Algebra · Paper 1
Worded Problems
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Section 1 of 10
Type 1: Linear — Number Puzzles
Question. When 3 is added to twice a number the result is 24. Find the number.
Number $= x$
$2x + 3 = 24$
$2x = 21$
$x = 10.5$
Question. When 5 is added to twice a number the result is 47. Find the number.
Number $= x$
$2x + 5 = 47$
$2x = 42$
$x = 21$
You try
When 7 is added to three times a number the result is 28. Find the number.
Number $= x$. Translate: $3x + 7 = 28$.
Number $= x$
$3x + 7 = 28$
$3x = 21$
$x = 7$
$x = 7$
You try
Twice a number, decreased by 5, equals 19. Find the number.
Number $= x$. Translate: $2x - 5 = 19$.
Number $= x$
$2x - 5 = 19$
$2x = 24$
$x = 12$
$x = 12$
Section 2 of 10
Linear — Age Problems
Question. Mary is 4 times older than John. In 5 years she will be 3 times his age. Find Mary's age now.
| John | Mary | |
|---|---|---|
| Now | $x$ | $4x$ |
| 5 yrs | $x + 5$ | $4x + 5$ |
$4x + 5 = 3(x + 5)$
$4x + 5 = 3x + 15$
$x = 10$
Mary's age now: $\;4x = 40$
Question. Mary is 5 times John's age. In 12 years she will be three times his age. Find Mary's age now.
| Now | 12 yrs | |
|---|---|---|
| Mary | $5x$ | $5x + 12$ |
| John | $x$ | $x + 12$ |
$5x + 12 = 3(x + 12)$
$5x + 12 = 3x + 36$
$2x = 24$
$x = 12$
Mary's age now: $\;5x = 60$
You try
Sarah is 3 times Tom's age. In 6 years she will be twice his age. Find Sarah's age now.
Now: Tom $= x$, Sarah $= 3x$. In 6 yrs: Tom $= x+6$, Sarah $= 3x+6$. She will be twice his age: $3x + 6 = 2(x+6)$.
Now: Tom $= x$, Sarah $= 3x$
In 6 yrs: Tom $= x + 6$, Sarah $= 3x + 6$
$3x + 6 = 2(x + 6)$
$3x + 6 = 2x + 12$
$x = 6$
Sarah now: $\;3x = 18$
Sarah is $18$ now.
You try
A father is 4 times his son's age. In 10 years he will be twice as old. Find both ages now.
Son $= x$, father $= 4x$. In 10 yrs: $4x + 10 = 2(x + 10)$.
Now: Son $= x$, Father $= 4x$
In 10 yrs: $4x + 10 = 2(x + 10)$
$4x + 10 = 2x + 20$
$2x = 10 \;\Rightarrow\; x = 5$
Son $= 5$, Father $= 20$
Son $5$, Father $20$.
Section 3 of 10
Linear — Rate Work
Question. A hot tap takes 4 minutes to fill a bath and a cold one takes $x$ minutes. Working together they take 3 minutes. Find $x$.
| Time | Fill | Per min | |
|---|---|---|---|
| Hot | $4$ | $1$ | $\dfrac{1}{4}$ |
| Cold | $x$ | $1$ | $\dfrac{1}{x}$ |
| Together | $\dfrac{1}{4} + \dfrac{1}{x} = \dfrac{x + 4}{4x}$ |
$\dfrac{1}{\;\dfrac{1}{4} + \dfrac{1}{x}\;} = 3$
$\dfrac{1}{\;\dfrac{x + 4}{4x}\;} = 3$
$\dfrac{4x}{x + 4} = 3$
$4x = 3x + 12$
$x = 12$
You try
A hot tap fills a bath in 6 minutes; a cold one takes $x$ minutes. Working together they take 2 minutes. Find $x$.
Hot per min $= \dfrac{1}{6}$, cold per min $= \dfrac{1}{x}$, together per min $= \dfrac{1}{2}$.
$\dfrac{1}{6} + \dfrac{1}{x} = \dfrac{1}{2}$
$\dfrac{x + 6}{6x} = \dfrac{1}{2}$
$2(x + 6) = 6x$
$2x + 12 = 6x$
$4x = 12$
$x = 3$ minutes
$x = 3$ minutes.
Section 4 of 10
Type 2: Quadratic
Question. One number is 4 larger than another. Product of numbers is 21. Find the numbers.
One $= x \qquad$ Other $= x + 4$
$x(x + 4) = 21$
$x^{2} + 4x = 21$
$x^{2} + 4x - 21 = 0$
$x^{2} + 7x - 3x - 21 = 0$
$x(x + 7) - 3(x + 7) = 0$
$(x + 7)(x - 3) = 0$
$x = 3 \qquad x = -7$
$3, \, 7 \;\;$ or $\;\; -7, \, -3$
Question. Find 2 consecutive even numbers which multiply to 48.
