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.

 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.

Let the three consecutive positive integers be n, n+1, n+2

Given that

that sum of square of the first and the product of the other two is 67

n² + (n+1)(n+2) = 67

n² + (n² + 2n + n + 2) = 67  

n² + n² + 3n + 2 = 67  

2n² + 3n + 2 = 67  

2n² + 3n + 2 - 67  = 0

2n² + 3n − 65 = 0

Solve by quadratic formula or factorisation

Factorisation method

Find two numbers that gives a product of 2 × (−65) = −130 and add to 3.

by trial and error the numbers13 and −10 satisfy the condition

use this to split the middle term of

2n² + 3n − 65 = 0

.2n² + 13n − 10n − 65 = 0

n(2n + 13) − 5(2n + 13) = 0

(n − 5)(2n + 13) = 0

solving

n − 5 = 0 ⇒ n = 5

2n + 13 = 0 ⇒ n = −13/2 (rejecct)

required positive integers are

n=5

n+1 =6

n+2=7

for a video of the solution

cbse 10th maths pervious years old question papers 2025 2026 standard mathematics

   quadratic equation word problem  formation and solution by splitting the middle term

∫ cos x / [(2 + sin x)(4 + sin x)] dx

 find ∫ cos x / [(2 + sin x)(4 + sin x)] dx 

integration by substitution and partial fractions

Substitution

Let t = sin x

Then dt = cos x dx

 ∫ cos x / [(2 + sin x)(4 + sin x)] dx =  ∫ 1 / [(2 + t)(4 + t)] dt

Method of partial fractions

 1 / [(t + 2(t + 4)] = A/(t + 2) + B/(t + 4)

multiply [t+2][t+4]

1 = A(t + 4) + B(t + 2)

Put t = −2

1 = A(2) 

⇒ A = ½

Put t = −4:

1 = B(−2) 

⇒ B = −½

1 / [(t + 2)(t + 4)] = ½[1/(t + 2) − 1/(t + 4)]

integrate

I = ½  ∫ [1/(t + 2)  − 1/(t + 4)] dt 

= ½[ln|t + 2| − ln|t + 4|] + C

= ½ ln|(t + 2)/(t + 4)| + C   use  t = sin x

=½ ln[(2 + sin x)/(4 + sin x)] + C

for  explanation watch this video 

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

The probability of guessing the correct answer of a certain test question is (x/12). If the probability of not guessing the correct answer is ( ⅚), then find the value of x.

 The probability of guessing the correct answer of a certain test question is (x/12). If the probability of not guessing the correct answer is ( ⅚), then find the value of x.

cbse 10 th math old board exam question paper probability question mathematics standard

watch the video for more

At first glance, it may appear simple, but this problem is an excellent example of a question that checks conceptual understanding, numerical accuracy, and knowledge of complementary events in probability. It can appear in numerous exams worldwide, including CBSE, ICSE, IGCSE, GCSE, IB, AP, SAT, ACT, GRE, GMAT, and SOA actuarial examinations.

Step-by-Step Solution

To solve the problem, we start by recalling one of the fundamental rules in probability:

The sum of the probability of an event and the probability of its complement is always equal to one.

Let’s denote:

• $P(\text{correct}) = x/12$
  

• $P(\text{not correct}) = 5/6$
  

According to the complementary rule:

$$P(\text{correct}) + P(\text{not correct}) = 1$$

Substituting the given values:

$$x/12 + 5/6 = 1$$

To solve for $x$ first express 5/6 as a fraction with denominator 12:

$$5/6 = 10/12$$

So the equation becomes:

$$x/12 + 10/12 = 1$$

Combine like terms:

$$(x + 10)/12 = 1$$

Multiply both sides by 12 to eliminate the denominator:

$$x + 10 = 12$$

Subtract 10 from both sides:

$$x=2$$

Therefore, the probability of guessing the correct answer is 2/12, which simplifies to 1/6. This satisfies the given probability of not guessing the correct answer (5/6), since $1/6 + 5/6 = 1$.

Understanding Complementary Probability

The problem provides a perfect example of the complementary rule. In probability theory, every event has a complement — the set of outcomes where the event does not occur. The sum of the probabilities of an event and its complement always equals one.

This principle is fundamental in mathematics curricula worldwide. It is introduced in CBSE Class 9 as part of Chapter 15: Probability, where students learn about experimental probability using dice, coins, and simple card experiments. ICSE Class 9 and 10 also emphasize complement probability in their algebraic and practical problem sections. In IGCSE and GCSE mathematics, complement rules are a standard part of both foundation and higher-tier probability questions. IB Diploma students encounter similar questions in Analysis and Approaches or Applications and Interpretation courses, where both theoretical and experimental probability are explored.

Even in professional contexts such as AP Statistics, GRE, and GMAT quantitative sections, the complementary rule forms the basis for more complex probability problems, including conditional probability, joint probability, and expected value calculations. SOA actuarial exams often use complement probability as a foundational step before progressing to advanced stochastic models and actuarial risk assessments.