Word problems using linear equations: Exercises

Intuition How to use this page

Each problem below is a rung on a ladder. L1 just asks you to recognise which pattern you're looking at. L5 asks you to build the whole equation from a tangled paragraph. Try each problem with the five-step method (listed just below) BEFORE opening the solution. Every solution is hidden inside a collapsible callout so this page doubles as a self-test.

How to read this page's boxes and marks:

A grey Solution box that you can click to expand is where the full worked answer lives — try the problem first, then open it.
A trap box after each level warns about the single most common mistake at that difficulty.
Text wrapped in double equals marks a key answer or term — in Obsidian this shows up highlighted; treat it as "this is the thing to remember."

Vocabulary used on this page (defined here so the page stands alone):

LCM = lowest common multiple: the smallest number that two numbers both divide into evenly. Example: the LCM of $6$ and $9$ is $18$ , because $18$ is the smallest number both $6$ and $9$ go into.
Rate = how much of a job (or distance) happens in ONE unit of time.
Ratio $a : b$ = a comparison telling you the two quantities can be written as $a$ parts and $b$ parts of the same size; you can build both from one scaling number $x$ as $a x$ and $b x$ .

Level 1 — Recognition

Here you only need to spot the pattern and write ONE short equation. No twist.

L1·Q1

A number increased by 7 gives 20. Find the number.

Recall Solution

Step 1 (unknown): let the number be $x$ . This is what the question asks for, so we point $x$ straight at it. Step 2: "increased by 7" means we add: $x + 7$ . Step 3 (equation): "gives 20" is the word "is", so: $x + 7 = 20$ Step 4 (solve): to isolate $x$ we must undo the $+ 7$ . The inverse of adding $7$ is subtracting $7$ — do it to both sides so the balance stays level: $x + 7 - 7 = 20 - 7 \Rightarrow x = 13$ Step 5 (verify): $13 + 7 = 20$ ✓ Answer: the number is 13.

L1·Q2

Five times a number is 45. Find the number.

Recall Solution

Step 1: let the number be $x$ . Step 2–3: "five times" means multiply by $5$ , so $5 x$ ; "is 45" gives: $5 x = 45$ Step 4: $x$ is multiplied by $5$ . The inverse of multiplying by $5$ is dividing by $5$ : $\frac{5 x}{5} = \frac{45}{5} \Rightarrow x = 9$ Step 5: $5 \times 9 = 45$ ✓ Answer: the number is 9.

Level 2 — Application

Now you must express a second quantity in terms of $x$ before building the equation.

L2·Q1

The sum of two consecutive integers is 57. Find them.

Recall Solution

Step 1: let the smaller integer be $x$ . Step 2: "consecutive" means the next one is one bigger: $x + 1$ . (See the number-line pattern in the parent note's Type 2.) Step 3: "sum ... is 57": $x + (x + 1) = 57$ Step 4: combine like terms (add the two $x$ 's): $2 x + 1 = 57$ . Undo $+ 1$ then undo $\times 2$ : $2 x = 56 \Rightarrow x = 28$ Step 5: integers are $28$ and $29$ ; $28 + 29 = 57$ ✓ Answer: 28 and 29.

L2·Q2

Rahul has ₹x. His sister has ₹40 more than he does. Together they have ₹210. How much does Rahul have?

Recall Solution

Step 1: Rahul's money $= x$ (already named for us). Step 2: sister $= x + 40$ ("more than" $\to +$ ). Step 3: "together they have ₹210": $x + (x + 40) = 210$ Step 4: $2 x + 40 = 210 \Rightarrow 2 x = 170 \Rightarrow x = 85$ . Step 5: Rahul ₹85, sister ₹125, total $85 + 125 = 210$ ✓ Answer: Rahul has ₹85.

L2·Q3

Pencils cost ₹4 each. Anita buys some pencils and a ₹15 sharpener, spending ₹51 in total. How many pencils did she buy?

