Q1. Solve: \( x^2 + \frac{1}{x^2} = 34 \), find \( x + \frac{1}{x}
\)
Answer:
Let \( a = x + \frac{1}{x} \)
Then, \( a^2 =
x^2 + \frac{1}{x^2} + 2 \)
So, \( a^2 = 34 + 2 = 36 \Rightarrow a = \pm6
\)
Thus, \( x + \frac{1}{x} = \pm6 \)
Q2. Find two numbers whose sum is 10 and the sum of their reciprocals
is \( \frac{5}{12} \).
Answer:
Let the numbers be \( x \) and
\( 10 - x \)
Then, \( \frac{1}{x} + \frac{1}{10 - x} = \frac{5}{12} \)
Solve:
\( \frac{10 - x + x}{x(10 - x)} = \frac{5}{12} \)
\( \frac{10}{x(10 - x)}
= \frac{5}{12} \)
Cross-multiply: \( 120 = 5x(10 - x) \Rightarrow 5x^2 -
50x + 120 = 0 \)
\( x^2 - 10x + 24 = 0 \Rightarrow x = 4 \) or \( 6 \)
The
numbers are 4 and 6.
Q3. A quadratic equation has one root double the other. If the equation
is \( x^2 + px + q = 0 \), find \( p \) and \( q \).
Answer:
Let roots be \( \alpha \) and \( 2\alpha \)
Sum: \( \alpha +
2\alpha = 3\alpha = -p \Rightarrow \alpha = -\frac{p}{3} \)
Product: \(
\alpha \cdot 2\alpha = 2\alpha^2 = q \)
Substitute \( \alpha \) into \( q
= 2(-\frac{p}{3})^2 = \frac{2p^2}{9} \)
Thus, \( p = p \), \( q =
\frac{2p^2}{9} \)
Q4. The product of two consecutive odd integers is 255. Find the
integers.
Answer:
Let the numbers be \( x \) and \( x + 2
\)
Then, \( x(x + 2) = 255 \Rightarrow x^2 + 2x - 255 = 0 \)
Solve:
\( x = 15 \) or \( -17 \)
Integers: 15 & 17 or -17 & -15
Q5. Area of a rectangle is 77 cm². Length is 4 cm more than breadth.
Find dimensions.
Answer:
Let breadth = \( x \), then length =
\( x + 4 \)
Area = \( x(x + 4) = 77 \Rightarrow x^2 + 4x - 77 = 0 \)
\(
x = 7 \), so breadth = 7 cm, length = 11 cm
Q6. A number whose square is 12 more than five times the number.
Answer:
Let number = \( x \)
Then, \( x^2 = 5x + 12 \Rightarrow x^2 - 5x -
12 = 0 \)
\( x = 6 \) or \( -2 \)
Q7. Solve for \( x \): \( \frac{1}{x} + \frac{1}{x - 2} = \frac{3}{4}
\)
Answer:
Take LCM: \( \frac{(x - 2) + x}{x(x - 2)} =
\frac{3}{4} \)
\( \frac{2x - 2}{x(x - 2)} = \frac{3}{4} \)
Cross-multiply:
\( 4(2x - 2) = 3x(x - 2) \)
\( \Rightarrow 8x - 8 = 3x^2 - 6x \)
\( 3x^2 - 14x + 8 = 0 \)
Solve:
\( x = 4 \), \( x = \frac{2}{3} \)
Reject \( x = 2 \) as not allowed.
Q8. The denominator is 2 more than numerator. Square of fraction is \(
\frac{25}{49} \). Find the fraction.
Answer:
Let numerator =
\( x \), denominator = \( x + 2 \)
Then, \( \left( \frac{x}{x + 2}
\right)^2 = \frac{25}{49} \)
Take square root: \( \frac{x}{x + 2} = \pm
\frac{5}{7} \)
Solve for \( x \): \( x = 5 \) or \( -10 \)
Fraction:
\( \frac{5}{7} \)
Q9. A car travels 150 km. If speed were 5 km/h more, time reduced by 1
hour. Find speed.
