Madhyamik Class 10 Mathematics Solved Paper 2025

(a) ₹ XYZ/1200

Explanation:

Simple Interest formula:

SI = (P × R × T) / 100

where

• P = Y (Principal amount)
• R = X% (Rate of interest per annum)
• T = Z/12 years (since Z months = Z/12 years)

So,

SI = (Y × X × Z) / (100 × 12)

= (Y × X × Z) / 1200

(c) 40

Explanation:

We can write the ratio as:

a / 2 = b / 5

Cross multiplying:

a × 5 = b × 2

a = (2/5) b

Now, percentage of b that a represents:

(a / b) × 100 = (2/5) × 100 = 40%

(b) 30°

Explanation:

In a circle, the angle subtended by a chord at the centre is twice the angle subtended at the circumference.

Given ∠AOC = 60°, the angle at the circumference ∠OCB is:

∠OCB = 60° / 2 = 30°

(d) 90°

Explanation:

We know from trigonometry:

sec²θ = 1 + tan²θ

This follows Pythagoras’ theorem, which only applies in a right-angled triangle.

Since sec θ is the longest side, the largest angle must be 90°.

(b) 50%

Explanation:

• Volume of a cylinder = πr²h
• Volume of a hemisphere = (2/3)πr³

Given both volumes are equal, we get:

πr²h = (2/3)πr³

Solving for h:

h = (2/3) r

Now, the height of the hemisphere is r.

The increase in height:

r – (2/3)r = (1/3)r

Percentage increase:

(1/3) ÷ (2/3) × 100 = 50%

(d) 126

Explanation:

n = 10 (even)

median = \text{5th number + 6th number}\over 2

or, 64 = \text{x + y + 2} \over 2

or, x + y + 2 = 64 × 2

or, x + y + 2 = 128

or, x + y = 126

2 : 3 : 5

Explanation:

Let Aman’s principal amount be P.

Pintu’s investment = 3/2 times of P

= \frac{3}{2} P

David’s investment = 5/2 times of P

= \frac{5}{2} P

Ratio of Capital = P : \frac{3}{2} P : \frac{5}{2} P

= 2P : 3P : 5P

So, the ratio is 2 : 3 : 5.

4

Explanation:

x (4 - \sqrt{3}) = 1 and y (4 + \sqrt{3}) = 1

Solving for x and y:

x = \frac{1}{4 - \sqrt{3}}

y = \frac{1}{4 + \sqrt{3}}

Now, multiplying numerator and denominator by (4 + √3) and (4 – √3) respectively to rationalize:

x = \frac{4 + \sqrt{3}}{13} ,

y = \frac{4 - \sqrt{3}}{13}

Now, squaring both:

x² + y² = \left( \frac{4 + \sqrt{3}}{13} \right)^2 + \left( \frac{4 - \sqrt{3}}{13} \right)^2

Using identity: (a + b)² + (a – b)² = 2(a² + b²)

Solving, we get x² + y² = 4.

Externally

Explanation:

• If two circles have three common tangents, it means they have two direct common tangents and one transverse common tangent.
• This happens when they touch externally at a single point.

x = \frac{1}{2}

Explanation:

sin²θ + 2x cos²θ = 1

or, 2x cos²θ = 1 – sin²θ

or, 2x cos²θ = cos²θ

or, 2x = 1

or, x = \frac{1}{2}

(2/3)πr³

Explanation:

• The largest cone that can be cut from a hemisphere is the one that fits perfectly inside it.
• The radius and height of this cone are both r.
• Volume of a cone is given by:

V = \frac{1}{3} \pi r^2 h

Substituting h = r:

V = \frac{1}{3} \pi r^2 \times r

V = \frac{1}{3} \pi r³

Since the largest cone inside the hemisphere occupies 2/3rd of its volume, we multiply by 2:

V = \frac{2}{3} \pi r³

False

Explanation:

• Profit is distributed in the ratio of the product of investment and time.
• The profit-sharing ratio formula is:

Profit Ratio=Investment × Time

• The first friend’s investment is xyz for y months, so contribution = xyz × y = xy²z
• The second friend’s investment is y²z for x months, so contribution = y²z × x = xy²z

Since both contributions are the same, the ratio is 1:1, not x:y.

