Madhyamik Class 10 Mathematics Solved Paper 2023

(a) ₹ \frac{\text{xz}}{\text{3x+y}}

Explanation

Ratio of the share = x : 2x : y

Total amount = x + 2x + y = 3x + y

Profit = z

A’s profit shares = \text{\text{x}}\over \text{\text{3x + y}} × z

= \frac{\text{xz}}{\text{3x+y}}

(b) 2

Explanation

x² =  x

or, x² –  x = 0

or, x(x – 1) = 0

or, x = 0 and 1

(b) 2

(a) 9

Explanation

5 + 4 sin θ [ax value of sin θ = 1]

= 5 + 4 × 1

= 9

(b) 9 : 4

(d) 4

Explanation

Mean = \frac{∑\text{fx}}{∑\text{f}}

or, 5.4 = \frac{4(p - 2) + 5(p + 1) + 7(p - 1)}{(p - 2) + (p + 1) + (p - 1)}

or, 5.4 = \frac{\text{6p - 10}}{\text{3p - 2}}

or, 16p – 10 = 5.4 (3p – 2)

or, 16p – 16.2p = -10.8 + 10

or, 0.2p = 0.8

or, p = 4

10 %

Explanation

Simple Interest = 198 – 180 = Rs 18

Rate (r) = 18 × 100\over 180 × 1 = 10 %

± 2abc

Explanation

\text{a²bc}\over \text{x}=\text{x} \over \text{4bc}

or, x² = 4a²b²c²

or, x = √4a²b²c² = ± 2abc

1\over√2.

Explanation

tan θ cos 60° = \frac{\sqrt{3}}{2}

or, tan θ × 1\over 2 = \frac{\sqrt{3}}{2}

or, tan θ = √3

or, tan θ = tan 60º

or, θ = 60º

sin(θ – 15°) = sin(60º – 15°)

= sin 45°

= 1\over√2

90º

11

Explanation

7, 8, 9, 10, 11, 11, 13, 15, 16

Here n = 9, which is odd

Median = n + 1\over 2 th Observation

= 9 + 1\over 2 th Observation

= 10\over 2 th Observation

= 5th Observation = 11

Cone and Cylinder

True

Explanation

In the compound interest if the rate of interest in the first three years is r₁,%, r₂%, 2r₃% respectively, then the amount for the principal P at the end of three years is  P \big(1+ \frac{ r_{1} }{100}) \big(1+ \frac{ r_{2} }{100}) \big(1+ \frac{ r_{3}}{100})

True

Explanation

sin 54°

= sin (90° – 36°)

= cos 36°

False

Explanation

This statement is false. As maximum of two tangents can be drawn on a circle from an external point.

True

Explanation

2ab : c², bc : a² and ca : 2b²

= (ab × bc × ca) : (c² × a² × b²)

= c² × a² × b² : c² × a² × b² = 1 : 1

True

Explanation

CSA of sphere = Volume of Sphere

4πr² = 4\over3πr³

or, 1 = 1\over3r

or, r = 3 units

False

Explanation

the mode is 2 as well as 5

SI = 2\over 5 of principal

time = 5 years

Rate = {2\over 5}P × 100\over P × 5 = 8%

Ratio of capital = \frac{1}{7} : \frac{1}{4}

= 4 : 7

Sum of ratio = 4 + 7 = 11

Profit of A = \frac{4}{11} × 11000 = ₹ 4000

Profit of B = \frac{7}{11} × 11000 = ₹ 7000

x² – x = k(2x – 1)

or, x² – x = 2kx – k

or, x² – (2k + 1)x + k = 0

Sum of roots = α + β = 2k + 1

ATP : 2k + 1 = 0

k = –\frac{1}{2}

b ∝ a³

or, b = ka³

b₁ : b₂ = ka₁³ : ka₂³

or, b₁ : b₂ = 2³ : 3³ = 8 : 27

Hence, b must increase in ratio 8 : 27

Given, AB and CD are two chords of a circle

ABCD is a cyclic quadrilateral,

So, the opposite angles in a cyclic quadrilateral is equal to 180°

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BAD = 180° – ∠BCD

⇒ ∠BAD = 180° – ∠PCB — (i) [∵ ∠BCD = ∠PCB]

From the fig, ∠PAD + ∠BAD = 180° (Linear pair)

