AHSEC Class 11 (SEBA Board) Mathematics Sample Papers, Chapterwise Weightage and Tips Read More »

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]]>The authorities have released the **Assam SEBA Board Class 11 Mathematics question paper 2022** for sebaonline.org. Students preparing for the upcoming Assam Board Class 11 Mathematics Exams can download the previous year’s class 11 Mathematics question papers released by the Central Board of Secondary Education. Students are advised to solve the AHSEC Class 11 Mathematics question paper to boost their preparation level. Moreover, it will help the candidates to understand the complete syllabus weightage in the exam.

Download Assam SEBA Board Class 11 Mathematics Question Sample Papers below

Class 11 Sample Paper 1 | |

Class 11 Sample Paper 2 | |

Class 11 Sample Paper 3 | |

Class 11 Sample Paper 4 |

Class 11 Mathematics Chapterwise Weightage

Unit |
Name of the Unit |
Chapters |
Allotted Marks |

1. | Sets and Functions | Sets Relations and Functions Trigonometric Functions |
23 |

2. | Algebra | Principles of Mathematical induction Complex Numbers and Quadratic Equations Linear Inequalities Permutation and Combinations Binomial Theorem Sequence and Series |
30 |

3. | Coordinate Geometry | Straight Lines Conic Sections Introduction to Three-dimensional Geometry |
10 |

4. | Calculus | Limits and Derivatives | 05 |

5. | Mathematical Reasoning | Mathematical Reasoning | 02 |

6. | Statistics and Probability | Statistics Probability |
10 |

Key Benefits Of Solving Assam SEBA Board Class 11 Mathematics Question Papers. There are a few advantages of this study work out:

- To foster an interest in arithmetic, for understudies who need to seek it for higher investigations
- It improves the speed of addressing questions and time usage abilities.
- It assists competitors with acquiring an understanding of the Assam SEBA Board Class 11 Mathematics test design.
- It assists understudies with acquiring information regarding the matter.
- Ace the fundamental ideas and abilities expected for Class 12.
- Figure out the kind of inquiries and checking plan.
- Helps in breaking down the readiness level.
- Gives a thought regarding the genuine test situation.
- Upgrades test demeanor and lift certainty.
- Tackling issues assist understudies with creating sensible thinking.
- Gives a thought regarding the themes significant for the assessment perspective.
- AHSEC 11 Mathematics Question papers help in understanding the Exam design and its degree of trouble.
- They can get basic comprehension with the assistance of visuals.

**Frequently Asked Questions**

**Q1: What are the chapters you can skip to pass Class 11 maths easily?**

Ans: If any students just want to pass the exam anyhow, then you must give more attention to the chapters like Sets, Complex Numbers, Probability, Mathematical Reasoning, and Limits & Derivatives which hold more than 25 Marks, and in the same way you can skip chapters like Trigonometry and Straight lines.

**Q2: Is class eleventh Math troublesome?**

Ans: Indeed, class 11 Maths is significantly more troublesome than class tenth Maths. New ideas like connection work, set hypothesis, Algebra, and Calculus are presented in this class. Understudies need to give additional consideration in homerooms to grasp these ideas.

**Q3: what number of sections are there in class 11 Maths?**

Ans: There are in each of the 16 sections in class 11 Maths. You can peruse the article above to know the names of these sections and their course contents.

Q4: Which guide is best for class 11 Maths?

Ans: RD Sharma and RS Aggarwal are considered the best class 11 Maths reference books. They show the ideas well as well as furnish understudies with a lot of training questions.

**Q5: Is NCERT enough for class 11 maths?**

Ans: NCERT Maths arrangements are viewed as the best as they furnish you with the best comprehension of the ideas.

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]]>APPLICATIONS OF DIGIT SUM METHOD Read More »

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]]>The digit-sum method has immense utility for practitioners of Numerology and other occult sciences. The knowledge that they can eliminate the 9’s and numbers that add up to 9 makes their task simpler.

For students giving competitive and other exams. this technique has a lot of utility. Many times they can check the digit-sum of each of the alternatives with the digit-sum of the question and try to arrive at the correct answer. This will eliminate the need for going through the whole calculation.

