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Year 10 Maths Advanced Algebra Review

Equation Problems

20 practice questions 1 video lesson Theory + worked examples

Equation problems turn a worded situation into an equation, solve it and then answer the question that was actually asked. In NSW Year 10 Advanced maths the situations take several steps to set up, from consecutive numbers and ages to speeds, percentages and areas, and the answer is checked against the story (NESA outcome MA5‑EQU‑C‑01).

You will choose a pronumeral, write expressions for each quantity, form and solve the equation, then use the answer, for example to find a product, an area or the largest angle. Problems include a hire bill with a free allowance, overtime pay at time-and-a-half, a tank filled from one fraction to another, a car catching a cyclist, a batting average and a drink topped up with juice. Some questions ask whether consecutive numbers can have a given sum. It is part of Year 10 Advanced Algebra Review.

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Theory

To solve an equation problem, choose a pronumeral for one unknown, write every other quantity in terms of it, and turn the key fact into an equation. This Year 10 Advanced maths page covers consecutive numbers, before-and-after problems, ages, trips at two speeds, percentages, two-digit numbers and mixtures.

A worded problem hides a linear equation. Your job is to find it. Start with a sentence such as "Let \(w\) be the width in cm." Choose the quantity that the others are described from: if the length is "\(4\) less than three times the width", let the width be \(w\) and the length is \(3w-4\).

Next, find the fact that gives the equation. It is usually a total (a sum, a perimeter, a bill), two things being equal (same distance, same amount after a swap) or a known difference.

Some problems lead to fractions (time \(=\) distance \(\div\) speed, an average, a fraction of a tank). Multiply every term by the LCD. Others give \(x^2\) terms on both sides, such as an area that changes, and these cancel to leave a linear equation.

Finish by answering the question that was asked, with units, and check the answer in the words of the problem. If the answer must be a whole number and it is not, the situation is impossible.

Translating words into algebraA two-column table. Consecutive whole numbers: n, n plus 1, n plus 2. Consecutive odd or even: n, n plus 2, n plus 4. 5 years ago or in 5 years: a minus 5 or a plus 5. 20 percent off price p: 0.8p. Two-digit number with digits t and u: 10t plus u. x km at 60 km/h takes x divided by 60 hours. In words In algebra consecutive whole numbers n, n + 1, n + 2 consecutive odd (or even) n, n + 2, n + 4 5 years ago / in 5 years a − 5 / a + 5 20% off the price p 0.8p two-digit number, digits t, u 10t + u x km at 60 km/h takes x ÷ 60 hours
Choose one pronumeral, then write every other quantity in terms of it.
Distance, speed and time triangleA triangle split into three parts: d at the top, s and t at the bottom. Cover the quantity you want: d equals s times t, t equals d divided by s, s equals d divided by t. d s t d = s × t t = d ÷ s s = d ÷ t
Cover the quantity you want. Time is distance over speed, so each part of a trip gives one fraction.

Facts that often give the equation:

SituationEquation comes from
Rectangle\(P = 2l+2w\), \(A = lw\)
Anglestriangle \(180^\circ\), quadrilateral \(360^\circ\)
Trips\(t = \dfrac{d}{s}\); add the times, or set two distances equal
Average\(\text{average} = \dfrac{\text{total}}{\text{number}}\), so total \(=\) average \(\times\) number
Price change\(r\%\) off: \(\left(1-\dfrac{r}{100}\right)p\); \(r\%\) on: \(\left(1+\dfrac{r}{100}\right)p\)
Pay with overtimenormal pay \(+\) \(1.5\times\) rate \(\times\) extra hours
Two-digit numbertens digit \(t\), units digit \(u\): value \(10t+u\), reversed \(10u+t\)
Mixtureamount of ingredient \(=\) percentage \(\times\) total volume
Catching up. When one traveller starts later, give them time \(t\) and the earlier one time \(t+h\), where \(h\) is the head start. They meet when their distances are equal.
Ages. If someone is \(a\) now, they were \(a-k\) years old \(k\) years ago and will be \(a+k\) in \(k\) years' time. Everyone ages by the same amount.

How to solve an equation problem

  1. Read to the end and note what is asked. Draw a quick diagram or bar model if it helps.
  2. Define one pronumeral in words, with units.
  3. Express every other quantity in terms of that pronumeral.
  4. Form the equation from the key fact: a total, a difference or two equal quantities.
  5. Solve it. Clear fractions with the LCD; let any \(x^2\) terms cancel.
  6. Answer the actual question in a sentence, with units. You may need one more step, such as finding the length after you find the width.
  7. Check in the original words, and that the answer makes sense (whole number of people, positive length).

Useful set-ups

  1. Fraction of what is left: if \(x\) mL is drunk first, a quarter of the rest is \(\dfrac{1}{4}(750-x)\).
  2. Same number added to top and bottom: \(\dfrac{a+n}{b+n}\) equals the new fraction; cross-multiply.
  3. Free allowance: only the amount over the limit is charged, so extra kilometres \(= d-\) limit.
  4. Is it possible? Solve anyway. A non-whole answer where a whole number is needed proves it is impossible (Example 1).
Example 1 — Consecutive numbers

(i) Three consecutive multiples of \(5\) have a sum of \(255\). Find them.
(ii) Can three consecutive even numbers have a sum of \(100\)? Justify your answer.
Solution

(i) Let the smallest be \(n\). The next two are \(n+5\) and \(n+10\).

