Madhyamik Class 10 Mathematics Solved Paper 2022

(a) P (1 + \text{r}\over100)ⁿ

(c) ₹ 3,000

Explanation:

Ratio of profit = 50 : 100 : 150

= 1 : 2 : 3

So, Capitals = 1x : 2x : 3x

Sum of ratio = 1x + 2x + 3x = 6x

or, 6x = 6000

or, x = 1000

∴ Smita’s capital = 3x = 3000

(d) 5 : 14

Explanation:

\text{A}\over \text{B} × \text{B}\over \text{C} × \text{C}\over \text{D}

= \frac{2 × 5 × 6}{3 × 8 × 7}

= 5 : 14

(d) 45°

Explanation:

Since PR = RQ, triangle PRQ is isosceles, and ∠PRQ = 90° (because PQ is the diameter).

So, ∠RPQ = ∠RQP = 45°.

(d) 4

Explanation:

If two circles are apart externally, then they have 2 external tangents and 2 internal tangents. Total = 4.

(a) 32\over 3r³ cm³

Explanation:

V = 4\over 3π(2r)³

= 4\over 3π × 2 × r³

= 32\over 3πr³

P(1 + r\over100)

Explanation:

After one year, amount = Principal × (1 + r\over100).

This amount becomes the 2nd year principal.

irrational number

Explanation:

√11 is irrational, and multiplying by 7 (a rational non-zero integer) keeps it irrational.

r³

Explanation:

Volume of a sphere = 4\over3πr³.

Hence v is directly proportional to r³.

in proportional

Explanation:

Condition of similarity: corresponding angles equal and corresponding sides in proportion.

cyclic quadrilateral

Explanation:

If ∠A + ∠C = 180° (or ∠B + ∠D = 180°), the quadrilateral can be inscribed in a circle.

cube

Explanation:

When all three dimensions of a cuboid are equal, it is called a cube.

False

Explanation:

A partnership can be formed with two or more persons. No rule requires at least 3.

True

Explanation:

Amount = Principal + Interest. Since Interest > 0,

Principal < Amount.

True

Explanation:

x² = 100

⇒ x = √100 = ± 10

So roots are + 10 and – 10.

False

Explanation:

Inverse variation means a ∝ 1\over b

⇒ ab = k (constant).

But a\over b is not constant; it varies with b.

False

Explanation:

Concentric circles (same center, different radii) have no common tangents. Tangents to one circle cannot be tangent to the other at the same time.

True

Explanation:

In a right circular cone:

l² = r² + h²

So (height h, radius r, slant height l) form a right-angled triangle.

Principal (P) = ₹ 690

Simple interest = 1\over16 × 690 = ₹ 43.125

Interest for 8 months = 8\over12 × 43.125

= ₹ 2\over3 × 43.125

= ₹ 28.75

Future population (V) = 17,280

Initial population (V_o) = 13,310

rate (r) = r %

Time (n) = 3 years

A = P × (1 + r/100)ⁿ

or, 17,280 = 13,310 × (1 + r/100)³

or, 17280\over 13310 = (1 + r/100)³

or, 1728\over 1331 = (1 + r/100)³

or, (12\over 11)³ = (1 + r/100)³

or,  1 + r/100 = 12\over 11

or, r/100 = 12\over 11 – 1

or, r = 1\over 11 × 100 = 9 1\over 11 %

The ratio of capitals of A, B and C = 1\over \text{x} : 1\over \text{y} : 1\over \text{z}

= 1\over \text{x} × xyz : 1\over \text{y} × xyz : 1\over \text{z} × xyz

= yz + xz + xy

The loss of C = \text{xy} \over \text{yz + xz + xy} × z

= \text{xyz} \over \text{yz + xz + xy}

Sum of roots (α + β) = 66\over 7

Produt of root (α β) = 27\over 7

Ratio of the sum and the product of two roots

= 66\over 7 : 27\over 7

= 66 : 27

= 22 : 9

= {12(\sqrt{15}+3)\over(\sqrt{15})^2-3^2}

= {12(\sqrt{15}+3)\over15-9}

= {12(\sqrt{15}+3)\over6}

= 2√15 + 6

Radius (r) = 13 cm

BM = 10\over 2 = 5 cm

Distance from the centre (OM) : OM² = 13² – 5²

or, OM² = 13² – 5² = 169 – 25 = 144

or, OM = √144 = 12 cm

Given:

• AOB is a diameter of the circle with center O.
• Point C lies on the circle.
• ∠OBC = 60°.