⇒ ∠PAD = 180° – ∠BAD

From the equation (i)

⇒ ∠PAD = 180° – (180° – ∠PCB)

∴ ∠PAD = ∠PCB (Proved)

AL = x – 2

AC = 2x + 3

BM = x – 3

BC = 2x

ΔMCL ∼ ΔBCA

\text{AL}\over \text{AC} = \text{BM}\over \text{BC}

or, \text{x - 2}\over \text{2x + 3} = \text{x - 3}\over \text{2x}

or, 2x² – 4x = 2x² – 6x + 3x – 9

or, – 4x + 6x – 3x =-9

or, x = 9 Ans

Given X and Y are two circles touch each other externally at C. AB is the common tangent to the circles X and Y at point A and B respectively.

To find: ∠ACB

Proof: Let P be a point on AB such that, PC is at right angles to the line joining the centers of the circles.

Note that, PC is a common tangent to both circles.

This is because the tangent is perpendicular to the radius at the point of contact for any circle.

Let ∠PAC = α and ∠PBC = β.

PA = PC [lengths of the tangents from an external point C]

In a triangle CAP, ∠PAC = ∠ACP = α

Similarly, PB = CP and ∠PCB = ∠CBP = β.

Now in the triangle ACB:

∠CAB + ∠CBA + ∠ACB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180° (Since ∠ACB = ∠ACP + ∠PCB = α + β).

2α + 2β = 180°

⇒  α + β = 90°

∴ ∠ACB = α + β = 90°

tan2A = cot(A − 30°),

We have tan2A = cot(A − 30°)

⇒ cot(90° − 2A) = cot(A − 30°),

⇒ (90° − 2A) = (A − 30°),

⇒ A = 40°

∴ Sec (A + 20°) = 40° + 20°

Sec (60°) = 2

tan θ = \frac{8}{15}

⇒ p = 8k and b = 15k

Pythagoras theorem: h² = p² + b²

or, h² = (8k)² + (15k)²

or, h² = 289k²

or, h = 17

∴ sin θ = 8/17

V = 1\over3πr²H

or, V = 1\over3AH   (∵ A = πr²)

or, 3 = \text{AH}\over \text{V}

or, \text{AH}\over \text{3V} = 1

Volume of circular cylinder = πr²h

or, Volume of cone = 1\over 3πr²h

Ratio = V_cylinder : V_cone

= πr²h : 1\over 3πr²h

= 1 : 1\over 3 = 3 : 1

Numbers in ascending order are 6, 8, 10, 12, 13, x

Mean = 6 + 8 + 10 + 12 + 13 + x\over 6 = 49 + x\over 6

No. of terms (n) (even)

Median = {\text{n}\over2} \text{term} +{\text{n}\over2} + 1 \text{term}\over 2 

= {6\over2} \text{term} +({6\over2} + 1) \text{term}\over 2 

= 3^{rd} \text{term} +4^{th} \text{term}\over 2 

= 10 + 12\over 2  = 11

ATP: Mean = Median

or, 49 + \text{x}\over 6 = 11

or, x = 17

Amount (A) = ₹ 22500

Principal = P

Rate (r) = 61\over 4 %

time (n) = 2

A = P (1 – \text{r}\over 100)ⁿ

or, 22500 = P(1 – 25\over 4 × 100)²

or, 22500 = P(15\over 16)²

or, 22500 = P15\over 16 × 15\over 16

or, P =  22500 × 16\over 15 × 16\over 15

or, p = 25600

Thus, 2 years ago there were 25600 smokers.

Let the capital of A = 6x

Let the capital of B = 4x

Let the capital of C = 3x

Half of 6x = 3x

Ratio of capital with respect to 1 month:

= [(6x × 4) + (3x × 8)]: (4x × 12): (3x × 12)

= (24x + 24x): 48x: 36x

= 48x: 48x: 36x

= 4:4:3

Total profit = ₹ 61,050

Profit of A = Share of A × total profit

= 4\over 11 × 61050 = ₹ 22200

Profit of A = ₹ 22200

Similarly, Profit of B = ₹ 22200 (since the ratio of B is same as A)

