Numbers Foundation Revision Card Answers | Maths Made Easy

Numbers Foundation Revision Card Answers

FN1 – Factors and Multiples.

QUESTION: List all the factors of 54.

ANSWER: We want to look for factor pairs, i.e. pairs of numbers that multiply to make 54. We get


Therefore, the full list of factors of 54 is: 1, 2, 3, 6, 9, 27, and 54.


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:


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.

FN3 – Types of Numbers – Prime Numbers

QUESTION: Which of the following numbers are prime numbers?

πŸ•, 𝟏𝟐, πŸπŸ—, πŸ‘πŸ–, πŸ“πŸ”, πŸ”πŸ•, πŸ–πŸ—

ANSWER: Looking at the table on the card we can see the following numbers are prime

πŸ•, 𝟏𝟐, πŸπŸ—, πŸ‘πŸ–, πŸ“πŸ”, πŸ”πŸ•, πŸ–πŸ—

FN4 – Types of Numbers – Squares and Cubes

QUESTION: Work out the value of 5^2\times 2^3.


Firstly, 5^2=5\times5=25.

Secondly, 2^3=2\times2\times2=8

Finally, 25\times8=200.

Final Answer = 200

FN5 – Types of Numbers – Square and Cube roots

QUESTION: Work out the value of \sqrt[2]{49}\times\sqrt[3]{27}.


First, \sqrt[2]{49}= 7

Next, \sqrt[3]{27} = 3

So finally, 7\times 3 = 21

So our final answer = 21

FN6 – 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


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


FN7 – LCM and HCF

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




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

FN8 – 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:


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


Now we can do the subtraction. We get:


FN9 – 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


Then, we do the multiplication:


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


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

FN10 – Fractions 3: Mixed Numbers and Fractions of Amounts

QUESTION: Find \dfrac{5}{6} of 96. Write your answer in its simplest form.

ANSWER: To find a fraction of a number, divide the number by the denominator and multiple by the numerator:

\dfrac{5}{6}\text{ of }96= (96 \div 6) \times 5

First,Β 96 \div 6 =16

Next,Β 16 \times 5 = 80

Final answer = 80

FN11 – Fractions, Decimals, and Percentages

QUESTION: Write the following fraction as a decimal and a percentage


ANSWER: Firstly, we we know that \dfrac{1}{4} = 25\%, so \dfrac{1}{8}= 12.5\%

Now we need to work outΒ \dfrac{5}{8}= 12.5\% \times 5 = 62.5\%

Last we need to convertΒ 62.5\% into a decimal. We do this by dividing by 100.

62.5 \div 100 = 0.625

Final answer

\dfrac{5}{8} = 62.5\% = 0.625

FN12 – Rounding – Decimal Places & Significant Figures


i) Round 7.789 to 1 decimal place.

ii) Round 0.0595 to 2 significant figures.


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)

FN13 – 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 β€œdistance = 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.

FN14 – Error Bounds

QUESTION: Bobby’s measures the mass, m, of his cat to be 6.74kg to 3 significant figures. Write down the error bound for the actual value of m.

ANSWER: The smallest value that, when rounded to 3sf, rounds up to 6.74 would be 6.735. The biggest value that, when rounded to 3sf, rounds down to 6.74 would be 6.745. Therefore, we get the error bound for m to be

6.735\leq m<6.745

FN15 – Standard Form


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

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


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. This is a big number, so its power is positive, and we get

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