However, there is one drawback with this technique. The drawback is that the digit-sum method can tell us only whether an answer is wrong or not. It cannot tell us with complete accuracy whether an answer is correct or not.

This sentence is so important that I would like to repeat it again.

The digit-sum method can only tell us whether an answer is wrong or not. It cannot tell us with complete accuracy whether an answer is correct or not.

Let me illustrate this with an example.

(Q) What is the product of 9993 multiplied by 9997.

Method: Assume that you have read the question and calculated the answer as 99900021. The digit sum of the question is 3 and the digit-sum of the answer is 3 hence we can assume that the answer is correct.

However, instead of 99900021 had your answer been 99900012 then too the digit-sum would have matched even though the answer is not correct. Or for that matter if your answer would have been 99990021 then too the digit-sum would have matched although this answer is incorrect too. Or in an extreme case, even if your answer would have been 888111021 then still the digit-sum would have matched although it has highly deviated from the correct answer!

Thus, even though the digit-sum of the answer matches that of the question, you cannot be 100% sure of its accuracy. You can be reasonably sure of its accuracy but cannot swear by it.

However, if the digit-sum of the answer does not match the digit sum of the question then you can be 100% sure that the answer is wrong.

In a nutshell,

- Then the answer is matched Then Most likely correct.
- If the digit-sums Do not match then Wrong.

For practitioners of numerology and other occult sciences, there is no question of checking answers and hence this can per se come to their aid.

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]]>Which one of the following is not a “Money Market Instrument”? Read More »

The post Which one of the following is not a “Money Market Instrument”? appeared first on Math Dart.

]]>A. Treasury bills

B. Commercial Paper

C. Certificate of deposit

D. Equity Shares

E. None of these

**Answer:** D: Equity Shares.

The reason is that equity shares are instruments that last for a long time therefore, they cannot be an instrument used to make money.

The money market is a market where financial institutions provide many investors and borrowers the chance to purchase and sell different types of short-term securities. There isn’t a physically-based “money market.” Instead, it’s an informal trading and banking network connected via fax, telephones machines, and computers. The money markets are present across the United States and abroad. Short-term loans and securities that are sold on the market – which are often referred to as market instruments have maturities ranging from a single day to a year and are highly liquid. Treasury bills and notes issued by federal agencies and certificates of deposit (CDs) and eurodollar deposits, commercial paper bankers’ acceptances Repurchases, and bankers’ acceptances are all examples of instruments. The money market instruments’ suppliers money market instruments are banks and individuals who have a preference for the most liquidity and lowest risk.

**What’s Money Market?**

Money Market is an investment market in which short-term financial assets with a liquidity of less than one year can be traded through stock exchanges. The trading bills or securities are very liquid. They also facilitate participants’ short-term borrowing requirements by trading bills. Participants in this financial market include typically large institutions, banks, and private investors.

There is a wide range of instruments that are traded on the market for money in both the stock exchanges NSE as well as BSE. They include Treasury bills as well as certificates of deposit commercial paper, purchase agreements etc. Since the securities that are traded are extremely liquid this market is considered to be a secure location to invest in.

**Specifics that are part of the Money Market:**

- The number of assets that are traded is usually very large.
- It’s still in the process of evolving. There is always the chance of adding new instruments.
- It meets the financial needs of the borrower. It also addresses the investments with an expiration date of one calendar year or less.

I hope the much-hyped question “Which one of the following is not a “Money Market Instrument?” is solved now.

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]]>Ram is twice as fast as Aman and Aman is thrice as fast as Rohit in doing a work. Read More »

The post Ram is twice as fast as Aman and Aman is thrice as fast as Rohit in doing a work. appeared first on Math Dart.

]]>**A.**15

**B.**20

**C.** 25

**D.** 27

**E.** 32

Answer: 25 Days

Solution:

Let the Rohit one day work be ‘x’

Aman’s one day work be ‘3x’

Ram’s one day work be ‘6x’

Therefore, three can do this particular work together = 1/10

\begin{array}{l} \therefore x+3 x+6 x=\frac{1}{10} \\ \Rightarrow 10 x=\frac{1}{10} \\ \therefore x=\frac{1}{100} \end{array}

So, Aman and Rohit can do this work is 3x+x=1/100+1/100

\begin{array}{l} \Rightarrow 3x+x=\frac{1}{100} \\ \Rightarrow 4x=\frac{1}{100} \\ \therefore x=25 \text { Days } \end{array}

Hence, Aman and Rohit can do this work in 25 days

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]]>What is consecutive prime number? Examples Read More »

The post What is consecutive prime number? Examples appeared first on Math Dart.

