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

(a) 1

Explanation

When two circles touch each other internally, a single common tangent can be drawn. This tangent touches both circles at their common point of contact.

(a) 9

Explanation

5 + 4 sin θ [ax value of sin θ = 1]

= 5 + 4 × 1

= 9

(b) 9 : 4

Explanation

Given ratio of volumes = 27 : 8

or, 4\over 3πr₁³ : 4\over 3πr₂³ = 27 : 8

or, r₁³ : r₂³ = 27 : 8

or, r₁ : r₂ = 3 : 2

Ratio of volumes = 4πr₁² : 4πr₂²

= r₁² : r₂²

= 3² : 2² = 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

P(1 + r\over100)

Explanation:

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

This amount becomes the 2nd year principal.

± 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

Explanation

Two angles are called complementary angles if the sum of their measures is 90 degrees.

So, by definition:

∠A and ∠B are complementary

⇒ ∠A + ∠B = 90°.

Example: If ∠A = 30°, then ∠B = 60°.

30° + 60° = 90°.

Therefore, the answer is 90°.

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

Let the principal be P

SI = 2\over 5 of principal = 2\over 5 P

time (t) = 5 years

Rate = \text{SI} × 100\over \text{P × t} = {2\over 5}\text{P} × 100\over \text{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}

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

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

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}^2z^2 + {k_2}^2 z^2 + z^2\over {k_1}{k_2}z({k_2}z) + k_2 z (z) + z (k_1k_2z)

= ({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

∴ a2 = b + c ; b2 = c + a ; c2 = a + b

\frac{1}{\text{1 + a}} + \frac{1}{\text{1 + b}} + \frac{1}{\text{1 + c}}

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

= \frac{\text{a}}{\text{a + b + c}} + \frac{\text{b}}{\text{b + c + a}} + \frac{\text{c}}{\text{c + a + b}}

= \frac{\text{a + b + c}}{\text{a + b + c}} = 1 (Proved)

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

The process of drawing an incircle of a right-angled triangle

1. Make a right-angled triangle (one angle should be 90°).
2. Cut two angles of the triangle into half using a compass (for example, angle at A and angle at C).
3. The two bisectors will meet at a point inside the triangle. This point is called the incenter.
4. From the incenter, draw a straight line to one side of the triangle so that it meets the side at 90°. The length of this line is the radius of the circle.
5. Place the compass on the incenter, set its length equal to the radius, and draw a circle.

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\over3πr³

Surface area of the sphere = 4πr²

Given that the volume is twice the surface area:

⇒ 4\over3πr³ = 2 × 4πr²

⇒ 4\over3r³ = 8r²

⇒ 4\over3r = 8

⇒ 4r = 24

⇒ 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.