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MCQEasyJEE 2025Measures of Central Tendency

JEE Mathematics 2025 Question with Solution

Marks obtained by all the students of class 1212 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 1414 with median class interval 12−1812-18 and median class frequency 1212. If the number of students whose marks are less than 1212 is 1818, then the total number of students is:

  • A

    4848

  • B

    4444

  • C

    4040

  • D

    5252

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Median = 1414, median class interval = 12−1812-18, median class frequency = 1212, and the number of students whose marks are less than 1212 is 1818.

Find: The total number of students, say NN.

For grouped data, the median formula is

Median=L+(N/2−Ff)×h\text{Median} = L + \left( \frac{N/2 - F}{f} \right) \times h

where L=12L=12, F=18F=18, f=12f=12, and class width h=18−12=6h=18-12=6.

Substituting the given values,

14=12+(N/2−1812)×614 = 12 + \left( \frac{N/2 - 18}{12} \right) \times 6

So,

14−12=(N/2−1812)×614 - 12 = \left( \frac{N/2 - 18}{12} \right) \times 6 2=(N/2−1812)×62 = \left( \frac{N/2 - 18}{12} \right) \times 6 26=N/2−1812\frac{2}{6} = \frac{N/2 - 18}{12} 13=N/2−1812\frac{1}{3} = \frac{N/2 - 18}{12}

Therefore,

12×13=N/2−1812 \times \frac{1}{3} = N/2 - 18 4=N/2−184 = N/2 - 18 N/2=22N/2 = 22 N=44N = 44

Therefore, the total number of students is 4444. The correct option is B.

Stepwise Algebra

Given: Median = 1414, lower limit of median class ℓ=12\ell = 12, cumulative frequency before the median class F=18F = 18, frequency of median class f=12f = 12, class width h=6h = 6.

Find: Total number of observations NN.

Using

Median=ℓ+(N2−Ff)×h\text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h

we get

14=12+(N2−1812)×614 = 12 + \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 2=(N2−1812)×62 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 2=N2−1822 = \frac{\frac{N}{2} - 18}{2} 4=N2−184 = \frac{N}{2} - 18 N2=22\frac{N}{2} = 22 N=44N = 44

Hence, the total number of students is 4444.

Common mistakes

  • Using the wrong cumulative frequency before the median class. Here, the number of students with marks less than 1212 is the cumulative frequency F=18F=18, not the median class frequency. Use FF as the frequency before the median class and ff as the frequency of the median class.

  • Taking the class width incorrectly. For the median class interval 12−1812-18, the class width is h=18−12=6h=18-12=6. Using any other value changes the equation and gives the wrong total number of students.

  • Confusing the lower limit of the median class with some other value. In the given formula, LL or ℓ\ell must be the lower limit of the median class, which is 1212 here. Substituting 1818 or the class mark would be incorrect.

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