Question

Given below are two statements
Statement I: The set of numbers (5, 6, 7, p, 6, 7, 8, q) has an arithmetic mean of 6 and mode (most frequently occurring number) of 7. Then p×q=16.
Statement II: Let p and q be two positive integers such that p+q+p×q=94. Then p+q=20.
In light of the above statements, choose the correct answer from the options given below

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is B

To solve the question, we must evaluate each statement separately for their validity.

Statement I

The given set of numbers is \( (5, 6, 7, p, 6, 7, 8, q) \), with an arithmetic mean of 6 and a mode of 7.

1. First, calculate the arithmetic mean:
   • The formula for the arithmetic mean of a set of numbers is: \[\text{Mean} = \frac{\text{Sum of all numbers}}{\text{Total number of numbers}}\]
   • Since the mean is given as 6: \[\frac{5 + 6 + 7 + p + 6 + 7 + 8 + q}{8} = 6\]
   • Simplifying, we have: \[39 + p + q = 48\]
   • This gives: \[p + q = 9\]
2. Now, consider the mode, which indicates the most frequently occurring number. In this set, the number 7 is repeated twice, which aligns with the given mode of 7.
3. Thus, the statement expects: \[p \times q = 16\]
4. Assuming possible values, if \( p = 1 \) & \( q = 8 \), or vice versa, \( p \times q = 8 \), which does not satisfy \( p \times q = 16 \).
5. Hence, Statement I is false.

Statement II

We are given \( p + q + p \times q = 94 \) and check if \( p + q = 20 \).

1. Given: \[ p + q + p \times q = 94\]
2. Substitute \( p + q = 20 \), then \[ 20 + p \times q = 94\], which implies:
3. \[ p \times q = 74\]
4. With \( p + q = 20 \) and \( p \times q = 74 \), solve the quadratic:
   • Use the equation: \[x^2 - 20x + 74 = 0\]
   • The discriminant is: \[\Delta = b^2 - 4ac = 400 - 296 = 104\]
   • \(\Delta\) is not a perfect square, so \( x \) is not an integer, hence \( p \) and \( q \) cannot be integers.
5. So, Statement II is also false.

As both statements are false, the correct answer is: Both Statement I and Statement II are false.