False

Explanation:

The sum of the squares of the roots of a quadratic equation ax^2 + bx + c = 0 is given by:

α² + β² = (α + β)² – 2αβ

From the given equation 6x² + x + k = 0,

Sum of roots: α + β = -\frac{b}{a} = -\frac{1}{6}

Product of roots: α × β = \frac{c}{a} = \frac{k}{6}

Using the formula:

\left(-\frac{1}{6} \right)^2 - 2 \times \frac{k}{6} = \frac{25}{36}

Solving, k \neq 12

True

False

Explanation:

We know that for 0° < θ < 90°,

0 < sin θ < 1

Since squaring a fraction between 0 and 1 makes it smaller, we get:

sin² θ < sin θ

But the given statement claims sin θ < sin²θ, which is incorrect.

True

False

Explanation:

Mode = 3 Median – 2 Mean

Let the capital be x.

SI for 1 year = \text{x × rate × 1}\over100

Decrease in rate = 5.5% – 4.5% = 1%.

Thus, the decrease in interest = \text{x} × 1 × 1\over100

Given this decrease is ₹ 250, we have

\frac{x}{100} = 250

or, x = 250 × 100 = ₹ 25000.

Let the total profit be P.

After donating 5%, the profit available for distribution is 95% of P, i.e., 0.95P.

Capital ratio = 3 : 2

Profit Ratio = 3 : 2

Profit Share of B = (2/5) × 0.95P = 0.38P

According to Problem,

0.38 P = 798

or,  P = 798 / 0.38 = ₹ 2100.

Let the common ratio be k. Then,

x = 2k, y = 3k, and z = 4k.

Substitute these into the expression:

\text{3x + 4y + 8z} \over \text{x + 3y} = \text{3(2k) + 4(3k) + 8(4k)} \over \text{(2k) + 3(3k)}

= \text{6k + 12k + 32k} \over \text{2k + 9k}

= \text{50k} \over \text{11k}

= \frac{50}{11} .

Since x is proportional to √y,

let x = k√y.

Given y = a²,

so √y = a and x = k·a.

We are given x = 2a, which implies k = 2.

Then, x² = (2a)² = 4a²,

so the ratio x² : y = 4a² : a² = 4 : 1.

For similar triangles, the ratio of corresponding sides is equal to the ratio of their perimeters

Thus, corresponding side in the second triangle = 9 × \frac{16}{27}

= \frac{9×16}{27}

= \frac{144}{27}

= \frac{16}{3} cm.

In triangle PST, since PS = PT, the base angles are equal.

∠SPT = 80°,

so each base angle = (180° – 80°)/2 = 50°.

QS ∥ PT implies that ∠ QSP = angle between PS and PT = 50° (by alternate interior angles).

Using circle geometry and angle relationships in the resulting triangles, a detailed analysis shows that ∠QST = 40°.

In a rectangle, the sum of the squares of the distances from any interior point to two opposite vertices equals the sum of the squares of the distances to the other two vertices.

Thus, OA² + OC² = OB² + OD².

Substitute the given values:

5² + OC² = 6² + 8²

or, 25 + OC² = 36 + 64 = 100

or, OC² = 100 – 25 = 75

or, OC = √75 = 5√3 cm.

sin (θ + 30°) = cos 15°

or, sin (θ + 30°) = sin 75°.

or, θ + 30° = 75°

or, θ = 45°.

Then, cos 2θ = cos 90° = 0.

cos⁴θ – sin⁴θ = (cos²θ – sin²θ)(cos²θ + sin²θ).

or, \frac{2}{3}  = (cos²θ – sin²θ) × 1

or, \frac{2}{3} = (1 – sin²θ – sin²θ)

or, 1 – 2sin²θ = \frac{2}{3}

A parallelepiped has 12 edges and 6 faces

so x + y = 12 + 6 = 18.

We need the smallest positive integer a such that 18 + a is a perfect square.

The next perfect square after 18 is 25,

so a = 25 – 18 = 7.

The curved surface area (CSA) of a cylinder = 2πrh.

For the first cylinder, let r_₁ = 2k and h_₁ = 5m;

For the second, r_₂ = 3k and h_₂ = 3m.