One $= x \qquad$ Other $= x + 2$
$x(x + 2) = 48$
$x^{2} + 2x - 48 = 0$
$x = -8 \qquad x = +6$
$-6, \, -8 \;\;$ or $\;\; 6, \, 8$
You try
Find two consecutive positive integers whose product is 132.
Let them be $x$ and $x + 1$. Then $x(x + 1) = 132$.
$x(x + 1) = 132$
$x^{2} + x - 132 = 0$
$x^{2} + 12x - 11x - 132 = 0$
$x(x + 12) - 11(x + 12) = 0$
$(x + 12)(x - 11) = 0$
$x = 11 \;$ (positive)
$11$ and $12$
$11$ and $12$.
You try
Find two consecutive odd numbers whose product is 63.
Let them be $x$ and $x + 2$. Then $x(x + 2) = 63$.
$x(x + 2) = 63$
$x^{2} + 2x - 63 = 0$
$(x + 9)(x - 7) = 0$
$x = 7 \;$ or $\; x = -9$
$7$ and $9$ (or $-9$ and $-7$)
$7$ and $9$.
Section 5 of 10
Quadratic — Rectangle
Question. One side of a rectangle is 3 m longer than the other. Given area is $38 \text{ m}^{2}$, find length to 1 decimal.
$A = \ell \times b$
$x(x + 3) = 38$
$x^{2} + 3x = 38$
$x^{2} + 3x - 38 = 0$
$x = \dfrac{-3 \pm \sqrt{9 - 4(-38)}}{2}$
$x = 4.8 \text{ m} \;\;$ or $\;\; -7.8$
Question. One side of a rectangle is 5 m longer than the other. Area is $66 \text{ m}^{2}$. Find sides.
$A = \ell \times b$
$x(x + 5) = 66$
$x^{2} + 5x - 66 = 0$
$x = -11 \qquad x = +6$
$6 \text{ m} \;\;$ and $\;\; 11 \text{ m}$
You try
One side of a rectangle is 4 m longer than the other. Area is $45 \text{ m}^{2}$. Find the sides.
Sides: $x$ and $x + 4$. Area: $x(x + 4) = 45$.
$x(x + 4) = 45$
$x^{2} + 4x - 45 = 0$
$(x + 9)(x - 5) = 0$
$x = 5 \;$ (positive)
$5 \text{ m}$ and $9 \text{ m}$
$5 \text{ m}$ and $9 \text{ m}$.
You try
The length of a rectangle is 2 m longer than the width. Area is $24 \text{ m}^{2}$. Find the dimensions.
Width $= x$, length $= x + 2$. Area: $x(x + 2) = 24$.
$x(x + 2) = 24$
$x^{2} + 2x - 24 = 0$
$(x + 6)(x - 4) = 0$
$x = 4 \;$ (positive)
Width $4 \text{ m}$, length $6 \text{ m}$
$4 \text{ m} \times 6 \text{ m}$.
Section 6 of 10
Quadratic — Sharing
Question. €$120$ is shared by $x$ people. If there were 5 more people each would receive €$4$ less. Find $x$.
| Money | People | Each | |
|---|---|---|---|
| Now | $120$ | $x$ | $\dfrac{120}{x}$ |
| If 5 more | $120$ | $x + 5$ | $\dfrac{120}{x + 5}$ |
$\dfrac{120}{x} - \dfrac{120}{x + 5} = 4$
$120(x + 5) - 120x = 4x(x + 5)$
$120x + 600 - 120x = 4x^{2} + 20x$
$4x^{2} + 20x - 600 = 0$
$x^{2} + 5x - 150 = 0$
$x^{2} + 15x - 10x - 150 = 0$
$x(x + 15) - 10(x + 15) = 0$
$(x + 15)(x - 10) = 0$
$x = -15 \qquad x = 10$
$x = 10$
Question. A cinema takes in €$400$ each time that all seats are sold. Next week, eight seats will be removed to make room for a new emergency exit. The price per seat will have to be increased by €$2.50$ in order to keep the takings at €$400$. Taking $x$ to be the number of seats now in the cinema, find the number of seats now and the price per seat now.
| Cost | Seats | Price per seat | |
|---|---|---|---|
| Before | $400$ | $x$ | $\dfrac{400}{x}$ |
| After | $400$ | $x - 8$ | $\dfrac{400}{x - 8}$ |
Difference $= 2.5$
$\dfrac{400}{x - 8} - \dfrac{400}{x} = \dfrac{2.5}{1}$
$400x - 400(x - 8) = 2.5x(x - 8)$
$400x - 400x + 3200 = 2.5x^{2} - 20x$
$2.5x^{2} - 20x - 3200 = 0$
$x^{2} - 8x - 1280 = 0$
$x^{2} - 40x + 32x - 1280 = 0$
$x(x - 40) + 32(x - 40) = 0$
$x = 40 \qquad x = -32$
$\dfrac{400}{40} = 10 \qquad \dfrac{400}{32} = 12.5$
$40$ seats at €$10$ now; €$12.50$ after.