\(n+(n+5)+(n+10)\)\(=\)\(255\)
\(3n+15\)\(=\)\(255\)
\(3n\)\(=\)\(240\)
\(n\)\(=\)\(80\)
3n+15=255⇒n=80

The numbers are \(\mathbf{80}\), \(\mathbf{85}\) and \(\mathbf{90}\). Check: \(80+85+90 = 255\).

(ii) Let them be \(n\), \(n+2\), \(n+4\).

\(3n+6\)\(=\)\(100\)
\(n\)\(=\)\(\dfrac{94}{3} = 31.33\ldots\)

\(n\) is not a whole number, so no. The equation proves it is impossible, not just hard to find.

Example 2 — Before and after
Kai has three times as many trading cards as Mia. After Kai gives Mia \(24\) cards, they have the same number. How many cards did each have at first?
Solution

Let Mia have \(m\) cards at first, so Kai has \(3m\). Write both amounts after the swap, then set them equal.

Bar model for Kai and MiaKai has three bars each equal to m; Mia has one bar m. Kai gives away 24 cards and Mia receives 24 cards. m m m m Kai Mia gives 24 +24
\(3m-24\)\(=\)\(m+24\)
\(2m\)\(=\)\(48\)
\(m\)\(=\)\(24\)
3m−24=m+24⇒m=24

Mia had \(\mathbf{24}\) cards and Kai had \(\mathbf{72}\). Check: after the swap both have \(48\).

Example 3 — A trip at two speeds
Jess rides to the beach at \(24\) km/h and rides home along the same road at \(16\) km/h. The round trip takes \(2\) hours \(30\) minutes of riding. How far is the beach?
Solution

Let the distance be \(d\) km. Each leg takes distance \(\div\) speed, and \(2\) h \(30\) min \(= 2.5\) h.

Trip to the beach and backHome and Beach are d km apart. The trip there is at 24 km/h and the trip back is at 16 km/h. Home Beach d km 24 km/h 16 km/h
\(\dfrac{d}{24}+\dfrac{d}{16}\)\(=\)\(2.5\)
\(2d+3d\)\(=\)\(120\)
\(5d\)\(=\)\(120\)
\(d\)\(=\)\(\mathbf{24}\text{ km}\)
d24+d16=2.5⇒d=24

The LCD of \(24\) and \(16\) is \(48\), and \(48\times 2.5 = 120\). Check: \(1\) h there and \(1.5\) h back.

Example 4 — Mixtures and percentages
A \(30\) L barrel of drink is \(20\%\) cordial. How much water must be added to make it \(12\%\) cordial?
Solution

Adding water does not change the amount of cordial. Let \(x\) L of water be added.

\(\text{cordial}\)\(=\)\(0.20\times 30 = 6\text{ L}\)
\(0.12(30+x)\)\(=\)\(6\)
\(30+x\)\(=\)\(50\)
\(x\)\(=\)\(\mathbf{20}\text{ L}\)
0.12(30+x)=6⇒x=20

Check: \(\dfrac{6}{50} = 0.12 = 12\%\). In a mixture problem, write an equation for the ingredient whose amount you can track.

Common pitfalls

Odd numbers go up by 2. Consecutive odd numbers are \(n\), \(n+2\), \(n+4\), not \(n\), \(n+1\), \(n+2\). The same is true of even numbers.
Change both amounts in a swap. If Kai gives \(24\) to Mia, Kai's amount goes down by \(24\) and Mia's goes up by \(24\). The gap closes by \(48\).
Match the units. With speeds in km/h, write \(45\) minutes as \(0.75\) h. With a price in dollars, write \(25\) cents as \(\$0.25\).
Answer the question asked. If you let \(w\) be the width but the question asks for the area, \(w = 12\) is not the final answer.
Watch out for percentages of the wrong total. A \(20\%\) discount leaves \(80\%\) of the original price, \(0.8p\). Taking \(20\%\) of the sale price instead gives the wrong equation.

Frequently asked questions

How do you turn a word problem into an equation?

Let a pronumeral stand for one unknown, write the other quantities in terms of it, then find the fact that links them, such as a total or two equal amounts. That fact written in symbols is the equation.

How do you write consecutive odd numbers in algebra?

Let the first be \(n\). The next odd numbers are \(n+2\), \(n+4\) and so on, because odd numbers are \(2\) apart. Even numbers work the same way.

How do you solve a distance speed time word problem?

Use \(t = \dfrac{d}{s}\) for each part of the trip. Add the times if the total time is given, or set the distances equal if one traveller catches another.

What if my answer is not a whole number?

Check your working first. If it is correct and the question needs a whole number, such as a number of people or an odd number, the situation is impossible, and your equation is the proof.

How do you do mixture problems?

Track one ingredient. Its amount is the percentage times the total volume. Write that amount before and after the change, and set up an equation. Adding water changes the total volume but not the amount of the other ingredient.