Learn how to identify one of the most common Abstract Reasoning patterns

In this article, we’re going to share the 5 step process for solving Ste A/Set B questions.

Students often find Abstract Reasoning to be the most challenging subtest of the UCAT.

However, if you are familiar with the **common patterns and rules**, it will be easier to identify them under time pressure.

Remember this **5 step strategy **for solving Set A/Set B questions for the Abstract Reasoning subtest:

- Concentrate on
**one of the sets**first. - Start with the
**simplest square**(if there is such a square). **Compare**the simplest square to the**next simplest square(s)**to identify the pattern.- If you succeed in seeing the pattern, check with the remaining squares to
**confirm the pattern**. If you cannot see the pattern, move on to the other set and repeat steps 1 to 4. - The pattern or rules for both sets will be different but
**of the same type**. Once you identify the pattern in one set, you can look for a pattern of the same type in the other set.

One of the most common types of Abstract Reasoning patterns is shown in this example.

Have a go at identifying the pattern and picking whether each of these 5 test shapes fits within **set A, set B or neither set**.

**Answer:** This is an example of an equivalence question where each of the shapes is equivalent to a certain number.

First, we start with the simplest square in Set A which is the middle right square.

This is the simplest square because there is only one type of shape in it. There are 7 pentagons, so we can assume that **pentagon = 1**.

Therefore, the **sum of each square must be 7**.

Then, we compare this square to the next simplest square which is the bottom left, because it contains pentagons and one other shape.

If we assume that the sum of each Set A square is 7, there are 4 pentagons so **triangle = 3**.

Lastly, we can apply the same logic to the top right square. The sum should be 7 and there are 3 pentagons so we can deduce that **circle = 2**.

If we assign the same values to the pentagon, circle and triangle in **set B**, we discover that the **sum of each square is 9**.

To summarise, pentagon = 1, circle = 2 and triangle = 3.

The sum of each square in set A is 7 and the sum of each square in set B is 9.

**Test shape 1:** There are 4 circles. Sum = 4 x 2 = 8; this fits in neither set.

**Test shape 2:** There is 1 circle and 2 triangles. Sum = 2 + 2 x 3 = 8; this fits in neither set.

**Test shape 3:** There are 9 pentagons. Sum = 9 x 1 = 9; this fits in set B.

**Test shape 4:** There are 2 pentagons, 1 circle and 1 triangle. Sum = 2 x 1 + 2 + 3 = 7; this fits in set A.

**Test shape 5:** There is 1 pentagon, 1 circle and 2 triangles. Sum = 1 + 2 + 2 x 3 = 9; this fits in set B.