Madhyamik Mathematics Board Paper 2026 | Full Solution | English Version

🔢 Madhyamik 2026 Board Question Paper - Mathematics

Prime Maths · Vinod Singh 9038126497

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📌 1. Choose the correct option in each case from the following questions:  [ 1 x 6 = 6 ]

Q1(i). If a principal becomes double in 10 years, the rate of simple interest per annum is:

1. 5%
2. 10%
3. 15%
4. 20%

📘 view stepwise solution

Step 1 – define variables: Let principal \(P\). Amount after 10 years \(A = 2P\).
Step 2 – interest earned: \(SI = A - P = 2P - P = P\).
Step 3 – apply simple interest formula: \(SI = \dfrac{P \times R \times T}{100}\) with \(T = 10\).
\[ P = \frac{P \times R \times 10}{100} \]
Step 4 – cancel \(P\) (since \(P \neq 0\)): \(1 = \dfrac{R \times 10}{100}\)
Step 5 – solve for \(R\): \(1 = \dfrac{R}{10} \implies R = 10\%\).
✅ correct option: (b) 10%

Q1(ii). Condition of two roots of \(ax^2+bx+c=0\;(a>0)\) equal in magnitude but opposite in sign:

1. \(b=0,c=0\)
2. \(b=0,c>0\)
3. \(b=0,c<0\)
4. \(b>0,c=0\)

📘 view stepwise solution

Step 1 – let the roots be \(\alpha\) and \(-\alpha\).
Step 2 – sum of roots: \(\alpha + (-\alpha) = 0 = -\dfrac{b}{a} \implies b = 0\) (since \(a\neq0\)).
Step 3 – product of roots: \(\alpha \cdot (-\alpha) = -\alpha^2 = \dfrac{c}{a}\).
Step 4 – sign analysis: \(\alpha^2 > 0\) (real roots), so \(-\alpha^2 < 0\). Hence \(\dfrac{c}{a} < 0\).
Step 5 – given \(a>0\), therefore \(c < 0\).
✅ correct option: (c) \(b=0,\;c<0\)

Q1(iii). If average of \(6,7,x,y,16\) is \(9\). Then:

1. \(x+y=21\)
2. \(x+y=16\)
3. \(x-y=21\)
4. \(x-y=19\)

📘 view stepwise solution

Step 1 – average formula: \(\dfrac{6+7+x+y+16}{5} = 9\).
Step 2 – sum of known numbers: \(6+7+16 = 29\). So \(\dfrac{29+x+y}{5} = 9\).
Step 3 – multiply by 5: \(29+x+y = 45\).
Step 4 – solve: \(x+y = 45-29 = 16\).
✅ correct option: (b) \(x+y=16\)

Q1(iv). Arc of length \(121\) cm of a circle makes \(77^\circ\) angle at the centre of the circle, then the radius of the circle will be. Radius = ?

1. 110 cm
2. 100 cm
3. 90 cm
4. 70 cm

📘 view stepwise solution

Step 1 – arc length formula: \(L = 2\pi r \times \dfrac{\theta}{360^\circ}\). Use \(\pi = \dfrac{22}{7}\).
Step 2 – substitute: \(121 = 2 \cdot \dfrac{22}{7} \cdot r \cdot \dfrac{77}{360}\).
Step 3 – simplify inside: \(2\cdot\dfrac{22}{7} = \dfrac{44}{7}\). Then \(\dfrac{44}{7} \times \dfrac{77}{360} = \dfrac{44 \times 77}{7 \times 360}\).
Step 4 – cancel 77 with 7: \(\dfrac{77}{7} = 11\), so expression becomes \(\dfrac{44 \times 11}{360} = \dfrac{484}{360}\).
Step 5 – equation: \(121 = r \cdot \dfrac{484}{360}\).
Step 6 – solve for \(r\): \(r = \dfrac{121 \times 360}{484}\). Note \(484 = 4 \times 121\). Thus \(r = \dfrac{360}{4} = 90\) cm.
✅ correct option: (c) 90 cm

Q1(v). Length of a side of a cube be \('a'\) unit and length of the diagonal be \('d'\) unit then the relation of a and d will be:

1. \(\sqrt{2}a = d\)
2. \(\sqrt{3}a = d\)
3. \(a = \sqrt{3}d\)
4. \(a = \sqrt{2}d\)

📘 view stepwise solution

Step 1 – space diagonal of a cube: \(d = \sqrt{a^2 + a^2 + a^2}\).
Step 2 – simplify: \(d = \sqrt{3a^2} = a\sqrt{3}\).
Step 3 – rewrite: \(\sqrt{3}\,a = d\).
✅ correct option: (b) \(\sqrt{3}a = d\)

Q1(vi). \(ABCD\) be a cyclic quadrilateral whose centre is \(O\). \(BC\) is extended upto \(E\). If \(\angle DCE = 96°\) then the value of \(\angle BOD\) will be:

1. 42°
2. 84°
3. 142°
4. 168°

📘 view stepwise solution

Step 1 – linear pair: \(\angle BCD + \angle DCE = 180^\circ\) (straight line BCE). So \(\angle BCD = 180^\circ - 96^\circ = 84^\circ\).
Step 2 – central angle theorem: Angle at centre (\(\angle BOD\)) is twice the angle at circumference subtended by the same arc BAD. Here \(\angle BCD\) subtends arc BAD.
Step 3 – calculate: \(\angle BOD = 2 \times \angle BCD = 2 \times 84^\circ = 168^\circ\).
✅ correct option: (d) 168°

📝 2. Fill in the blanks (any five) [ 1 x 5 = 5]

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(i) If the ratio of principal and yearly amount be 8:9, then yearly rate of interest is _________.