Profit of C = 3 × 5550 = ₹ 16650

\text{x² - 6x + 9 - (x² + 6x + 9)}\over x² - 9 + 48\over 7 = 0

or, – \text{12x}\over \text{x² - 9} + 48\over 7 = 0

or, -84x + 48x² – 432 = 0

or, -7x + 4x² – 36 = 0

or, 4x² – 7x – 36 = 0

or, 4x² – (16x – 9x) – 36 = 0

or, 4x² – 16x + 9x – 36 = 0

or, 4x(x – 4) + 9(x – 4) = 0

or, (x – 4)(4x + 9) = 0

x = 4, -9/4

Let the price of 1 dozen pen at present is ₹ x

∴ In ₹ at present 12\over \text{x} × 30 pens = 360\over \text{x} pens are got

If the price is reduced by ₹ 6 per dozen, then it becomes ₹ (x – 6)

Then for Rs 30 we get 12\over \text{x - 6} × 30 pens = 360\over x - 6

As per question, 360\over \text{x - 6} – 360\over \text{x} = 3

or, 360(1\over \text{x - 6} – 1\over \text{x}) = 3

or, 120 {\text{x - x + 6}\over \text{(x - 6)x}} = 1

or, 120 {6\over \text{(x - 6)x}} = 1

or, 720\over \text{x² - 6x} = 1

or, x² – 6x – 720 = 0

or, x² – 6x – 720 = 0

or, x² – (30 – 24)x – 720 = 0

or, x² – 30x + 24x – 720 = 0

or, x(x – 30) + 24(x – 30) = 0

or, (x – 30)(x + 24) = 0

either x – 30 = 0 or x + 24 = 0
=> x = 30 or x = -24

But the price of pens cannot be negative, so x ≠ -24, hence x = 30

Hence, before the reduction of prices, the price of 1 dozen pens was Rs 30.

x = \frac{1}{2-\sqrt{3}}

or, x + 1 = \frac{1}{2-\sqrt{3}} + 1

= \frac{1 + 2-\sqrt{3}}{2-\sqrt{3}}

= \frac{3 -\sqrt{3}}{2-\sqrt{3}} × \frac{2 +\sqrt{3}}{2 + \sqrt{3}}

= \frac{6 + 3\sqrt{3} - 2\sqrt{3} - 3}{4 - 3}

= 3 + √3

y = \frac{1}{2+\sqrt{3}}

y + 1 = \frac{1}{2+\sqrt{3}} + 1 = 3 – √3

Now, \frac{1}{x+1} + \frac{1}{y+1}

= \frac{1}{3 + √3} + \frac{1}{3 - √3}

= \frac{3 - √3 + 3 + √3}{9 - 3}

= \frac{6}{6} = 1

y ∝ z ⇒ y = k₂z — (i)

x ∝ y ⇒ x = k₁y ⇒x = k₁k₂z — (ii)

\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy} \propto \frac{1}{x} + \frac{1}{y} + \frac{1}{z}

or, \frac{\text{x² + y² + z²}}{xyz} ∝ \frac{\text{yz + xz + xy}}{\text{xyz}}

or, \frac{\text{x² + y² + z²}}{\text{yz + xz + xy}} = k

To prove the proportionality for the given expression, the value of \frac{\text{x² + y² + z²}}{\text{yz + xz + xy}} should be non-zero constant.

Now substitute the values of x and y from (i) and (ii),

= ({k_1}^2{k_2}^2 + {k_2}^2  + 1)z^2\over {k_1}{k_2}^2 + k_2  + (k_1k_2)z^2

= ({k_1}^2{k_2}^2 + {k_2}^2  + 1)\over {k_1}{k_2}^2 + k_2  + (k_1k_2)

= Non – zero constant

\frac{\text{a²}}{\text{b + c}} = \frac{\text{b²}}{\text{c + a}} = \frac{\text{c²}}{\text{a + b}} = 1

Let the continued proportion be a, ak, ak², ak³, ak⁴

Given: ak³ = 54 --- (1) and  ak⁴ = 162--- (2)

Dividing (2) and (1)

k = 3

Put k in eq (1)

ak³ = 54

or, a × 3³ = 54

or, a × 27 = 54

or, a = 2

So, first number will be ‘2'.

Given: ABCD is a cyclic quadrilateral

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

Construction: Two diagonals AC and BD are drawn.