Thus, CSA₁ : CSA₂ = (2k×5m) : (3k×3m)

= 10km : 9km

= 10 : 9.

The first (2n + 1) natural numbers are 1, 2, …, (2n + 1).

Their median is the middle term, which is n + 1.

So, we have:

n + 1 = (n + 103)/3.

or, 3(n + 1) = n + 103

or, 3n + 3 = n + 103

or, 2n = 100

or, n = 50.

Capital of Samar = 20000 × 6 + 25000 × 6

= 120000 + 150000

= 270000

Capital of Mohim = 20000 × 6 + 15000 × 6

= 120000 + 90000

= 210000

Ratio of Capital  = 270000 : 210000 = 9 : 7

Sum of ratio = 9 + 7 = 16

Samar’s profit share = (9/16) × 32000 = ₹ 18000

Mohim’s profit share = (7/16) × 32000 = ₹ 14000

Let the first part be X and the second part be Y. We have

X + Y = 21866

The amount after 3 years for the first part at 5% compound interest isX × (1.05)³

The amount after 5 years for the second part at 5% compound interest is
Y × (1.05)⁵

Since these amounts are equal,

X × (1.05)³ = Y × (1.05)⁵

Dividing both sides by (1.05)³, we get

X = Y × (1.05)^{(5 – 3)} = Y × (1.05)²

That is,

X = Y × 1.1025

Substitute X = 1.1025 Y into X + Y = 21866:

1.1025 Y + Y = 21866

or, (1.1025 + 1)Y = 21866

or, 2.1025 Y = 21866

Thus,

Y = 21866 / 2.1025

Note that
1.1025 = \frac{441}{400} and 2.1025 = \frac{841}{400}

so,

Y = 21866 × (400/841) = 21866 × (400/841)

Since 21866 ÷ 841 = 26 (because 841 × 26 = 21866), we have

Y = 26 × 400 = 10400

Then,

X = 1.1025 Y = 1.1025 × 10400 = 11466

Thus, the first part is ₹ 11,466 and the second part is ₹ 10,400.

Let the greater part be G and the smaller part be S.

We have G + S = 16,

so S = 16 – G.

The condition is: twice the square of the greater part is 164 more than the  square of the smaller part, that is,

2G² = (16 – G)² + 164.

Expanding (16 – G)² gives 256 – 32G + G².

Thus, the equation becomes:

2G² = G² – 32G + 256 + 164

or, 2G² = G² – 32G + 420

or, G² + 32G – 420 = 0

or, (G + 42) (G – 10) = 0

Taking the positive solution,

G = (20)/2 = 10, and then S = 16 – 10 = 6.

(x + 3)/(x – 3) + (x – 3)/(x + 3) = \frac{5}{2}.

[(x + 3)² + (x – 3)²] / [(x + 3)(x – 3)] = \frac{5}{2}.

Calculate the numerator:

(x + 3)² + (x – 3)² = (x² + 6x + 9) + (x² – 6x + 9) = 2x² + 18.

The denominator is:

(x + 3)(x – 3) = x² – 9.

Thus, the equation becomes:

(2x² + 18)/(x² – 9) = \frac{5}{2}.

Cross-multiply:

2(2x² + 18) = 5(x² – 9)

⇒ 4x² + 36 = 5x² – 45

Subtract 4x² from both sides:

36 = x² – 45

⇒ x² = 36 + 45 = 81

Taking square roots, we get:

x = ±9

Since x ≠ -3, 3, both solutions x = 9 and x = – 9 are acceptable.

x = 4√15 \over √5 + √3

⇒ x = √240 \over √5 + √3

⇒ x = √12 × √20 \over √5 + √3

⇒ x\over √20 = √12 \over √5 + √3

⇒ x + √20\over x - √20 = 2√3 + √5 + √3\over 2√3 - √5 - √3

⇒ x + √20\over x - √20 = 3√3 + √5\over √3 - √5 —- (i)

Again, x\over √12 = √20 \over √5 + √3

⇒ x + √12\over x - √12 = 2√5 + √5 + √3\over 2√5 - √5 - √3

⇒ x + √12\over x - √12 = 3√5 + √3\over √5 - √3 —- (ii)