Answer:
Let speed = \( x \), time = \(
\frac{150}{x} \)
Faster speed = \( x + 5 \), time = \( \frac{150}{x+5}
\)
Equation: \( \frac{150}{x} - \frac{150}{x+5} = 1 \)
Solve: \(
150(x + 5 - x) = x(x + 5) \)
\( \Rightarrow 750 = x^2 + 5x \)
\( x^2 + 5x - 750 = 0 \Rightarrow x = 25
\)
Original speed = 25 km/h
Q10. Solve: \( \sqrt{x + 7} - \sqrt{x - 1} = 2 \), \( x > 1 \)
Answer:
Let: \( \sqrt{x + 7} = a \), \( \sqrt{x - 1} = b \), so \( a - b = 2
\)
Then, \( a = b + 2 \)
So, \( (b + 2)^2 = x + 7 \) and \( b^2 = x
- 1 \)
Subtract: \( (b + 2)^2 - b^2 = 8 \)
\( b^2 + 4b + 4 - b^2 = 8
\Rightarrow 4b = 4 \Rightarrow b = 1 \)
Then, \( \sqrt{x - 1} = 1
\Rightarrow x = 2 \)
Q11. Prove: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A \)
Answer:
Using identities:
\( \tan^2 A + 1 = \sec^2 A \), \( \cot^2 A + 1 =
\csc^2 A \)
So, LHS = \( \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2
A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \)
Hence proved.
Q12. Prove: \( \frac{1 - \cos A}{\sin A} + \frac{1 + \cos A}{\sin A} =
2 \csc A \)
Answer:
LHS = \( \frac{(1 - \cos A + 1 + \cos
A)}{\sin A} = \frac{2}{\sin A} = 2 \csc A \)
Hence proved.
Q13. Evaluate: \( \frac{\sin^2 60^\circ + \cos^2 30^\circ - \tan^2
45^\circ}{\cot 45^\circ + \sec 60^\circ - \csc 30^\circ} \)
Answer:
Values: \( \sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 30^\circ =
\frac{\sqrt{3}}{2}, \tan 45^\circ = 1 \)
Numerator = \( \left(
\frac{3}{4} + \frac{3}{4} - 1 \right) = \frac{3}{2} - 1 = \frac{1}{2} \)
Denominator
= \( 1 + 2 - 2 = 1 \)
Result = \( \frac{1}{2} \)
Q14. If \( \sec \theta = \frac{13}{12} \), find all other ratios.
Answer:
Use triangle: \( \sec \theta = \frac{\text{hyp}}{\text{adj}} =
\frac{13}{12} \)
Then opp = \( \sqrt{13^2 - 12^2} = 5 \)
So:
\(
\sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}, \)
\( \tan \theta = \frac{5}{12}, \csc \theta = \frac{13}{5}, \cot \theta =
\frac{12}{5} \)
Q15. If \( \tan A = \frac{3}{4} \), and A is acute, evaluate \( \sin A
+ \cos A + \frac{1}{\sin A} + \frac{1}{\cos A} \)
Answer:
Let
perpendicular = 3, base = 4 → hypotenuse = 5
So: \( \sin A = \frac{3}{5},
\cos A = \frac{4}{5} \)
Expression = \( \frac{3}{5} + \frac{4}{5} +
\frac{5}{3} + \frac{5}{4} = \frac{7}{5} + \frac{20}{12} + \frac{25}{20} \)
LCM
approach: Final result = \( \frac{7}{5} + \frac{5}{3} + \frac{5}{4} = \frac{84
+ 100 + 75}{60} = \frac{259}{60} \)
Q16.From a point on the ground, the angle of elevation of the top of a
tower is 60°. From another point 20 m closer, the angle is 75°. Find the
height of the tower.
Answer:
Let height = \( h \), original
distance = \( x \)
\( \tan 60^\circ = \frac{h}{x} \Rightarrow \sqrt{3} =
\frac{h}{x} \Rightarrow h = x\sqrt{3} \)
\( \tan 75^\circ = \frac{h}{x -
20} \Rightarrow 3.732 = \frac{x\sqrt{3}}{x - 20} \)
Solve: Approximate
value → \( h ≈ 51.96 \) m
Q17. A man observes the top of a building at an angle of elevation of
30°. On moving 50 m closer, the angle becomes 60°. Find the height of the
building.
