## 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 182cm 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 175cm.

## 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.

**ANSWER:**

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

**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

**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.