JEE Advanced 2025 · Paper 1 · Mathematics Q14

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MathematicsList match+4 / -1Question 14 of 48

Consider the following frequency distribution:

Value	4	5	8	9	6	12	11
Frequency	5	$f_1$	$f_2$	2	1	1	3

Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.

For the given frequency distribution, let $\alpha$ denote the mean deviation about the mean, $\beta$ denote the mean deviation about the median, and $\sigma^2$ denote the variance.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I

P
$7f_1 + 9f_2$ is equal to
Q
$19\alpha$ is equal to
R
$19\beta$ is equal to
S
$19\sigma^2$ is equal to

List-II

1
$146$
2
$47$
3
$48$
4
$145$
5
$55$

P → 5; Q → 3; R → 2; S → 4

P → 5; Q → 2; R → 3; S → 1

P → 5; Q → 3; R → 2; S → 1

P → 3; Q → 2; R → 5; S → 4

Answer and worked solution

Official answer: C

Frequencies. $5+f_1+f_2+2+1+1+3=19$ , so $f_1+f_2=7$ .

Median. In order the values are 4 (×5), 5 (× $f_1$ ), 6 (×1), 8 (× $f_2$ ), 9, 11, 12. The median is the 10th value. For it to be 6 we need $5+f_1<10\le6+f_1$ , so $f_1=4$ and $f_2=3$ .

(P) $7\cdot4+9\cdot3=55$ → (5).

Mean. $\frac{20+20+6+24+18+33+12}{19}=\frac{133}{19}=7$ .

(Q)
$19\alpha=\sum f|x-7|=5\cdot3+4\cdot2+1+3\cdot1+2\cdot2+3\cdot4+5=48 \;\to\; (3).$

(R)
$19\beta=\sum f|x-6|=5\cdot2+4\cdot1+0+3\cdot2+2\cdot3+3\cdot5+6=47 \;\to\; (2).$

(S)
$19\sigma^2=\sum f(x-7)^2=45+16+1+3+8+48+25=146 \;\to\; (1).$

The match is P→5, Q→3, R→2, S→1, which is option (C).

Topics:Variance and standard deviation Measures of central tendency

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JEE Advanced 2025 · Paper 1, Mathematics question 14