JEE Main20262 April 2026Morning ShiftMathematicsStatisticsActual
If the mean of the data Class 5-10 10-15 15-20 20-25 25-30 30-35 Frequency 2 k 28 54 k+1 5 is 21 , then k is one of the roots of the equation :
Options
- A2x^2 - 23x - 10 = 0
- B4x^2 - 35x + 24 = 0
- C2x^2 - 19x - 10 = 0
- D2x^2 - 35x + 98 = 0
Correct answer
C. 2x^2 - 19x - 10 = 0
Step-by-step solution
The class marks (mid-points) x_i for the given classes are 7.5, 12.5, 17.5, 22.5, 27.5, 32.5 . The sum of the frequencies is: f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 2k + 90 The sum of the products f_i x_i is: f_i x_i = 2(7.5) + k(12.5) + 28(17.5) + 54(22.5) + (k + 1)(27.5) + 5(32.5) f_i x_i = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 f_i x_i = 40k + 1910 The mean is given as 21 . Using the formula for the mean x = f_i x_i f_i : 40k + 1910 2k + 90 = 21 40k + 1910 = 21(2k + 90) 40k + 1910 = 42k + 1890 2k = 20 k =