 30 = \boxed{90}\) centimeters. Writing the second length in terms of the first keeps the whole problem on one variable.
That dictionary is the whole vocabulary. Every word problem is built from those same few phrases, so no new translation rules are coming. What changes is how many phrases stack into one story and which quantity the question finally asks for. The next problems run on the same five steps in fuller settings, including counts of two different kinds of thing, a choice between two pricing plans, ages a few years apart, and runs of consecutive whole numbers.
Problem
A puzzle box holds \(30\) tokens, each a square with \(4\) corners or a triangle with \(3\) corners, and all the corners together count \(104\). Let \(s\) be the number of squares, so the triangles are \(30 - s\), the key move being that the second category is the total minus the first. The question asks for triangles. How many triangle tokens are in the box?
Show a hint
- You know two totals here, the number of tokens and the number of corners. Pick a letter for one of the counts. Let \(s\) be the number of squares. Since the two kinds add up to \(30\) tokens, how can you write the number of triangles using \(s\)? Then total the corners, \(4\) for each square and \(3\) for each triangle.
- The triangles must be \(30 - s\), the total minus the squares. Counting corners gives \(4s + 3(30 - s) = 104\). Distribute and combine like terms to get \(s + 90 = 104\), so \(s = 14\). That is the number of squares, so finish by computing the triangles the question wants, which is \(30 - 14\).
Show the full solution
Let \(s\) be the squares, so the triangles are \(30 - s\). Counting corners gives \(4s + 3(30 - s) = 104\). Distribute to get \(4s + 90 - 3s = 104\), so \(s + 90 = 104\) and \(s = 14\) squares. The triangles are \(30 - 14 = \boxed{16}\). Check it. \(14 + 16 = 30\) tokens, and \(4 \times 14 + 3 \times 16 = 104\) corners.
Problem
A street magician's riddle gives two descriptions of the same spot. Half her secret number with four shaved off lands exactly where a third of her number with three piled on lands, so \(\tfrac{1}{2}x - 4 = \tfrac{1}{3}x + 3\). Clear the fractions, gather the variable, and solve. What is her secret number?
Show a hint
- Let \(x\) stand for the secret number. Now translate each half of the sentence on its own. "Half of my number and shave off four" becomes \(\tfrac{1}{2}x - 4\), and "a third of my number and pile on three" becomes \(\tfrac{1}{3}x + 3\). The phrase "land in exactly the same spot" is the equals sign, so set the two pieces equal.
- You have \(\tfrac{1}{2}x - 4 = \tfrac{1}{3}x + 3\), with the variable on both sides and fractions in the way. Clear the fractions by multiplying every single term by \(6\). That turns it into \(3x - 24 = 2x + 18\). Now gather the \(x\) terms on one side and the numbers on the other, and solve.
Show the full solution
Both descriptions land on the same spot, so \(\tfrac{1}{2}x - 4 = \tfrac{1}{3}x + 3\). Multiply every term by \(6\) to clear the fractions, giving \(3x - 24 = 2x + 18\). Subtract \(2x\), then add \(24\), so her secret number is \(\boxed{42}\). Check it. Half of \(42\) minus four is \(17\), and a third of \(42\) plus three is \(17\) too.
Problem
A climbing gym offers two plans. The Steady plan charges a flat \(\$12\) per month plus \(\$0.20\) per session, and the Casual plan charges a flat \(\$4\) per month plus \(\$0.30\) per session. For few sessions Casual is cheaper, for many Steady wins, so somewhere they cost the same. How many sessions in one month make the two plans cost the same amount?
Show a hint
- Give the unknown a name. Let \(s\) be the number of sessions in the month. Now write the monthly cost of each plan as an expression in \(s\), and remember that "cost the same" is just an equals sign waiting to happen. To keep everything in whole numbers you can work in cents, so \(\$0.20\) is \(20\) cents and the flat \(\$12\) is \(1200\) cents.
