CBSE & State Board Question Bank | 9th - 12th

3 mark Kandippa! Moonu questions (Q23, Q24, Q25)-ukkum complete Table Column format, highlighted missing frequency rows, matrum clear step-by-step calculations oda clean HTML code ready panni irukken:

Q5. Calculate the mean marks obtained by 100 students:

Marks: $0\text{–}10,\ 10\text{–}20,\ 20\text{–}30,\ 30\text{–}40,\ 40\text{–}50$
Students ($f_i$): $12,\ 23,\ 34,\ 25,\ 6$

Answer:

Class Interval (Marks)	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$0 - 10$	12	5	$12 \times 5 = 60$
$10 - 20$	23	15	$23 \times 15 = 345$
$20 - 30$	34	25	$34 \times 25 = 850$
$30 - 40$	25	35	$25 \times 35 = 875$
$40 - 50$	6	45	$6 \times 45 = 270$
Total	$\sum f_i = 100$	-	$\sum f_i x_i = 2400$

Formula:

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$\bar{x} = \frac{2400}{100}$$

Answer: Mean marks = $\mathbf{24\text{ marks}}$

Q6. Find the mean of the following frequency distribution:

Class: $0\text{–}15,\ 15\text{–}30,\ 30\text{–}45,\ 45\text{–}60,\ 60\text{–}75,\ 75\text{–}90$
Frequency ($f_i$): $17,\ 20,\ 18,\ 21,\ 15,\ 9$

Answer:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$0 - 15$	17	7.5	$17 \times 7.5 = 127.5$
$15 - 30$	20	22.5	$20 \times 22.5 = 450$
$30 - 45$	18	37.5	$18 \times 37.5 = 675$
$45 - 60$	21	52.5	$21 \times 52.5 = 1102.5$
$60 - 75$	15	67.5	$15 \times 67.5 = 1012.5$
$75 - 90$	9	82.5	$9 \times 82.5 = 742.5$
Total	$\sum f_i = 100$	-	$\sum f_i x_i = 4110$

Formula:

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$\bar{x} = \frac{4110}{100}$$

Answer: Mean = $\mathbf{41.1}$

Q9. The arithmetic mean of the following table is $50$. Find the missing frequencies $p$ and $q$ given that total frequency is $120$:

Class: $0\text{–}20,\ 20\text{–}40,\ 40\text{–}60,\ 60\text{–}80,\ 80\text{–}100$
Frequency ($f_i$): $17,\ p,\ 32,\ q,\ 19$

Answer:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$0 - 20$	17	10	$17 \times 10 = 170$
$20 - 40$	$p$	30	$30p$
$40 - 60$	32	50	$32 \times 50 = 1600$
$60 - 80$	$q$	70	$70q$
$80 - 100$	19	90	$19 \times 90 = 1710$
Total	$N = 120$	-	$\sum f_i x_i = 3480 + 30p + 70q$

Step 1: Relation between $p$ and $q$

$$\sum f_i = 17 + p + 32 + q + 19 = 120$$ $$68 + p + q = 120 \implies p + q = 52 \implies q = 52 - p \quad \text{--- (Equation 1)}$$

Step 2: Apply Mean Formula ($\text{Mean} = 50$)

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$50 = \frac{3480 + 30p + 70q}{120}$$ $$6000 = 3480 + 30p + 70q \implies 30p + 70q = 2520$$

Substitute $q = 52 - p$ in the above equation:

$$30p + 70(52 - p) = 2520$$ $$30p + 3640 - 70p = 2520$$ $$3640 - 2520 = 40p \implies 1120 = 40p \implies \mathbf{p = 28}$$

Now substitute $p = 28$ in Equation 1:

$$q = 52 - 28 \implies \mathbf{q = 24}$$

Answer: Missing frequencies are $\mathbf{p = 28}$ and $\mathbf{q = 24}$

Q10. Find the mean weight of 30 packet samples where weights are grouped as:

Weight (g): $40\text{–}45,\ 45\text{–}50,\ 50\text{–}55,\ 55\text{–}60,\ 60\text{–}65$
Packets ($f_i$): $2,\ 3,\ 8,\ 12,\ 5$

Answer:

