1. **Problem:** Write the sample space and sample points for each experiment. **a) Three coins are tossed simultaneously.** - Each coin has two outcomes: Head (H) or Tail (T). - Total outcomes = $2^3 = 8$. - Sample space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$. - Each element is a sample point. **b) A die is rolled.** - A die has 6 faces numbered 1 to 6. - Sample space $S = \{1, 2, 3, 4, 5, 6\}$. - Each number is a sample point. 2. **Problem:** Find the sum of the first ‘n’ odd natural numbers. - The first $n$ odd natural numbers form an arithmetic progression (A.P.) with first term $t_1=1$ and common difference $d=2$. - Sum of $n$ terms of an A.P. is: $$S_n = \frac{n}{2} [2t_1 + (n-1)d]$$ - Substitute $t_1=1$ and $d=2$: $$S_n = \frac{n}{2} [2 \times 1 + (n-1) \times 2] = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} \times 2n = n^2$$ - **Answer:** The sum of the first $n$ odd natural numbers is $S_n = n^2$. 3. **Problem:** Find the values of $y$ for the equation $5x - 3y = 1$ at given $x$ values. - Rearrange the equation to solve for $y$: $$5x - 3y = 1 \implies -3y = 1 - 5x \implies y = \frac{5x - 1}{3}$$ - Calculate $y$ for each $x$: 1. For $x=0$: $$y = \frac{5 \times 0 - 1}{3} = \frac{-1}{3} = -\frac{1}{3}$$ 2. For $y=0$ (find $x$): $$5x - 3 \times 0 = 1 \implies 5x = 1 \implies x = \frac{1}{5}$$ 3. For $x=1$: $$y = \frac{5 \times 1 - 1}{3} = \frac{4}{3}$$ 4. For $x=-2$: $$y = \frac{5 \times (-2) - 1}{3} = \frac{-10 - 1}{3} = \frac{-11}{3}$$ - Ordered pairs: $$(0, -\frac{1}{3}), (\frac{1}{5}, 0), (1, \frac{4}{3}), (-2, -\frac{11}{3})$$ 4. **Problem:** Complete the cumulative frequency table. | Class | Frequency | Cumulative Frequency (Less than type) | |----------|-----------|---------------------------------------| | 200-210 | 4 | 4 | | 210-220 | 14 | 4 + 14 = 18 | | 220-230 | 26 | 18 + 26 = 44 | | 230-240 | 10 | 44 + 10 = 54 | 5. **Problem:** Probability questions for spinning arrow on numbers 2,4,8,16,32,64,128,256. - Total outcomes $n(S) = 8$. **a) Number which is either a perfect square or perfect cube.** - Perfect squares in set: $4 = 2^2$, $16 = 4^2$, $64 = 8^2$. - Perfect cubes in set: $8 = 2^3$, $64 = 4^3$. - Numbers either perfect square or cube: $\{4, 8, 16, 64\}$. - Number of favorable outcomes $n(E) = 4$. - Probability: $$P = \frac{n(E)}{n(S)} = \frac{4}{8} = \frac{1}{2}$$ **b) Number which is both perfect square and perfect cube.** - Numbers that are both perfect square and cube are perfect sixth powers. - In set: $64 = 2^6$. - Number of favorable outcomes $n(E) = 1$. - Probability: $$P = \frac{1}{8}$$ 6. **Problem:** Find the nth term of the A.P. $4, \frac{14}{3}, \frac{16}{3}, 6, \ldots$ - First term $a = 4$. - Common difference $d = \frac{14}{3} - 4 = \frac{14}{3} - \frac{12}{3} = \frac{2}{3}$. - General term: $$t_n = a + (n-1)d = 4 + (n-1) \times \frac{2}{3} = \frac{12}{3} + \frac{2}{3}(n-1) = \frac{12 + 2n - 2}{3} = \frac{2n + 10}{3}$$ 7. **Problem:** Length and breadth of rectangular storehouse. - Let breadth = $x$ m. - Length = $x + 7$ m. - Diagonal = length + 1 = $x + 7 + 1 = x + 8$ m. - By Pythagoras theorem: $$\text{diagonal}^2 = \text{length}^2 + \text{breadth}^2$$ $$ (x + 8)^2 = (x + 7)^2 + x^2$$ Expand: $$x^2 + 16x + 64 = x^2 + 14x + 49 + x^2$$ Simplify: $$x^2 + 16x + 64 = 2x^2 + 14x + 49$$ Bring all terms to one side: $$0 = 2x^2 + 14x + 49 - x^2 - 16x - 64 = x^2 - 2x - 15$$ Solve quadratic: $$x^2 - 2x - 15 = 0$$ $$x = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}$$ - $x = \frac{2 + 8}{2} = 5$ (valid, length and breadth positive) - $x = \frac{2 - 8}{2} = -3$ (discard) - Breadth = 5 m, Length = $5 + 7 = 12$ m. 8. **Problem:** Solve system of equations: $$\frac{4}{x} + \frac{5}{y} = 7$$ $$\frac{3}{x} + \frac{4}{y} = 6$$ Let $p = \frac{1}{x}$ and $q = \frac{1}{y}$. Rewrite: $$4p + 5q = 7$$ $$3p + 4q = 6$$ Multiply second equation by 4 and first by 3: $$12p + 15q = 21$$ $$12p + 16q = 24$$ Subtract: $$(12p + 16q) - (12p + 15q) = 24 - 21 \implies q = 3$$ Substitute $q=3$ into first equation: $$4p + 5 \times 3 = 7 \implies 4p + 15 = 7 \implies 4p = -8 \implies p = -2$$ Recall $p = \frac{1}{x} = -2 \implies x = -\frac{1}{2}$. Recall $q = \frac{1}{y} = 3 \implies y = \frac{1}{3}$. **Final answers:** - Sample spaces for coins and die. - Sum of first $n$ odd numbers is $n^2$. - Points on line $5x - 3y = 1$ are $(0,-\frac{1}{3}), (\frac{1}{5},0), (1, \frac{4}{3}), (-2, -\frac{11}{3})$. - Cumulative frequencies: 4, 18, 44, 54. - Probability perfect square or cube = $\frac{1}{2}$; both square and cube = $\frac{1}{8}$. - $n$th term of A.P. = $\frac{2n + 10}{3}$. - Storehouse dimensions: length = 12 m, breadth = 5 m. - Solution to system: $x = -\frac{1}{2}$, $y = \frac{1}{3}$.