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.

 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.

Let numerator = x, denominator = y. Fraction = x⁄y

Given

sum of numerator and denominator of a fraction is 4 less than twice the denominator

 x + y = 2y − 4

⇒ x = y − 4  --- (1)

Given

If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃

 (x − 1)⁄(y − 1) = ¹⁄₃

⇒ 3(x − 1) = y − 1

⇒ 3x − 3 = y − 1

⇒ 3x = y + 2  --- (2)

Solve using substitution method

Substitute (1) into (2):

3(y − 4) = y + 2

3y − 12 = y + 2

2y = 14

y = 7 

 Then 

x = y − 4 

x = 7 -4

x= 3  

x =3 

y=7

 The fraction is ³⁄₇

linear equations in two unknowns

formation of linear equations and solution by substitution method

cbse 10th maths previous years question papers 

2025 2026

for a detailed video of the solution using elimination method

\refer  

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

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

given

cos A + sin A = (√2) cos A

sin A =  (√2) cos A - cos A

sin A = (√2 - 1 ) cos A

sin A/ (√2 - 1 )  = cos A

rationalising

(√2 + 1 ) sin A/ [ (√2 - 1 ) (√2 + 1 ) ]  = cos A

(√2 + 1 ) sin A / [2-1] = cosA

(√2 + 1 ) sin A / [1] = cosA

(√2 + 1 ) sin A = cosA

apply distribution law

(√2) sin A + 1 sinA = cosA

(√2) sin A + sin A = cosA  

(√2) sin A  = cosA - sin A

OR

cos A − sin A = (√2) sin A

cbse 10th maths old board exam question paper 2025 2026 trigonometry

video of an alternate method 

AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the length PA.

 AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the length PA.

Let O be the centre of the circle and Q the midpoint of chord /AB

Given: OA = OB = 15 cm, AB = 24 cm

Q is midpoint of AB

AQ = QB = 24/2 = 12 cm,

also OQ ⊥ AB  

In right triangle  ΔOQA:

OA² = OQ² + AQ²

15² = OQ² + 12²

225 = OQ² + 144

OQ² = 81 

OQ = 9 cm

  ∠OAP = 90° since radius ⊥ tangent

so ΔOAQ ~ ΔOPA

using corresponding sides

OA/OP = OQ/OA

OA² = OQ × OP

15² = 9 × OP

225 = 9 × OP 

 OP = 25 cm 

In right triangle  ΔOPA:

 PA² = OP² − OA²

PA² = 25² − 15² = 625 − 225 = 400

PA = 20 cm  

cbse 10th maths old board exam question paper previous years 2025 2026 stansdard mathematics

circles tangents to circles

a detailed video on an alternative method to solve the problem

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).

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)

Given

P(x, y) is equidistant from A and B, 

 PA = PB

squaring both sides

PA² = PB²

Using distance formula to find PA and PB

(x - 5)² + (y - 3)² = (x - 1)² + (y - 7

Expand both sides using identity

x² - 10x + 25 + y² - 6y + 9 = x² - 2x + 1 + y² - 14y + 49

-10x - 6y + 34 = -2x - 14y + 50

-10x + 2x - 6y + 14y = 50 - 34

-8x + 8y = 16

or

-x + y = 2

or 

y = x + 2

watch this video for more details

cbse 10th maths 2025 2026 old board exam question paper coordinate geometry distance formula