🔍 detailed solution

Let \(P = 8k\), \(A = 9k\). Interest for 1 year \(= A-P = k\). Simple interest formula: \(k = \dfrac{8k \times R \times 1}{100}\). Cancel \(k\): \(1 = \dfrac{8R}{100} \Rightarrow R = \dfrac{100}{8} = 12.5\%\).
\(\Rightarrow\) \(12\frac{1}{2}\%\).

(ii) Conjugate surd of \(\sqrt{3} - 5\) is ______.

🔍 detailed solution

Write as \(-5 + \sqrt{3}\). Conjugate: change sign of the irrational part \(\rightarrow -5 - \sqrt{3}\).

(iii) Two tangents at the end point of a diameter of a circle are mutually _________.

🔍 detailed solution

Parallel (radius ⟂ tangent, both radii are collinear, so tangents are perpendicular to the same line → parallel).

(iv) If \(x = a\sec\theta,\; y = b\cot\theta\), then \(\dfrac{x^2}{a^2} - \dfrac{b^2}{y^2} =\) ______.

🔍 detailed solution

\(\dfrac{x}{a} = \sec\theta,\; \dfrac{y}{b} = \cot\theta \Rightarrow \dfrac{b}{y} = \tan\theta\). Then \(\dfrac{x^2}{a^2} - \dfrac{b^2}{y^2} = \sec^2\theta - \tan^2\theta = 1\).

(v) The radius of a solid hemisphere is \(3r\) unit, the area of total surface is _________.

🔍 detailed solution

TSA of hemisphere (solid) = \(3\pi R^2\). Here \(R = 3r\) → \(3\pi (3r)^2 = 3\pi \cdot 9r^2 = 27\pi r^2\) sq units.

(vi) The frequency of \(1, 2, 3, 4, 5\) are respectively \(1, 2, 3, 4, f\) and their arithmetic mean is \(4\) then value of \(f\) is _________.

🔍 detailed solution

\(\sum fx = 1\cdot1 + 2\cdot2 + 3\cdot3 + 4\cdot4 + 5f = 1+4+9+16+5f = 30+5f\).
\(\sum f = 1+2+3+4+f = 10+f\). Mean = \(\frac{30+5f}{10+f} = 4\).
Cross multiply: \(30+5f = 40+4f \Rightarrow f = 10\).

🔎 3. True or False (any five) [ 1 x 5 = 5 ]

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(i) \(\sin^2\theta = (\sin\theta)^2\) for \(0° \theta<90°\).

explanation

True – conventional notation: \(\sin^2\theta\) means \((\sin\theta)^2\).

(ii) The length of a side of a largest cube be \(4\sqrt{2}\) cm inscribed in a sphere of radius 4 cm.

explanation

False. Sphere diameter = 8 cm = space diagonal of cube \(\Rightarrow \sqrt{3}\,a = 8 \Rightarrow a = 8/\sqrt{3} \approx 4.618\) cm. \(4\sqrt{2}\approx5.657\) cm (that would be face diagonal).

(iii) The angle in the segment of a circle which is greater than semicircle is an obtuse angle.

explanation

False. Major segment ( > semicircle) contains an acute angle; obtuse angle lies in minor segment.

(iv) If the Arithmetic mean of \(x-3, x-1, 7, x, 2x-1, 3x-5\) be \(7.5\), then their median will be \(3\).

explanation

Sum = \((x-3)+(x-1)+7+x+(2x-1)+(3x-5) = 8x -3\). Mean = \(\frac{8x-3}{6}=7.5 \Rightarrow 8x-3=45 \Rightarrow x=6\).
Data: 3,5,7,6,11,13 → sorted 3,5,6,7,11,13 → median = \(\frac{6+7}{2}=6.5 \neq 3\). False.

(v) If \(x \propto 1/y\) then \((xy)^{10}\) is a constant.

explanation

True. \(xy = k\) (constant) ⇒ \((xy)^{10}=k^{10}\) constant.

(vi) In a business, the ratio of capital of Raju and Asif is \(5:4\). If Raju got Rs. 80 of total profit then Asif got Rs. 100.

explanation

False. Profit ratio = capital ratio = 5:4. So Asif = \(\frac{4}{5}\times 80 = 64\).

✍️ 4. Answer any ten (2 marks each) [ 2 x 10 = 20]

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(i) A and B started a business by investing Rs. 15,000 and Rs. 45,000 respectively. After 6 months B got a profit of Rs. 3,030, what is the profit of A?

stepwise solution

Capital ratio A:B = 15000:45000 = 1:3. Profits shared in same ratio (no time adjustment as period same).
Let A's profit = \(x\), then B's profit = \(3x\). Given \(3x = 3030 \Rightarrow x = 1010\). So A gets ₹1010.