- In cents the Steady plan costs \(1200 + 20s\) and the Casual plan costs \(400 + 30s\). Set them equal to get \(1200 + 20s = 400 + 30s\). Get the variable on one side by subtracting \(20s\) from both sides, then subtract \(400\) from both sides, and finish by dividing.
Show the full solution
Let \(s\) be the sessions in the month, and work in cents to stay exact. Steady costs \(1200 + 20s\) and Casual costs \(400 + 30s\), so they match when \(1200 + 20s = 400 + 30s\). Subtract \(20s\) to get \(1200 = 400 + 10s\), then \(800 = 10s\), so the plans tie at \(\boxed{80}\) sessions. At \(80\) sessions both come to \(2800\) cents, which is \(\$28\).
Problem
Marisol notices that six years from now her tortoise Pebble will be exactly seven sixths of its current age. Let \(t\) be Pebble's age now, so its future age is \(t + 6\), and that equals \(\tfrac{7}{6}t\). Clear the fraction and solve. How old is Pebble right now, in years?
Show a hint
- Pick a letter for the thing you do not know yet, the age right now, and call it \(t\). A future age is just the current age plus however many years pass, so in six years Pebble will be \(t + 6\). The words say that future age equals seven sixths of the age now, so set \(t + 6\) equal to \(\tfrac{7}{6}t\).
- Start from \(t + 6 = \tfrac{7}{6}t\). The fraction is annoying, so multiply every single term by \(6\) to clear it, which gives \(6t + 36 = 7t\). Now it is an ordinary equation with the variable on both sides, so subtract \(6t\) from each side and read off \(t\).
Show the full solution
Let \(t\) be Pebble's age now. Six years from now the age is \(t + 6\), and that equals \(\tfrac{7}{6}t\), so \(t + 6 = \tfrac{7}{6}t\). Multiply every term by \(6\) to get \(6t + 36 = 7t\), then subtract \(6t\), so Pebble is \(\boxed{36}\) years old. Check it. \(36 + 6 = 42\), and \(\tfrac{7}{6}\cdot 36 = 42\).
Problem
Three steel rings have weights that are consecutive even numbers of kilograms and total \(96\) kg. What does the heaviest ring weigh, in kilograms?
Show a hint
- Let \(n\) stand for the weight of the lightest ring in kilograms. Because the rings step up by \(2\) each time, the middle ring is \(n + 2\) and the heaviest is \(n + 4\). The three weights add to \(96\), so write that sum as an equation in \(n\).
- Your equation is \(n + (n + 2) + (n + 4) = 96\). Combine the like terms on the left to get \(3n + 6 = 96\), then subtract \(6\) and divide by \(3\) to find \(n\). Once you have \(n\), the question wants the heaviest ring, which is \(n + 4\), not \(n\) itself.
Show the full solution
Let \(n\) be the lightest ring in kilograms. Consecutive even weights step up by \(2\), so the three rings are \(n\), \(n + 2\), and \(n + 4\), and they total \(96\). That gives \(3n + 6 = 96\), so \(3n = 90\) and \(n = 30\). The weights are \(30\), \(32\), and \(34\), so the heaviest ring is \(\boxed{34}\) kilograms. The variable landed on the lightest ring, but the question wanted the heaviest, so take the extra step.
Problem
A baker fills exactly \(40\) boxes, each a small box holding \(6\) cupcakes or a large box holding \(10\), for a total of \(312\) cupcakes. Let \(s\) be the small boxes, so the large boxes are \(40 - s\). Count the cupcakes, solve, then answer for the large boxes. How many large boxes did she fill?
Show a hint
- You have two unknowns but they are tied together. Let \(s\) be the number of small boxes. Since every box is small or large and there are \(40\) boxes in total, the number of large boxes must be \(40 - s\). Now count cupcakes, the small boxes contribute \(6s\) and the large boxes contribute \(10\) of them each.
- Write the cupcake count as one equation, \(6s + 10(40 - s) = 312\). Distribute and combine like terms to get \(-4s + 400 = 312\). Solve for \(s\), then remember the question asks for large boxes, which is \(40 - s\), not \(s\) itself.
