Miraculous world of Numbers: March 2023

Sunday, March 5, 2023

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.

The blogs on NCERT 10th-class mathematics solutions are published.

#NCERT #10thStd #Mathematics

ANIL SATPUTE

Friday, March 3, 2023

138-How to check the acceptance of Jeevan pramaan by pension authority

Click here for ⇨ previous blog on Life Certificate.

To check whether the Pension Distributing Agency has accepted your Digital Life Certificate (DLC) or not, you can follow the steps given below:

1) Visit the Jeevan Pramaan website

https://jeevanpramaan.gov.in/ppouser/login

2) Enter your Pramaan ID, and the captcha code displayed on the screen.
3) Click on the "Generate OTP" button.

4) Enter OTP and click on "Login" button.

Note: This disclaimer will be as per your pension distribution unit.

5) It depends on the pension distrubution unit when to accept the DLC (Jeevan

pramaan). Normally we can the status of our DLC, at the end of month we can check it at any number of times. If the DLC is accepted, then the bottom remark will change.

6) Repeat the steps 1 to 5 given above. see the remark at the bottom of your

certificate.

Note: This remark will be as per your pension distribution unit.

7) Now we can download the DLC (Jeevan Pranaam) for your record.

8) This DLC (Jeevan Pranaam) is for one year.

9) We can use the same blog for the next year.

ANIL SATPUTE

Thursday, March 2, 2023

137-Jeevan pramaan by mobile app (Life Certificate by mobile app)

Jeevan Pramaan is a digital life certificate that is issued to pensioners in India. It was launched by the Government of India in 2014 as part of its Digital India initiative.

The Jeevan Pramaan certificate is used to provide proof of life to pensioners so that they can continue to receive their pension without having to physically visit the pension disbursing agency. This is especially beneficial for elderly pensioners who may have mobility issues or live in remote areas.

The Jeevan Pramaan certificate is issued based on the pensioner's biometric authentication, which includes fingerprint or iris scanning. The pensioner can provide their biometric details at a Jeevan Pramaan center or through a mobile app. Once the authentication is successful, the digital life certificate is generated and sent to the pension disbursing agency.

The Jeevan Pramaan certificate is valid for one year and can be renewed annually. This digital certificate has simplified the pension verification process, reduced paperwork, and eliminated the need for physical visits by pensioners to government offices.

Yes, Jeevan Pramaan can also be generated through a mobile app. The Jeevan Pramaan mobile app is available for both Android and iOS devices and can be downloaded from the respective app stores.

To generate the Jeevan Pramaan certificate through the mobile app, the pensioner needs to first download and install the app on their mobile device. They will then need to register by providing their Aadhaar number and other details.

Once the registration is complete, the pensioner can provide their biometric details, through the mobile app. The app will then generate the digital life certificate, which can be downloaded and saved for future use.

It is important to note that to use the Jeevan Pramaan mobile app, the pensioner's mobile device must have a working internet connection.

Additionally, the pensioner must have linked their Aadhaar number with their bank account and pension account to ensure successful authentication and disbursal of pension.

Now we will see step-by-step procedures to use a mobile app for generating a digital life certificate.