(ii) In \(\triangle ABC\), line parallel to BC intersects AB, AC at P, Q. AP=4 cm, QC=9 cm, PB=AQ. Find PB.

stepwise solution

By Basic Proportionality Theorem: \(\dfrac{AP}{PB} = \dfrac{AQ}{QC}\). Set \(PB = AQ = x\). Then \(\dfrac{4}{x} = \dfrac{x}{9} \Rightarrow x^2 = 36 \Rightarrow x = 6\) (positive). PB = 6 cm.

(iii) Two Chords AB and CD are equidistant from centre O. If ∠AOB = 60° and CD = 6 cm, find radius.

stepwise solution

Equidistant chords ⇒ equal lengths: AB = CD = 6 cm. In ΔAOB, OA=OB (radii), ∠AOB=60° ⇒ triangle is equilateral (isosceles with vertex 60° ⇒ base angles each 60°). Hence radius = OA = AB = 6 cm.

(iv) If \(\tanθ + \cotθ = 2\) then find \(\tan^7θ + \cot^7θ\).

stepwise solution

\(\tanθ + \frac{1}{\tanθ}=2\). Multiply by \(\tanθ\): \(\tan^2θ + 1 = 2\tanθ \Rightarrow \tan^2θ -2\tanθ +1=0 \Rightarrow (\tanθ-1)^2=0 \Rightarrow \tanθ=1\). So \(\cotθ=1\). Then \(\tan^7θ+\cot^7θ = 1+1 = 2\).

(v)  If \(x\) and \(y\) are positive real numbers then \(\sec \theta = \frac{x}{y}\) is correct or not? Give reason.

stepwise solution

No. For real θ between \(0^{\circ}\) and \(90^{\circ}\), \(0 \leq \cos \theta \leq 1\). As  \(\sec \theta = \frac{x}{y}\) which means \(\cos \theta = \frac{y}{x}\) and for \(y > x \), \( \cos \theta > 1 \) which is not possible.

(vi) For two right circular cylinders, if ratio of their heights be \(1:2\) and ratio of the circumference of the base be \(3:4\) then find the ratio of their volume.

stepwise solution

Let \(h_1:h_2 = 1:2\). Circumference \(2πr_1 : 2πr_2 = 3:4 \Rightarrow r_1:r_2 = 3:4\). Volume = \(πr^2h\). Ratio = \(\left(\frac{r_1}{r_2}\right)^2 \times \frac{h_1}{h_2} = \left(\frac{3}{4}\right)^2 \times \frac{1}{2} = \frac{9}{16}\times\frac12 = \frac{9}{32}\).

(vii) Arithmetic mean of \(x_1,x_2,......,x_n\) is \(\bar x\). Prove that \(\sum (x_i-\bar{x})^2 = \sum x_i^2 - n\bar{x}^2 \).

stepwise proof

\(\sum (x_i-\bar{x})^2 = \sum (x_i^2 - 2x_i\bar{x} + \bar{x}^2) = \sum x_i^2 - 2\bar{x}\sum x_i + \sum \bar{x}^2\).
Now \(\sum x_i = n\bar{x}\) and \(\sum \bar{x}^2 = n\bar{x}^2\). Hence expression = \(\sum x_i^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2 = \sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 = \sum x_i^2 - n\bar{x}^2\).

(viii)If the rate of interest increases from 5.5% to 6% then the yearly interest increases by ₹49.50. Find the capital.

stepwise solution

Increase in rate = \(0.5\%\). This increase applies to the whole capital: \(0.5\% \times \text{capital} = 49.50\). So \(\frac{0.5}{100} \times C = 49.50 \Rightarrow C = 49.50 \times \frac{100}{0.5} = 49.50 \times 200 = 9900\).

(ix)If the sum of the roots of the equation \(x^2-4x = K(x-1)-5\) is 7. Find the value of \(K\).

stepwise solution

Expand RHS: \(Kx - K -5\). Bring all terms: \(x^2 -4x - Kx + K +5 =0 \Rightarrow x^2 -(4+K)x + (K+5)=0\). Sum of roots = \(4+K\). Given \(4+K = 7 \Rightarrow K=3\).

(x) If \((a+b):\sqrt{ab}=2:1\) then find \(a:b\).

stepwise solution

\(\frac{a+b}{\sqrt{ab}} = 2 \Rightarrow a+b = 2\sqrt{ab}\). Square both sides: \((a+b)^2 = 4ab \Rightarrow a^2+2ab+b^2 = 4ab \Rightarrow a^2-2ab+b^2=0 \Rightarrow (a-b)^2=0 \Rightarrow a=b\). Ratio \(1:1\).

(xi) If the radius of the sphere is increased by 50% find the percentage increase of the volume.

stepwise solution

New radius = \(1.5r\). Volume ∝ \(r^3\). New volume = \((1.5)^3 = 3.375\) times original. Increase = \(3.375-1 = 2.375\) → \(237.5\%\).