∠ACB = 90° (angle subtended by a diameter at any point on the circle is a right angle)

Triangle OBC is isosceles (OB = OC, both are radii of the circle)

Since ∠OBC = 60°, we also have ∠OCB = 60°.

∠OCA + ∠OCB + ∠ACB = 180° (The sum of angles in any triangle is 180°)

or, ∠OCA + 150° = 180°.

or, ∠OCA = 180° – 150° = 30°.

Let the radius of the circle be r.

The line from the center O to the point of tangency T (where the tangent meets the circle) is perpendicular to the tangent line PT at the point of tangency. Thus, we have a right-angled triangle OTP, where:

• OT = r (the radius of the circle),
• PT = 10 cm (the length of the tangent),
• OP = 26 cm (the distance from the point P to the center O).

Using the Pythagorean theorem:

OP² = OT² + PT²

Substitute the known values:

26² = r² + 10²

Simplify the equation:

676 = r² + 100

Now, solve for r²:

r² = 676 – 100 = 576

Finally, take the square root of both sides:

r = √576 = 24 cm

Thus, the length of the radius of the circle is 24 cm.

Given

• AD = 5 cm
• DB = 6 cm
• AE = 7.5 cm

AB = AD + DB

or, AB = 5 cm + 6 cm

or, AB = 11 cm

(AD / AB) = (AE / AC)   [Since DE || BC]

or, (5 / 11) = (7.5 / AC)

or, 5 × AC = 7.5 × 11

or, AC = (7.5 × 11) / 5

or, AC = 82.5 / 5

or, AC = 16.5 cm

h₁ : h₂ = 1 : 2 — (i)

Perimeters of the base = 3 : 4

C₁ : C₂ = 3 : 4

or, 2πr1 : 2πr2 = 3 : 4

or, r₁ : r₂ = 3 : 4 — (ii)

Ratio of their volumes = V₁ : V₂

= πr₁²h₁ : πr₂²h₂

= r₁²h₁ : r₂²h₂ 

= {r_1}²\over {r_1}² × {h_1}\over {h_2}

= ({3\over 4})² × {1\over 2}

= {9\over 16} × {1\over 2}

= 9 : 32

Original radius = r (let)

New radius = 1.5 r

Original Surface Area = 4πr²

New Surface Area = 4π(1.5r)² = 9πr²

Increase % = \text{9πr² − 4πr²}\over \text{4πr²} × 100% = = \text{5πr²}\over \text{4πr²} × 100% = 125%

Diagonal = 4√3

⇒ a√3 = 4√3 (where a = side)

⇒ a = 4

TSA = 6a² = 6 × 16 = 96 cm²

Let Principal = P, rate = r% per annum.

Amount after 7 years = 7,100

⇒ P + I₇ = 7,100 —- (i)

Amount after 4 years = 6,200

⇒ P + I₄ = 6,200 —- (ii)

Substracting (i) – (ii)

Interest for 3 years (I₃) = 7,100 – 6,200 = 900

Interest for 1 years (I₁) = 900 ÷ 3 = Rs. 300

Interest for 4 years (I₄) = Rs. 300 × 4 = Rs. 1,200

Substituting I₄ in equation (ii)

Principal = Amount − Interest = 6,200 − 1,200 = Rs. 5,000

Rate % = \frac{300}{5000} × 100 = 6%

∴ Principal = ₹ 5,000; Rate = 6% p.a.

Capitals Ratio = ₹8,000 : ₹10,000 : ₹12,000

= 4 : 5 : 6

Sum of Ratio = 4x + 5x + 6x = 15x

Total profit = ₹13,400

Bank installment = ₹5,000

Profit to share = ₹13,400 − ₹5,000 = ₹ 8,400

Profit share of Partner 1 = \frac{4x}{15x} × ₹ 8,400 = ₹ 2,240

Profit share of Partner 2 = \frac{5x}{15x} × ₹ 8,400 = ₹2,800

Profit share of Partner 3 = \frac{6x}{15x} × ₹ 8,400 = ₹3,360

Principal (P) = ₹ 20,000

Rate (r) = 5% p.a.

