1. \[ \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \] This is a mathematical equation involving rational expressions. Here's the English translation of the statement: **"The sum of the following three fractions equals one-sixth:"** - The first fraction is: one divided by \((x - 1)(x - 2)\) - The second fraction is: one divided by \((x - 2)(x - 3)\) - The third fraction is: one divided by \((x - 3)(x - 4)\) And their total is equal to \(\frac{1}{6}\).

2. Doubling the principal means Principal \((p) =\) Interest \((I) = x\) taka Rate of interest \((r) = 6\frac{1}{4}\% = \frac{25}{4}\%\) \[ \therefore \text{Time } (t) = \frac{100 \times I}{p \times r} \text{ years} = \frac{100 \times x}{x \times \frac{25}{4}} = \frac{100 \times 4}{25} = 16 \text{ years} \] \[ \therefore \text{At an annual simple interest rate of } 6\frac{1}{4}\%, \text{ any principal will double in 16 years.} \]

3.

4. Let the length of the space diagonal of the cube be \(x\), the length of a face diagonal be \(y\), and the edge of the cube be \(a\). Then, \[ x = a\sqrt{3} \Rightarrow a = \cfrac{x}{\sqrt{3}} \quad \text{— (i)} \] And, \[ y = a\sqrt{2} \Rightarrow a = \cfrac{y}{\sqrt{2}} \quad \text{— (ii)} \] Comparing equations (i) and (ii): \[ \cfrac{x}{\sqrt{3}} = \cfrac{y}{\sqrt{2}} \Rightarrow x = \cfrac{\sqrt{3}}{\sqrt{2}} \cdot y \Rightarrow x = \cfrac{\sqrt{6}}{2} \cdot y \] \(\therefore\) The length of the space diagonal of a cube is \(\cfrac{\sqrt{6}}{2}\) times the length of its face diagonal.

(a) undefined (b) positive (c) negative (d) zero

5. If \[ \cfrac{x}{xa + yb + zc} = \cfrac{y}{ya + zb + xc} = \cfrac{z}{za + xb + yc} \] and \(x + y + z \ne 0\), then show that each of the above ratios is equal to \(\cfrac{1}{a + b + c}\).

6. If \(a\) is a positive number and \[ a : \cfrac{27}{64} = \cfrac{3}{4} : a, \] then find the value of \(a\).

(a) \(\cfrac{81}{256}\) (b) 9 (c) \(\cfrac{9}{16}\) (d) \(\cfrac{16}{9}\)

7. If \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} \] then prove that each of these ratios is either \(\frac{1}{2}\) or \(-1\).

8. In right-angled triangle \( \triangle ABC \), where \( \angle A = 90^\circ \), a perpendicular \( AD \) is drawn from point \( A \) to the hypotenuse \( BC \). Prove that: \[ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ACD} = \frac{BC^2}{AC^2} \]

9. Let \(a\) be a positive number, and given: \[ a : \frac{27}{64} = \frac{3}{4} : a \] Find the value of \(a\).

10. Given: \[ x \cos \theta = 3 \quad \text{and} \quad 4 \tan \theta = y \] Find the relation between \(x\) and \(y\) eliminating \(\theta\).

11. If \[ x \cos 60^\circ = \frac{2 \tan 45^\circ}{1 + \tan^2 45^\circ} - \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} \] then find the value of \(x\).

12. If \[ \frac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} + \frac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} = 14 \] then determine the possible values of \(x\).

13. In \( \triangle ABC \), the vertex \( A \) has a perpendicular \( AD \) drawn to the side \( BC \). If \[ \frac{BD}{DA} = \frac{DA}{DC} \] then prove that \( \triangle ABC \) is a right-angled triangle.

14. If \[ \cfrac{ax + by}{a} = \cfrac{bx - ay}{b} \] then prove that each ratio is equal to \(x\).

15. If \[ \frac{x}{y + z} = \frac{y}{z + x} = \frac{z}{x + y} \] then prove that the value of each ratio is either \(\frac{1}{2}\) or \(-1\).

