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 :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q8 – JEE-MAIN 2026

← Back

Home Maths StatisticsQuestion #424

If the mean of the data

\begin{array}{ccccccc} Class & 5 - 10 & 10 - 15 & 15 - 20 & 20 - 25 & 25 - 30 & 30 - 35 \\ Frequency & 2 & k & 28 & 54 & k + 1 & 5 \end{array}

is 21, then

k

is one of the roots of the equation :

Clue 1 of 3: Understand Grouped Data

Recall the formula for the mean of grouped data, which involves class marks and frequencies.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate Class Marks and Sums of Frequencies and Products✦ Active

First, determine the class mark ( $x_{i}$ ) for each class interval. The class mark is the midpoint of the interval. Then, calculate the sum of frequencies ( $\sum f_{i}$ ) and the sum of the products of frequency and class mark ( $\sum f_{i} x_{i}$ ). The class marks are:

x_{i} : 7.5, 12.5, 17.5, 22.5, 27.5, 32.5

The sum of frequencies is:

\sum f_{i} = 2 + k + 28 + 54 + (k + 1) + 5 = 90 + 2 k

The sum of $f_{i} x_{i}$ is:

\sum f_{i} x_{i} = (2 \times 7.5) + (k \times 12.5) + (28 \times 17.5) + (54 \times 22.5) + ((k + 1) \times 27.5) + (5 \times 32.5)

= 15 + 12.5 k + 490 + 1215 + 27.5 k + 27.5 + 162.5

= (15 + 490 + 1215 + 27.5 + 162.5) + (12.5 k + 27.5 k)

= 1910 + 40 k

Step 2: Calculate the Value of k○ Expand

Given that the mean of the data is 21, use the formula for the mean of grouped data:

Mean = \frac{\sum f_{i} x_{i}}{\sum f_{i}}

21 = \frac{1910 + 40 k}{90 + 2 k}

21 (90 + 2 k) = 1910 + 40 k

1890 + 42 k = 1910 + 40 k

42 k - 40 k = 1910 - 1890

2 k = 20

k = 10

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid errors when solving for $k$ .

Step 3: Identify the Correct Quadratic Equation○ Expand

Substitute $k = 10$ into each of the given quadratic equations to find which one has $k$ as a root. Let $x = 10$ :

1. $2 x^{2} - 23 x - 10 = 2 (10)^{2} - 23 (10) - 10 = 200 - 230 - 10 = - 40 \neq 0$

2. $4 x^{2} - 35 x + 24 = 4 (10)^{2} - 35 (10) + 24 = 400 - 350 + 24 = 74 \neq 0$

3. $2 x^{2} - 19 x - 10 = 2 (10)^{2} - 19 (10) - 10 = 200 - 190 - 10 = 0$

4. $2 x^{2} - 35 x + 98 = 2 (10)^{2} - 35 (10) + 98 = 200 - 350 + 98 = - 52 \neq 0$

The equation $2 x^{2} - 19 x - 10 = 0$ is satisfied when $x = 10$ . Therefore, $k = 10$ is a root of this equation.

💡 Teacher's Secret Hint

Remember to substitute the calculated value of $k$ into the variable $x$ in the quadratic equations provided in the options.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 8 – JEE-MAIN 2026

Clue 1 of 3: Understand Grouped Data

🧠 Full Solution Path