# Solving the toughest 2023 PSLE Maths question on Ratios and Fractions

- Tags:
- Primary Maths

Mathematics can be a challenging subject, and one of the most formidable aspects for many students is understanding and working with ratios and fractions. In the 2023 PSLE exam, students encountered a particularly tough maths question that put their knowledge to the test. In this article, we'll explore the intricacies of ratios and fractions and provide you with a step-by-step guide to conquer this challenging question.

## Understanding Ratios And Fractions

Ratios are a way to compare two or more quantities, expressing their proportional relationship. Fractions, on the other hand, represent a part of a whole. In many maths problems, including the 2023 PSLE question we will discuss below, these concepts intertwine, and a solid grasp of them is essential.

### Tackling the complex 2023 PSLE Maths Question on Ratios and Fractions

#### Question:

In the figure, \(\text{ADGE}\) is a rectangle where \(\mathrm{AB = BC = CD}\). \(\text{CE}\) intersects \(\text{BG}\) at \(\text{F}\). Given that the area ratio of \(\mathrm{\triangle AEC}\) to \(\mathrm{\triangle FBC}\) is \(8:1\), what fraction of rectangle \(\text{ADGE}\) is shaded?

#### Solution:

Let's look at the question,

We're told that \(\mathrm{\triangle AEC : \triangle FBC = 8 : 1}\)

So, if the area of \(\mathrm{\triangle FBC = 1}\) unit,

Then the area of \(\text{AEFB}\) would be \(8 - 1 = 7\) units.

Since \(\text{ADGE}\) is a rectangle, \(\text{AE}\) would be the same height as \(\text{GD}\).

This means that,

Area of \(\text{AEFB}\) = Area of \(\text{GFCD = 7}\) units.

So, if we add the area of \(\text{AEFB}\), area of \(\text{GFCD}\), and area of \(\mathrm{\triangle FBC}\), we get the area of the shaded area.

Therefore, the area of shaded area = \(7+7+1=15\) units.

Now, if we just knew the area of rectangle \(\text{ADGE}\), we'd be able to find the exact area of the shaded area.

How do we find that?

We know that \(\mathrm{AB = BC = CD}\) and that the area of \(\mathrm{\triangle AEC : \triangle FBC = 8 : 1}\).

Let's imagine that there was a straight line that went up straight from point \(\text{B}\). And let the point at which that line would have touched line \(\text{EG}\) be \(\text{K}\).

So, the rectangle \(\mathrm{KBDG}\) would be \(\displaystyle{\frac{2}{3}}\)rd of the rectangle \(\mathrm{AEGD}\), and \(\mathrm{\triangle KGB}\) would be 8 units.

This means that the area of \(\displaystyle{\frac{2}{3}}\)rd of the rectangle is \(8+1+7 = 16\) units.

And \(\displaystyle{1 \over 3}\)rd of the rectangle would be \(\displaystyle{\frac{16}{2}= 8}\) units.

This also means that the area of \(\displaystyle{3 \over 3}\) of the rectangle will be \(\displaystyle{8 \times 3 = 24}\) units.

We already know that the area of the shaded figure is 15 units.

The fraction of the shaded figure \(\displaystyle{={15 \over 24} = {5\over 8}}\).

**Answer:**

\(\displaystyle{\frac{5}{8}}\)

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