SOLUTION OF SECTION D - solvedpapers.net

Mathematics Standard 2025| Code (30-1-1) | CBSE Class 10 Board Solved Papers

📌 SECTION D

LA · 4 Questions · 5 marks each

Long Answer type – internal choice in Q33 & Q34

Q32.
💰 Simple Interest · Investment

Vijay invested in schemes A (8% p.a.) and B (9% p.a.). Total annual interest = ₹1860. If amounts interchanged, interest would be ₹20 more. Find investments in each scheme.

🔍 VIEW SOLUTION

✅ ANSWER: Scheme A = ₹12000, Scheme B = ₹10000

Let investment in A = ₹x, in B = ₹y.
\(\frac{8x}{100} + \frac{9y}{100} = 1860\) \(8x + 9y = 186000 (i) \quad\)
Interchanged: \(\frac{9x}{100} + \frac{8y}{100} = 1880\) \(9x + 8y = 188000 (ii) \quad\)
Add (i) & (ii): \(17x + 17y = 374000\) \(x+y = 22000\)
Subtract (ii)-(i): \(x – y = 2000\)
Solving: \(x = 12000,\)\( y = 10000\).

Q33.(a)
🔷 Parallelogram · Similarity

The diagonal BD of parallelogram ABCD intersects line segment AE at F, where E is on BC. Prove that \(DF \times EF = FB \times FA\).

Triangle

🔍 VIEW PROOF

In \(\triangle ADF\) and \(\triangle EBF\):
\(\angle DFA = \angle EFB\) (vert. opp.)
\(\angle ADF = \angle FBE\) (alt. int. angles, AD ∥ BC)
\( \triangle ADF \sim \triangle EBF\) (AA)
\(\frac{DF}{FB} = \frac{FA}{EF}\) ⇒ \(DF \times EF = FB \times FA\).

OR 33.(b)
📐 Right Triangle Proof

In \(\triangle ABC\), \(AD \perp BC\) and \(AD^2 = BD \times DC\). Prove that \(\angle BAC = 90^\circ\).

Right Triangle

🔍 VIEW PROOF

\(AD^2 = BD \times DC\) \(\Rightarrow \frac{AD}{DC} = \frac{BD}{AD}\).
\(\angle ADB = \angle ADC = 90^\circ\).
\(\triangle DBA \sim \triangle DAC\) (SAS).
So, these two triangles will be equiangular.\[
\angle 1 = \angle C \quad \text{and} \quad \angle 2 = \angle B
\]\[
\angle 1 + \angle 2 = \angle B + \angle C
\]\[
\angle A = \angle B + \angle C \quad \ldots(i)
\]By angle sum property,\[
\angle A + \angle B + \angle C = 180^\circ
\]

\[
\angle A + \angle A = 180^\circ \quad \ldots
\]

From (i) we get

\[
2\angle A = 180^\circ
\]

\[
\angle A = \angle BAC = 90^\circ
\]

Q34.(a)
📐 Right Triangle

Perimeter of right triangle = 60 cm, hypotenuse = 25 cm. Find lengths of other two sides.

🔍 VIEW SOLUTION

✅ Sides: 15 cm, 20 cm

Let sides = x, y. \(x+y+25=60\) \(\Rightarrow y=35-x\).
Pythagoras: \(x^2 + y^2 = 625\) ⇒ \(x^2 + (35-x)^2 = 625\)
⇒ \(x^2 + 1225 -70x + x^2 = 625\) ⇒ \(2x^2 -70x +600=0\) ⇒ \(x^2-35x+300=0\)
⇒ \((x-20)(x-15)=0\) ⇒ x = 20 or 15. Sides: 20 & 15 cm.

OR 34.(b)
🚆 Train · Speed

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.

🔍 VIEW SOLUTION

✅ Speed = 40 km/h

Let speed = \(x\) km/h. Time = \(\frac{480}{x}\) h. New speed = \(x-8\) km/h, time = \(\frac{480}{x-8}\) h.
\(\frac{480}{x-8} – \frac{480}{x} = 3\) ⇒ \(480\left(\frac{x – (x-8)}{x(x-8)}\right) = 3\) ⇒ \(\frac{3840}{x(x-8)} = 3\)
⇒ \(x(x-8) = 1280\) ⇒ \(x^2 -8x -1280 = 0\) ⇒ \((x-40)(x+32)=0\) ⇒ \(x=40\) km/h.

Q35.
📊 Mean & Mode

Find missing frequency ‘f’ if mean = 18. Hence find the mode.

Daily Allowance	Number of Children
11-13	7
13-15	6
15-17	9
17-19	13
19-21	f
21-23	5
23-25	4

🔍 VIEW SOLUTION

✅ f = 20, Mode ≈ 19.95

Class	xᵢ	fᵢ	fᵢxᵢ
11-13	12	7	84
13-15	14	6	84
15-17	16	9	144
17-19	18	13	234
19-21	20	f	20f
21-23	22	5	110
23-25	24	4	96
Total		44+f	752+20f

Mean = \(\frac{752+20f}{44+f} = 18\) ⇒ \(752+20f = 792 + 18f\) ⇒ \(2f = 40\) ⇒ \(f=20\).
Modal class = 19-21 (highest frequency 20).
Mode = \(19 + \frac{20-13}{22} \times 3\) \(= 19 + \frac{7}{22} \times 3\) \( = 19 + 0.9545 \approx 19.95\).

🏁 END OF SECTION D · LA (5 marks each) 🏁
✅ 4 Long Answer questions – fully solved, click to reveal