]]>A few instances of indivisible numbers are 5, 7, 11, 13, and 17. Involving these numbers in a grouping, for example, 11, 13, 17 methods would be continuous, as there are no indivisible numbers between any of these three numbers.

Then again, arrangements 7, 13, 17 wouldn’t address sequential indivisible numbers on the grounds that the indivisible number 11 can go between the 7 and the 13.

Since 2 is the main prime considerably number, It’s conceivable on the grounds that the following significant number, 4, is a composite number, similar to each and every much number after that since they are for the most part uniformly detachable by 2. As a result of all the even numbers beginning with 4 being composite, it’s difficult to have two more prime consecutive. Or then again one more method for saying it is that when you recognize an indivisible number, it’s surefire that the number promptly going before it, as well as the number succeeding it, will be composite.

**Q: Are 1, 2, and 3 the main continuous primes?**

As numerous different responses have been brought up, 1 isn’t prime and 2 and 3 are the main two continuous indivisible numbers (as any remaining even numbers are separable by 2 and subsequently not prime). In any case, I thought I’d pause for a minute to give some thinking with regards to why 1 is certifiably not an indivisible number, despite the fact that this is truly somewhat of a beside your unique inquiry.

One reason primes are so helpful/intriguing is that each and every other number can be communicated as a result of primes. For instance,

60=2^2⋅3⋅5.

Additionally, this articulation is one of a kind.

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]]>Trivial Solution Meaning Read More »

The post Trivial Solution Meaning appeared first on Math Dart.

]]>Ans: connecting with or being the numerically least difficult case.

In maths, the modifier trivial is frequently used to allude to a case or a case that can be promptly acquired from a setting, or an item that has a straightforward construction. The thing technicality, as a rule, alludes to a basic specialized part of some verification or definition.

Let us take a system of 3 equations

a1x+b1y+c1z=0

a2x+b2y+c2z=0

a3x+b3y+c3z=0

and coefficient matrix “A”. So I have been told that the solution of this matrix will be non-trivial if |A|=0 and trivial in any other case. As far as I know, a nontrivial solution means solutions are not equal to zero but in any case, x,y,z=0 will satisfy given equations regardless of the value of the determinant. So, why do we call it a “non-trivial” solution?

If x=y=z=0 then trivial solution And if |A|=0 then non-trivial solution that is the determinant of the coefficients of x,y,z must be equal to zero for the existence of a non-trivial solution.

For example, the equation x+10y=0 has the trivial solution x=0,y=0. Nontrivial solutions include x=10,y=–1 and x=–2,y=0.2.

**Trivial Meaning in Hindi**: मामूली

We wrote trivial solutions in Mathematics higher education. But we never get into depth, sure you’d learn something today.

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]]>The post Assam SLET Mathematics Previous Year Paper: 2012-2022 Download PDF appeared first on Math Dart.

]]>

2012 | Download |

2013 | Download |

2014 | Download |

2015 | Download |

2016 | NA |

2017 | NA |

2018 | NA |

2019 | Download |

2020 | Not Held |

2021 | Download |

Assam SLET Mathematics 2022 PDF |
Download |

“Assam SLET Mathematics 2022 Previous Year Paper Download PDF”

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]]>Pythagorean Values & Triplets Read More »

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]]>We know that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. For example, if the sides of the right-angled triangle are 3, 4, and 5 then the square of 5 equals the square of 3 plus the square of 4. The values 3, 4, and 5 together form a Pythagorean triplet. Other examples of triplets are (6-8-10), (9-40-41), (1160-61), etc. We are going to study a method by which we can form triplets out of a given value. This technique will be of great help to students dealing with geometric figures in competitive exams.

