34. In a class of 60 students, 50 offer Biology and 40 offer Physics. If 5 do not offer the two subjects, how many students offer both subjects? A. 10 B. 20 C 35 D. 45 35. A chord, 7 cm long, is Jrawn in a circle with radius 3.7 cm . Calculate the distance of the chord from the centre of the circle. A. 0.7 cm (B. $$ 1.2 \mathrm{~cm} \mathrm{C} $$.2.0 cm D. 2.5 cm 36. A chord subtends an angle of $$ 120^{3} $$ at the centre of a circle of radius 35 cm . Find the perimeter of the minor sector containing the chord. $$ \left[ ight. $$ Take $$ \left.\pi=\frac{22}{7} ight] $$ A. $$ 14 \frac{1}{3} \mathrm{~cm} $$ B. $$ 12 \frac{5}{5} \mathrm{~cm} \quad $$ C. $$ B \frac{1}{7} \mathrm{~cm} $$ D. $$ 7 \frac{1}{3} \mathrm{~cm} $$ 37. If $$ \sin x=\frac{5}{13} $$ and $$ 0^{0} \leq x \leq 90^{0} $$. find the value of $$ (\cos x- an x) $$. A. $$ \frac{7}{13} $$ B. $$ \frac{12}{13} $$ C) $$ \frac{79}{156} $$ D. $$ \frac{209}{156} $$ 38. One of the roots of the equation $$ x^{2}+9=0 $$ is (8) 3 B. -3 C. -9 D. None of these

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34. In a class of 60 students, 50 offer Biology and 40 offer Physics. If 5 do not offer the two subjects, how many students offer both subjects? A. 10 B. 20 C 35 D. 45 35. A chord, 7 cm long, is Jrawn in a circle with radius 3.7 cm . Calculate the distance of the chord from the centre of the circle. A. 0.7 cm (B. $$ 1.2 \mathrm{~cm} \mathrm{C} $$.2.0 cm D. 2.5 cm 36. A chord subtends an angle of $$ 120^{3} $$ at the centre of a circle of radius 35 cm . Find the perimeter of the minor sector containing the chord. $$ \left[ight. $$ Take $$ \left.\pi=\frac{22}{7}ight] $$ A. $$ 14 \frac{1}{3} \mathrm{~cm} $$ B. $$ 12 \frac{5}{5} \mathrm{~cm} \quad $$ C. $$ B \frac{1}{7} \mathrm{~cm} $$ D. $$ 7 \frac{1}{3} \mathrm{~cm} $$ 37. If $$ \sin x=\frac{5}{13} $$ and $$ 0^{0} \leq x \leq 90^{0} $$. find the value of $$ (\cos x-	an x) $$. A. $$ \frac{7}{13} $$ B. $$ \frac{12}{13} $$ C) $$ \frac{79}{156} $$ D. $$ \frac{209}{156} $$ 38. One of the roots of the equation $$ x^{2}+9=0 $$ is (8) 3 B. -3 C. -9 D. None of these

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### Question 34
**Given:**
- Total students = 60
- Students offering Biology = 50
- Students offering Physics = 40
- Students not offering either = 5

**To find:** Students offering both subjects.

**Step 1: Calculate students offering at least one subject.**
Total students offering at least one subject = Total students - Students not offering either = $$60 - 5 = 55$$.

**Step 2: Use the principle of inclusion-exclusion.**
Let $$x$$ be the number of students offering both subjects.
$$
\text{Students offering at least one subject} = \text{Students offering Biology} + \text{Students offering Physics} - \text{Students offering both}
$$
$$
55 = 50 + 40 - x
$$
$$
55 = 90 - x
$$
$$
x = 90 - 55 = 35
$$
35 students offer both subjects.

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