Show the full solution
Let \(s\) be the small boxes, so the large boxes are \(40 - s\). Counting cupcakes gives \(6s + 10(40 - s) = 312\). Distribute to get \(6s + 400 - 10s = 312\), so \(-4s + 400 = 312\), then \(-4s = -88\) and \(s = 22\) small boxes. The large boxes are \(40 - 22 = \boxed{18}\). Check it. \(22 + 18 = 40\) boxes, and \(6 \cdot 22 + 10 \cdot 18 = 312\) cupcakes.
Practice these ideas
Practice
A magician guessing the number on your card says, four more than triple your number lands exactly on thirty-one. What number is on the card?
Show the solution
Let \(x\) be the number on the card. Triple it and add four to land on thirty-one, so \(3x + 4 = 31\). Subtract 4 from both sides for \(3x = 27\), then divide by 3, so the card reads \(\boxed{9}\). Four more than triple means the tripling happens first and the four gets added after.
Practice
A lighthouse keeper logs ships each night. Wednesday's count was eight less than Tuesday's and came to \(23\), and eight less than Tuesday's number is \(x - 8\). How many ships passed on Tuesday?
Show the solution
Let \(x\) be Tuesday's count. Eight less than that number is \(x - 8\), and it came to 23, so \(x - 8 = 23\). Add 8 to both sides, so \(\boxed{31}\) ships passed on Tuesday. Less than flips the order you hear. It is \(x - 8\), never \(8 - x\).
Practice
A museum desk starts the morning with \(50\) maps and \(12\) remain by lunch. The number handed out subtracted from \(50\) leaves \(12\). How many maps were handed out?
Show the solution
Let \(x\) be the maps handed out. They came off the pile of fifty and twelve remain, so \(50 - x = 12\). Add \(x\) and subtract \(12\) from both sides to get \(50 - 12 = x\), so the desk handed out \(\boxed{38}\) maps. Subtracted from fifty puts the fifty first, so it is \(50 - x\) and not \(x - 50\).
Practice
Doubling the count of new birdhouses and adding \(7\) gives \(33\). Solve for the count, then find how many would remain if five were taken down. How many birdhouses would be left?
Show the solution
Let \(x\) be the birdhouses the crew hung. Doubling and adding seven gives \(2x + 7 = 33\), so \(2x = 26\) and \(x = 13\). Taking five down leaves \(13 - 5 = \boxed{8}\) birdhouses. The variable was the number hung, not the number left, so that last subtraction is part of the answer.
Practice
Two bakery trays hold \(45\) croissants together, and the window tray holds \(9\) fewer than the kitchen tray. How many croissants are on the kitchen tray?
Show the solution
Let \(k\) be the croissants on the kitchen tray, so the window tray holds \(k - 9\). The two trays hold 45 together, so \(k + (k - 9) = 45\). That is \(2k - 9 = 45\), so \(2k = 54\) and the kitchen tray has \(\boxed{27}\). Check it. The window tray has \(27 - 9 = 18\), and \(27 + 18 = 45\).
Practice
A llama drinks four times as much water as a goat, and together they drink \(40\) liters. How many liters did the llama drink?
Show the solution
Let \(g\) be the goat's liters, so the llama drank \(4g\). Together that is \(g + 4g = 40\), so \(5g = 40\) and \(g = 8\). The question asks about the llama, which is \(4 \times 8 = \boxed{32}\) liters. Check it. \(8 + 32 = 40\), and \(32\) is four times \(8\).
Practice
A robotics club builds \(26\) machines, each a roller with \(4\) wheels or a hopper with \(2\) wheels, totaling \(84\) wheels. How many hoppers did they build?
Show the solution
Let \(c\) be the rollers, so the hoppers are \(26 - c\). Counting wheels gives \(4c + 2(26 - c) = 84\), which expands to \(4c + 52 - 2c = 84\), so \(2c + 52 = 84\) and \(c = 16\). The hoppers are \(26 - 16 = \boxed{10}\). Check it. \(16 + 10 = 26\) machines, and \(4 \times 16 + 2 \times 10 = 84\) wheels.
