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

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 present them in a systematic way 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:

a_{1}= a = 3, a_{2}= 15, a_{3}= 27, a_{4}= 39.

d = a_{2}- a_{1}

d = 15 - 3

d = 12 --------- equation 1

_{n}= a + (n – 1) d,

a_{n}= a + (n – 1) d

a_{54}= 3 + 12 (54 – 1)

a_{54}= 3 + 12 (53)

a_{54}= 3 + 636

a_{54}= 639 --------- equation 2

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

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:

ax^{3}+ bx^{2}+ cx + d, then

π° + π + πΈ = - b/a = [- (coeficient of x^{2})/(coeficient of x^{3})]

π°π + ππΈ + πΈπ° = c/a = [(coeficient of x)/(coeficient of x^{3})]

π° x π x πΈ = - d/a = [- (constant)/(coeficient of x^{3})]

p(x) = ax^{3}+ bx^{2}+ cx + d, then p(π°) = p(π) = p(πΈ) = 0.

### Solution:

^{3}+ bx

^{2}+ cx + d be our cubic polynomial.

π° + π + πΈ = - b/a

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

π°π + ππΈ + πΈπ° = c/a

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

π° x π x πΈ = - d/a-14/1 = -d/a ------------- therefore a = 1, and c = 14.

4) So, our cubic polynomial will be x^{3} - 2x^{2} - 7x + 14.

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

**Q2. Prove that 3 + 2√5 is irrational.**

### Explanation:

### Solution:

3 + 2√5 = p/q^{ }2√5 = (p/q) - 32√5 = [(p-3q)/q]√5 = (p-3q)/2q

**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) = 415p + 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.**

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

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