You try
€$100$ is shared equally by $x$ people. If there were 5 more people, each would receive €$1$ less. Find $x$.
Each gets $\dfrac{100}{x}$ now and $\dfrac{100}{x + 5}$ after. Difference $= 1$.
$\dfrac{100}{x} - \dfrac{100}{x + 5} = 1$
$100(x + 5) - 100x = x(x + 5)$
$500 = x^{2} + 5x$
$x^{2} + 5x - 500 = 0$
$(x + 25)(x - 20) = 0$
$x = 20 \;$ (positive)
$x = 20$ people
$x = 20$ people.
Section 7 of 10
Quadratic — Speed / Rate
Question. Water flows at $1 \text{ km/h}$. I travel at $x \text{ km/h}$ for $12 \text{ km}$ up and down stream. Total time is $9$ hours. Find $x$.
Speed down $= x + 1$, speed up $= x - 1$.
| Distance | Speed | Time | |
|---|---|---|---|
| Down | $12$ | $x + 1$ | $\dfrac{12}{x + 1}$ |
| Up | $12$ | $x - 1$ | $\dfrac{12}{x - 1}$ |
$\dfrac{12}{x + 1} + \dfrac{12}{x - 1} = \dfrac{9}{1}$
$\dfrac{12(x - 1) + 12(x + 1)}{(x + 1)(x - 1)} = 9$
$12x - 12 + 12x + 12 = 9(x^{2} - 1)$
$9x^{2} - 24x - 9 = 0$
$3x^{2} - 8x - 3 = 0$
$3x^{2} - 9x + x - 3 = 0$
$3x(x - 3) + 1(x - 3) = 0$
$(x - 3)(3x + 1) = 0$
$x = 3 \qquad x = -\dfrac{1}{3}$
$x = 3 \text{ km/h}$
Question. A farmer must feed bales of hay to his cattle for a total of 90 days. He feeds the cattle 540 bales of hay over a number of days. Their average consumption is $x$ bales per day. If the average consumption is increased by 1 bale per day, then the cattle consume 300 bales in the remaining days. Find $x$ and the number of days taken to consume the first 540 bales.
| Bales | Each | Days | |
|---|---|---|---|
| First | $540$ | $x$ | $\dfrac{540}{x}$ |
| Remaining | $300$ | $x + 1$ | $\dfrac{300}{x + 1}$ |
$\dfrac{540}{x} + \dfrac{300}{x + 1} = \dfrac{90}{1}$
$540(x + 1) + 300x = 90x(x + 1)$
$54(x + 1) + 30x = 9x(x + 1)$
$54x + 54 + 30x = 9x^{2} + 9x$
$9x^{2} - 75x - 54 = 0$
$3x^{2} - 25x - 18 = 0$
$3x^{2} - 27x + 2x - 18 = 0$
$3x(x - 9) + 2(x - 9) = 0$
$(x - 9)(3x + 2) = 0$
$x = 9 \qquad x = -\dfrac{2}{3}$
Days for first 540 bales: $\;\dfrac{540}{9} = 60$
$x = 9$ bales/day; 60 days for the first 540 bales.
You try
A train takes 1 hour less than a car to travel $240 \text{ km}$. The train is $20 \text{ km/h}$ faster than the car. Find the car's speed.
Car speed $= x$, train $= x + 20$. Car time $= \dfrac{240}{x}$, train $= \dfrac{240}{x + 20}$. Difference $= 1$ hour.
$\dfrac{240}{x} - \dfrac{240}{x + 20} = 1$
$240(x + 20) - 240x = x(x + 20)$
$4800 = x^{2} + 20x$
$x^{2} + 20x - 4800 = 0$
$(x + 80)(x - 60) = 0$
$x = 60 \;$ (positive)
Car: $\;60 \text{ km/h}$
$60 \text{ km/h}$.
Section 8 of 10
Optimisation — Farmer's Field
Question. A farmer has $12 \text{ m}$ of wire to make a rectangular field. One side is $x$. Show area is $A = 6x - x^{2}$. Graph $f(x) = 6x - x^{2}$, hence find (i) maximum area (ii) dimensions for area of $4 \text{ m}^{2}$. The domain of graph is $0 \le x \le 6$.
