A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

 A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

Given:

 P(Success) = 90% = 0.9 = 9/10

 P(Failure) = 1 − 0.9 = 0.1 = 1/10

let S debite Success,

 F denote Failure

Probability that exactly one surgery is successful

possibilities SFF, FSF, FFS

P(SFF) = 0.9 × 0.1 × 0.1 = 0.009

P(FSF) = 0.1 × 0.9 × 0.1 = 0.009

P(FFS) = 0.1 × 0.1 × 0.9 = 0.009

  P(exactly one auccess) = 0.009 + 0.009 + 0.009 = 0.027 = 27/1000 

 Probability that at most two surgeries are successful

At most two successes means 0 or 1 or 2 successes

so

 Probability that at most two surgeries are successful 

= 1 − P(all three successful)

= 1 - P[ SSS]

= 1 - [0.9 × 0.9 × 0.9]

 = 1 -  0.729 

= 0.271 

= 271/1000

see this video for more explanation 

basic probability , independent events , cbse 12th maths old board exam question paper 20252026

From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

 From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

Cylinder and conical cavity have same radius and height

r = 2.1 cm, h = 2.8 cm

.Volume of remaining solid

 Volume of cylinder − Volume of cone 

Volume of cylinder = πr²h = π × (2.1)² × 2.8 = π × 4.41 × 2.8 = 12.348π cm³  

Volume of cone = (1/3)πr²h = (1/3) × 12.348π = 4.116π cm³

Remaining volume = 12.348π − 4.116π = 8.232π cm³  

Using π = 22/7

Remaining volume  = 8.232 × 22/7 = 25.872 cm³ ≈ 25.87 cm³

. Total surface area of remaining solid

Surfaces left after hollowing:  

Bottom circular base of cylinder = πr²  

Curved surface of cylinder = 2πrh  

Curved surface of cone = πrl  

Slant height of cone: l = √(r² + h²) = √(2.1² + 2.8²) = √(4.41 + 7.84) = √12.25 = 3.5 cm 

Required surface area = πr² + 2πrh + πrl

 = π(2.1)² + 2π(2.1)(2.8) + π(2.1)(3.5)

 = 4.41π + 11.76π + 7.35π = 23.52π cm²  Using π = 22/7:

 = 23.52 × 22/7 = 73.92 cm²  

for more details watch this video 

mensuration,   surface area and volume of solids , cbse th maths old board exam question paper 2025 2026

cbse 10th maths questions

 cbse 10th maths questions

Linear equations in two unknowns 

 The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

solution

Quadratic equations

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

solution

Coordinate Geometry

Find a relation between x and y such that the point P(x, y) is equidistant from the

 points A(5, 3) and B(1, 7)

solution

Trigonometry

 If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

solution

Mensuration surface area and volume of solids

 From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

solution

While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola. Based on the above information answer the following questions : (a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ? (b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola. (c) Find the expression for the above polynomial. OR (c) If the zeroes of the polynomial are −5 and 3, find its expression.

 While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola.

Based on the above information answer the following questions :

(a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ?

(b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola.

(c) Find the expression for the above polynomial.

(d) If the zeroes of the polynomial are −5 and 3, find its expression.

[a] A parabola is the graph of a quadratic polynomial

Given

parabola, intersects the x-axis at −2 and 3 

therefore zeroes are [-2] and 3

Polynomial with roots α and β is given by

f(x) = k(x − α)(x − β)

or

f(x) = k[x² − (α + β)x + αβ]

where k ≠ 0 is any real number.

given that zeroes are [-2] and 3

Take them as

α = [-2]

β = 3

 f(x) = k(x + 2)(x − 3)

use distribution law

f(x) =k(x² − x − 6)

to find k use the condition that parabola cuts y-axis at −3

which means it passes through (0,-3)

so Use f(0) = −3 

k[0-0-6] = [-3]

we get k =(1/2)

and  f(x) = (1/2)(x² − x − 6)

in the last part the zeroes are given to be  −5 and 3

so g(x) = k(x + 5)(x − 3) =kx² + 2x − 15

choose k=1

g(x)  = x² + 2x − 15

see this video for more details 

cbse 10th maths old board exam question papers 2025 2026 to find the equation of a polynomial whose zeroes are given