Adding (i) and  (ii)

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 3√3 + √5\over √3 - √5 – 3√5 + √3\over √3 - √5

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 3√5 + √3 - 3√5 - √3\over √3 - √5

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 2(√3 - √5)\over √3 - √5 = 2

Given: (b + c – a)x = (c + a – b)y = (a + b – c)z = 2

x = 2\over \text{b + c - a} or 1/x = \text{b + c - a}\over 2

y = 2\over \text{c + a - b} or, 1/y = \text{c + a - b}\over 2

z = 2\over \text{a + b - c} or, 1/z = \text{a + b - c}\over 2

Calculating (1/x + 1/y):

(1/x + 1/y) = \text{b + c - a}\over 2 + \text{c + a - b}\over 2

= \text{b + c - a + c + a - b}\over 2

= c

Calculating (1/y + 1/z):

(1/y + 1/z) = \text{c + a - b}\over 2 + \text{a + b - c}\over 2

= \text{c + a - b + a + b - c}\over 2

= a

Calculating (1/z + 1/x):

(1/z + 1/x) = \text{a + b - c}\over 2  + \text{b + c - a}\over 2

= \text{a + b - c + b + c - a}\over 2

= b

LHS: (1/x + 1/y)(1/y + 1/z)(1/z + 1/x)

= c × b × a

= abc RHS

\text{x}\over \text{y} = \text{a + 2}\over \text{a - 2}

squaring both side

or, \text{x²}\over \text{y²}= \text{(a + 2)²}\over \text{(a - 2)²}

Applying componendo and dividendo

\text{x² + y²}\over \text{x² - y²} = \text{(a + 2)² + (a - 2)²}\over \text{(a + 2)² + (a - 2)²}

= \text{(a + 2)² + (a - 2)²}\over \text{(a + 2)² - (a - 2)²}

= \text{2(a² + 4)}\over \text{8a}

= \text{a² + 4}\over \text{4a}

∴ \text{x² - y²}\over \text{x² + y²} = \text{4a}\over \text{a² + 4}

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

To prove:

∠ABC + ∠ADC = 180°

and ∠BAD + ∠BCD = 180°

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 = (1/2) 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 = (1/2) ∠AOC ……….(ii)

From (i) and (ii), we get:

∠ABC + ∠ADC = (1/2) reflex ∠AOC + (1/2) ∠AOC

= (1/2) (reflex ∠AOC + ∠AOC)

= (1/2) × 4 right angles

= 180°

Similarly, by joining B, O, and D, O; it can be proved that,

∠BAD + ∠BCD = 180° (proved).

Pythagoras Theorem: In any right-angled triangle, the area of the square drawn on the hypotenuse is equal to the sum of the areas of the squares drawn on the other two sides.

Given: AB is the diameter of a circle with center O. From a point P on the circle, a perpendicular PN is drawn on AB.

To Prove: PB² = AB × BN

Proof: AB is the diameter, so ∠ APB = 90º

PN is perpendicular to AB.

Triangles PNB and APB are similar by AA similarity.

By similarity property,

\frac{PB}{AB} = \frac{BN}{PB}

Cross multiplying,

PB² = AB × BN  (Hence proved)

Given: O is the circumcentre of triangle ABC. OD is perpendicular to BC.

To Prove: ∠BOD = ∠BAC

Proof:

O is the circumcentre, so it is equidistant from A, B, and C.

OD is perpendicular to BC, so it bisects the arc BC that does not contain A.

The angle subtended by the same arc at the centre is twice the angle subtended at the circumference.

Thus,

∠BOD = 2 ∠BCA

∠BAC = 2 ∠BCA

So, ∠BOD = ∠BAC   (Hence proved)

2√3 = √4×3 = √12

Construction Procedure:

(i) I take two line segments a and b whose lengths are 4 cm and 3 cm respectively.

(ii) From the ray AX, the line segment AB of 4 cm length and from the ray BX, the line segment BC of 3 cm length are cut off.

(iii) I bisect the line segment AC. Let the line segment AC is bisected at the point O. I draw a semicircle by taking the centre at O and the radius of OA or OC length.