Answer:
Let height = \( h \), initial distance = \(
x \)
\( \tan 30^\circ = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} =
\frac{h}{x} \Rightarrow h = \frac{x}{\sqrt{3}} \)
\( \tan 60^\circ =
\frac{h}{x - 50} \Rightarrow \sqrt{3} = \frac{h}{x - 50} \)
Substitute
and solve: \( h ≈ 43.3 \) m
Q18. A passenger of on aero plane observes that Howrah station is at
one side of the plane and Saheed minor Is just opposite side. The angle of
depression of Howrah station and Shaheed minar from the passenger of aero
plane are 60° and 30° respectively. If the aero plane is at a height of 545 √3
meters at that time, let us find the distance between Howrah station
Answer:
Height = \( 545\sqrt{3} \)
Let horizontal distances: Howrah = \(
x_1 \), Minar = \( x_2 \)
\( \tan 60^\circ = \frac{545\sqrt{3}}{x_1}
\Rightarrow x_1 = 545 \)
\( \tan 30^\circ = \frac{545\sqrt{3}}{x_2}
\Rightarrow x_2 = 1635 \)
Distance = \( x_2 + x_1 = 2180 \) m
Q19. Prove: \( \frac{1 + \sin A}{\cos A} + \frac{1 - \sin A}{\cos A} =
2 \sec A \)
Answer:
LHS = \( \frac{1 + \sin A + 1 - \sin
A}{\cos A} = \frac{2}{\cos A} = 2 \sec A \)
Hence proved.
Q20. Find the value of \( A \), if \( \frac{1 + \cos^2 A}{\sin^2 A} = 1 + \cot^2 A \)
Answer:
We know, \( \cot^2 A = \frac{\cos^2 A}{\sin^2 A} \)
So, \( \frac{1 + \cos^2 A}{\sin^2 A} = 1 + \frac{\cos^2 A}{\sin^2 A} \)
⇒ \( \frac{1 + \cos^2 A}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A} \)
⇒ \( \frac{1 + \cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A} \)
⇒ \( 1 + \cos^2 A = 1 \)
⇒ \( \cos^2 A = 0 \)
⇒ \( \cos A = 0 \)
⇒ \( A = 90^\circ \)
Final Answer: \( \boxed{A = 90^\circ} \)
Q21. If \( \theta = 30^\circ \), evaluate: \( \tan(90^\circ - \theta) +
\cot \theta \)
Answer:
\( \tan(90^\circ - 30^\circ) + \cot
30^\circ \)
\( = \cot 30^\circ + \cot 30^\circ \)
\(= 2 \cot
30^\circ = 2\sqrt{3} \)
Q22. Prove: \( \sin(90^\circ - A) = \cos A \)
Answer:
From complementary angle identity, \( \sin(90^\circ - A) = \cos A \).
Hence proved.
Q23. If \( \tan A = 1 \), verify: \( \sin^2 A + \cos^2 A = 1 \)
Answer:
\( \tan A = 1 \Rightarrow A = 45^\circ \)
\( \sin 45^\circ = \cos
45^\circ = \frac{1}{\sqrt{2}} \)
So, \( \sin^2 A + \cos^2 A = \frac{1}{2}
+ \frac{1}{2} = 1 \). Verified.
Q24. Prove: \( \cot(90^\circ - \theta) = \tan \theta \)
Answer:
From complementary angle identity, \( \cot(90^\circ - \theta) = \tan
\theta \). Hence proved.
