5 study habits to prepare effectively for Board Exam

Here are 5 study-habits to prep-up for your upcoming examinations. Now that the practical-examinations have started across many schools in India, the theory papers are just a month away. This is the time to focus all your energy towards your preparation for best results in your examinations.

Scholr brings to you these 5 study-habits to take-up, to take the best out-of you.

STUDY HABITS

  1. Revision Notes– It is very important to make revision notes which are short and very informative. They are just points which takes you back to the entire context related to the topic. So make catchy flash-cards, flow-charts, formula-sheets and all that you might need at the very last minute. Sometimes all you remember in the exam is where you read the topic in your notes, if you remember these points, you will get the answer right!
  2. Don’t Mock the Mocks– Get sample-tests for yourself, and practice them like a real exam. Time yourself, keep all study-materials away, and write as if you were writing the final paper. Post sample-paper, evaluate the answer-sheet or take help from your mentor. Evaluate yourself seriously to know where you are losing on marks.
  3. Routine-Up: Past, Present and Future- Analyse syllabus for all subjects and allot time for the coming month. Look back on your preparations to see, which topic or subject is stronger and requires relatively less time. Accordingly, divide time left for the final exams and routine up!
  4. Take Breaks and Reward Yourself: It is very important to keep taking breaks in between. Maybe you can take a Sunday off, this will kick-start your preparations when you resume them. Also, for every sample paper where you cross your target score, treat yourself. Maybe, go watch a movie or order pizza. The idea is to not over-stress yourself as it reduces efficiency
  5. Pre-plan the Answer Structure: And finally, depending upon the blue-print of the question-paper, structure answers for each topic. Include the must-haves of the topic, any diagrams or flow-charts that might give your answer a more complete look!

So, use these 5-study habits and you are ready to go, just stay confident and you will rule your board-examinations!

Also read: Roll-numbers for Class Xth CBSE Board Examinations are now out!

CBSE BOARD EXAMINATIONS: ROLL-NUMBERS ARE OUT!

CBSE Boards

The roll-numbers are out!

The Central Board of Secondary Education has released the roll numbers for aspirants appearing for 10th and 12th Board Examinations this year. Follow the steps given below to get your roll-numbers.Roll-Numbers

Step I- Visit CBSE’s Official Website at cbse.nic.in

Step II- Scroll down and look for the In-Focus Tab on your right. Under the tab find ‘Roll No. LOC for Class X/XII Exam,2019’

Step III- The authentication page will pop-up and you will require the following credentials: a. User ID b. Password. Also, type the security key given to proceed further.

Step IV- Almost there! Now simply download your roll numbers.

Who knows this number-combination might actually turn lucky for you! Your lucky number is now just a few clicks away.

Also, are you appearing from the private boards? Your admit-cards are out too! Let us guide you to them too! Just reach here and begin with your credentials. Select your region and keep the following credentials handy

  1. Application No.
  2. Previous Roll. No and Year
  3. Candidate Name

Now you can simply download your admit card and get it printed! Keep it safely and remember to keep a digital copy of it. For rest of the students, wait until your admit-cards are released and we will guide you through it.

CBSE 10th Board Examinations preparation strategy

Board exam Xth

Boost Up, your 10th Board Examinations will be here in no time. Excited or Nervous? Beat the exam blues and here are 8 tips to boost your exam preparations. 

Boost Up. Get, Set, Go!

Boost Up. Get, Set, Go!