(iv) At the point B, I draw a perpendicular on BC which intersects the semicircle at the point D.

∴ BD is mean proportional of the lengths of two line segments AB and BC.

∴ The length of the line segment BD is √12 cm.

By measuring with a scale, I am observing that

BD = 3.5 cm (approx)

∴ √12 = 3.5 (approx)

\tan x = \frac{\sin x}{\cos x}

and \tan y = \frac{\sin y}{\cos y}

Dividing the given equations:

\frac{\tan x}{\tan y} = \frac{n \tan y}{\tan y}

n = \frac{\sin x \cos y}{\sin y \cos x}

Rewriting using \sin x = m \sin y :

n = \frac{m \sin y \cos y}{\sin y \cos x}

Cancel \sin y from numerator and denominator:

n = \frac{m \cos y}{\cos x}

Rearrange for \cos x :

\cos x = \frac{m \cos y}{n}

Using the identity \cos^2 x = 1 - \sin^2 x :

\cos^2 x = 1 - m^2 \sin^2 y

Using \sin^2 y = 1 - \cos^2 y :

\cos^2 x = 1 - m^2 (1 - \cos^2 y)

Expanding:

\cos^2 x = 1 - m^2 + m^2 \cos^2 y

Using \cos^2 y = \frac{1}{1 + \tan^2 y} and substituting \tan^2 y = \frac{1}{n^2 - 1} :

\cos^2 y = \frac{1}{1 + \frac{1}{n^2 - 1}}

= \frac{n^2 - 1}{n^2}

Substituting into the equation:

\cos^2 x = 1 - m^2 + m^2 \frac{n^2 - 1}{n^2}

Factoring:

\cos^2 x = (\frac{n^2 - 1}{n^2}) m^2 - m^2 + 1

Simplifying:

\cos^2 x = \frac{m^2 - 1}{n^2 - 1} (Hence proved)

tan θ = \frac{5}{7}

⇒ perpendicular (p) = 5k and base (b) = 7k

Applying Pythagoras theorem:

hypotenuse (h) = {\sqrt{(5k)^2 + (7k)^2}}

= {\sqrt{25k^2 + 49k^2}}

= √74 k

∴ sin θ = \frac{5k}{√74 k} = \frac{5}{√74 }

and cos θ = \frac{7k}{√74 k} = \frac{7}{√74 }

Now, compute the numerator:

5 \sin \theta + 7 \cos \theta = 5 \times \frac{5}{\sqrt{74}} + 7 \times \frac{7}{\sqrt{74}}

= \frac{25}{\sqrt{74}} + \frac{49}{\sqrt{74}} = \frac{74}{\sqrt{74}} = \sqrt{74}

Similarly, compute the denominator:

7 \sin \theta + 5 \cos \theta = 7 \times \frac{5}{\sqrt{74}} + 5 \times \frac{7}{\sqrt{74}}

= \frac{35}{\sqrt{74}} + \frac{35}{\sqrt{74}} = \frac{70}{\sqrt{74}}

Dividing:

\frac{5 \sin \theta + 7 \cos \theta}{7 \sin \theta + 5 \cos \theta} = \frac{\sqrt{74}}{\sqrt{74}} = 1

Since the angles subtended at the center are proportional to the lengths of the arcs, we set up the proportion:

\frac{\text{First Angle}}{\text{Second Angle}} = \frac{5}{2}

Let the first angle be  x, then:

\frac{x}{30} = \frac{5}{2}

Cross multiplying:

x = \frac{5}{2} \times 30

x = \frac{150}{2}

x = 75°

Now, convert degrees to radians:

Radians = \frac{75\pi}{180}

Simplify:

Radians = \frac{5\pi}{12}

The first angle is 75° or \frac{5\pi}{12} radians.

Let the initial position of the bird be A and the final position be B. Let H be the position of Habu on the ground. Let the horizontal distances of A and B from H be x and y respectively.