(xii) \(ABCD\) is a cyclic quadrilateral. If \(AD=AB\), ∠DAC=60°, ∠BDC=50° then find the ∠ACD.

stepwise solution

1. Angles in same segment: \(\angle BAC = \angle BDC = 50^\circ\) (subtended by arc BC).
2. So \(\angle DAB = \angle DAC + \angle CAB = 60^\circ+50^\circ=110^\circ\).
3. In \(\triangle ADB\), \(AD=AB\) ⇒ base angles equal: \(\angle ABD = \angle ADB = \frac{180^\circ-110^\circ}{2}=35^\circ\).
4. Again, \(\angle ACD = \angle ABD\) (angles subtended by arc AD). Hence \(\angle ACD = 35^\circ\).

📊 5. Arithmetic (answer any one)

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(i) If the rate of compound interest be 4% in the 1st year and 5% in the 2nd year, then find the interest of Rs. 25,000 for two years.

stepwise solution

Amount = \(25000 \times \left(1+\frac{4}{100}\right) \times \left(1+\frac{5}{100}\right) = 25000 \times 1.04 \times 1.05\).
\(1.04\times1.05 = 1.092\). So \(A = 25000\times1.092 = 27300\). Interest = \(27300-25000 = 2300\).

(ii) Three friends invest Rs. 4,800, Rs. 6,600 and Rs. 9,600. respectively in a business. 1^st person received 1/8 th of the profit as salary for looking after the business and the remaining profit  was distributed among them in ratio of their capitals. If after one year 1^st person received Rs. 780 find the amount recieved by other two.

stepwise solution

Capitals ratio = 48:66:96 = 8:11:16 (sum 35). Let total profit = \(P\). Salary = \(P/8\). Remaining = \(7P/8\) divided in ratio 8:11:16.
1st's extra = \(\frac{8}{35}\times\frac{7P}{8} = \frac{P}{5}\). So 1st total = \(\frac{P}{8} + \frac{P}{5} = \frac{5P+8P}{40} = \frac{13P}{40}\). Given \(\frac{13P}{40}=780 \Rightarrow P = 2400\).
Remaining after salary = \(2400 - 2400/8 = 2400-300 = 2100\).
2nd gets \(\frac{11}{35}\times2100 = 660\); 3rd gets \(\frac{16}{35}\times2100 = 960\).

🧩 6. Algebra – quadratic eq. (any one)

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(i) Solve: \(b(c-a)x^2 + c(a-b)x + a(b-c)=0\).

stepwise solution

Check sum of coefficients: \(b(c-a)+c(a-b)+a(b-c) = bc - ab + ac - bc + ab - ac = 0\). Hence one root is \(1\).
Product of roots = \(\frac{a(b-c)}{b(c-a)}\). Thus other root = \(\frac{a(b-c)}{b(c-a)}\).
Alternatively: \(b(c-a)x^2 + c(a-b)x + a(b-c)=0 \) \(\implies b(c-a)x^2 -(b(c-a)+a(b-c))x + a(b-c)=0\) \(\implies b(c-a)x^2 -b(c-a)x-a(b-c)x + a(b-c)=0\) \(\implies b(c-a)x(x-1)-a(b-c)(x-1)=0\) Now complete!

(ii) Digit in ten's place of a two digit number is less by 3 than the digit in the unit place. Product of digits is less than the number by 15. Find the number.

stepwise solution

Let units = \(y\), tens = \(x = y-3\). Number = \(10x+y = 10(y-3)+y = 11y-30\).
Product \(xy = (y-3)y = y^2-3y\). Given \(xy = (11y-30) - 15\) → \(y^2-3y = 11y-45\).
Rearr: \(y^2 -14y +45=0\) ⇒ \((y-5)(y-9)=0\) ⇒ \(y=5\) or \(9\). Then numbers: \(25\) or \(69\).

⚡ 7. Algebra – variation / surds (any one)

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(i) If \((x^3+y^3)\propto (x^3-y^3)\), then prove that \((x^2+y^2)\propto xy\).

stepwise proof

\(\frac{x^3+y^3}{x^3-y^3}=k\) (constant). Apply componendo-dividendo: \(\frac{(x^3+y^3)+(x^3-y^3)}{(x^3+y^3)-(x^3-y^3)} = \frac{k+1}{k-1}\) ⇒ \(\frac{2x^3}{2y^3} = \frac{k+1}{k-1}\) ⇒ \(\left(\frac{x}{y}\right)^3 = \text{constant}\). Hence \(\frac{x}{y}=c\) (constant). Then \(\frac{x^2+y^2}{xy} = \frac{c^2+1}{c}\) = constant ⇒ \((x^2+y^2)\propto xy\).

(ii) If \(x(2-\sqrt3)=y(2+\sqrt3)=1\) then find the value of  \(3x^2-5xy+3y^2\).

stepwise solution

\(x = \frac{1}{2-\sqrt3}\). Rationalise: multiply numerator and denominator by \(2+\sqrt3\) ⇒ \(x = 2+\sqrt3\).
\(y = \frac{1}{2+\sqrt3} = 2-\sqrt3\). Then \(x+y=4,\; xy=1\).
\(3x^2-5xy+3y^2 = 3(x^2+y^2)-5\). Now \(x^2+y^2 = (x+y)^2 - 2xy = 16-2=14\). So expression = \(3\times14-5 = 42-5=37\).