We will divide our study into two parts. In the first part, we will express the square of a number as the sum of two squared numbers. In the second part, we will express a given number as the difference between two squared numbers. The first part is further divided into two cases – the first case dealing with odd numbers and the second dealing with even numbers. a

(A) Expressing the square of a number as the sum of two squared numbers

Case 1: Odd Numbers

Rule: The square of an odd number is also odd. This square is the sum of two consecutive middle digits.

Example: (a) 3^2 = 9 = 4 + 5 (b) 5^2 = 25 = 12 + 13 (c) 9^2 = 81 = 40 +41

We see that the square of 3 (odd) is 9 (odd), which is the sum of two consecutive numbers, viz. 4 and 5. Similarly, 25 is the sum of two consecutive numbers, 12 and 13.

But how do we find the consecutive numbers? The answer is very simple. To find the consecutive numbers, just divide the square by 2 and round it off.

In the first example, the square of 3 is 9. When 9 is divided by 2 the answer is 4.5. When 4.5 is rounded off to its next higher and lower numbers we have 4 and 5! Thus, (3-4-5) form a triplet where the square of the highest number is equal to the sum of the square of the other two numbers.

In the second example, the square of 5 is 25. When 25 is divided by 2 we get 12.5. When 12.5 is rounded off to its next higher and lower numbers, we have 12 and 13. Thus, (5-12-13) forms a triplet where the square of 13 equals the square of 5 plus the square of 12.

In the last example, the square of 9 is 81. When 81 is divided by 2 and rounded off it gives 40 and 41. Thus, (9-40-41) is a triplet where the square of the biggest number (41) is equal to the sum of the square of the other two numbers (9 and 40)

Similarly, 35^2= 1225 (1225/2 = 612.5)

Thus, 35, 612, and 613 form a triplet where 613 units are the length of the hypotenuse and 35 units and 612 units are the lengths of the other two sides of the triangle.

Case 2: Even Numbers

The square of an even number is an even number. Therefore, we cannot have two middle digits on dividing it by 2. Thus, we divide the even number by 2, 4, 8, 16, etc. (powers of 2) until we get an odd number. Once we get an odd number we continue the procedure as done in the case of odd numbers.

If you have divided the even number by 2, 4, 8, etc. the final answer will have to be multiplied by the same number to get the triplet.

Q. One value of a Pythagorean triplet is 6; find the other two values.

We divide 6 by 2 to get an odd number 3. Next, we follow the same procedure as explained in the case of odd numbers to form a triplet (3-4-5). Now, since we have divided the number by 2 we multiply all the values of (3-4-5) by 2 to form the triplet (6-8-10).

Thus, the other two values are 8 and 10.

Q. One value of a triplet is 20; find the other two values.

We divide 20 by 4 to get an odd number 5. Next, we follow the same procedure as explained before to form a triplet (5-1213). Now, since we have divided the number by 4, we multiply all values of the triplet (5-12-13) by 4 and make it (20-48-52).

Thus the other two values are 48 and 52

(B) Expressing a given number as a difference of two squares

We express a given number ‘n’ as a product of two numbers ‘a’ and ‘b’ and then express it as:

\mathrm{n}=[(\mathrm{a}+\mathrm{b}) / 2]^{2}-[(\mathrm{a}-\mathrm{b}) / 2]^{2}Q. Express 15 as a difference of two squared numbers. We know that 15 = 5 x 3. Thus, the value of ‘a’ is 5 and ‘b’ is 3.

Thus,

\begin{aligned} 15 &=[(5+3) / 2]^{2}-[(5-3) / 2]^{2} \\ &=(8 / 2)^{2}-(2 / 2)^{2} \\ &=(4)^{2}-(1)^{2} \end{aligned}

(We can verify that 16 – 1 = 15)

From the above, it can be proved that if one side of the triangle is \sqrt{15} then the other side is 1 unit and the hypotenuse is 4 units. This can be checked with the expression \sqrt{15}^2 + 12 = 42

Q. In A ABC, the value of side angle B is 90 degrees and side BC is 6 units. Find a few possible values of side AB and side AC.

Since the triangle is a right-angled triangle, BC = ACP – AB2. The value of BC2 = 36. Thus, 36 = AC2 – AB?. In other words, we have to express 36 as a difference between the two squares. We will apply the same rule that we used in the previous example.