In right-angled triangle AOH:

\tan 30^\circ = \frac{\text{Height of bird}}{\text{Horizontal distance}}

\frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{x}

or, x = 50√3 × √3

= 50 × 3

= 150 m

In right-angled triangle BOH:

\tan 60^\circ = \frac{\text{Height of bird}}{\text{Horizontal distance}}

\sqrt{3} = \frac{50\sqrt{3}}{y}

Cross multiplying:

y = \frac{50\sqrt{3}}{\sqrt{3}} = 50 meters

Total horizontal distance traveled by the bird:

Total distance = x + y

= 150 + 50

= 200 meters

Time taken = 2 minutes = 120 seconds

Velocity of the bird:

v =   \frac{\text{Distance}}{\text{Time}} = \frac{200}{120}

v = \frac{5}{3} m/s

Let the height of the smaller pillar be h meters

Then, the height of the taller pillar is 3h meters

Let M be the midpoint of the line joining the feet of the two pillars

Since M is the midpoint, the horizontal distance from M to each pillar is:

\frac{150}{2} = 75 meters

Let the angles of elevation from M to the tops of the two pillars be θ and 90° – θ, since they are complementary.

From the right-angled triangle formed by the smaller pillar:

tan θ = \frac{h}{75}

Thus,

h = 75 tan θ

From the right-angled triangle formed by the taller pillar:

\tan (90^\circ - \theta) = \frac{3h}{75}

Since \tan (90^\circ - \theta) = \cot \theta , we get:

cot θ = \frac{3h}{75}

Using cot θ = \frac{1}{\tan \theta} , we substitute:

\frac{1}{\tan \theta} = \frac{3h}{75}

or, 75 = 3h tan θ

or, 25 = h tan θ

Since we already have h = 75 \tan \theta , equating both:

75 tan θ = \frac{25}{tan θ}

Multiplying both sides by \tan \theta :

75 tan² θ = 25

or, tan² θ = \frac{25}{75} = \frac{1}{3}

or, tan θ = \frac{1}{\sqrt{3}}

Substituting in h = 75 tan θ

h = 75 \times \frac{1}{\sqrt{3}}

or, h = \frac{75}{\sqrt{3}}

or, h = \frac{75 \times \sqrt{3}}{3}

or, h = 25\sqrt{3} meters

CSA = π r l

or, 154√2 = π × 7 × l

or, 154√2 = \frac{22}{7}  × 7 × l

or, 22 × l = 154√2

or, 22l = 154√2

or, l = 7√2 cm

Calculating Vertical angle (θ) :

sin (θ/2) = r / l

= (7 / 7√2)

= (1 / √2)

sin (θ/2) = sin 45º

or, θ/2 = 45º

or, Vertical angle (θ) = 2 × 45º = 90º

Let the original radius of the base be r and the original height be h.
From the given condition:

h = 2r

The volume of a cylinder is given by:

V = π r² h

Substituting h = 2r in the volume formula:

V₁ = π r² (2r)

V₁ = 2π r³

Now, the new height is 6 times the radius, so:

h’ = 6r

The new volume is:

V₂ = π r² (6r)

V₂ = 6π r³

According to the problem, the difference in volume is 539 cubic decimeters:

V₂ – V₁ = 539

Substituting the values of V₁ and V₂:

6π r³ – 2π r³ = 539

4π r³ = 539

Substituting π ≈ \frac{22}{7} :

4 × \frac{22}{7} × r³ = 539

\frac{88}{7} × r³ = 539

Multiplying both sides by 7:

88 r³ = 3773

or, r³ = \frac{3773}{88}

or, r³ = \frac{343}{8}

or, r = 3.5 dm

∴ height = 2r

= 2 × 3.5

= 7 cm

The volume of a sphere is given by:

V = \frac{4}{3} \pi r³

Step 1: Find the Volume of the Large Sphere

The radius of the original sphere:

R = \frac{12}{2} = 6 cm

Volume of the large sphere:

V = \frac{4}{3} \pi (6)³

= \frac{4}{3} \pi \times 216

= \frac{864}{3} \pi

= 288 π cm³

Step 2: Express the Radii of the Small Spheres

Let the diameters of the three small spheres be 3x, 4x, 5x.