Recall Solution

Step 1: let number of pencils $= x$ . Step 2: cost of pencils $= 4 x$ (rate × quantity, the Type 3 money pattern). Step 3: pencils plus sharpener equals total: $4 x + 15 = 51$ Step 4: undo $+ 15$ : $4 x = 36$ ; undo $\times 4$ : $x = 9$ . Step 5: $4 (9) + 15 = 36 + 15 = 51$ ✓ Answer: 9 pencils.

Level 3 — Analysis

Here the equation hides behind a time shift, a fraction, or a subtraction that can flip a sign.

L3·Q1 (Age)

A mother is 4 times as old as her daughter. In 20 years she will be twice as old as her daughter. Find their present ages.

Recall Solution

Step 1: daughter's present age $= x$ . Step 2: mother now $= 4 x$ . In 20 years, both age by 20: daughter $= x + 20$ , mother $= 4 x + 20$ . Step 3: "she will be twice as old as her daughter" (in 20 years): $4 x + 20 = 2 (x + 20)$ Step 4: expand the right side: $4 x + 20 = 2 x + 40$ . Collect $x$ 's on the left, numbers on the right: $4 x - 2 x = 40 - 20 \Rightarrow 2 x = 20 \Rightarrow x = 10$ Step 5: daughter 10, mother 40; in 20 yrs $\to$ 30 and 60, and $60 = 2 \times 30$ ✓ Answer: daughter 10, mother 40.

L3·Q2 (Number / fraction)

When 3 is added to a number and the result is doubled, we get 26. Find the number.

Recall Solution

Step 1: let the number $= x$ . Step 2: "3 added" $\to x + 3$ ; "the result doubled" wraps the whole thing: $2 (x + 3)$ . Step 3: "we get 26": $2 (x + 3) = 26$ Step 4: divide both sides by $2$ first (undo the outermost operation): $x + 3 = 13$ , then $x = 10$ . Step 5: $2 (10 + 3) = 2 (13) = 26$ ✓ Answer: the number is 10.

L3·Q3 (Money with sign flip)

A jar holds ₹96 in ₹2 and ₹5 coins. There are 30 coins in all. How many of each?

Recall Solution

Step 1: let number of ₹2 coins $= x$ . Step 2: number of ₹5 coins $= 30 - x$ (the rest of the 30 coins). Step 3: total value is the independent constraint: $2 x + 5 (30 - x) = 96$ Step 4: distribute the $5$ : $2 x + 150 - 5 x = 96$ . Combine the two $x$ -terms: $- 3 x + 150 = 96$ . Now we want $x$ alone, so we peel off the $+ 150$ . Why move the 150? It's stuck to the $x$ -term by addition, so to undo an addition we subtract $150$ from both sides (keeping the equation balanced): $- 3 x + 150 - 150 = 96 - 150 \Rightarrow - 3 x = - 54$ Finally $x$ is multiplied by $- 3$ , so we divide both sides by $- 3$ . A negative divided by a negative is positive: $x = \frac{- 54}{- 3} = 18$ ₹5 coins $= 30 - 18 = 12$ . Step 5: coins $18 + 12 = 30$ ✓; value $2 (18) + 5 (12) = 36 + 60 = 96$ ✓ Answer: 18 coins of ₹2 and 12 coins of ₹5.

Level 4 — Synthesis

Now the rate–time–work and distance machinery joins in; you assemble several relationships at once.

L4·Q1 (Work)

Tap P fills a tank in 10 hours; tap Q fills it in 15 hours. Both are opened together. How long to fill the tank?