16. A circle is drawn with diameter \(AB\) and center \(O\). From a point \(P\) on the circle, a perpendicular is drawn to the diameter \(AB\), intersecting it at point \(N\). Prove that: \[ PB^2 = AB \cdot BN \]

17. For the following data of handicraft scores awarded to 45 girl students in our village, calculate the **median** of the frequency distribution:

18. If a planet's gravitational force on its satellite is directly proportional to the planet's mass (M) and inversely proportional to the square of their distance (D), and the square of the satellite's orbital period (T) is directly proportional to the cube of the distance and inversely proportional to the gravitational force, then for two sets of values \(m_1, d_1, t_1\) and \(m_2, d_2, t_2\) corresponding to M, D, and T respectively, prove that: \[ m_1 t_1^2 d_2^3 = m_2 t_2^2 d_1^3 \]

19. If \[ 1\frac{1}{2} \cdot \frac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} + \frac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} = 14 \] then find the value of \(x\).

20. A person bought \(y\) pencils for ₹\(x\). If each pencil had cost ₹1 less, then with the same amount ₹\(x\), he would have received one additional pencil. Prove that: \[ 2y = \sqrt{1 + 4x} - 1 \]

21. If \(a = 3 + 2\sqrt{2}\), then what is the value of \[ \frac{a^6 + a^4 + a^2 + 1}{a^3}? \]

(a) 100 (b) 200 (c) 204 (d) 250

22. The modal class of the above frequency distribution is 15–20. So, the mode is calculated as: \[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \] Where: - \(l = 15\) (lower boundary of modal class) - \(f_1 = 28\) (frequency of modal class) - \(f_0 = 18\) (frequency of class before modal class) - \(f_2 = 17\) (frequency of class after modal class) - \(h = 5\) (class width) Substituting the values: \[ = 15 + \left(\frac{28 - 18}{2 \times 28 - 18 - 17}\right) \times 5 = 15 + \frac{10}{21} \times 5 = 15 + \frac{50}{21} = 15 + 2.38 = 17.38 \quad \text{(approx)} \] ✅ Therefore, the mode is approximately **17.38**.

23. Left-hand side (LHS): \[ 1 + \cfrac{\tan A}{\tan B} = 1 + \cfrac{\tan(90^\circ - B)}{\tan B} = 1 + \cfrac{\cot B}{\cot B} = 1 + \cot^2 B = \csc^2 B \] Right-hand side (RHS): \[ \tan^2 A \cdot \sec^2 B = \tan^2(90^\circ - B) \cdot \sec^2 B = \cot^2 B \cdot \sec^2 B = \cfrac{\cos^2 B}{\sin^2 B} \cdot \cfrac{1}{\cos^2 B} = \cfrac{1}{\sin^2 B} = \csc^2 B \] \(\therefore\) LHS = RHS (Proved)

24. If \[ \tan \theta \cdot \cos 60^\circ = \frac{\sqrt{3}}{2} \] then find the value of \[ \sin(\theta - 15^\circ) \] given that \(0^\circ < \theta < 90^\circ\).

25. If \[ \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = 2\frac{51}{79} \] then find the value of \(\sin\theta\).

26. If the total surface area of a cube is \(s\) square units and the length of its diagonal is \(d\) units, then the relationship between \(s\) and \(d\) is \[ s = 6d^2 \]

27. If \[ x \cot\left(\frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{3}\right) + \frac{3}{4} \sec^2\left(\frac{\pi}{4}\right) + 4 \sin\left(\frac{\pi}{6}\right) \] then find the value of \(x\).

28. Prove that the quadratic equation \[ 2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0 \] will have no real roots if \(a \ne b\).

29. If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \[ 5x^2 + 2x - 3 = 0, \] find the value of \(\alpha^2 + \beta^2\).

30. The compound ratio of \((x + y) : (x - y),\ (x^2 + y^2) : (x + y)^2,\ (x^2 - y^2)^2 : (x^4 - y^4)\) is: \[ = \frac{(x + y)(x^2 + y^2)(x^2 - y^2)^2}{(x - y)(x + y)^2(x^4 - y^4)} = \frac{1}{1} \]