Question

If the mean of the following grouped data is \(21\): then \(k\) is one of the roots of the equation:

• A. \(2x^2-23x-10=0\)
• B. \(4x^2-35x+24=0\)
• C. \(2x^2-19x-10=0\)
• D. \(2x^2-35x+98=0\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For grouped data, the mean is given by \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\), where \(x_i\) is the mid-point of the class intervals. We first find \(k\) and then check which quadratic equation it satisfies.
Step 2: Key Formula or Approach:
1. Midpoints (\(x_i\)): 7.5, 12.5, 17.5, 22.5, 27.5, 32.5.
2. Frequencies (\(f_i\)): 2, \(k\), 28, 54, \(k+1\), 5.
Step 3: Detailed Explanation:
Calculate \(\sum f_i\):
\(\sum f_i = 2 + k + 28 + 54 + (k+1) + 5 = 2k + 90\).
Calculate \(\sum f_i x_i\):
\(\sum f_i x_i = (7.5 \times 2) + (12.5 \times k) + (17.5 \times 28) + (22.5 \times 54) + 27.5(k+1) + (32.5 \times 5)\)
\(\sum f_i x_i = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5\)
\(\sum f_i x_i = 40k + 1910\).
Set mean to 21:
\[ \frac{40k + 1910}{2k + 90} = 21 \]
\[ 40k + 1910 = 21(2k + 90) \]
\[ 40k + 1910 = 42k + 1890 \]
\[ 2k = 20 \Rightarrow k = 10 \].
Now, test \(k=10\) in the options:
(C) \(2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0\). (Correct)
Step 4: Final Answer:
The value of \(k\) is 10, which is a root of the equation \(2x^2 - 19x - 10 = 0\).