📌 SECTION D
Questions 35 to 40 · Internal choice in Q35, Q36, Q38, Q39
A train travels 480 km at uniform speed. If speed were 8 km/h less, it would take 3 hours more. Find speed of train.
OR
Find value of K so that \(kx(x-2)+6=0\) has equal roots.
🔍 VIEW ANSWER
✅ Speed = 40 km/h | OR k = 6
Main: \(\frac{480}{x-8} – \frac{480}{x} = 3\) \(⇒ 480\left(\frac{8}{x(x-8)}\right)=3\) \(⇒ x^2-8x-1280=0\) \(⇒ (x-40)(x+32)=0\) \(⇒ x=40\).
OR:
\(kx^2-2kx+6=0\). For equal roots, D=0 \(⇒ 4k^2-24k=0\) \(⇒ 4k(k-6)=0\) \(⇒ k=6\) (k≠0).
A solid iron pole: cylinder height 220 cm, diameter 24 cm, surmounted by another cylinder height 60 cm, radius 8 cm. Find mass (1 cm³ = 8g, π=3.14).
OR
From a solid cylinder (h=2.4 cm, d=1.4 cm), a conical cavity of same height/diameter is hollowed out. Find total surface area of remaining solid.
🔍 VIEW ANSWER
✅ Mass ≈ 892.3 kg | OR TSA ≈ 18 cm²
Main: 
V₁ = π×12²×220
= 31680π,
V₂ = π×8²×60
= 3840π,
Total = 35520π cm³.
Mass = 8×35520×3.14
= 892262.4 g ≈ 892.3 kg.
OR:
r=0.7 cm, h=2.4 cm,
l=√(h²+r²)=2.5 cm.
CSA cone = πrl = 5.5 cm²,
CSA cylinder = 2πrh = 10.56 cm²,
base area = πr² = 1.54 cm².
Total = 10.56+1.54+5.5 = 17.6 ≈ 18 cm².
From top of 7 m building, angle of elevation of top of cable tower is 60° and angle of depression of foot is 45°. Find height of tower.
🔍 VIEW ANSWER
✅ Height = 7(√3+1) m
Let building
AB = 7 m.
In right ΔABC,
tan45° = AB/BC
⇒ BC = 7 m = AD.
In right ΔADE,
tan60° = DE/AD
⇒ DE = 7√3 m.
Tower height = DE + CD
= 7√3 + 7
= 7(√3+1) m.
Evaluate: \(\frac{5\cos^2 60° + 4\sec^2 30° – \tan^2 45°}{\sin^2 30° + \cos^2 30°}\)
OR
Prove: \(\frac{\sin\theta – 2\sin^3\theta}{2\cos^3\theta – \cos\theta} = \tan\theta\),
🔍 VIEW ANSWER
✅ Value = \(\frac{67}{12}\)
Main: \(\frac{5×\frac14 + 4×\frac43 – 1}{\frac14 + \frac34} \)\(= \frac{\frac54 + \frac{16}{3} – 1}{1} \)\(= \frac{15+64-12}{12}\)\( = \frac{67}{12}\). OR:
\(\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}\)\( = \tan\theta \cdot \frac{\cos^2\theta – \sin^2\theta}{\cos^2\theta – \sin^2\theta} \)\(= \tan\theta\).
If a line intersects sides AB and AC of ΔABC at D and E respectively and is parallel to BC, prove that \(\frac{AD}{AB} = \frac{AE}{AC}\).
OR
A vertical pole 6 m casts a shadow 4 m. At same time, a tower casts a shadow 28 m. Find height of tower.
🔍 VIEW ANSWER
✅ Proof | OR Tower height = 42 m
Main: DE ∥ BC \(⇒\frac{AD}{DB}\)\( = \frac{AE}{EC}\) \(⇒ \frac{DB}{AD}\)\( = \frac{EC}{AE}\) \(⇒\frac{DB}{AD}+1\)\( = \frac{EC}{AE}+1\) \(⇒ \frac{AB}{AD} \)\(= \frac{AC}{AE}\) \(⇒ \frac{AD}{AB}\)\( = \frac{AE}{AC}\).
OR:
By similarity, \(\frac{6}{h}\)\( = \frac{4}{28}\) \(⇒ h = 6×7\) \(= 42 m.\)
If median of distribution is 28.5, find x and y:
🔍 VIEW ANSWER
✅ x = 8, y = 7
45 + x + y = 60
⇒ x + y = 15.
Median = 28.5
⇒ median class 20-30. l=20,
cf = 5+x, f=20, h=10, n/2=30. 28.5 = 20 + \(\frac{30-(5+x)}{20}×10\)
⇒ 8.5 = \(\frac{25-x}{2}\)
⇒ 17 = 25-x
⇒ x=8.
Then
y = 15-8 = 7.