⚖️ 8. Ratio & proportion (any one)

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(i) If \(\frac{a+b-c}{a+b}=\frac{b+c-a}{b+c}=\frac{c+a-b}{c+a}\) and \(a+b+c \neq 0\), then prove that \(a=b=c\).

stepwise proof

Subtract each fraction from 1: \(1-\frac{a+b-c}{a+b} = \frac{c}{a+b}\); similarly \(\frac{a}{b+c}\) and \(\frac{b}{c+a}\). So \(\frac{c}{a+b}=\frac{a}{b+c}=\frac{b}{c+a}\). Add 1 to each: \(\frac{a+b+c}{a+b}=\frac{a+b+c}{b+c}=\frac{a+b+c}{c+a}\). Assuming \(a+b+c\neq0\), denominators equal ⇒ \(a+b=b+c=c+a\) ⇒ \(a=b=c\).

(ii) If \(x=\frac{8ab}{a+b}\), then find the value of \(\frac{x+4a}{x-4a}+\frac{x+4b}{x-4b}\).

stepwise solution

\(\frac{x}{4a} = \frac{2b}{a+b}\). Apply componendo-dividendo: \(\frac{x+4a}{x-4a} = \frac{2b+(a+b)}{2b-(a+b)} = \frac{a+3b}{b-a}\).
Similarly \(\frac{x}{4b} = \frac{2a}{a+b}\) ⇒ \(\frac{x+4b}{x-4b} = \frac{2a+(a+b)}{2a-(a+b)} = \frac{3a+b}{a-b} = -\frac{3a+b}{b-a}\).
Sum = \(\frac{a+3b - (3a+b)}{b-a} = \frac{-2a+2b}{b-a} = 2\).

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🧠 For more problems keep visiting Prime Maths

📐 9. Geometry theorems (any one)

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(i) Prove that the angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle.

📘 view stepwise proof

Given: Circle with centre \(O\). Arc \(AB\) subtends \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part.
To prove: \(\angle AOB = 2\angle ACB\).
Construction: Join \(CO\) and extend it to point \(D\).
Proof:

• In \(\triangle AOC\), \(OA = OC\) (radii) ⇒ \(\angle OAC = \angle OCA\).
• Exterior angle \(\angle AOD = \angle OAC + \angle OCA = 2\angle OCA\).
• In \(\triangle BOC\), \(OB = OC\) ⇒ \(\angle OBC = \angle OCB\).
• Exterior angle \(\angle BOD = \angle OBC + \angle OCB = 2\angle OCB\).
• Adding: \(\angle AOD + \angle BOD = 2(\angle OCA + \angle OCB)\).
• But \(\angle AOD + \angle BOD = \angle AOB\) and \(\angle OCA + \angle OCB = \angle ACB\).
• Hence \(\angle AOB = 2\angle ACB\). ∎

(ii) If two circles touch each other, prove that the point of contact lies on the straight line joining their centres.

📘 view stepwise proof

Given: Two circles with centres \(A\) and \(B\) touching at point \(P\).
To prove: \(A\), \(P\), \(B\) are collinear.
Construction: Draw a common tangent \(ST\) at \(P\).
Proof:

• Radius \(AP\) is perpendicular to tangent \(ST\) ⇒ \(\angle APT = 90^\circ\).
• Radius \(BP\) is perpendicular to tangent \(ST\) ⇒ \(\angle BPT = 90^\circ\).
• Thus \(\angle APT + \angle BPT = 180^\circ\).
• Angles \(\angle APT\) and \(\angle BPT\) form a linear pair on line \(ST\).
• Therefore, \(AP\) and \(BP\) are along the same straight line ⇒ \(A\), \(P\), \(B\) are collinear. ∎

🔧 10. Geometry applications (any one)

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(i) In an isosceles \(\triangle ABC\), \(\angle B = 90^\circ\). The bisector of \(\angle BAC\) meets \(BC\) at \(D\). Prove that \(CD^2 = 2\,BD^2\).

📘 view stepwise solution

Given: \(AB = BC\) (isosceles right‑angled at \(B\)). \(AD\) bisects \(\angle BAC\).
To prove: \(CD^2 = 2 BD^2\).
Proof:

• Let \(AB = BC = a\). Then hypotenuse \(AC = \sqrt{a^2 + a^2} = a\sqrt{2}\).
• By angle‑bisector theorem in \(\triangle ABC\): \[ \frac{BD}{CD} = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}. \]
• Hence \(CD = \sqrt{2}\, BD\).
• Squaring both sides: \(CD^2 = 2 BD^2\). ∎

(ii) \(O\) is any point inside rectangle \(ABCD\). Prove that \(OA^2 + OC^2 = OB^2 + OD^2\).

📘 view stepwise solution

Construction: Through \(O\) draw lines parallel to the sides, meeting \(AB, BC, CD, DA\) at \(P, Q, R, S\) respectively.
Proof:

• By Pythagoras in right triangles: \[ \begin{aligned} OA^2 &= AP^2 + OP^2,\\ OC^2 &= CQ^2 + OQ^2,\\ OB^2 &= BP^2 + OP^2,\\ OD^2 &= DQ^2 + OQ^2. \end{aligned} \]
• Because of the parallel lines, \(AP = BQ\) and \(CQ = PD\).
• Adding the first two: \(OA^2 + OC^2 = AP^2 + OP^2 + CQ^2 + OQ^2\).
• Adding the last two: \(OB^2 + OD^2 = BP^2 + OP^2 + DQ^2 + OQ^2\).
• But \(AP^2 = BQ^2\) and \(CQ^2 = DQ^2\). Hence both sums are equal. ∎

🛠️ 11. Construction (any one)

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(i) Construct the circumcircle of \(\triangle ABC\) where \(BC = 6\ \text{cm}\), \(\angle ABC = 60^\circ\) and \(AB = 8\ \text{cm}\). (Write the main steps.)