Time (t) = 2 years

Simple Interest (SI) = \text{P × r × t} \over 100

= 20,000 × 5 × 2 \over 100

= ₹ 2,000

Compound Interest (annual compounding) = P[(1 + {r\over100})² − 1]

= 20,000[(1 + {5\over100})² − 1]

= 20,000[({105\over100})² − 1]

= 20,000 × (11025\over10000 − 1)

= 20,000 × (11025 – 10000\over10000)

= 20,000 × (1025\over10000)

= ₹ 2,050

∴ Difference (CI − SI) = 2,050 − 2,000 = ₹ 50

Given: \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x} with x ≠ 0, – (a + b)

or, \frac{1}{a+b+x} – \frac{1}{x} = \frac{1}{a}+\frac{1}{b}

or, \frac{x – a – b – x}{(a+b+x)x} = \frac{a + b}{ab}

or, – \frac{(a + b)}{(a+b+x)x} = \frac{a + b}{ab}

or, – \frac{1}{(a+b+x)x} = \frac{1}{ab}

or, – ab = (a+b+x)x

or, x² + ax + bx + ab = 0

or, x (x + a) + b (x + a) = 0

or, (x + a)(x + b) = 0

x = -a or -b

Sum of roots = −4 + 3 = −1

Product of roots = (−4) × 3 = −12

Standard monic quadratic: x² − (sum)x + (product) = 0

⇒ x² − (−1)x + (−12) = 0

⇒ x² + x − 12 = 0

Given: m + 1\over m = √3

(a) m² + 1\over m² = (m + 1\over m )² – 2

= (√3)² – 2

= 3 – 2 = 1

(b) m³ + 1\over m³ = (m + 1\over m )³ – 3 (m + 1\over m )

= (√3)³ – 3 (√3)

= 3√3 – 3 √3

= 0

\frac{√5}{√3 + √2} = \frac{√5}{√3 + √2} × \frac{√3 – √2}{√3 – √2}

= \frac{√5(√3 – √2)}{(√3)² – (√2)²}

= \frac{√5(√3 – √2)}{3 – 2}

= √5(√3 – √2)

= √15 – √10

Given: a = \frac{√5 + 1}{√5 – 1}

⇒ a = \frac{√5 + 1}{√5 – 1} × \frac{√5 + 1}{√5 + 1}

⇒ a = \frac{(√5 + 1)²}{(√5)² – 1²}

⇒ a = \frac{(√5)² + 2√5 + (1)²}{(√5)² – 1²}

⇒ a = \frac{5 + 2√5 + 1}{5 – 1}

⇒ a = \frac{6 + 2√5}{4}

⇒ a = \frac{3 + √5}{2}

Now, ab = 1

⇒ b = \frac{1}{\text{a}}

⇒ b = \frac{2}{3 + √5}

⇒ b = \frac{2}{3 + √5} × \frac{3 – √5}{3 – √5}

⇒ b = \frac{2(3 – √5)}{(3)² – (√5)²}

⇒ b = \frac{2(3 – √5)}{4}

⇒ b = \frac{(3 – √5)}{2}

a + b = \frac{(3 + √5)}{2} + \frac{(3 – √5)}{2}

a + b = \frac{(3 + √5) + (3 – √5)}{2} = 3

a² + b² = (a + b)² – 2ab = 3² – 2 × 1 = 9

Now, (\frac{\text{a}}{\text{b}} + \frac{\text{b}}{\text{a}}) = (\frac{\text{a² + b²}}{\text{ab}})

= \frac{9}{1}

= 9

No. of farmers = A, no. of days = B & area of land = C

No. of days is in inverse variation with no. of farmers, when area of land remains constant

i.e., B ∝ \frac{1}{A} when C is constant

Again, No. of days is indirect variation with area of land; when No. of farmers remains constant

∴ B ∝ C when A is Constant.

According to the theorem on joint variation,

B ∝ \frac{C}{A} when C & A both vary

∴ B = K \frac{C}{A} where K is a constant of variation.

Given A = 15, B = 5, & C = 18.

5 = K \frac{18}{15}

or, K =   \frac{15 \times 5}{18} = \frac{25}{6}

⇒ B = K \frac{C}{A}

⇒ B = \frac{25}{6} \times \frac{12}{10} = 5

∴ No. of days = 5.

Given: a : b = b : c ⇒ ac = b²

abc = (ac)·b = b²·b = b³ — (i)

ab + bc + ca = b(a + c) + ac

ab + bc + ca = b(a + c) + b²

ab + bc + ca = b(a + b + c) — (ii)

Now, \frac{abc(a+b+c)^3}{(ab+bc+ca)^3} = \frac{b^3(a+b+c)^3}{\big(b(a+b+c)\big)^3} = 1

\dfrac{a}{1-a}+\dfrac{b}{1-b}+\dfrac{c}{1-c} = 1

or, \dfrac{a}{1-a} + 1 +\dfrac{b}{1-b} + 1 +\dfrac{c}{1-c} + 1 = 1 + 1 + 1 + 1

or, \dfrac{a + 1 – a}{1-a} + \dfrac{b + 1 – b}{1-b} + \dfrac{c + 1 – c}{1-c} = 4

or, \dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c} = 4

Given: ABCD is a cyclic quadrilateral of a circle with centre O.

