In your science and maths problems, you will often meet very large or very small numbers. Writing all those zeros is tedious and can lead to mistakes. This is where Standard Form comes in. It is a neat way to write numbers, especially the big and small ones, using powers of 10.

Let's convert a large number. Take 93,000,000.

First, we find 'A'. Place a decimal point after the first non-zero digit. Here, that gives us 9.3.

Next, we find 'n'. Count how many places you moved the decimal point from its original position (at the end of the number) to its new position. We moved it 7 places to the left. So, $n = 7$.

Putting it together: $93,000,000 = 9.3 \times 10^7$.

Now for a small number, like 0.0052.

Again, find 'A' by placing the decimal after the first non-zero digit: 5.2.

To find 'n', count how many places you moved the decimal point to the right. We moved it 3 places. Because we are moving it to the right for a small number, the power will be negative. So, $n = -3$.

This gives us: $0.0052 = 5.2 \times 10^{-3}$.

We are used to counting in base ten, using digits from 0 to 9. The binary number system works in base two. It only uses two digits: 0 and 1. This system is the foundation of all modern computers.

In the BECE, you will be asked to convert numbers between base ten and base two.

To convert a base ten number to base two, we use repeated division by 2. Let’s convert $25_{10}$ to base two.

We divide 25 by 2 and record the remainder each time. We continue until the result is 0.

Now, read the remainders from the bottom up to get the answer: $11001_2$. So, $25_{10} = 11001_2$.

To convert from base two back to base ten, we use place values, which are powers of 2. Let's convert $1011_2$ to base ten.

Now we calculate the value: $(1 \times 8) + (0 \times 4) + (1 \times 2) + (1 \times 1)$ $= 8 + 0 + 2 + 1$ $= 11$

So, $1011_2 = 11_{10}$.

Ratios and percentages are used everywhere, from sharing items to calculating discounts in a shop. The BECE loves to test these with word problems.

A ratio compares quantities. If a recipe needs 2 cups of flour for every 1 cup of sugar, the ratio of flour to sugar is 2:1. The total parts in this ratio is $2 + 1 = 3$.

Let's say you need to share $90 between Kofi and Ama in the ratio 2:3. First, find the total number of parts: $2 + 3 = 5$ parts. Then, find the value of one part: $\text{\textdollar}90 \div 5 = \text{\textdollar}18$. Now, calculate each share: Kofi's share (2 parts): $2 \times \text{\textdollar}18 = \text{\textdollar}36$. Ama's share (3 parts): $3 \times \text{\textdollar}18 = \text{\textdollar}72$. Wait, that's not right. $3 \times \text{\textdollar}18 = \text{\textdollar}54$. Always double check your multiplication! $36 + 54 = 90$. Perfect.

Percentages are just a special type of ratio where the total is always 100. You will often face problems about profit, loss, and discounts.

• Profit = Selling Price (SP) - Cost Price (CP)
• Loss = Cost Price (CP) - Selling Price (SP)
• Percentage Profit or Loss = $(Profit\ or\ Loss \div Cost\ Price) \times 100\%$

Example: A trader buys a phone for $500 and sells it for $650. What is the percentage profit?

Profit = $650 - 500 = \text{\textdollar}150$. Percentage Profit = $(150 \div 500) \times 100\%$.

To solve this without a calculator, we can simplify the fraction: $150/500 = 15/50 = 3/10$. Now, $(3/10) \times 100 = 30$. So the percentage profit is 30%.

The Highest Common Factor (HCF) and Lowest Common Multiple (LCM) are important for working with fractions and solving certain types of problems. A reliable way to find them is with prime factorization.

Every whole number greater than 1 is either a prime number or can be written as a unique product of prime numbers. This is a very powerful idea in mathematics. To find the prime factors of a number, we break it down using a factor tree or repeated division.

Let's find the prime factors of 36 and 48. For 36: $36 = 2 \times 18 = 2 \times 2 \times 9 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$. For 48: $48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1$.

To find the HCF, we take the lowest power of each common prime factor and multiply them.

The common prime factors are 2 and 3. The lowest power of 2 is $2^2$. The lowest power of 3 is $3^1$. HCF = $2^2 \times 3^1 = 4 \times 3 = 12$.

To find the LCM, we take the highest power of every prime factor present in either number and multiply them.

The prime factors present are 2 and 3. The highest power of 2 is $2^4$. The highest power of 3 is $3^2$. LCM = $2^4 \times 3^2 = 16 \times 9 = 144$.

Now, let's test what you've learned. These questions are similar in style to what you will see in Section A of the exam.

Mastering these topics is a big step towards success in your BECE maths paper. Practice these methods until they become second nature.