📘 view construction steps

• Draw a line segment \(BC = 6\ \text{cm}\).
• At \(B\), construct \(\angle CBA = 60^\circ\) and cut \(AB = 8\ \text{cm}\) from the ray.
• Join \(AC\) to complete \(\triangle ABC\).
• Draw the perpendicular bisectors of sides \(AB\) and \(BC\). Let them intersect at \(O\).
• With centre \(O\) and radius \(OA\) (or \(OB\) or \(OC\)), draw the circle – this is the required circumcircle.

(ii) Construct a square equal in area to an equilateral triangle of side \(6\ \text{cm}\). (Write the main steps.)

📘 view construction steps

• Draw an equilateral triangle \(ABC\) with side \(6\ \text{cm}\).
• Compute (or measure) its altitude: \(h = \sqrt{6^2 - 3^2} = \sqrt{36-9} = \sqrt{27} = 3\sqrt{3}\ \text{cm}\).
• Area of triangle = \(\frac{1}{2}\times 6 \times 3\sqrt{3} = 9\sqrt{3}\ \text{cm}^2\).
• Construct a rectangle with length \(6\ \text{cm}\) and width \(3\sqrt{3}\ \text{cm}\) (using a right triangle to obtain \(\sqrt{3}\)).
• Now construct a square with area equal to that rectangle using the mean‑proportional method:
  • Extend the rectangle’s length \(L\) by its width \(W\).
  • Draw a semicircle on the total segment as diameter.
  • Erect a perpendicular at the junction; its length to the circle gives the side of the square.

📐 12. Trigonometry (any two)

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(i) If the ratio of three angles of a triangle is \(2:3:4\) then determine the circular value of the greatest angle.

📘 view stepwise solution

Step 1: Let the angles be \(2x,\ 3x,\ 4x\). Sum = \(2x+3x+4x = 9x\).
Step 2: In a triangle, \(9x = 180^\circ\) ⇒ \(x = 20^\circ\).
Step 3: Greatest angle = \(4x = 80^\circ\).
Step 4: Convert to radians: \(80^\circ \times \frac{\pi}{180} = \frac{4\pi}{9}\).
✅ Greatest angle = \(\frac{4\pi}{9}\) radians

(ii) If \(\tan\theta = \frac{4}{3}\), find \(\sin\theta + \cos\theta\).

📘 view stepwise solution

Step 1: Construct a right triangle with opposite = \(4\), adjacent = \(3\). Hypotenuse = \(\sqrt{4^2+3^2} = \sqrt{16+9}=5\).
Step 2: \(\sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{4}{5}\), \(\cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{3}{5}\).
Step 3: \(\sin\theta + \cos\theta = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}\).
✅ \(\sin\theta + \cos\theta = \frac{7}{5}\)

(iii) If \(A\) and \(B\) are complementary angles, prove that \((\sin A + \cos B)^2 = 1 + 2\sin A \sin B\).

📘 view stepwise proof

Given: \(A + B = 90^\circ\). Hence \(\cos B = \sin A\) and \(\sin B = \cos A\).
Proof: \[ \begin{aligned} \text{LHS} &= (\sin A + \cos B)^2 \\ &= (\sin A + \sin A)^2 \quad (\because \cos B = \sin A)\\ &= (2\sin A)^2 = 4\sin^2 A. \end{aligned} \] Wrong Problem!!. Usually the intended statement is: \((\sin A + \sin B)^2 = 1 + 2\sin A \sin B\) because \(\sin B = \cos A\). Let's verify the standard form: \[ (\sin A + \sin B)^2 = \sin^2 A + \sin^2 B + 2\sin A\sin B. \] Since \(\sin^2 A + \sin^2 B = \sin^2 A + \cos^2 A = 1\), we get \(1 + 2\sin A\sin B\).
If the problem statement uses \(\cos B\) instead of \(\sin B\), then note \(\cos B = \sin A\) and the expression becomes \((\sin A + \sin A)^2 = 4\sin^2 A\), which is not generally equal to \(1+2\sin A\sin B\).
Assuming the intended identity uses \(\sin B\) (common in complementary angle problems):
\[ (\sin A + \sin B)^2 = 1 + 2\sin A \sin B \quad\text{(proved above).} \] ✅ Full marks on attempt.

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📏 13. Height and Distance (any one)

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(i) From the roof of a building, the angles of depression of the top and foot of a lamp post are \(30^\circ\) and \(\theta^\circ\) respectively. The ratio of the heights of the building and the lamp post is \(3:2\), then find the value of \(\theta\).

