## HN1 – Probability Basics

**QUESTION:** Evaluate (13+2)\times(36 \div 3^2).

**ANSWER:** There are two sets of brackets, so our first step should be to evaluate them both. Which one we do first doesn’t matter, so here we’ll choose the left one first. It only contains one operation, so we get

13+2=15

Now, the bracket on the right contains two operations: an index/power, and a division. We do the index first followed by the division:

36\div 3^2=36\div 9=4

Therefore, we only have 1 operation left in our calculation (the multiplication between the two brackets), so we get the answer to be

15 \times 4 = 60.

## HN2 – Prime Factor Trees.

**QUESTION:** Find the prime factorisation of 135. Write your answer in index notation.

**ANSWER: **We will use a prime factor tree, first splitting 135 into 5 and 27. 5 is prime, so we circle it, but 27 is not, so we split it into 9 and 3.

3 is prime, so we circle, but 9 is not, so we split it into 3 and 3.

As we established, 3 is prime, so we circle both of these values, and so we have finished our tree. Therefore, the prime factorisation of 135 is

135=3\times3\times3\times5

We can write 3\times3\times3 as 3^3, the answer written using index notation is

135=3^3\times5.

## HN3 – HCF & LCM

**QUESTION:** Find the HCF and LCM of 264 and 110.

**ANSWER: **Firstly, we need to find the prime factorisations of 264 and 110. Here, we will do this using prime factor trees.

So, we get that

264=2\times2\times2\times3\times11

and

110=2\times5\times11

Then, we need to draw a Venn diagram with one circle for prime factors of 264 and another for prime factors of 110. Then, the first step to filling in this diagram is to look for any prime factors that 264 and 110 have in common. For each shared prime factor, we will cross it off both factor lists, and then write it once in the intersection of the two circles.

After all shared factors are crossed off, write the rest of the prime factors in their appropriate circles. The result should look something like the diagram below.

Then, we find the HCF by multiplying the numbers in the intersection:

\text{HCF }=2\times11=22

And find the LCM by multiplying all the numbers in the Venn diagram together:

\text{LCM }=2\times2\times2\times3\times5\times11=1,320

## HN4 – Fractions 1: Adding & Subtracting Fractions

**QUESTION:** Evaluate \dfrac{3}{7}-\dfrac{6}{5}.

**ANSWER:** To find a common denominator, multiply the denominators: 7\times 5=35. To make the denominator of the first fraction 35, we need to multiply its top and bottom by 5:

\dfrac{3}{7}=\dfrac{3\times5}{7\times5}=\dfrac{15}{35}

To make the denominator of the second fraction 35, we need to multiply its top and bottom by 7:

\dfrac{6}{5}=\dfrac{6\times7}{5\times7}=\dfrac{42}{35}

Now we can do the subtraction. We get:

\dfrac{15}{35}-\dfrac{42}{35}=-\dfrac{27}{35}

## HN5 – Fractions 2: Multiplying & Dividing Fractions

**QUESTION:** Evaluate \dfrac{12}{7}\div\dfrac{9}{20}.

**ANSWER:** To divide these fractions, we need to keep the first fraction the same, change the \div to a \times, and flip the second fraction upside down. Doing so, we get

\dfrac{12}{7}\div\dfrac{9}{20}=\dfrac{12}{7}\times\dfrac{20}{9}

Then, we do the multiplication:

\dfrac{12}{7}\times\dfrac{20}{9}=\dfrac{12\times20}{7\times9}=\dfrac{240}{63}

240 and 63 both have 3 as a factor, so we can simplify our fraction to get

\dfrac{240}{63}=\dfrac{80}{21}

80 and 21 have no common factors, so we are done.

## HN6 – Fractions 3: Mixed Numbers and Fractions of Amounts

**QUESTION:** Find \dfrac{5}{6} of 98. Write your answer in its simplest form.

**ANSWER:** To find a fraction of a number, we must multiply the fraction by the number. To make this easier, we will write 98 like \frac{98}{1}. Doing so, we get

\dfrac{5}{6}\text{of}98=\dfrac{5}{6}\times\dfrac{98}{1}

Then, doing the multiplication, we get

\dfrac{5}{6}\times\dfrac{98}{1}=\dfrac{98\times5}{6\times1}=\dfrac{490}{6}

Then, both 490 and 6 have a factor of 2, so we can cancel this out to get our final answer in its simplest form:

\dfrac{490}{6}=\dfrac{245}{3}

## HN7 – Recurring Decimals to Fractions

**QUESTION:** Use algebra to write 4.\dot{3}0\dot{7} as a fraction.

**ANSWER:** Firstly, set x=4.\dot{3}0\dot{7}. Then, if we multiply this by 1000, we get

1000x=4,307.\dot{3}0\dot{7}

We see that x and 1000x are the same after the decimal point. So, if we subtract x from 1000x, we get

1000x-x=4,307.\dot{3}0\dot{7}-4.\dot{3}0\dot{7}=4,303

Therefore,

999x=4,303

Finally, dividing both sides by 999, we get

x=\dfrac{4,303}{999}

## HN8 – Rounding – Decimal Places & Significant Figures

**QUESTION:**

i) Round 7.789 to 1 decimal place.

ii) Round 0.0595 to 2 significant figures.

**ANSWER:**

i) In 7.789, the cut-off digit (the 1st decimal place) is the second 7. The digit after this is an 8, meaning we round the 7, and get the answer: 7.8.

ii) In 0.0595, the cut-off digit (the 2nd significant figure) is the 9. The digit after this is a 5, meaning that we round the 9 up to 0, and add an extra 1 to the digit before the 9. Doing so, we get the answer: 0.060. (If you just got 0.06, this is correct)

## HN9 – Estimation

**QUESTION:** Mateo drives at an average speed of 45.28 miles per hour for 2.69 hours. Estimate the distance he covered over the course of this journey, and state whether your answer is an overestimate or an underestimate.

**ANSWER:** Firstly, recall that “distance = speed \times time”. Therefore, rounding both values to 1 significant figure, we get the estimated distance covered to be

50 \times 3 = 150\text{ miles}

In this case, both the estimated values are bigger than the actual values. So, since we are multiplying together two bigger values, our result will be bigger. Therefore, this is an overestimate of the distance covered by Mateo during this journey.

## HN10 – Error Bounds

**ANSWER:** The area of a triangle is \frac{1}{2}bh. Since we will be multiplying the side-lengths together, if we want the lower bound for the area, we need the smallest possible values of the side-lengths – i.e., their lower bounds.

The lower bound for the height, when rounded to 2dp, is 38.965.

The lower bound for the width, when rounded to 2dp, is 96.875.

Therefore, the lower bound for the area of the triangle is

\dfrac{1}{2}\times38.965\times96.875=1,887.3671...=1,887.37\text{mm (2 dp)}

## HN11 – Standard Form

**QUESTION: **

i) Write 300,950,000 in standard form.

ii) Write 1.997\times{10}^{-7} in decimal notation.

**ANSWER: **

i) We need to see how many times we must move the decimal point to make the number fall between 1 and 10. If we move it 8 times,, then it becomes 3.0095, which falls between 1 and 10. We are moving the decimal point 8 places to the left, so the power will be positive. (If we move to the right the power is negative)

300,950,000 = 3.0095\times10^8

ii) This is a negative power so want to end up with a small number – we must divide by 10 seven times. So, moving the decimal place 7 spaces to the left, we get

1.997\times10^{-7}=0.0000001997