139-NCERT-10-Mathematics-All-Topics-Solutions will be published soon

Very soon, the blogs on NCERT 10th-class mathematics solutions will be published. 

1) Developing a strong foundation in mathematics is important for students as it can help them in various fields and in their daily lives. Solving problems systematically and presenting them in an organized manner can help students understand the concepts better and remember them for a longer time.

2) Using easy and efficient methods for calculations can also save time and help students to solve problems quickly and accurately.

3) Solving every problem of all the exercises in NCERT 10th standard mathematics systematically can help students build a strong foundation in the subject and develop problem-solving skills.

4) By solving problems systematically, students can learn how to approach a problem step by step and break it down into manageable parts. This can help them to understand the problem better and find the solution more easily. We have taken the time to solve every problem in the exercises of the textbook and systematically present them for students to learn from. This can be a valuable resource for students who are studying this subject.

5) We are planning to publish such blogs, as they can be a valuable resource for students who are studying 10th-standard mathematics.

6) See the following problems and their solutions in a step-by-step pattern:

1) Exercise 5.2 - Arithmetic Progressions: problem no: 11

Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution:

1) Here,

a₁ = a = 3, a₂ = 15, a₃ = 27, a₄ = 39.

2) Here.

d = a₂ - a₁

d = 15 - 3
d = 12 --------- equation 1

3) Using the formula a_n = a + (n – 1) d,

a_n = a + (n – 1) d 

a₅₄ = 3 + 12 (54 – 1)

a₅₄ = 3 + 12 (53)

a₅₄ = 3 + 636

a₅₄ = 639 --------- equation 2

4) Now we will find n for the term which is more by 132 than the 54th term.

a

a_n = 639 + 132
a_n = 771 --------- equation 3

5) Now we will find the value of n. 

a_n = a + (n – 1) d 

771 = 3 + 12 (n – 1)

12 (n – 1) = 771 - 3

12 (n – 1) = 768

(n – 1) = 768/12

(n – 1) = 64

n = 64 + 1

n = 65 --------- equation 4

6) So, 65th term is 132 more than 54th term.

7) Second method: our term is 132 more than the 54th term. As the difference is 12,

132 will give us 132/12 = 11. So our term is 54 + 11 = 65. So, 65th term is 132 more than 54th term.

2) Exercise 2.4 - Polynomials: problem no: 11

Q2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. 

Explanation:

1) We know that if 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial 

ax³ + bx² + cx + d, then 

𝝰 + 𝛃 + 𝜸 = - b/a = [- (coeficient of x²)/(coeficient of x³)]

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a = [(coeficient of x)/(coeficient of x³)]
𝝰 x 𝛃 x 𝜸 = - d/a = [- (constant)/(coeficient of x³)]

2)  If 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial

p(x) = ax³ + bx² + cx + d, then p(𝝰) = p(𝛃) = p(𝜸) = 0.

Solution:

1) Let ax³ + bx² + cx + d be our cubic polynomial.

2) Here 𝝰 + 𝛃 + 𝜸 = 2, 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = - 7, 𝝰 x 𝛃 x 𝜸 = - 14.

3) We know that 

𝝰 + 𝛃 + 𝜸 = - b/a

2/1 = - b/a ------------- therefore a = 1, and b = -2.

 

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a

-7/1 = c/a ------------- therefore a = 1, and c = - 7.

 

𝝰 x 𝛃 x 𝜸 = - d/a

-14/1 = -d/a ------------- therefore a = 1, and c = 14.

4) our cubic polynomial will be x³ - 2x² - 7x + 14.  

3) Exercise 1.3 - Real Numbers: problem no: 2

Q2. Prove that 3 + 2√5 is irrational.

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let if possible, consider that 3 + 2√5 be a rational number.

2) So, 3 + 2√5 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

3 + 2√5 = p/q 

2√5 = (p/q) - 3

2√5 = [(p-3q)/q]

√5 = (p-3q)/2q

3) As p and q are co-primes, (p-3q)/2q is rational, so √5 is also rational which contradicts the actual fact that √5 is irrational.

4) So, 3 + 2√5 is an irrational number.

4) Exercise 3.6 - Pair of Linear Equations in Two Variables: problem no: 2-(iii)

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

1) Let the speed of the train be x km/h.

2) Let the speed of the bus be y km/h.

3) Total distance traveled by Roohi is 300 km.

4) According to the first condition, she traveled 60 km by train and 240 km by bus in

4 hrs, so, we have,

60/x + 240/y = 4, put 1/x = p and 1/y = q, we have,

60p + 240q = 4

4(15p + 60q) = 4

15p + 60q = 1

15(p + 4q) = 1

p + 4q = 1/15
p = ((1/15) - 4q) -------------------------- equation 1.

5) According to the second condition, she traveled 100 km by train and 200 km by

bus in 4 hrs and 10 minutes i.e. 4 + (10/60) = 4 + (1/6) = 25/6 hrs, so, we have,

100/x + 200/y = 25/6, put 1/x = p and 1/y = q, we have,

100p + 200q = 25/6

25(4p + 8q) = 25/6

(4p + 8q) = 1/6 -------------------------- equation 2.

6) Put the value of p = ((1/15) - 4q) from equation 1 in equation 2, we get,

(4p + 8q) = 1/6

(4((1/15) - 4q) + 8q) = 1/6

(4/15 - 16q + 8q) = 1/6

(4/15 - 8q) = 1/6 

8q = 4/15 - 1/6 

8q = (8/30) - (5/30)

8q = 3/30

8q = 1/10

q = 1/(10(8)) 

q = 1/80 -------------------------- equation 3.

7) Put the value of q = 1/80 from equation 3 in equation 1, we get,

p = ((1/15) - 4q)

p = ((1/15) - 4(1/80))

p = ((1/15) - 1/20))

p = ((4/60) - 3/60))

p = (4 - 3)/60

p = 1/60  -------------------------- equation 4.

8) The value of p = 1/60 and the value of q = 1/80.

9) As, p = 1/x and p = 1/60, we have x = 60.

10) As, q = 1/y and q = 1/80, we have y = 80.

11) The speed of the train is 60 km/h. 

12) The speed of the bus is 80 km/h.

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