Q25. Simplify: \( \frac{\sec^2 A - \tan^2 A}{1 + \cot^2 A} \)
Answer:
Numerator = \( \sec^2 A - \tan^2 A = 1 \) (by identity)
Denominator
= \( 1 + \cot^2 A = \csc^2 A \)
So, the result = \( \frac{1}{\csc^2 A} =
\sin^2 A \)
Q26. Find the mean of the following data:
| Class | Frequency (f) |
|---|---|
| 0–10 | 4 |
| 10–20 | 8 |
| 20–30 | 10 |
| 30–40 | 12 |
| 40–50 | 6 |
Midpoints (x): 5, 15, 25, 35, 45
Using formula: \( \bar{x} = \frac{\sum fx}{\sum f} \)
\( \sum fx = 4×5 + 8×15 + 10×25 + \) \( 12×35
+ 6×45 = 20 + 120 + 250 \) \( + 420 + 270 = 1080 \)
\( \sum f = 4 + 8 + 10 + 12 + 6 = 40 \)
Mean = \( \frac{1080}{40} = 27 \)
Q27. Find the median from cumulative frequency table:
| Marks ≤ x | Cumulative Frequency |
|---|---|
| 10 | 5 |
| 20 | 15 |
| 30 | 35 |
| 40 | 60 |
| 50 | 80 |
Total frequency = 80, Median class = 30–40 (since \( \frac{80}{2} = 40 \))
Use formula:
\( \text{Median} = L + \frac{\frac{N}{2} - CF}{f} \times h \)
L = 30, CF = 35, f = 25, h = 10
\( \text{Median} = 30 + \frac{40 - 35}{25} \times 10 = 30 + 2 = 32 \)
Q28. Find the mode:
| Marks | Frequency |
|---|---|
| 0–20 | 5 |
| 20–40 | 12 |
| 40–60 | 15 |
| 60–80 | 6 |
| 80–100 | 2 |
Modal class = 40–60
Formula:
\( \text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
L = 40, \( f_1 = 15, f_0 = 12, f_2 = 6 \), h = 20
\( \text{Mode} = 40 + \frac{15 - 12}{30 - 12 - 6} \times 20 \) \( = 40 + \frac{3}{12} \times 20 \) \(= 40 + 5 = 45 \)
Q29. Missing frequency x. Mean = 42.5, total freq = 50
| Class | Frequency |
|---|---|
| 0–10 | 4 |
| 10–20 | 6 |
| 20–30 | 10 |
| 30–40 | x |
| 40–50 | 15 |
| 50–60 | 5 |
Midpoints: 5, 15, 25, 35, 45, 55
\( \sum fx = 4×5 + 6×15 + 10×25 + x×35 \) \( + 15×45 + 5×55 = 20 + 90 + 250 \) \( + 35x + 675 + 275 = 1310 + 35x \)
Mean = 42.5 = \( \frac{1310 + 35x}{50} \)
Solve: \( 2125 = 1310 + 35x \) \( \Rightarrow x = \frac{815}{35} = 23.28 \)
Hence, x ≈ 23 (rounded)
Q30. Find mode of grouped data:
| Class | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 10 |
| 30–40 | 25 |
| 40–50 | 15 |
| 50–60 | 8 |
Modal class = 30–40, L = 30, \( f_1 = 25, f_0 = 10, f_2 = 15, h = 10 \)
\( \text{Mode} = 30 + \frac{25 - 10}{2×25 - 10 - 15} × 10 \) \(= 30 + \frac{15}{25} × 10 = 30 + 6 = 36 \)
Q31. A cone and a cylinder have the same base and height. Find the
ratio of their volumes.
Answer:
\( \text{Volume of Cone} = \frac{1}{3}\pi r^2 h, \) \( \text{Volume of
Cylinder} = \pi r^2 h \)
\( \text{Ratio} = \frac{1}{3}\pi r^2 h : \pi r^2 h = 1:3 \)
Q32. A solid metal cone of height 24 cm and radius 7 cm is melted and
recast into a cylinder of same radius. Find the height of the cylinder.
Answer:
\( \text{Volume of cone} = \frac{1}{3} \pi r^2 h \) \( = \frac{1}{3} \pi \cdot
7^2 \cdot 24 = 392\pi \)
\( \pi r^2 H = 392\pi \) \( \Rightarrow 49H = 392 \Rightarrow H = 8 \text{ cm}
\)
Q33. A solid metal cone of height 24 cm and radius 7 cm is melted and
recast into a cylinder of same radius. Find the height of the cylinder.
