32. Out of 70 schools 42 of them can be attended by boys and 35 can be attended by girls, If a pupil is selected at random from these schools, find the probability that he/she is from a mixed school. (A) $$ \frac{1}{1} $$ (B) $$ \frac{1}{10} $$ (C) $$ \frac{1}{6} $$ (D) $$ \frac{1}{5} $$ 33. Given that $$ a=i-3 j, b=2 i+5 j $$ and $$ c=3 i-11 j $$, calculate $$ |a+b+c| $$. (A) $$ \sqrt{13} $$ (B) $$ 3 \sqrt{13} $$ (C) $$ 9 \sqrt{13} $$ (D) $$ 6 \sqrt{13} $$ 34. What is the probability of obtaining a head and a six when a fair coin and a die are tossed together? (A) $$ \frac{1}{12} $$ (B) $$ \frac{1}{3} $$ (C) $$ \frac{1}{2} $$ (D) $$ \frac{2}{3} $$ 35. If $$ \rightarrow=\binom{-7}{6} $$ and $$ \rightarrow=\binom{16}{-11} $$, find $$ \underset{Y X}{ } $$. (A) $$ \binom{9}{-5} $$ (B) $$ \binom{-23}{-5} $$ (C) $$ \binom{9}{17} $$ (D) $$ \binom{-23}{17} $$ 36. Factorize $$ 10 x^{2}+12 x+3 $$ (A) $$ (5 x+1)(2 x+3) $$ (B) $$ (2 x-3)(5 x+1) $$ (C) $$ (6 x+5)(2 x+3) $$ $$ (5 x-1)(2 x-3) $$ 37. Solve the equation $$ x^{2}-64=0 $$. (A) $$ \pm 8 $$

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UpStudy AI · Accepted Answer

Let's solve each of the given questions step by step:

---

### **Question 32**

**Out of 70 schools, 42 can be attended by boys and 35 can be attended by girls. Find the probability that a randomly selected pupil is from a mixed school.**

**Options:**
- (A) $$ \frac{1}{1} $$
- (B) $$ \frac{1}{10} $$
- (C) $$ \frac{1}{6} $$
- (D) $$ \frac{1}{5} $$

**Solution:**

To find the number of mixed schools (schools attended by both boys and girls), we use the principle of inclusion:

$$
\text{Number of mixed schools} = (\text{Schools for boys}) + (\text{Schools for girls}) - \text{Total schools}
$$

$$
= 42 + 35 - 70 = 7
$$

**Probability:**

$$
P(\text{Mixed school}) = \frac{\text{Number of mixed schools}}{\text{Total schools}} = \frac{7}{70} = \frac{1}{10}
$$

**Answer:** 
**Option (B) $$ \frac{1}{10} $$**

---

### **Question 33**

**Given vectors $$ \vec{a} = \mathbf{i} - 3\mathbf{j} $$, $$ \vec{b} = 2\mathbf{i} + 5\mathbf{j} $$, and $$ \vec{c} = 3\mathbf{i} - 11\mathbf{j} $$, calculate $$ |\vec{a} + \vec{b} + \vec{c}| $$.**

**Options:**
- (A) $$ \sqrt{13} $$
- (B) $$ 3\sqrt{13} $$
- (C) $$ 9\sqrt{13} $$
- (D) $$ 6\sqrt{13} $$

**Solution:**

First, find $$ \vec{a} + \vec{b} + \vec{c} $$:

$$
\vec{a} + \vec{b} + \vec{c} = (\mathbf{i} - 3\mathbf{j}) + (2\mathbf{i} + 5\mathbf{j}) + (3\mathbf{i} - 11\mathbf{j}) = 6\mathbf{i} - 9\mathbf{j}
$$

Now, compute the magnitude:

$$
|\vec{a} + \vec{b} + \vec{c}| = \sqrt{6^2 + (-9)^2} = \sqrt{36 + 81} = \sqrt{117} = 3\sqrt{13}
$$