If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx

 If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx.

differential calculus topic of logarthimic differentiation

given

(sin x)ʸ = yᶜᵒˢˣ  [power contains variables]

Take ln both sides

ln[(sin x)ʸ] = ln[yᶜᵒˢˣ]  [use properties of logarithms]

 y. ln(sin x) = cos x . ln y

 Differentiate w.r.t. x

[use product rule and chain rule]

  [dy/dx] ln(sin x) + y cot x  =  −sin x · ln y + cos x · (1/y) ·[ dy/dx]

[dy/dx ] ln(sin x) + y cot x = −sin x ln y + (cos x / y) [dy/dx]

 Collect dy/dx terms

 [dy/dx] ln(sin x) − (cos x / y)[ dy/dx ]= −sin x ln y − y cot x

[ dy/dx] [ln(sin x) − cos x / y] = −[sin x ln y + y cot x]

dy/dx = −[sin x ln y + y cot x] / [ln(sin x) − cos x / y]

  Multiply numerator and denominator by y:

dy/dx = −y[sin x ln y + y cot x] / [y ln(sin x) − cos x]

see this video for more explanation 

cbse 12th maths old board exam question paper

2025 2026 differentiation

logarthmic differentiation

 

Find a particular solution of the differential equation (x + 1) dy/dx = 2 e⁻ʸ − 1, given that y = 0 when x = 0

 Find a particular solution of the differential equation

(x + 1) dy/dx = 2 e⁻ʸ − 1,

given that y = 0 when x = 0

variable separable differential equation with initial condition

given

(x + 1) dy/dx = 2 e⁻ʸ − 1,

separating

 dy / (2 e⁻ʸ − 1) = dx / (x + 1)

multiply numerator and denominator of LHS with  eʸ 

eʸ dy / (2 − eʸ) = dx / (x + 1)

integrating

∫ eʸ / (2 − eʸ) dy = ∫ 1 / (x + 1) dx

for the LHS use the substitution

u = 2 − eʸ 

⇒ du = −eʸ dy

eʸ dy =-du

we get

 ∫ −du / u = ∫ 1 / (x + 1)  dx

 −ln|u| = ln|x + 1| + C

re substitute the value of u = 2 − eʸ 

 −ln|2 − eʸ| = ln|x + 1| + C

 ln|2 − eʸ| +ln|x + 1| = -C

using properties of logarithms,

ln|(2 − eʸ)(x + 1)| = -C

using definition of logarithm

 (x + 1)(2 − eʸ) = A-----(1)

, where A is another arbitrary constant

use the initial condition   y = 0 when x = 0 to solve for A

we get A =1

(1)⇒ (x + 1)(2 − eʸ) =1

or

2 − eʸ = 1 / (x + 1)

eʸ = 2 − [ 1 / (x + 1)]

eʸ =  (2x + 1) / (x + 1)

using definition of log

y = ln[(2x + 1) / (x + 1)]

for more details watch this video 

cbse 12th past year question papers 2025 2026 differential equations

variable separable differential equation dy/dx = (2 − y) / (x + 1)

 Solve the following variable separable differential equation

dy/dx = (2 − y) / (x + 1)

dy/dx = (2 − y) / (x + 1)

clearly it is variable separable type

dy / (2 − y) = dx / (x + 1)

integrating both sides

∫ dy / (2 − y) = ∫ dx / (x + 1)

− ln|2 − y| = ln|x + 1| + lnC [ note the ( -1), coefficient of y when integrating]

0 = ln|2 − y| + ln|x + 1| + lnC

use the property of logarithms

0=ln [ |2 − y| |x + 1| C ]

or

[ |2 − y| |x + 1| C ] = 1

for more details refer to this video

cbse  12th applied mathematics 2025 2026 old board exam question papers applied mathematics

variable sepaprable differential equation