We have to express the number 36 as a product of 2 numbers ‘a’ and ‘b’.

(a) 36 = 18 x2 = [(18+2)/2]^2 – [(18-2)/2]^2 = [(20)/2]^2 – [(16)/2]^2 = 10^2 – 8^2 = 10,8

(b) 36 = 9 x4 = [(13)/2]^2 – [(5)/2]^2 = 6.5^2 – 2.5^2 = 6.5, 2.5

(c) 36 = 12 x 3 = [(15)/2]^2 – [(9)/2]^2 = 7.5^2 – 4.5^2 = 7.5, 4.5

In example (a) the value of sides AC and AB is 10 units and 8 units.

In example (b) the value of sides AC and AB is 6.5 units and 2.5 units.

In example (c) the value of sides AC and AB is 7.5 units and 4.5 units.

We can verify:

6^2= 10^2 – 8^2

6^2= 6.5^2 – 2.5^2

6^2= 7.5^2 – 4.5^2

From the examples, we can conclude that if the length of side BC is 6 units, then the lengths of sides AC and AB can be either 10 and 8 or 6.5 and 2.5 or 7.5 and 4.5 units. We have thus formed three triplets (6,8,10) (2.5, 6, 6.5) and (4.5, 6, 7.5).

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]]>What is cpct in maths ? Read More »

The post What is cpct in maths ? appeared first on Math Dart.

]]>This implies that at least two triangles are harmonious, then, at that point, their comparing points as a whole and sides are compatible also. Comparing portions of harmonious triangles or **cpct in maths** is utilized to indicate the connection between the sides and the points of two consistent triangles.

Assuming there are two triangles that are harmonious to one another by any of the accompanying guidelines of congruency, then, at that point, their relating sides, as well as points, should be equivalent to one another.

- SSS (Side-Side-Side)
- SAS (Side-Angle-Side)
- ASA or AAS (Angle-Side-Angle)
- RHS (Right point Hypotenuse-Side)

Allow us to take a model if two triangles ΔABC ≅ ΔDEF, by any of the previously mentioned congruency rules, then, at that point

**Corresponding Vertices:**A and D, B and E, C and F**Corresponding Sides:**AB and PQ, BC and QR, AC and PR**Corresponding Angles:**∠A and ∠D ∠ B and ∠E, ∠C and ∠F

Hope you are clear What is cpct in maths ? Please share this with your maths nerds.

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]]>How to predict how much money a person has in his pocket Read More »

The post How to predict how much money a person has in his pocket appeared first on Math Dart.

]]>STEPS

(a) Ask him to take the amount he has in his pocket (just the

rupees, ignore the paisa)

(b) Next ask him to add 5 to it (C) Multiply the answer by 5 (d) Double the answer so obtained

(e) Finally, ask him to add his favourite one digit number (any number from 0 to 9) (f) Add 10

After the steps are over ask them to tell you the final answer. And just by listening to the final answer you will come to know the amount he has in his pocket!

SECRET

• Ignore the digit in the unit’s place.

• From the remaining number, subtract 6 and you will come to know the amount he has in this pocket.

Example:

Let us suppose a person has 20 Rupees in his pocket. He would work out the steps as given below:

(a) Take the amount in your pocket 20

(b) Add 5 25

(c) Multiply by 5: 125

(d) Double the answer: 250

(e) Add your favourite single-digit number (say, 7) 257 (f) Add 1 267

Thus his final answer would have been 267. Now let us see how we can find out the amount he has from the final answer. As mentioned earlier, we will ignore the digit in the unit’s place (in this case it is 7). Now the remaining number is 26. From 26, we subtract 6 to get 20. Thus our answer is confirmed. Similarly if the total was 1062, 63 and 170 the amount would

be 100, 0 and 11 respectively.

(From 1062, we ignore the last digit 2 and take only 106. From 106, we subtract 6 and get the answer as 100 and so on…)

I use this technique in my seminars to give the audience some relief from complex calculations. There are many ways by which we can predict the amount a person has in his pocket but I prefer this technique because it neither involves any complex calculations nor is it too obvious for the audience to guess the secret.

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