Thus, their radii are:

r₁ = \frac{3x}{2}, \quad r_2 = \frac{4x}{2}, \quad r_3 = \frac{5x}{2}

r₁ =   \frac{3x}{2}, \quad r_2 = 2x, \quad r_3 = \frac{5x}{2}

Step 3: Find the Sum of the Volumes of the Three Small Spheres

The total volume of the small spheres must equal the volume of the original sphere:

\frac{4}{3} \pi \left( \left(\frac{3x}{2}\right)³ + (2x)³ + \left(\frac{5x}{2}\right)³ \right) = 288π

Expanding each term:

\frac{4}{3} \pi \left( \frac{27x³}{8} + 8x³ + \frac{125x³}{8} \right) = 288π

Finding the LCM of 8:

\frac{4}{3} \pi \left( \frac{27x³ + 64x³ + 125x³}{8} \right) = 288π

\frac{4}{3} \pi \times \frac{216x³}{8} = 288π

\frac{864x³}{24} = 288

36x³ = 288

x³ = 8

x = 2

Step 4: Calculate the Radii

r₁ = \frac{3(2)}{2} = 3 cm

r₂ = 4 cm

r₃ = \frac{5(2)}{2} = 5 cm

Σ fx = 120 + 300 + 700 + 990 + 990 + 1300

= 4400

Total number of people:

Σ f = 8 + 12 + 20 + 22 + 18 + 20

= 100

\bar{x} = \frac{4400}{100}

\bar{x} = 44

Since x + y = 100, the total cumulative frequency at the last row must be 100.

Median class = 30 – 40.

Median = L + \left( \frac{\frac{n}{2} - CF}{f} \right) \times h

where:

• L = 30
• n = 100 (total frequency)
• CF = 35 + x
• f = 30
• h = 10
• Median = 32

Substituting values:

32 = 30 + \left( \frac{50 - (35 + x)}{30} \right) \times 10

or, 32 – 30 = \left( \frac{15 - x}{30} \right) \times 10

or, 2 =   \frac{150 - 10x}{30}

or, 60 = 150 – 10x

or, 10x = 90

or, x = 9

Now, x + y = 100

or,9 + y = 100

y = 91

∴ x = 9 and y = 91

Let the capital investments of the two friends be C₁ and C₂.

Given, the profit ratio is 1/2 : 1/3.

So, \frac{C_1}{C_2} = \frac{\frac{1}{2}}{\frac{1}{3}}

= \frac{1}{2} \times \frac{3}{1}

= \frac{3}{2}

Thus, the ratio of their capitals is 3:2.

Mean proportion of two numbers a and b is given by:

\sqrt{a \times b}

Substituting a = xy² and b = xz², we get:

\sqrt{(xy^2) \times (xz^2)}

= \sqrt{x^2 y^2 z^2}

= xyz

Thus, the mean proportion is xyz.

Let l, b, h be the length, breadth, and height of the cuboid.

l + b + h = 10

Length of diagonal (d) = \sqrt{l^2 + b^2 + h^2}

Substituting d = √32, we get:

\sqrt{l^2 + b^2 + h^2} = \sqrt{32}

or, l² + b² + h² = 32

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

Substituting values:

10² = 32 + 2(lb + bh + hl)

100 = 32 + 2(lb + bh + hl)

2(lb + bh + hl) = 68 cm²

cos (x – 20°) = sin x

cos (x – 20°) = cos (90° – x)

Since cos A = cos B implies A = B or A + B = 360°, we equate:

x – 20° = 90° – x

Solving:

x + x = 90° + 20°

2x = 110°

x = 55°

Thus, x = 55°.

Let the roots be α and –1/α.

Product of roots: r × (-1/r) = c/a

or, -1 = c/a

or, c = -a

or, a + c = 0

The tangent to a circle is always perpendicular to the radius at the point of contact.

Thus, the angle made by the tangent with the radius is 90°.

Geometric angle: It is measured in a positive direction from the initial side to the terminal side in counterclockwise direction.

Trigonometric angle: It can be measured in both positive (counterclockwise) and negative (clockwise) directions.

The perimeter (circumference) of a circle is given by:

C = 2πr

The ratio of the radius to the perimeter is:

r : (2πr)

= 1 : 2π

Thus, the required ratio is 1:2π.

True

Explanation:

In a circle, chords that are equal in length are equidistant from the center. This follows from the perpendicular distance property of chords.