Question

A box contains 6 cricket balls, 5 tennis balls and 4 rubber balls. Of these, some balls are defective. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than the proportion of defective rubber balls.
Moreover, the overall proportion of defective balls is twice the proportion of defective tennis balls. What BEST can be said about the number of defective rubber balls in the box?

• A. It is either 2 or 3
• B. It is exactly 2
• C. It is exactly 3
• D. It is either 3 or 4
• E. It is either 0 or 1

Answer 1

The Correct Option is C

Given the problem, let's define the variables:
Let \( x \) be the number of defective cricket balls, 
\( y \) be the number of defective tennis balls,
\( z \) be the number of defective rubber balls.
We have the following counts:
Cricket balls: 6,
Tennis balls: 5,
Rubber balls: 4.

Total balls = 6 + 5 + 4 = 15.

The conditions given are:

1. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than that of defective rubber balls:
   \[\frac{x}{6} > \frac{y}{5} \] and \[\frac{x}{6} < \frac{z}{4}\]
2. The overall proportion of defective balls is twice the proportion of defective tennis balls:
   \[\frac{x+y+z}{15} = 2 \times \frac{y}{5}\]

From the second condition:
\[\frac{x+y+z}{15} = \frac{2y}{5}\]
Simplifying, we get:
\[x+y+z = 6y\]
Rearranging gives us:
\[x+z = 5y \tag{1}\]

Analyzing Equation (1):
\(x\) and \(y\) must be integers as they represent counts of balls. We also need to satisfy the conditions:
\[\frac{x}{6} > \frac{y}{5}\] and \[\frac{x}{6} < \frac{z}{4}\].
Substitute expressions for \(z\) from \(x+z=5y\) into inequalities:

• Using \(\frac{x}{6} > \frac{y}{5}\), multiply by 30 to remove fractions:
  \[5x > 6y \tag{2}\]
• Substituting \(z = 5y-x\) in \[\frac{x}{6} < \frac{5y-x}{4}\]
  \[4x < 6(5y - x)\]
• Rearranging gives:
  \[4x < 30y - 6x\]
  \[10x < 30y\]
  \[x < 3y \tag{3}\]

Considering integer solutions that satisfy constraints from (2) and (3), particularly \(5x > 6y\) and \(x < 3y\), we test possible \(y\):
 

• For \(y=1\):
  inequalities give no valid \(x\) since \(5x > 6\) and \(x < 3\).
• For \(y=2\):
  \(5x > 12\), therefore \(x \geq 3\)
  \(x < 6\)
  Giving \(x = 3\) as solution.

Thus if \(y=2\) then \(x=3\), \(z=5y-x=10-3=7\), doesn't exist since \(z\) should be 4 or less.

• For \(y=3\):
  \(5x > 18\), i.e., \(x \geq 4\)
  Also \(x < 9\) gives possibility at \(x=4\)
  So, \(z=5y-x=15-4=11\) is invalid.

The only solution satisfying the constraints is \(y=1\), \(x=2\) and \(z=3\). Thus, the number of defective rubber balls is exactly 3.