**Answer:** 
**Option (B) $$ 3\sqrt{13} $$**

---

### **Question 34**

**What is the probability of obtaining a head and a six when a fair coin and a die are tossed together?**

**Options:**
- (A) $$ \frac{1}{12} $$
- (B) $$ \frac{1}{3} $$
- (C) $$ \frac{1}{2} $$
- (D) $$ \frac{2}{3} $$

**Solution:**

- **Total possible outcomes:** $$ 2 $$ (coin) $$ \times $$ $$ 6 $$ (die) $$ = 12 $$
- **Favorable outcome:** Getting a head **and** a six $$ = 1 $$

**Probability:**

$$
P(\text{Head and Six}) = \frac{1}{12}
$$

**Answer:** 
**Option (A) $$ \frac{1}{12} $$**

---

### **Question 35**

**If $$ \vec{X} = \binom{-7}{6} $$ and $$ \vec{Y} = \binom{16}{-11} $$, find $$ \vec{Y} - \vec{X} $$.**

**Options:**
- (A) $$ \binom{9}{-5} $$
- (B) $$ \binom{-23}{-5} $$
- (C) $$ \binom{9}{17} $$
- (D) $$ \binom{-23}{17} $$

**Solution:**

Calculate $$ \vec{Y} - \vec{X} $$:

$$
\vec{Y} - \vec{X} = \binom{16}{-11} - \binom{-7}{6} = \binom{16 - (-7)}{-11 - 6} = \binom{23}{-17}
$$

However, this exact option isn't listed. It appears the question might be asking for $$ \vec{X} - \vec{Y} $$:

$$
\vec{X} - \vec{Y} = \binom{-7}{6} - \binom{16}{-11} = \binom{-23}{17}
$$

**Answer:** 
**Option (D) $$ \binom{-23}{17} $$**

---

### **Question 36**

**Factorize $$ 10x^2 + 12x + 3 $$.**

**Options:**
- (A) $$ (5x + 1)(2x + 3) $$
- (B) $$ (2x - 3)(5x + 1) $$
- (C) $$ (6x + 5)(2x + 3) $$
- (D) $$ (5x - 1)(2x - 3) $$

**Solution:**

We need to find two binomials that multiply to $$ 10x^2 + 12x + 3 $$.

Let's test **Option (A)**:

$$
(5x + 1)(2x + 3) = 10x^2 + 15x + 2x + 3 = 10x^2 + 17x + 3 \neq 10x^2 + 12x + 3
$$

Next, **Option (B)**:

$$
(2x - 3)(5x + 1) = 10x^2 + 2x - 15x - 3 = 10x^2 - 13x - 3 \neq 10x^2 + 12x + 3
$$

Continue with **Option (C)**:

$$
(6x + 5)(2x + 3) = 12x^2 + 18x + 10x + 15 = 12x^2 + 28x + 15 \neq 10x^2 + 12x + 3
$$

Finally, **Option (D)**:

$$
(5x - 1)(2x - 3) = 10x^2 - 15x - 2x + 3 = 10x^2 - 17x + 3 \neq 10x^2 + 12x + 3
$$

**Conclusion:**
The polynomial $$ 10x^2 + 12x + 3 $$ does not factor neatly into integer coefficients based on the provided options. However, none of the given options correctly factorize the polynomial. It appears there might be an error in the options provided.

**Answer:** 
**None of the provided options correctly factorize the polynomial.**

---

### **Question 37**

**Solve the equation $$ x^2 - 64 = 0 $$.**

**Options:**
- (A) $$ \pm8 $$

**Solution:**

$$
x^2 - 64 = 0 \implies x^2 = 64 \implies x = \pm8
$$

**Answer:** 
**Option (A) $$ \pm8 $$**

---
**Solutions:** 1. **Question 32:** - **Answer:** $$ \frac{1}{10} $$ - **Option:** (B) 2. **Question 33:** - **Answer:** $$ 3\sqrt{13} $$ - **Option:** (B) 3. **Question 34:** - **Answer:** $$ \frac{1}{12} $$ - **Option:** (A) 4. **Question 35:** - **Answer:** $$ \binom{-23}{17} $$ - **Option:** (D) 5. **Question 36:** - **Answer:** The polynomial does not factor neatly with the given options. - **Option:** None of the provided options are correct. 6. **Question 37:** - **Answer:** $$ \pm8 $$ - **Option:** (A)

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