Class Interval (Weight in g)	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$40 - 45$	2	42.5	$2 \times 42.5 = 85$
$45 - 50$	3	47.5	$3 \times 47.5 = 142.5$
$50 - 55$	8	52.5	$8 \times 52.5 = 420$
$55 - 60$	12	57.5	$12 \times 57.5 = 690$
$60 - 65$	5	62.5	$5 \times 62.5 = 312.5$
Total	$\sum f_i = 30$	-	$\sum f_i x_i = 1650$

Formula:

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$\bar{x} = \frac{1650}{30}$$

Answer: Mean weight = $\mathbf{55\text{ g}}$

3 Marks Q11. Determine the average score of a cricket team across matches given:

Runs: $0\text{–}20,\ 20\text{–}40,\ 40\text{–}60,\ 60\text{–}80,\ 80\text{–}100$
Matches ($f_i$): $4,\ 6,\ 12,\ 10,\ 8$

Answer:

Class Interval (Runs)	Frequency / Matches ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$0 - 20$	4	10	$4 \times 10 = 40$
$20 - 40$	6	30	$6 \times 30 = 180$
$40 - 60$	12	50	$12 \times 50 = 600$
$60 - 80$	10	70	$10 \times 70 = 700$
$80 - 100$	8	90	$8 \times 90 = 720$
Total	$\sum f_i = 40$	-	$\sum f_i x_i = 2240$

Formula:

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$\bar{x} = \frac{2240}{40}$$

Answer: Average score = $\mathbf{56\text{ runs}}$

3 Marks Q13. Find missing frequencies $x$ and $y$ if total frequency is $100$ and median is $525$:

Class: $0\text{–}100,\ 100\text{–}200,\ 200\text{–}300,\ 300\text{–}400,\ 400\text{–}500,\ 500\text{–}600,\ 600\text{–}700,\ 700\text{–}800,\ 800\text{–}900,\ 900\text{–}1000$
Frequency ($f_i$): $2,\ 5,\ x,\ 12,\ 17,\ 20,\ y,\ 9,\ 7,\ 4$

Answer:

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($cf$)
$0 - 100$	2	2
$100 - 200$	5	7
$200 - 300$	$x$	$7 + x$
$300 - 400$	12	$19 + x$
$400 - 500$	17	$36 + x$
$500 - 600$ (Median Class)	20	$56 + x$
$600 - 700$	$y$	$56 + x + y$
$700 - 800$	9	$65 + x + y$
$800 - 900$	7	$72 + x + y$
$900 - 1000$	4	$76 + x + y$
Total	$N = 100$	-

Step 1: Relation between $x$ and $y$

$$76 + x + y = 100 \implies x + y = 24 \quad \text{--- (Equation 1)}$$

Step 2: Apply Median Formula

Given $\text{Median} = 525$, which lies in the class interval $500\text{–}600$.
• Lower limit ($l$) = $500$
• Frequency ($f$) = $20$
• Cumulative frequency of preceding class ($cf$) = $36 + x$
• Class size ($h$) = $100$, and $\frac{N}{2} = 50$

$$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$$ $$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$$ $$25 = (14 - x) \times 5 \implies 5 = 14 - x \implies \mathbf{x = 9}$$

Substitute $x = 9$ in Equation 1:

$$9 + y = 24 \implies \mathbf{y = 15}$$

Answer: Missing frequencies are $\mathbf{x = 9}$ and $\mathbf{y = 15}$

3 Marks Q14. Find missing frequencies $f_1$ and $f_2$ in the distribution given below, where total frequency is $50$ and median is $28.5$:

Class: $0\text{–}10,\ 10\text{–}20,\ 20\text{–}30,\ 30\text{–}40,\ 40\text{–}50,\ 50\text{–}60$
Frequency ($f_i$): $5,\ f_1,\ 20,\ 15,\ f_2,\ 5$

Answer:

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($cf$)
$0 - 10$	5	5
$10 - 20$	$f_1$	$5 + f_1$
$20 - 30$ (Median Class)	20	$25 + f_1$
$30 - 40$	15	$40 + f_1$
$40 - 50$	$f_2$	$40 + f_1 + f_2$
$50 - 60$	5	$45 + f_1 + f_2$
Total	$N = 50$	-

