To solve this problem, we need to determine the number of balls in each of the three bags, I, II, and III, given that together they contain a total of 10 balls. Let's define:
Let \( x \) be the number of balls in Bag I,
\( y \) be the number of balls in Bag II,
\( z \) be the number of balls in Bag III.
The relationship between these is \( x + y + z = 10 \).
Statement (1): Bag I contains five balls more than Bag III.
This can be expressed as \( x = z + 5 \).
Substituting into the total equation:
\( (z + 5) + y + z = 10 \)
Simplifying gives \( 2z + y + 5 = 10 \) or \( 2z + y = 5 \).
With this equation alone, we cannot uniquely determine the values of \( x \), \( y \), and \( z \) because different combinations of \( z \) and \( y \) will satisfy the equation.
Therefore, statement 1 alone is insufficient.
Statement (2): Bag II contains half as many balls as Bag I.
This relationship can be expressed as \( y = \frac{x}{2} \).
Substituting into the total equation:
\( x + \frac{x}{2} + z = 10 \)
This simplifies to \( \frac{3x}{2} + z = 10 \) or \( 3x + 2z = 20 \).
With this equation alone, similar to statement 1, we have multiple solutions because different pairs of \( x \) and \( z \) can satisfy this equation.
Thus, statement 2 alone is insufficient.
Combining Statements (1) and (2):
From statement (1): \( x = z + 5 \)
From statement (2): \( 3x + 2z = 20 \)
Substitute \( x = z + 5 \) from statement (1) into statement (2):
\( 3(z + 5) + 2z = 20 \)
Simplifying, \( 3z + 15 + 2z = 20 \)
This simplifies to \( 5z = 5 \), yielding \( z = 1 \).
Substitute back \( z = 1 \) into \( x = z + 5 \), thus \( x = 6 \).
Substituting \( x = 6 \) into \( y = \frac{x}{2} \), gives \( y = 3 \).
Hence, using both statements together, we can determine the exact number of balls in each bag as:
Bag I: 6 balls, Bag II: 3 balls, Bag III: 1 ball.
The correct option is: BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.