Figure — Word problems using linear equations

Recall Solution

Step 1: let the time together $= t$ hours. Step 2 (why rates?): we cannot add "10 hours" and "15 hours" directly — hours don't add. What adds is how much of the tank each tap fills per hour — its rate. This is exactly what the figure shows. The top bar (labelled "P: 1/10") is P's one-hour contribution, $\frac{1}{10}$ of the tank; the middle bar (labelled "Q: 1/15") is Q's one-hour contribution, $\frac{1}{15}$ of the tank. Stacking those two bars gives the bottom bar (labelled "together") — how much both taps fill in one hour. Step 3: their bars stack, so combined they fill $\frac{1}{10} + \frac{1}{15}$ per hour. Over $t$ hours that must equal one whole tank (the vertical dashed line at value 1 on the figure marks "1 whole tank"): $t (\frac{1}{10} + \frac{1}{15}) = 1$ Step 4: add the fractions using their lowest common multiple, $LCM (10, 15) = 30$ (the smallest number both $10$ and $15$ divide into): $\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$ So $t \cdot \frac{1}{6} = 1$ . To free $t$ we divide both sides by $\frac{1}{6}$ . Why does dividing by a fraction mean multiplying by its reciprocal? Because dividing by $\frac{1}{6}$ asks "how many sixths fit into the thing?" — and there are $6$ sixths in every $1$ whole, so dividing by $\frac{1}{6}$ multiplies by $6$ . Hence $t = 1 \times 6 = 6$ . Step 5: in 6 h, P fills $6 \times \frac{1}{10} = 0.6$ , Q fills $6 \times \frac{1}{15} = 0.4$ ; total $= 1$ tank ✓ Answer: 6 hours.

L4·Q2 (Distance–rate–time)

Two cars start from the same point and drive in opposite directions. One goes at 40 km/h, the other at 60 km/h. After how many hours are they 250 km apart?

Recall Solution

Step 1: let the time $= t$ hours. Step 2: using $Distance = Rate \times Time$ , car 1 covers $40 t$ and car 2 covers $60 t$ . The figure makes the key point visible: the two arrows leave the same "start" dot pointing in opposite directions (the left-pointing arrow labelled "40t km" and the right-pointing arrow labelled "60t km"), so the total gap between the cars — the double-headed arrow labelled "gap" spanning both — is the sum of the two distances, not their difference. Step 3: $40 t + 60 t = 250$ Step 4: combine: $100 t = 250 \Rightarrow t = 2.5$ . Step 5: in 2.5 h, distances are $100$ km and $150$ km; $100 + 150 = 250$ ✓ Answer: 2.5 hours.

L4·Q3 (Mixture / ratio)

A 20-litre mixture of juice and water is $\frac{3}{5}$ juice. How many litres of pure water must be added so the mixture becomes half juice?

Recall Solution

Step 1: let litres of water added $= x$ . Step 2 (key idea — juice is conserved): adding water changes the total but not the amount of juice. Juice present $= \frac{3}{5} \times 20 = 12$ litres, always. New total volume $= 20 + x$ . Step 3: "becomes half juice" means juice is $\frac{1}{2}$ of the new total: $12 = \frac{1}{2} (20 + x)$ Step 4: multiply both sides by 2: $24 = 20 + x \Rightarrow x = 4$ . Step 5: new total $= 24$ L, half of that is 12 L juice ✓ Answer: add 4 litres of water. (This is a ratio idea — see Ratio and Proportion — because "half juice" compares juice to total.)

Level 5 — Mastery

A single tangled paragraph. You must extract every relationship yourself, then build one clean equation.

L5·Q1

Ravi is thinking of a number. He says: "If I multiply my number by 3, then subtract 4, I get the same result as adding 14 to my number." What is Ravi's number?

Recall Solution

Step 1: Ravi's number $= x$ . Step 2: left description: "multiply by 3 then subtract 4" $\to 3 x - 4$ . Right description: "add 14" $\to x + 14$ . Step 3: "I get the same result as" is the equals sign: $3 x - 4 = x + 14$ Step 4: move $x$ 's left, numbers right: $3 x - x = 14 + 4 \Rightarrow 2 x = 18 \Rightarrow x = 9$ . Step 5: left $= 3 (9) - 4 = 23$ ; right $= 9 + 14 = 23$ ✓ Answer: the number is 9.

