Turkish wrote:

A box contains a total of 56 balls of red and blue color in the ratio of 3: 5. What is the least number of red balls that should be removed from the box, such that the ratio of red and blue balls is less than 2 to 7?

a.9

b. 10

c.11

d.12

e.13

In questions like this, it helps to represent the given information visually in a rough diagram.

Let's first represent the initial situation visually:

Since the Red: Blue balls are in the ratio 3:5, we can write their number as 3x and 5x respectively.

So, total number of balls = 3x + 5x = 56 (given)

=> 8x = 56

=> x = 7

So, initial number of red balls = 3x = 3*7 = 21

And, initial number of blue balls = 5x = 5*7 = 35

Let's now look at the final situation:

Let y balls be removed from the red balls.

So, the final situation looks like this:

So, the final ratio of Red: Blue balls =\(\frac{21-y}{35}\)

The question requires that:

\(\frac{21-y}{35}\) < \(\frac{2}{7}\)

Multiplying both sides of this inequality with 35 (a positive number) will not change the sign of inequality. We'll get:

21-y < 10

Adding y - 10 to both sides of the inequality, we get:

11 < y

That is, y > 11

Since y denotes

the number of balls, y must be a positive integer.

Therefore, the smallest possible value of y is 12.

The

takeaway that I hope you draw from this alternate solution is:

In questions that involve multiple quantities and changes done to those quantities, representing the given information visually helps in getting a better understanding of how to solve the question.Hope you found this little tip useful!

Best Regards

Japinder

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