📘 view stepwise solution

Step 1 – assign variables: Let height of building \(H = 3x\), height of lamp post \(h = 2x\).
Step 2 – vertical difference: The top of the lamp is \(3x - 2x = x\) below the roof.
Step 3 – horizontal distance \(d\): From roof, angle of depression \(30^\circ\) to top of lamp ⇒ \(\tan 30^\circ = \frac{x}{d}\).
\[ \frac{1}{\sqrt{3}} = \frac{x}{d} \quad\Rightarrow\quad d = x\sqrt{3}. \] Step 4 – angle \(\theta\) to foot of lamp: From roof, foot is at depth \(3x\) below, same horizontal distance \(d\).
\[ \tan\theta = \frac{3x}{d} = \frac{3x}{x\sqrt{3}} = \sqrt{3}. \] Step 5 – solve for \(\theta\): \(\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ\).
✅ \(\theta = 60^\circ\)

(ii) From the foot of a Tilla the angle of elevation of the top is \(45^\circ\). By moving \(100\) m towards the Tilla along a slope of \(30^\circ\), the angle of elevation of the top becomes\(60^\circ\). Find the height of the Tilla.

📘 view stepwise solution

Step 1 – Main triangle: The distance of the point from the foot of the Tilla is \(h\), since \(\tan 45^{\circ}=1=\frac{h}{distance}\)
Step 2 – movement along slope: From the starting point walk \(100\) m at \(30^\circ\) upward slope (towards the Tilla). The displacement components: Horizontal distance \( = 100\cos30^\circ = 100 \times \frac{\sqrt3}{2} = 50\sqrt3\), Vertical distance \(= 100\sin30^\circ = 50\). So the new position is at a horizontal distance of \(h-50\sqrt{3}\) from the Tilla and its vertical distance from the top of the Tilla is \(h-50\)
Step 3 – elevation from \(B\): Angle of elevation to top of Tilla is \(60^\circ\). \[ \tan60^\circ = \frac{h-50}{h - 50\sqrt3}. \] Step 4 – solve for \(h\): \(\tan60^\circ = \sqrt3\): \[ \sqrt3 = \frac{h-50}{h - 50\sqrt3} \implies \sqrt3\,(h - 50\sqrt3) = h - 50 \implies \sqrt3\,h - 150 = h - 50. \] \[ (\sqrt3 - 1)h = 100 \quad\Rightarrow\quad h = \frac{100}{\sqrt3 - 1}. \] Step 5 – rationalise: Multiply numerator and denominator by \(\sqrt3 + 1\): \[ h = \frac{100(\sqrt3+1)}{3-1} = 50(\sqrt3 + 1)\ \text{m}. \] ✅ Height \(= 50(\sqrt3 + 1)\) m

🧊 14. Mensuration (any two)

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(i) The ratio of the length, breadth and height of a solid rectangular parallelopiped is \(4:3:2\) and area of the whole surface is \(468\ \text{cm}^2\). Find the volume of the parallelopiped.

📘 view stepwise solution

Step 1 – let dimensions: Let length \(l = 4x\), breadth \(b = 3x\), height \(h = 2x\).
Step 2 – total surface area (TSA): \[ TSA = 2(lb + bh + hl) = 2\bigl(4x\cdot3x + 3x\cdot2x + 2x\cdot4x\bigr) = 2(12x^2 + 6x^2 + 8x^2) = 2 \times 26x^2 = 52x^2. \] Step 3 – given TSA = 468: \[ 52x^2 = 468 \quad\Rightarrow\quad x^2 = 9 \quad\Rightarrow\quad x = 3. \] Step 4 – actual dimensions: \(l = 12\ \text{cm},\ b = 9\ \text{cm},\ h = 6\ \text{cm}\).
Step 5 – volume: \[ V = l \times b \times h = 12 \times 9 \times 6 = 648\ \text{cm}^3. \] ✅ Volume = \(648\ \text{cm}^3\)

(ii) The internal and external radius of a hollow cylinder of height \(20\ \text{cm}\) are \(4\ \text{cm}\) and \(5\ \text{cm}\) respectively. By meltingthis cylinder a solid cone of height equal to one‑third the height of the cylinder is formed. Find the diameter of the base of the cone.

📘 view stepwise solution

Step 1 – volume of hollow cylinder: \[ V_{\text{cyl}} = \pi h_{\text{cyl}} (R^2 - r^2) = \pi \times 20 \times (5^2 - 4^2) = 20\pi (25-16) = 20\pi \times 9 = 180\pi\ \text{cm}^3. \] Step 2 – height of cone: \(h_{\text{cone}} = \frac{1}{3}\times 20 = \frac{20}{3}\ \text{cm}\).
Step 3 – volume of cone: \(V_{\text{cone}} = \frac13 \pi r_{\text{cone}}^2 h_{\text{cone}}\). Equate volumes: \[ \frac13 \pi r_{\text{cone}}^2 \times \frac{20}{3} = 180\pi. \] Cancel \(\pi\) and simplify: \[ \frac{20}{9} r_{\text{cone}}^2 = 180 \quad\Rightarrow\quad r_{\text{cone}}^2 = 180 \times \frac{9}{20} = 81. \] Step 4 – radius and diameter: \(r_{\text{cone}} = 9\ \text{cm}\) (positive). Diameter = \(2\times 9 = 18\ \text{cm}\).
✅ Base diameter = \(18\ \text{cm}\)

(iii) A hemispherical bowl of internal radius \(9 \text{cm}\) is full of water. How many cylindrical bottles each of diameter \(3\ \text{cm}\) and height \(4\ \text{cm}\) are required to fill this water?

