# Ratio Higher Revision Card Answers

## HR1 – Ratio

QUESTION: The ratio of the ages of Kemah compared to Bob and Deborah is 1∶2∶4. Deborah is 21 years older than Kemah. Work out the ages of Kemah, Bob, and Deborah.

ANSWER: We know that difference between Deborah’s and Kemah’s ages is 21. Looking at the ratio, Kemah has 1 part and Deborah has 4, meaning that the difference between them (21 years) constitutes 3 parts in the ratio. Therefore, we get that

$\text{1 part }=21\div 3=7$

Kemah, Bob, and Deborah have 1, 2, and 4 parts in the ratio respectively. So, we get that

$\text{Kemah’s age }=1\times 7=7$

$\text{Bob’s age }=2\times 7=14$

$\text{Deborah’s age }=4\times 7=28$

## HR2 – Direct Proportion

QUESTION: The force, $F$, acting on a car is directly proportional to the acceleration of the car, $a$. When a force of $7,500$ N is acting on a car, it accelerates at $6\text{m/s}^2$.

a) Express $F$ in terms of $a$.

b) Find the acceleration achieved when $12,000$ N of force is acting on this car.

ANSWER: a) If $F$ is directly proportional to $a$ then we can write

$F\propto a$

which becomes the equation:

$F=ka$

Now, to find $k$, we will substitute in the values given in the question. Doing so, we get

$7,500=k\times 6$

Dividing both sides by $6$, we get

$k=7,500 \div 6=1,250$

Therefore, our equation for $F$ in terms of $a$ is

$F=1,250a$

b) To find the acceleration when $F=12,000$, we must substitute this into the equation:

$12,000=1,250a$

Then, dividing both sides by $1,250$, we get the acceleration to be

$a=12,000\div1,250=9.6\text{ m/s}^2$

## HR3 – Inverse Proportion

QUESTION: The force, $F$, due to gravity is inversely proportional to the square of the distance, $d$, between objects. Two objects $200$ km apart result in a gravitational force of $4,500$ N.

a) Express $F$ in terms of $d$.

b) Work out the distance these two objects must be apart to result in a gravitational force of $2,000$ N.

ANSWER: $F$ is inversely proportional to the square of $d$, so we get

$F\propto\dfrac{1}{d^2}$

which becomes the equation

$F=\dfrac{k}{d^2}$

Now, to find $k$, we will substitute in the values given in the question. Doing so, we get

$4,500=\dfrac{k}{(200)^2}=\dfrac{k}{40,000}$

Multiplying both sides by $40,000$, we get

$k=4,500 \times 40,000=180,000,000$

Therefore, our equation for $F$ in terms of $d$ is

$F=\dfrac{180,000,000}{d^2}$

b) To find the distance when $F=2,000$, we must substitute this into the equation:

$2,000=\dfrac{180,000,000}{d^2}$

Firstly, multiplying both sides by $d^2$ gives us

$2,000d^2=180,000,000$

Then, divide by $2,000$ to get

$d^2=180,000,000\div2,000=90,000$

Finally, we get the answer, by square rooting both sides, to be

$d=\sqrt{90,000}=300\text{ km}$

## HR4 – Percentage Change

QUESTION: Matt buys a TV for $\pounds 550$, and a year later sells it to his friend Dave for $32\%$ less. Calculate how much Dave purchased the TV for.

ANSWER: This is a $32\%$ decrease, so the multiplier for a $32\%$ decrease is

$1-\dfrac{32}{100}=0.68$

Therefore, multiplying this by the original value Matt bought the TV for, we get the price that Dave purchased it for to be

$550\times 0.68=\pounds 374$

## HR5 – Reverse Percentage

QUESTION: Tom measures himself to be $182$cm tall and calculates that this new height is a $4\%$ increase on his height 2 years ago. Work out how tall Tom was 2 years ago.

ANSWER: We need to consider how we would calculate a $4\%$ increase. We know that $4\%=0.04$, so we get the multiplier for a $4\%$ increase to be

$1+0.04=1.04$

Let $H$ be Tom’s height from two years ago. We know that the result of multiplying $H$ by $1.04$ must be $182$. We can write this as an equation:

$H\times 1.04=182$

Then, if we divide both sides by $1.04$ we get

$H=182\div 1.04=175$

So, Tom’s height two years ago was $175$cm.

## HR6 – Growth & Decay

QUESTION: Bacteria are being grown in a lab. Initially, there are $480$ bacteria in a dish, and the number increases by $40\%$ every day. The scientist estimates that after one week, there will be over $5,000$ bacteria in the dish. Show that the scientist’s estimate is correct.

ANSWER: This is a case of compound growth. Firstly, the multiplier for a $40\%$ increase is

$1+0.40=1.4$

We are looking at the number of bacteria after one week, which means SEVEN $40\%$ increases. Therefore, our calculation is

$480\times (1.4)^7=5,060\text{ (to nearest whole number)}$

$5,060$ is clearly bigger than $5,000$, so the scientist’s estimate is correct.

## HR7 – Speed, Distance, Time

QUESTION: An aeroplane travels at an average speed of $230$ kilometres per hour for a total journey time of $414$ minutes. Work out the total distance covered by this aeroplane.

If we cover up $d$ in the triangle, we see that we will have to multiply $s$ by $t$ to get our answer. However, the units don’t match up – we need to convert the minutes to hours, which we will do by dividing it by $60$.

$t=414\div 60=6.9\text{ hours}$

Now, we can do the multiplication:

$\text{distance covered }=d=230\times 6.9=1,587\text{ km}$

## HR8 – Pressure & Density

QUESTION: A pressure of $40 N/M^2$ is exerted evenly on the area of the triangle shown here. Calculate the force being applied to exert this amount of pressure.

ANSWER: Covering up $F$ on the triangle,

we see that we need to multiply $p$ (pressure) by $A$ (area). To do this, we need to find the area of the triangle above. Before that, however, notice that the sides of the triangle are measured in centimetres which doesn’t match up with the “Newtons per metres squared” in the question. So, we will convert the dimensions of the triangle into metres by dividing by 100:

$\text{height }=80\div 100=0.8\text{ m}$

$\text{base }=150\div 100=1.5\text{ m}$

Now we can calculate the area of the triangle:

$\text{area }=\dfrac{1}{2}bh=\dfrac{1}{2}\times0.8\times1.5=0.6\text{ m}^2$

Therefore, we get that the force being applied is

$F=p\times A=40\times 0.6=24\text{ N}$

## HR9 – Rates of Change

QUESTION: For the function sketched on the adjacent graph, find the instantaneous rate of change of $y$ with respect to $x$ at the point $x=3$.

ANSWER: We must draw a tangent to the graph at the point $x=3$ and find its gradient. Firstly, that tangent looks like

Calculating the gradient of the tangent line (orange), we get

$\text{gradient }=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{3-5}{4.8-2.6}=-0.91\text{ (2dp)}$

Therefore, the instantaneous rate of change of $y$ with respect to $x$ at $x=3$ is $-0.91$, to 2dp.

Note: your answer may vary slightly – anything between $-0.8$ and $-1$ is acceptable.