Proof: ∠ADB = ∠ACB [angles in the same segment of the circle]

Again, ∠BAC = ∠BDC [angles in the same segment of the circle]

Again, ∠ADC = ∠ADB + ∠BDC

= ∠ACB + ∠BAC

∴ ∠ADC + ∠ABC = ∠ACB + ∠BAC + ∠ABC

∴ ∠ADC + ∠ABC = 2 right angles [∴ sum of three angles of a triangle is 2 right angles]

Similarly we can prove that, ∠BAD + ∠BCD = 2 right angles

Given: AB is a tangent at the point P of a circle with center O and OP is a radius through the point P. To prove: OP and AB are perpendicular to each other i.e. OP ⊥ AB.

Construction: Any other point Q is taken on the tangent AB, O, Q are joined.

Proof: Any other point on AB except P is outside the circle; ∴ OQ intersects the circle at a point. Let R be the point of intersection.

∴ OR < OQ [∴ R is a point between O, Q]

Again, OR = OP [∴ radii of the same circle]

∴ OP < OQ

∴ The point Q is any point on AB,

∴ OP is the least of all the line segments drawn from the center O to the tangent AB.

Again, the least distance is perpendicular distance.

∴ OP ⊥ AB (proved)

Given:  The bisector of ∠DAB and ∠BCD intersect the circle at the points X and Y.

To Find : ∠XOY

The angles ∠YAB and ∠YCB subtended by the minor arc YB are on the same segment of the circle.

∴ ∠YAB = ∠YCB = 1\over2 ZBCD ---- (1) [ ∴ CY is bisector of ∠BCD]

Again, ∠XAY = ∠XAB + ∠YAB

= 1\over2 ∠BAD + 1\over2 ∠BCD [From (1) we get, ∴ AX is bisector of ∠DAB]

= 1\over2 (∠BAD + ∠BCD)

= 1\over2 × 180° [∴ ABCD is a cyclic quadrilateral]

= 90° ∴ ∠XAY is a semicircular angle.

∴ XY is a diameter and ∠XOY = 180°

ABCD is a cyclic trapezium of which AD || BC

AB = DC or ABCD is a rectangle and AC = BD.

∠ADC + ∠DCB = 180° [∠ AD || BC and DC is transversal]

Again, ∠BAD + ∠DCB = 180° [∠ ABCD is a cyclic quadrilateral]

∴ ∠ADC + ∠DCB = ∠BAD + ∠DCB ∴ ∠ADC = ∠BAD ..

In ∆BAD and ∆ADC, ∠BAD = ∠ADC [From (1) we get]

∠ABD = ∠DCA [∴ Angles in the same segment]

AD is common side

∴ ∆BAD = ∆ADC [A-A-S congruence property]

∴ AB = DC .. ABCD is an isosceles trapezium or a rectangle and AC = BD (Similar part of congruent triangle) [Proved]

Steps of construction:

1. Draw a line BC of 6 cm.
2. At point B draw a right angle.
3. Take a distance of 5 cm and cut an arc from the point B. This will give the point
4. Join A to C. This is the right angle triangle ABC.
5. Draw angle bisector of any two angles say ∠B and ∠C of △ABC and let these intersect at a point say O.
6. Taking O as centre and OM as radius, draw a circle.
7. The circle touches the other two sides of triangle. This will be the required in circle of the triangle.

Cos θ = \text{x}\over \sqrt{\text{x² + y²}}

base (b) = x

hypotenuse (h) = \sqrt{\text{x² + y²}}

Pythagoras Theorem:  p² = h² - b²

or, p² = x² + y² - x²

or, p² = y²

or, p = y

LHS: x sin θ = x \text{y}\over \sqrt{\text{x² + y²}}

= y \text{x}\over \sqrt{\text{x² + y²}}

= y cos θ RHS 

Hence, x sin θ = y cos θ proved

Length of arc = 5.5 cm

Radius of the circle = 7 cm

Angle substend by arc (θ) = 5.5\over 7 = 11\over 14 radians

or, Angle substended by arc at the centre

= 11\over 14 × 180\over π

= 11\over 14 × 180 × 7\over 22

= 45º or π/4

LHS: \text{tan θ + sec θ - 1}\over \text{tan θ - sec θ + 1}

= \text{(tan θ + sec θ) - (sec² θ - tan² θ)}\over \text{tan θ - sec θ + 1}

= \text{(sec θ + tan θ) - (sec θ + tan θ)(sec θ - tan θ)}\over \text{tan θ - sec θ + 1}