To prove: ∠ABC + ∠ADC = 2 right angles and ∠BAD + ∠BCD = 2 right angles

Construction: A, O and C, O are joined.

Proof: The reflex angle ∠AOC at the centre and the angle ∠ABC on the circle are formed with the circular arc ADC.

∴ Reflex ∠AOC = 2∠ABC

∴ ∠ABC = ½ reflex ∠AOC ………(i)

Again, ∠AOC is the angle at the centre and ∠ADC is the angle on the circle formed with the circular arc ABC.

∴ ∠AOC = 2∠ADC

∴ ∠ADC = ½ ∠AOC ………(ii)

∴ From (i) and (ii), we get

∠ABC + ∠ADC = ½ reflex ∠AOC + ½ ∠AOC

= ½ (reflex ∠AOC + ∠AOC)

= ½ × 4 right angles = 2 right angles

Similarly, by joining B, O and D, O; it can be proved that, ∠BAD + ∠BCD = 2 right angles (Proved)

Given : AB is a chord of the circle with its centre at O, which is not a diameter, and OD, is perpendicular on the chord AB.

To prove : OD, bisects the chord AB i.e., AD = DB.

Construction : O, A and O, B are joined.

Proof : In △ODA and △ODB,

∠ODA = ∠ODB (Each is right-angle)

OA = OB [radii of same circle] and

OD = OD (common side)

∴ △ODA ≅ △ODB [By R-H -S axiom of congruency]

∴ AD = DB [Corresponding sides of the congruent triangles] [proved]

Two angles in circle produced by the arc BC are ∠BAC and ∠BDC.

∴ ∠BAC = ∠BDC …… (1)

Again, two angles in circle produced by the arc CE are ∠CAE and ∠CDE.

∴ ∠CAE = ∠CDE …… (2)

Now, ∠FDE + ∠CDE + ∠BDC = 1 straight angle = 180°.

or, ∠CDE + ∠CDE + ∠BDC = 180° [∵ DE is bisector of ∠FDC : ∠FDE = ∠CDE]

or, 2∠CDE + ∠BDC = 180° …… (3)

Again, (∠GAE + ∠CAE + ∠BAC) = 1 straight angle = 180° …… (4)

From (3) and (4) we get, 2∠CDE + ∠BDC = ∠GAE + ∠CAE + ∠BAC

or, 2∠CDE = ∠GAE + ∠CAE (from (1))

or, 2∠CAE = ∠GAE + ∠CAE (by the figure)

or, ∠CAE = ½ ∠GAC

∴ AE is the bisector of ∠GAC. Hence AE is the external bisector of ∠BAC.

Let AB and CD be two chords of a circle with centre O. When produced, AB and CD intersect at P.

To prove: ∠AOC − ∠BOD = 2 ∠BPC

Construction: Join B and C.

Proof: The central angle produced by arc AC is ∠AOC, and the angle in the circle subtended by the same arc is ∠ABC.

Therefore, ∠AOC = 2 ∠ABC …….. (1) [by theorem]

Similarly, the central angle produced by arc BD is ∠BOD, and the angle in the circle subtended by the same arc is ∠BCD.

Therefore, ∠BOD = 2 ∠BCD …….. (2)

Now, subtracting (2) from (1), we get:

∠AOC − ∠BOD = 2 ∠ABC − 2 ∠BCD …….. (3)

Again, in triangle BPC, external angle ∠ABC = ∠BPC + ∠BCP.

But here, ∠BCP = ∠BCD.

So, ∠ABC = ∠BPC + ∠BCD.

Multiplying both sides by 2,

2 ∠ABC = 2 ∠BPC + 2 ∠BCD …….. (4)

From (3) and (4),

∠AOC − ∠BOD = 2 ∠BPC (Hence proved)

Length = 2.1 m

Breadth = 1.5 m

Base area = 2.1 × 1.5 = 3.15 m²

Volume poured = 630 L = \frac{630}{1000} m³ = 0.630 m³

Increase in height = \frac{0.630}{3.15} m

= 0.2 m

= 20 cm

The water level rises by 20 cm.

Given: height = 2r; if height = 6r, volume increases by 539 dm³.