Answer:
\( \text{Volume of big sphere} = \frac{4}{3} \pi (7)^3 \) \( = \frac{4}{3} \pi
\cdot 343 = \frac{1372}{3} \pi \)
\( \text{Volume of one small sphere} = \frac{4}{3} \pi (1)^3 \) \( =
\frac{4}{3} \pi \)
\( \text{Number of small spheres} = \frac{1372}{4} = 343 \)
Q34. Find the total surface area of a cone whose radius is 5 cm and
slant height is 13 cm.
Answer:
\( \text{TSA} = \pi r(l + r) = \pi \cdot 5(13 + 5) \) \( = \pi \cdot 5 \cdot
18 = 90\pi \)
\( \text{TSA} = 282.74\ \text{cm}^2 \quad (\pi \approx 3.14) \)
Q35. If the ratio of the curved surface areas of a solid cone and a
solid right circular cylinder having some base radii and same height is 5:8,
then let us determine the ratio of their base radii and heights.
Answer:
\( \text{CSA of cone} = \pi r l, \quad \) \( \text{CSA of cylinder} = 2\pi r h
\)
\( \frac{\pi r l}{2\pi r h} = \frac{l}{2h} = \frac{5}{8} \Rightarrow l:h = 5:4
\)
Q36. A hemisphere is cut from the top of a cylinder. Radius is 7 cm,
height is 21 cm. Find surface area of the remaining solid.
Answer:
\( \text{CSA of cylinder} = 2\pi r h \) \( = 2\pi \cdot 7 \cdot 21 \) \( =
294\pi \)
\( \text{CSA of hemisphere} = 2\pi r^2 \) \( = 2\pi \cdot 49 \) \( = 98\pi
\)
\( \text{Total Surface Area} = 294\pi + 98\pi \) \( = 392\pi \) \( \approx
1231.5\ \text{cm}^2 \)
Q37. The base radius of a solid right circular cone is equal to the
length of the radius of a solid sphere. If the volume of the sphere is twice
of that of the cone, the let us write by calculating the ratio of the height
and base radius of the cone
Answer:
\( \frac{4}{3}\pi r^3 = 2 \cdot \frac{1}{3}\pi r^2 h \) \( \Rightarrow 4r = 2h
\Rightarrow h = 2r \Rightarrow h:r = 2:1 \)
Q38. A cone of height 24 cm and radius 7 cm is cut to form a frustum by
removing a cone of height 10 cm. Find the frustum volume.
Answer:
\( R = 7, H = 24, h = 10, \) \( r = \frac{7}{24} \times 10 = 2.92 \text{ cm}
\)
\( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \) \( \approx \frac{1}{3}\pi \cdot 14
(49 + 8.5 + 20.44) \) \( \approx 1031.25 \text{ cm}^3 \)
Q39. A tent is shaped like a cylinder surmounted by a cone. Radius is
10 m, cylinder height 6 m, cone slant height 13 m. Find canvas needed.
Answer:
\( \text{CSA of cylinder} = 2\pi r h \) \( = 2\pi \cdot 10 \cdot 6 \) \( =
120\pi \)
\( \text{CSA of cone} = \pi r l = \pi \cdot 10 \cdot 13 = 130\pi \)
\( \text{Total canvas} = 250\pi \approx 785\ m^2 \)
Q40.A toy is made by attaching a hemisphere to a cone. Radius is 3.5
cm, height of cone is 6 cm. Find the total surface area of the toy.
Answer:
\( l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 6^2} \) \( = \sqrt{12.25 + 36} =
\sqrt{48.25} \approx 6.95 \)
\( \text{TSA} = \pi r l + 2\pi r^2 \) \( = \pi \cdot 3.5 \cdot 6.95 + 2\pi
\cdot 3.5^2 \) \( = 24.325\pi + 24.5\pi \) \( = 48.825\pi \approx 153.3\
\text{cm}^2 \)