$P = 12$
$2x + 2y = 12$
$x + y = 6$
$y = 6 - x$
$A = \ell \times b$
$\phantom{A} = x(6 - x)$
$A = 6x - x^{2}$
Graphing $\;f(x) = 6x - x^{2}\;$ on $\;0 \le x \le 6$:
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|---|
| $f(x)$ | $0$ | $5$ | $8$ | $9$ | $8$ | $5$ | $0$ |
(i) Maximum area $= 9 \text{ m}^{2}$ (at $x = 3$)
(ii) For area $= 4$: $x = 0.7 \;\;$ or $\;\; x = 5.3$
You try
A farmer has $20 \text{ m}$ of wire to make a rectangular field. One side is $x$. Show that $A = 10x - x^{2}$, hence find the maximum area and the dimensions.
Perimeter $= 20$ gives $y = 10 - x$. Area $= x(10 - x)$. The parabola peaks at $x = 5$.
$P = 20 \;\Rightarrow\; 2x + 2y = 20 \;\Rightarrow\; y = 10 - x$
$A = x(10 - x) = 10x - x^{2}$
Complete the square: $\;A = -(x - 5)^{2} + 25$
Maximum at $\;x = 5, \;\; y = 5$
Maximum area $= 25 \text{ m}^{2}$; field is $\;5 \text{ m} \times 5 \text{ m}$ (a square)
Max area $25 \text{ m}^{2}$ at $5 \text{ m} \times 5 \text{ m}$.
Section 9 of 10
Parabola — Boat Bow
Question. The bow of a boat is parabolic in shape. This parabola is expressed by $y = -x^{2} + 6x - 8$. If the length of the bow is $9 \text{ m}$, calculate its width, $w \text{ m}$, at the widest point.
$y = -x^{2} + 6x - 8$
$-x^{2} + 6x - 8 = 0$
$x^{2} - 6x + 8 = 0$
$(x - 2)(x - 4) = 0$
$x = 2 \qquad x = 4$
Complete the square:
$-y = x^{2} - 6x + 8$
$-y = x^{2} - 6x + 9 + 8 - 9$
$-y = (x - 3)^{2} - 1$
$y = 1 - (x - 3)^{2} \qquad$ vertex $(3, 1)$
Bow length is $9 \text{ m}$, so the base is at $\;y = 1 - 9 = -8$.
$-x^{2} + 6x - 8 = -8$
$-x^{2} + 6x = 0$
$x^{2} - 6x = 0$
$x(x - 6) = 0$
$x = 0 \qquad x = 6$
$w = 6 \text{ m}$
You try
A parabolic arch is given by $\;y = -x^{2} + 4x$. Find where it meets the ground ($y = 0$), and its maximum height.
Roots come from $y = 0$. Maximum height comes from completing the square.
$y = 0: \;\; -x^{2} + 4x = 0$
$x(4 - x) = 0$
$x = 0 \;\;$ or $\;\; x = 4$ (arch is $4 \text{ m}$ wide)
Complete the square: $\;-y = x^{2} - 4x = (x - 2)^{2} - 4$
$y = 4 - (x - 2)^{2}$ vertex $(2, 4)$
Meets ground at $x = 0$ and $x = 4$; max height $= 4$
Ground: $x = 0, 4$. Max height $= 4$.
Section 10 of 10
Simultaneous Equations
Question. A farm has pigs and hens. There are $32$ heads in total and $104$ legs. How many of each?
Number of Pigs $= x \qquad$ hens $= y$
Heads: $\;x + y = 32$
Legs: $\;4x + 2y = 104$
$\begin{aligned} -x - y &= -32 \\ \phantom{-}2x + y &= \phantom{-}52 \end{aligned}$ (first × −1, second ÷ 2; add)
$x = 20 \qquad y = 12$
$20$ pigs and $12$ hens.
Question. I have $25$ coins. Either €$1$ or €$2$ with total value of €$42$. Find number of each.
€$1 = x \qquad$ €$2 = y$
Coins: $\;x + y = 25$
Value: $\;x + 2y = 42$
Subtract: $\;y = 17$
$x = 25 - 17 = 8$
$8$ coins of €$1$ and $17$ coins of €$2$.
You try
A farm has rabbits and chickens. There are $20$ heads in total and $56$ legs. How many of each?
Rabbits $= x$ (4 legs), chickens $= y$ (2 legs). Two equations.
Heads: $\;x + y = 20$
Legs: $\;4x + 2y = 56 \;\Rightarrow\; 2x + y = 28$
Subtract: $\;x = 8 \;\Rightarrow\; y = 12$
$8$ rabbits and $12$ chickens
$8$ rabbits, $12$ chickens.
You try
I have $15$ coins, either €$1$ or €$5$, totalling €$43$. Find how many of each.
€$1$ coins $= x$, €$5$ coins $= y$. Two equations.
Coins: $\;x + y = 15$
Value: $\;x + 5y = 43$
Subtract: $\;4y = 28 \;\Rightarrow\; y = 7$
$x = 15 - 7 = 8$
$8$ coins of €$1$ and $7$ coins of €$5$
$8$ × €$1$ and $7$ × €$5$.