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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