Practice
A satellite finds that a quarter of the ice crystal count plus \(5\) equals a sixth of the count plus \(9\). How many ice crystals are in the cloud?
Show the solution
Let \(x\) be the ice crystals. A quarter of the count plus five equals a sixth plus nine, so \(\frac{1}{4}x + 5 = \frac{1}{6}x + 9\). Multiply every term by \(12\) to clear the fractions, giving \(3x + 60 = 2x + 108\). Subtract \(2x\), then subtract \(60\), so \(x = \boxed{48}\). Check it. Both sides come out to \(17\).
Practice
A club buys \(12\) glow items, each a stick at \(\$3\) or a lantern at \(\$5\), spending \(\$44\) in all. How many glow lanterns did they buy?
Show the solution
Let \(p\) be the glow sticks, so the lanterns are \(12 - p\). The spending gives \(3p + 5(12 - p) = 44\), which expands to \(3p + 60 - 5p = 44\), so \(-2p + 60 = 44\), then \(-2p = -16\) and \(p = 8\). The lanterns are \(12 - 8 = \boxed{4}\). Check it. \(8 + 4 = 12\) items, and \(3 \cdot 8 + 5 \cdot 4 = 44\) dollars.
Practice
Four years ago a boat was two thirds of its current age, so with \(t\) its age now, \(t - 4 = \tfrac{2}{3}t\). How old is the boat now, in years?
Show the solution
Let \(t\) be the boat's age now. Four years ago it was \(t - 4\), which equals two thirds of its current age, so \(t - 4 = \tfrac{2}{3}t\). Multiply every term by \(3\) for \(3t - 12 = 2t\), then subtract \(2t\), so the boat is \(\boxed{12}\) years old. Check it. Four years ago it was \(8\), and \(\tfrac{2}{3}\times 12 = 8\).
Practice
Priya and Dev's ages add to \(44\) years, and Priya is exactly three times as old as Dev. How old is Priya, in years?
Show the solution
Let \(y\) be Dev's age, so Priya's age is \(3y\). Their ages add to \(44\), so \(y + 3y = 44\), giving \(4y = 44\) and \(y = 11\). The question asks for Priya, whose age is \(3 \cdot 11 = \boxed{33}\) years. Check it. \(11 + 33 = 44\), and \(33\) is three times \(11\).
Practice
Three hallway panels have consecutive integer widths in tiles that sum to \(72\). How many tiles wide is the widest panel?
Show the solution
Let \(n\) be the narrowest panel in tiles, so the three widths are \(n\), \(n+1\), and \(n+2\). They sum to 72, so \(3n + 3 = 72\), giving \(3n = 69\) and \(n = 23\). The widths are 23, 24, and 25, so the widest panel is \(\boxed{25}\) tiles. Solving gives you the narrowest panel, so take one more step to reach the widest.
Practice
Two padlock codes are consecutive odd numbers that sum to \(56\). What is the larger of the two codes?
Show the solution
Let \(n\) be the smaller code. Consecutive odd numbers sit two apart, so the larger is \(n + 2\) and \(n + (n + 2) = 56\). That is \(2n + 2 = 56\), so \(2n = 54\) and \(n = 27\). The larger code is \(27 + 2 = \boxed{29}\). Check it. \(27\) and \(29\) are both odd and add to \(56\).
Practice
Garage A charges \(\$5\) plus \(\$2\) per hour, and Garage B charges \(\$11\) plus \(\$1\) per hour. At how many hours of parking do the two garages cost the same?
Show the solution
Let \(h\) be the hours parked. Garage A costs \(5 + 2h\) dollars and Garage B costs \(11 + h\), and they charge the same when \(5 + 2h = 11 + h\). Subtract \(h\) for \(5 + h = 11\), then subtract \(5\), so the costs match at \(\boxed{6}\) hours. At \(6\) hours both garages come to \(\$17\).