= \text{(sec θ + tan θ)(1 - sec θ + tan θ)}\over \text{tan θ - sec θ + 1}

= sec θ + tan θ

= 1\over \text{cos θ} + \text{sin θ}\over \text{cos θ}

= 1 + \text{sin θ}\over \text{cos θ}

In ΔABO

tan 30° = \text{AB}\over \text{OA}

or, 1\over √3 = \text{AB}\over 50

or, AB = 50\over √3 = 28.86 m

In ΔAOC,

tan 45° = \text{AC}\over \text{OA}

or, AC = 50 m

Height of the of tower increased = 50 m - 28.86 m

= 21.13 m

In ΔEDC, tan 30° = \text{ED}\over \text{DC}

or, 1\over √3 = \text{AE - AD}\over \text{CD} --- (1)

In ΔEAB,

tan 60° = \text{AE}\over \text{AB}

or, √3 = \text{AE}\over \text{CD} --- (2)

Divide (1) by (2),

\text{AE - AD}\over \text{AE} = 1\over √3 × 1\over √3

or, {\text{AE}\over \text{AE}} - {\text{AD}\over \text{AE}} = 1\over 3

or, 1 - {\text{AD}\over \text{AE}} = 1\over 3

or, {\text{AD}\over \text{AE}} = 1 - 1\over 3

or, {\text{AD}\over \text{AE}} = 2\over 3

or, {\text{BC}\over \text{AE}} = 2\over 3

or, {\text{AE}\over \text{BC}} = 2\over 3

The ratio of the heights of building and lamp post = 3 : 2

Volume of sphere = 4\over3πr³

Total volume of two sphere = 4\over3π(1³ + 6³)

Let internal radius of hollow sphere = r cm

Volume of the iron of this sphere = 4\over3π(9³ - r³)

According to the question,

4\over3π(1³ + 6³) = 4\over3π(9³ - r³)

or, (1³ + 6³) = (9³ - r³)

or, r³ = 9³ - 1³ - 6³ = 512

or, r³ = 8³

or, r = 8 cm

Let, radius of cylinder = r and the height of cylinder = 2r

If its height be 7 times its diameter, new height of cylinder = 14r

Case - 1: Volume = 1\over 3 2πr²h

= 1\over 3 × 2 × 22\over 7 × r² × 2r

= 44r³\over 21

Case - 2:  Radius = r, Height = 14r

Volume = 1\over 3 πr²h

= 1\over 3 × 2 × 22\over 7 × r² × 14r

= 308r³\over 21

According to the Question,

14\over 3 × πr³ - 2\over 3 × πr³ = 539

or, 4πr³ = 539

or, 4 × 22\over 7 × r³ = 539

or, r³ = 539 × 7\over 22 × 4

or, r³ = 7 × 7 × 7 \over 2 × 2 × 2

or, r = 7\over 2 = 3.5 cm

Given Height is twice of the radius,

H = r × 2 = 3.5 × 2 = 7 cm

Given:

• Curved surface area = 440 sq. decimeters
• Density of the wood = 3 kg per cubic decimeter
• Weight of the log = 18.48 quintals (1 quintal = 100 kg)

Convert the weight of the log to kilograms:

Weight of the log = 18.48 × 100 = 1848 kg

Formula for the curved surface area (CSA) of a cylinder:

CSA = 2πrh

Given that CSA = 440:

2πrh = 440 --- (1)

Volume of the cylinder:

Volume = πr²h

Weight of the log = Volume × Density

1848 = πr²h × 3

πr²h = 616 --- (2)

Divide the second equation by the first:

πr²h \over 2πrh = 616 \over 440

r \over 2 = 14 \over 10

r = 2.8 decimeters

Diameter = 2r = 2 × 2.8 = 5.6 decimeters

Therefore, the diameter of the log is 5.6 decimeters.

Σf = 120

or, 68 + f₁ + f₂ = 120

or, f₁ + f₂ = 120 - 68

or, f₁ + f₂ = 52 --- (1)

Σfx = mean × Σf

or, 3480 + 30f₁ + 70f₂ = 50 × 120

or, 30f₁ + 70f₂ = 2520

or, 3f₁ + 7f₂ = 252 --- (2)

Solving (1) and (2), we get

f₁ = 28

and  f₂ = 24

Hence, the values of f₁ and f₂ are 28 and 24.