L5·Q2

The length of a rectangle is 3 cm more than twice its width. Its perimeter is 60 cm. Find its length and width. (Recall: perimeter $= 2 (length + width)$ .)

Recall Solution

Step 1: width $= x$ (the simplest quantity everything else is built from). Step 2: length $= 2 x + 3$ ("3 more than twice the width"). Step 3: perimeter formula: $2 ((2 x + 3) + x) = 60$ Step 4: inside the bracket: $3 x + 3$ . So $2 (3 x + 3) = 60 \Rightarrow 6 x + 6 = 60 \Rightarrow 6 x = 54 \Rightarrow x = 9$ . Length $= 2 (9) + 3 = 21$ . Step 5: perimeter $= 2 (21 + 9) = 2 (30) = 60$ ✓ Answer: width 9 cm, length 21 cm.

L5·Q3

A father's present age is 6 years more than three times his son's age. Five years ago, the father was 5 times as old as his son was then. Find their present ages.

Recall Solution

Step 1: son's present age $= x$ . Step 2: father now $= 3 x + 6$ ("6 more than three times"). Five years ago (subtract 5 from each age — do the subtraction once, on the full expression): son $= x - 5$ , father $= (3 x + 6) - 5 = 3 x + 1$ . Step 3: "five years ago the father was 5 times as old as his son": $3 x + 1 = 5 (x - 5)$ Step 4: expand the right side: $3 x + 1 = 5 x - 25$ . Collect $x$ 's on one side, numbers on the other: $1 + 25 = 5 x - 3 x \Rightarrow 26 = 2 x \Rightarrow x = 13$ Father now $= 3 (13) + 6 = 45$ . Step 5: son 13, father 45. Five years ago: son 8, father 40, and $40 = 5 \times 8$ ✓ Answer: son 13, father 45.

L5·Q4

Two numbers are in the ratio 4 : 7. If 8 is added to each, the new ratio becomes 3 : 5. Find the original numbers.

Recall Solution

Step 1 (why one letter for a ratio?): numbers in ratio $4 : 7$ can be written $4 x$ and $7 x$ — one unknown scaling factor $x$ builds both while keeping the ratio exact (this is the "one scaling number" idea from the vocabulary list at the top of the page). Step 2: after adding 8: $4 x + 8$ and $7 x + 8$ . Step 3: the new ratio equals $\frac{3}{5}$ : $\frac{4 x + 8}{7 x + 8} = \frac{3}{5}$ Step 4: cross-multiply — multiply both sides by $5$ and by $(7 x + 8)$ to clear the fractions, which multiplies the top-left by the bottom-right and sets it equal to the top-right times the bottom-left: $5 (4 x + 8) = 3 (7 x + 8)$ Expand both sides: $20 x + 40 = 21 x + 24$ Collect $x$ 's on one side and numbers on the other: $40 - 24 = 21 x - 20 x \Rightarrow 16 = x$ Original numbers: $4 x = 4 (16) = 64$ and $7 x = 7 (16) = 112$ . Step 5: $64 + 8 = 72$ and $112 + 8 = 120$ , and $72 : 120 = 3 : 5$ (divide both by $24$ ) ✓ Answer: the numbers are 64 and 112.

Recall Self-test recall

Two taps fill in 10 h and 15 h — do you add the times or the rates? ::: Add the rates ( $\frac{1}{10} + \frac{1}{15}$ ); times never add. Cars going opposite ways at 40 and 60 km/h — distance apart after $t$ hours? ::: $40 t + 60 t = 100 t$ (distances add for opposite directions). Ratio $4 : 7$ with one unknown — how do you write the two numbers? ::: As $4 x$ and $7 x$ . Adding water to juice — which quantity stays constant? ::: The amount of juice; only the total volume grows.

Related: Word problems using linear equations · Linear Equations in One Variable · Algebraic Expressions · Inverse Operations · Applications of Algebra · Simultaneous Linear Equations