🔢 Madhyamik 2026 Board Question Paper - Mathematics
Prime Maths · Vinod Singh 9038126497
📌 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:
- 5%
- 10%
- 15%
- 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:
- \(b=0,c=0\)
- \(b=0,c>0\)
- \(b=0,c<0\)
- \(b>0,c=0\)
📘 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:
- \(x+y=21\)
- \(x+y=16\)
- \(x-y=21\)
- \(x-y=19\)
📘 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 = ?
- 110 cm
- 100 cm
- 90 cm
- 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:
- \(\sqrt{2}a = d\)
- \(\sqrt{3}a = d\)
- \(a = \sqrt{3}d\)
- \(a = \sqrt{2}d\)
📘 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:
- 42°
- 84°
- 142°
- 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]
(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 ]
(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]
(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)
(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.
1st 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 1st 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)
(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)
(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)
(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 = \(\frac{a+3b - (3a+b)}{b-a} = \frac{-2a+2b}{b-a} = 2\).
🧠 For more problems keep visiting Prime Maths
📐 9. Geometry theorems (any one)
(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)
(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)
(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)
(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)
(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 \(H-h = 3x - 2x = x\) 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)
(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)
(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)
📈 Prime Maths