Modal class = 79.5 - 89.5

• $L$ = 79.5
• f₁​ =  50
• f₀​ = 40
• f₂​ = 30
• $h$ =  10

Mode = 79.5 + (50 - 40\over 2×50 - 40 - 30) × 10

= 79.5 + (10\over 30) × 10

= 79.5 + (100\over 30)

= 82.833

To draw an incircle of a right-angled triangle, follow these steps:

1 Draw a right-angled triangle: Start by drawing a triangle with one angle of 90 degrees,

which means that one of the sides must be vertical, and another horizontal.

2 Find the Incenter: To draw the incircle, you first need to find the incenter of the triangle. The incenter is the point where the angle bisectors of the triangle intersect. To find the incenter, draw the angle bisectors of any two angles of the triangle. These two lines will intersect at a single point, which is the incenter.

3 Draw the Incircle: With the incenter located, you can now draw the incircle. The incircle is a circle that is tangent to all three sides of the triangle. To draw the incircle, use a compass to draw a circle with the incenter as the center, and a radius equal to the distance between the incenter and any of the sides of the triangle.

Construction Procedure:

(i) I drew a triangle ABC, of which AB, BC, and CA are 7 cm, 6 cm, and 3 cm, respectively.

(ii) I drew a rectangle EFCG whose area is equal to the area of triangle ABC.

(iii) Now from the extended EG, I cut off GK, which is equal to GC.

(iv) Now I drew a semicircle by taking the line segment EK as the diameter.

(v) I extended CG, which intersects the semicircle at the point.

(vi) I drew a squared figure HGJI by taking the side GH.
HGJI is the required square whose area is equal to the area of triangle ABC.

x ∝ y implies x = k₁ y.

y ∝ z implies y = k₂ z.

z ∝ x implies z = k₃ x.

Substitute y = k₂ z into x = k₁ y:

x = k₁ (k₂ z) = k₁ k₂ z.

Now substitute x = k₁ k₂ z into z = k₃ x:

z = k₃ (k₁ k₂ z).

Simplifying:

z = k₁ k₂ k₃ z.

For this to hold true, k₁ k₂ k₃ = 1.

Thus, the relation between the constants is:

k₁ k₂ k₃ = 1.

Capiatal ratio of A and B = \frac{3}{2} : 1 = 3 : 2

Sum of ratio = 3 + 2= 5

Profit share of B = ₹ 1500

Let P be the total profit

or, \frac{2}{5} × P = ₹ 1500

or, P = ₹ 1500 × \frac{5}{2} = ₹ 3750

Profit share of A = \frac{3}{5} × 3750 = ₹ 2250

x + √(x² - 9) = 9

or, 9 - x = √(x² - 9)

squaring both sides

(9 - x)² = {√(x² - 9)}²

or, 81 - 18x + x² = x² - 9

or, 81 - 18x = - 9

or, 18x = 90

or, x = 5

Now, x - √(x² - 9) =  5 - √(5² - 9)

= 5 - √16

= 5 - 4 = 1

Volume of the sphere = (4/3)πr³

Surface area of the sphere = 4πr²

Given that the volume is twice the surface area:

(4/3)πr³ = 2 × 4πr²

Simplifying:

(4/3)r³ = 8r²

(4/3)r = 8

4r = 24

Now, divide by 4:

r = 6

Thus, the radius of the sphere is 6 units.

x = √7 - √2

1\over \text{x} = 1\over √7 - √2

= √7 + √2\over 5

y = √8 - √3

1\over \text{y} = 1\over √8 - √3

= √8 + √3\over 5

Clearly, 1\over \text{x} > 1\over \text{y}

or, y > x

or, √8 - √3 > √7 - √2

the condition for the quadratic equation ax² + bx + c=0 to have one zero root is:

b ≠ 0 and c = 0.

If the lengths of the corresponding sides of two triangles are in proportion, then the two triangles are similar triangles.

Principal = P

rate = 6 \frac{1}{4}% = \frac{25}{4}%

Amount = 4P

Simple Interest (SI) = 4P - P = 3P

Time (t) = \frac{SI × 100}{\text {P × r}}

or, Time (t) = \frac{3p × 100}{\text {P × 25/4}} = 16 years

The front angle formed at the center of a circle by an arc is twice the angle formed by the same arc at any point on the circle.