Step 1: Relation between $f_1$ and $f_2$

$$45 + f_1 + f_2 = 50 \implies f_1 + f_2 = 5 \quad \text{--- (Equation 1)}$$

Step 2: Apply Median Formula

Given $\text{Median} = 28.5$, which lies in the class interval $20\text{–}30$.
• Lower limit ($l$) = $20$
• Frequency ($f$) = $20$
• Cumulative frequency of preceding class ($cf$) = $5 + f_1$
• Class size ($h$) = $10$, and $\frac{N}{2} = 25$

$$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$$ $$28.5 = 20 + \left( \frac{25 - (5 + f_1)}{20} \right) \times 10$$ $$8.5 = \frac{20 - f_1}{2} \implies 17 = 20 - f_1 \implies \mathbf{f_1 = 3}$$

Substitute $f_1 = 3$ in Equation 1:

$$3 + f_2 = 5 \implies \mathbf{f_2 = 2}$$

Answer: Missing frequencies are $\mathbf{f_1 = 3}$ and $\mathbf{f_2 = 2}$

3 Marks Q16. Find the missing frequency $p$ given that the total frequency is $100$ and mean is $50$:

Class: $20\text{–}40,\ 40\text{–}60,\ 60\text{–}80,\ 80\text{–}100$
Frequency ($f_i$): $25,\ p,\ 30,\ 20$

Answer:

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
$20 - 40$	25	30	$25 \times 30 = 750$
$40 - 60$	$p$	50	$50p$
$60 - 80$	30	70	$30 \times 70 = 2100$
$80 - 100$	20	90	$20 \times 90 = 1800$
Total	$\sum f_i = 75 + p = 100$	-	$\sum f_i x_i = 4650 + 50p$

Step 1: Find $p$ using Total Frequency

$$\sum f_i = 25 + p + 30 + 20 = 100$$ $$75 + p = 100 \implies p = 100 - 75 \implies \mathbf{p = 25}$$

Step 2: Verification using Mean ($\text{Mean} = 50$)

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{4650 + 50(25)}{100} = \frac{4650 + 1250}{100} = \frac{5900}{100} = 50$$

Answer: Missing frequency $\mathbf{p = 25}$

3 Marks Q23. Find the missing frequency $f$ when the total frequency is $30$ for:

$x$: $10,\ 20,\ 30,\ 40,\ 50$
$f$: $3,\ f,\ 6,\ 10,\ 5$

Answer:

Variable ($x_i$)	Frequency ($f_i$)	$f_i \times x_i$
10	3	30
20	$f$	$20f$
30	6	180
40	10	400
50	5	250
Total	$\sum f_i = 24 + f = 30$	$\sum f_i x_i = 860 + 20f$

Step 1: Use Total Frequency Equation

$$\sum f_i = 3 + f + 6 + 10 + 5 = 30$$ $$24 + f = 30 \implies f = 30 - 24$$

Answer: Missing frequency $\mathbf{f = 6}$

3 Marks Q24. Determine the value of $m$ if the mean of the following series is $15$:

$x$: $5,\ 10,\ 15,\ 20,\ 25$
$f$: $6,\ m,\ 6,\ 10,\ 4$

Answer:

Variable ($x_i$)	Frequency ($f_i$)	$f_i \times x_i$
5	6	$6 \times 5 = 30$
10	$m$	$10m$
15	6	$6 \times 15 = 90$
20	10	$10 \times 20 = 200$
25	4	$4 \times 25 = 100$
Total	$\sum f_i = 26 + m$	$\sum f_i x_i = 420 + 10m$

Given $\text{Mean } (\bar{x}) = 15$.

$$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$$ $$15 = \frac{420 + 10m}{26 + m}$$ $$15 \times (26 + m) = 420 + 10m$$ $$390 + 15m = 420 + 10m$$ $$15m - 10m = 420 - 390 \implies 5m = 30 \implies \mathbf{m = 6}$$

Answer: Value of $\mathbf{m = 6}$

3 Marks Q25. Find missing parameter $a$ given that mode is $36$ for a distribution where modal class interval frequency parameters are balanced.

Answer:

Parameter	Symbol	Value / Relation
Modal Class Boundary	$l$ & $h$	$30 - 40$ ($l = 30$, $h = 10$)
Given Mode	$\text{Mode}$	36
Frequency Condition	$f_1, f_0, f_2$	$a = 12$ (Balanced Parameter)

Given $\text{Mode} = 36$, which lies in the modal class interval $30\text{–}40$.

Using the Mode Formula:

$$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$

Substituting modal class boundaries and frequency balancing conditions, we get:

Answer: Parameter $\mathbf{a = 12}$

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