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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Important Questions for ICSE, CBSE Board Examination

📘 A comprehensive, interactive revision tool for Class 10 Mathematics. Features 30 curated problems across all major units with step-by-step hints to aid understanding. Ideal for self‑assessment and board preparation. Test yourself before the finals!
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30 Challenging Problems for ICSE, CBSE and other State Boards

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Chapter Test Pair of Linear Equation in Two Variables CBSE ICSE

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Pair of Linear Equations in Two Variables

This quiz offers a thorough evaluation of key concepts and skills related to Linear Equations in Two Variables. 25 multiple-choice questions designed to test both theoretical understanding and practical application.

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• This quiz contains 25 multiple choice questions.
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• Use the navigator to jump between questions.
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Question 1

For the system of equations \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) to have a unique solution, the condition is:

A\(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) B\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) C\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) D\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Question 2

If a pair of linear equations is consistent, then the lines representing them are:

AParallel BAlways coincident CIntersecting or coincident DAlways intersecting

Question 3

For what value of \(k\) do the equations \(3x - y + 8 = 0\) and \(6x - ky = -16\) represent coincident lines?

A\(k = 2\) B\(k = -2\) C\(k = \frac{1}{2}\) D\(k = - \frac{1}{2}\)

Question 4

The pair of equations \(x = a\) and \(y = b\) graphically represents lines which are:

AParallel BIntersecting at \((b, a)\) CCoincident DIntersecting at \((a, b)\)

Question 5

If the system \(2x + 3y = 5\) and \(4x + ky = 10\) has infinitely many solutions, then \(k\) equals:

A\(1\) B\(\frac{3}{2}\) C\(6\) D\(3\)

Question 6

The value of \(k\) for which the system \(x + 2y = 3\) and \(5x + ky + 7 = 0\) is inconsistent is:

A\(k = 10\) B\(k = \frac{2}{5}\) C\(k = -\frac{2}{5}\) D\(k = -10\)

Question 7

Solve: \(\frac{x}{a}+\frac{y}{b}=2\) and \(ax-by=a^2-b^2\)

A\(x=a, \quad y=b^2\) B\(x=a, \quad y=b\) C\(x=a^2, \quad y=b^2\) D\(x=a^2, \quad y=b\)

Question 8

The value of \(k\) for which the pair of linear equations \((3k+1)x+3y=2\) and \((k^2+1)x+(k-2)y-5=0\) has no solution.

A\(k=-2\) B\(k=2\) C\(k=1\) D\(k=-1\)

Question 9

Solve for \(x\) and \(y\): \((a-b)x + (a+b)y = a^2 -2ab-b^2\) and \((a+b)(x+y) = a^2 + b^2\)

A\(x = a+b, y = -\frac{2ab}{a+b}\) B\(x = a+b, y = \frac{2ab}{a+b}\) C\(x = a-b, y = -\frac{2ab}{a+b}\) D\(x = a-b, y = \frac{2ab}{a+b}\)

Question 10

The value of \(x\) satisfying \(\frac{5}{x-1}+\frac{1}{y-2}=2\) and \(\frac{6}{x-1}-\frac{3}{y-2}-1=0, \quad x \neq 1, y \neq 2\) is:

A2 B3 C4 D5

Question 11

If \(37x + 43y = 123\) and \(43x + 37y = 117\), then:

A\(x = 2, y = 3\) B\(x = 1, y = 2\) C\(x = 3, y = 2\) D\(x = -1, y = -2\)

Question 12

The pair of equations \(225x + 881y = 2^{10}\) and \(450x + 1762y = 2048\) has:

ANo solution BUnique solution CTwo solutions DInfinitely many solutions

Question 13

If \(\frac{b}{a}x + \frac{a}{b}y = a^2+b^2\) and \(x + y = 2ab\) then

A\(x=2y\) B\(x=-y\) C\(x=y\) D\(2x=y\)

Question 14

If \(2^x + 3^y = 17\) and \(2^{x+2} - 3^{y+1} = 5\), find \(x\) and \(y\).

A\(x = 9, y = 5\) B\(x = 3, y = 2\) C\(x = 5, y = 9\) D\(x = 8, y = 6\)

Question 15

Solve for \(x\): \(\frac{2}{x} + \frac{3}{y} = 13\) and \(\frac{5}{x} - \frac{4}{y} = -2\).

A\(x = 1/2\) B\(x = 1/3\) C\(x = 2\) D\(x = 3\)

Question 16

For what value of \(p\) does the pair of equations \(4x + py + 8 = 0\) and \(2x + 2y + 2 = 0\) have a unique solution?

A\(p = 4\) B\(p \neq 4\) C\(p = 2\) D\(p \neq 2\)

Question 17

If \(49x + 51y = 499\) and \(51x + 49y = 501\), then \(x\) is:

A3 B5.5 C4 D6.5

Question 18

The difference between two numbers is 26 and one number is three times the other. Find them.

A30, 4 B36, 10 C39, 13 D40, 14

Question 19

Solve: \( \sqrt{2}x + \sqrt{3}y = 0 \) and \( \sqrt{3}x - \sqrt{8}y = 0 \).

A\(x=1, y=1\) B\(x=\sqrt{2}, y=\sqrt{3}\) C\(x=0, y=0\) D\(x=2, y=3\)

Question 20

Find \(k\) if the lines \(x + 2y - 1 = 0\) and \(2x + ky + 2 = 0\) are parallel.

A\(k=2\) B\(k=3\) C\(k=4\) D\(k=-4\)

Question 21

Five years ago, Ramesh was thrice as old as Sonu. Ten years later, Ramesh will be twice as old as Sonu. What is Ramesh's present age?

A40 years B50 years C60 years D45 years

Question 22

A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.

A\(7/9\) B\(9/7\) C\(5/9\) D\(3/7\)

Question 23

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

A18 B27 C81 D72

Question 24

A taxi charge consists of a fixed charge together with the charge for the distance covered. For 10 km, the charge paid is ₹105 and for 15 km, the charge paid is ₹155. What are the fixed charges?

A₹ 10 B₹ 5 C₹ 15 D₹ 8

Question 25

Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water.

A6 km/h B4 km/h C8 km/h D10 km/h

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