Hello,

Would someone be able to help me with the following:

For the first one, I selected A as the answer and used the choosing numbers approach. I first did x = 3 and y = -1 and then x = 3 and y = -2. In both cases A was greater than B.

For the second one, I actually just couldn’t figure out how to approach it.

F

Try a value of y close to 0.

Put it this way: show that he can take 24 pencils and not have a single pencil of the fifth colour.

ohhhh that makes a lot of sense now. In general, when choosing numbers should we be looking at extremes? For example, a very big x value and a very small y value etc?

I am sorry I am still not following

For a QC question, that’s often a good option.

Let’s look at the question again. We want to take pencils from the box. As with the previous problem, this is again a question where we need to look at *extremes*.

Suppose he takes

- 6 pencils of the first colour
- 6 pencils of the second colour
- 6 pencils of the third colour
- 6 pencils of the fourth colour

Notice that

- he has take 24 colours in total, and
- we haven’t used a single pencil of the fifth colour

Can you extend this to solve the original problem?