📘 view stepwise solution

Step 1 – volume of hemispherical bowl: \[ V_{\text{bowl}} = \frac23 \pi R^3 = \frac23 \pi (9)^3 = \frac23 \pi \times 729 = 486\pi\ \text{cm}^3. \] Step 2 – volume of one cylindrical bottle: radius \(r = \frac{3}{2} = 1.5\ \text{cm}\), height \(h = 4\ \text{cm}\). \[ V_{\text{bottle}} = \pi r^2 h = \pi (1.5)^2 \times 4 = \pi \times 2.25 \times 4 = 9\pi\ \text{cm}^3. \] Step 3 – number of bottles: \[ n = \frac{V_{\text{bowl}}}{V_{\text{bottle}}} = \frac{486\pi}{9\pi} = 54. \] ✅ 54 bottles are required

📊 15. Statistics (any two)

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(i) Find the arithmetic mean for the following frequency distribution:

Class	5-14	15-24	25-34	35-44	45-54	55-64
Frequency	3	6	18	20	10	3

📘 view stepwise solution

Step 1 – find midpoints (class marks): \[ x_1 = \frac{5+14}{2} = 9.5,\; x_2 = 19.5,\; x_3 = 29.5,\; x_4 = 39.5,\; x_5 = 49.5,\; x_6 = 59.5. \] Step 2 – compute \(f_i x_i\): \[ \begin{aligned} f_1x_1 &= 3 \times 9.5 = 28.5,\\ f_2x_2 &= 6 \times 19.5 = 117,\\ f_3x_3 &= 18 \times 29.5 = 531,\\ f_4x_4 &= 20 \times 39.5 = 790,\\ f_5x_5 &= 10 \times 49.5 = 495,\\ f_6x_6 &= 3 \times 59.5 = 178.5. \end{aligned} \] Step 3 – total frequency and sum of products: \[ \sum f = 3+6+18+20+10+3 = 60, \] \[ \sum fx = 28.5 + 117 + 531 + 790 + 495 + 178.5 = 2140. \] Step 4 – arithmetic mean: \[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2140}{60} \approx 35.666\ldots \approx 35.67. \] ✅ Mean \(\approx 35.67\)

(ii) For the following frequency distribution, construct a 'greater than' cumulative frequency table and draw the ogive (more than type).

Class	100-120	120-140	140-160	160-180	180-200
Frequency	8	14	10	12	4

📘 view stepwise solution

Step 1 – compute total frequency: \[ N = 8 + 14 + 10 + 12 + 4 = 48. \] Step 2 – obtain 'greater than' cumulative frequencies:

• More than 100 : \(48\)
• More than 120 : \(48 - 8 = 40\)
• More than 140 : \(40 - 14 = 26\)
• More than 160 : \(26 - 10 = 16\)
• More than 180 : \(16 - 12 = 4\)
• More than 200 : \(4 - 4 = 0\) (optional)

Step 3 – points to plot for the ogive: \[ (100,\,48),\quad (120,\,40),\quad (140,\,26),\quad (160,\,16),\quad (180,\,4). \] Step 4 – drawing the ogive: On a graph sheet, take the upper limits (or the lower limits for more‑than) on the x‑axis and cumulative frequencies on the y‑axis. Plot the points and join them with a smooth freehand curve. The resulting graph is the 'more than' ogive.
✅ Table and points ready for plotting.

(iii) Find the mode of the following frequency distribution:

Marks obtained	less than 10	less than 20	less than 30	less than 40	less than 50	less than 60
Number of students	8	15	29	42	60	70

📘 view stepwise solution

Step 1 – convert to ordinary frequency distribution: The given cumulative frequencies are for "less than" limits. Subtract successive cumulatives to obtain class frequencies.

• \(0-10\) : \(8\)
• \(10-20\): \(15 - 8 = 7\)
• \(20-30\): \(29 - 15 = 14\)
• \(30-40\): \(42 - 29 = 13\)  (this is \(f_0\))
• \(40-50\): \(60 - 42 = 18\)  (this is \(f_1\) – modal class)
• \(50-60\): \(70 - 60 = 10\)  (this is \(f_2\))

Step 2 – identify modal class: The highest frequency is \(18\) in the class \(40-50\). So modal class = \(40-50\) with \[ l = 40,\quad f_1 = 18,\quad f_0 = 13,\quad f_2 = 10,\quad h = 10. \] Step 3 – apply mode formula: \[ \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 40 + \frac{18 - 13}{2\times18 - 13 - 10} \times 10. \] Step 4 – simplify: \[ 2f_1 - f_0 - f_2 = 36 - 13 - 10 = 13,\quad f_1-f_0 = 5. \] \[ \text{Mode} = 40 + \frac{5}{13} \times 10 = 40 + \frac{50}{13} \approx 40 + 3.846 = 43.846. \] ✅ Mode \(\approx 43.85\) (rounded to two decimals)

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📈 Prime Maths

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