Initial volume = πr²h = πr²(2r) = 2πr³

New volume = πr²h = πr²(6r) = 6πr³

Increase in volume = 6πr³ – 2πr³ = 4πr³

ATP: Increase in volume = 539

⇒ 4πr³ = 539

⇒ r³ = 539 × {7\over 22 × 4}

⇒ r³ = {343\over 8}

⇒ r = 3.5 dm

Height (actual) = 2r = 7 dm

Base area needed for 11 person = 11 × 4 = 44 m²

Air (volume) needed for 11 person = 11 × 20 = 220 m³

For a cone:

base area = πr² = 44

⇒ r² = \frac{44}{\pi}

Volume = 220 m³

⇒ \frac{1}{3}πr²h = 220

⇒ \frac{1}{3} × π × \frac{44}{\pi} × h = 220

⇒ 44 × \frac{h}{3} = 220

⇒ h = \frac{220 \times 3}{44} = 15 m

Radius of big sphere = 8 cm

Radius of each marble = 1 cm

Formula: Volume of a sphere = \frac{4}{3}πr³

Number of marbles = Volume (big) ÷ Volume (one marble)

= \frac{\tfrac{4}{3}\pi\cdot 8^3}{\tfrac{4}{3}\pi\cdot 1^3}

= 8³ = 512

Inner dimensions: 7.5 dm × 6 dm × 5.4 dm

Volume of tea box = 7.5 × 6 × 5.4 = 243 dm³

Weight filled = 52 kg 350 g = 52.35 kg

Empty weight = 3.75 kg

Weight of tea = 52.35 − 3.75 = 48.60 kg

Weight per 1 dm³ = \frac{48.6}{243} kg

= 0.2 kg

= 200 g

∴ 0.2 kg (200 g) per cubic dm

Given: The two legs (sides containing the right angle) have known lengths.

Construction:

1. Draw AB equal to one given leg.
2. At A, construct a line perpendicular to AB. On this perpendicular, mark AC equal to the other given leg.
3. Join BC to form right △ABC with the right angle at A.
4. Construct the perpendicular bisector of BC to locate M, the midpoint of the hypotenuse BC.
5. With center M and radius MB (or MC), draw a circle. This circle passes through A, B, and C — the circumcircle.

Construction:

1. Draw the given circle with center O and mark the external point P.
2. Join OP.
3. Find M, the midpoint of OP (use the perpendicular bisector of OP).
4. With center M and radius OM, draw a circle having diameter OP.
5. Let this circle cut the given circle at T.
6. Join PT. PT is the required tangent (if you need only one, use this PT; the second tangent comes from the other intersection).

Substitute in 2x² + ax + 8 = 0:

2(1)² + a(1) + 8 = 0

2 + a + 8 = 0

a + 10 = 0

a = −10

Given: In 10 years, interest = \frac{2}{5} of the amount.

Let principal = P, rate = r% p.a., interest in 10 years = I.

Relation with amount:

I = \frac{2}{5}(P + I)

⇒ 5I = 2P + 2I

⇒ 3I = 2P

⇒ I = \frac{2}{3}P

Simple interest formula for 10 years:

I = P × r × 10 / 100

So, P × r × 10 / 100 = \frac{2}{3}P

⇒ r × 10 / 100 = \frac{2}{3}

⇒ r = \frac{2}{3}\times \frac{100}{10}=\frac{20}{3}

√8 = 2√2

√18 = 3√2

√27 = 3√3

√72 = 6√2

Not similar: √27 (the others are multiples of √2).

Diameter = 20 cm ⇒ radius (r) = 10 cm

Slant height (l) = 25 cm

Total Surface Area (TSA) of cone = πrl + πr²

= π × 10 × 25 + π × 10²

= 250π + 100π

= 350π cm²

TSA = 350 × \frac{22}{7} = 1100 cm²

For time = 1 year (same rate, same principal):

Simple interest = Pr/100

Compound interest = Pr/100

So, CI = SI for 1 year.

90°

Explanation: An angle standing on a semicircle (i.e., subtended by a diameter) is a right angle.

The arc is 180°, and an inscribed angle is half of its arc ⇒ 180° ÷ 2 = 90°.

Let the third Proportion be x

5, 10, x ⇒ 5, 10, 10, x

Product of mean = 10 × 10 = 100

Product of extreme = 5 × x = 5x

Now, Product of extreme = Product of mean

⇒ 5x = 100

⇒ x = 20

∴ The third Proportion is 20.

Sphere

Explanation: A sphere has one continuous curved surface with no edges or vertices—just a single surface all around.

They are supplementary (sum = 180°).

Explanation:

Each pair of opposite angles subtends arcs that together make the full circle (360°); an inscribed angle is half the arc it subtends, so each opposite pair adds to 180°.