Today, we’re diving into an intriguing mathematical problem from the Merchant Taylors’ School Specimen Paper 1. The question might seem simple at first, but it requires a bit of thoughtful consideration. So, let’s break it down and solve it together.

## The Problem: Calculating 90% Red Balls

In this problem, we have a bag containing 50 balls—49 red balls and 1 blue ball. The goal is to determine how many red balls need to be removed to have 90% of the remaining balls be red.

**Given:**

- 49 red balls
- 1 blue ball

The question posed is: “How many red balls should be removed from the bag to contain 90% red balls?”

## Common Misconception: The Initial Reaction

At first glance, it might seem straightforward. Here’s how some might initially approach it:

- We know that 90% red balls means 90 out of 100. Since we don’t have 100 balls, another approach is to think of 90% of 50.
- If we consider 50 as our total number, 90% of 50 is 45.

So, if we want 90% of the balls to be red, we need 45 out of 50 to be red. This might lead some to think that removing 4 balls (49 – 45 = 4) would work. But let’s examine why this isn’t the correct approach.

## Why Removing Four Balls Doesn’t Work

Let’s break it down:

- If we remove 4 red balls, we’re left with 45 red balls and 1 blue ball, making a total of 46 balls.

However, 45 out of 46 is not 90%. It actually turns out to be approximately 97.83%.

“If you wrote four for this question, you wouldn’t get the marks.”

## Correct Approach: Fractions and Percentages

We need to go deeper into understanding fractions and percentages. We know that 90 out of 100 is proportional to 9 out of 10. This fraction (9/10) must remain consistent when we have fewer total balls.

Given:

- ( x ) red balls remaining
- 1 blue ball remaining

We need:[ \frac{x}{x+1} = 0.90 ]

This equation arises because if ( x ) is the number of red balls, then ( x + 1 ) is the total number of balls (including the 1 blue ball).

## Solving the Equation

Let’s solve for ( x ):

[ \frac{x}{x+1} = 0.90 ]

To clear the fraction, we use cross multiplication:

[ x = 0.90(x + 1) ]

[ x = 0.90x + 0.90 ]

Subtract ( 0.90x ) from both sides:

[ 0.10x = 0.90 ]

Divide both sides by 0.10:

[ x = 9 ]

This tells us that ( x = 9 ). So, for the percentage of red balls to be 90%, there must be exactly 9 red balls remaining.

## Conclusion: Removing the Right Number of Red Balls

Initially, we had 49 red balls. We need 9 red balls remaining. Therefore, the number of red balls to be removed:

**49 – 9 = 40**

So, **we need to remove 40 red balls** to achieve the goal of having 90% red balls in the bag.

By carefully considering the percentage and understanding the relationship between fractions and total quantities, we can solve this problem accurately.

## Final Thoughts

The key takeaway here is to step beyond the initial reaction and delve deeper into the problem’s requirements. It’s not merely about subtracting balls but understanding the underlying math.

Thanks for joining us in this mathematical adventure! If you have any questions or different approaches, feel free to discuss them in the comments.