  1. Gather yourself– You have come this far and it is about to get over. Just calm yourself and stay focused. Keep treating yourself so that you stay motivated!
  2. Practice: Practice is very important for any subject, whether they are chemistry formulas or history timeliness. Therefore, Go over and over again the topics, so that they do not slip out.
  3. Take Breaks– However, do not over-stress yourself, keep taking small breaks between your study-hours. Talk to people around you, especially friends who are sailing in the same boat.
  4. Group-Study: Also, studying with a buddy can do wonders. Talk to your close friends about syllabus and preparation tricks, learn from them, and help them study too.
  5. Take Mocks: More importantly, solving sample questions is a very good way to prepare and do not consider it just a mock. Time yourself and asses your strengths and weakness. After every paper list down what went wrong and how you could use time optimally.
  6. Doubts: On going through the syllabus over and over again, identify your doubts. Which is the one formula you always forget? Or the one mistake you tend to repeat? Is there a numerical that you cannot understand? List your doubts and get them solved. Friends, Parents or Teachers- just reach-out. Unsolved doubts can be dangerous!
  7. Understand the Concepts: Considering the vast syllabus, differentiate between what can be cleared at a conceptual level and what needs memorizing. DO NOT TRY TO MEMORIZE EVERYTHING. For concept based questions, conceptual clarity is must!
  8. Revision Notes: And finally, Make revision notes for lengthy topics. Revision Notes like flowcharts and flash cards will help you in last-minute revision. So, make handy and catchy notes, for last-minute help.

That’s all from us in this post. Do try to include the above points in your board examinations preparation strategy. All the best!

Meanwhile, you can also read about the newly introduced Know Your Aptitude Exam. 

Sequence and Series

Sequences Series

Arithmetic Progression

An arithmetic progression can be defined as a sequence of numbers wherein every term is derived from the preceding term by adding or subtracting a fixed number called the common difference “d”

For instance, the sequence 10, 6, 2, -2, …. is an arithmetic progression which has -4 as the common difference. The progression -2, 0, 2 , 4, 6 is an Arithmetic Progression (AP) with 2 as the common difference.

nth term of an AP series is Tn = a + (n – 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn – Tn-1.

The sum of first n terms of an AP: S =(n/2)[2a + (n- 1)d]
S =(n/2)[a + l] where l is the last term

Tn = Sn – Sn-1 , where Tn = nth term

If a, b and c are three terms in AP then arithmetic mean b = (a+c)/2

Geometric Progression

In a geometric sequence, every term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. For example, the sequence 9, 3, 1, 1/3,…. is a Geometric Progression (GP) for which 1/3 is the common ratio.

The general form of a GP is a, ar, ar2, ar3 and so on.

The nth term of a GP series is Tn = arn-1, where a = first term and r = common ratio = Tn/T(n-1) .

The formula applied to calculate the sum of first n terms of a GP:
a(1−rn)/1−r   where  r<1    OR     a(rn-1)/r-1         where  r > 1

When there are three quantities present in GP, the middle one is called as the geometric mean of the other two. If a, b and c are three quantities in GP and b is the geometric mean of a and c i.e. b =√ac

The sum of infinite terms of a GP series S∞= a/(1-r) where 0< r<1.

Harmonic Progression

A series of terms is called as an HP or HS series when their reciprocals are in arithmetic progression.

Example: 1/a, 1/(a+d), 1/(a+2d), and so on are in HP because a, a + d, a + 2d are in AP.

The nth term of a HP series is Tn =1/ [a + (n -1) d].

For solving a problem on Harmonic Sequence, you need to make the corresponding AP series and then solve the problem.

nth term of H.P. = 1/(nth term of corresponding A.P.)

If there are three terms a, b, c in HP, then b =2ac/(a+c).

Solved Examples:

1.
9+99+999+9999+99999+……. n terms
Let S= 9+99+999+9999+99999+……. n terms
S= (10–1)+(100–1)+(1000–1)+(1000–1)…..n terms
S= {(10+100+1000+…..n terms)-(1+1+1+1+…..n terms)}
S= {(10+100+1000+…..n terms)-n}
Therefore first bracket becomes a GP.
S= 10(10n – 1)/(10–1) – n
S=[ 10(10n – 1) – 9n ] / 9

 

2.
S=0.5+0.55+0.555+…..n terms
Taking 5 as common we get
S=5(0.1+0.11+0.111+…nterms)
Multiply rhs with 9/9
S=(5×9/9)[0.1+0.11+0.111+…..n terms]
S=5/9[0.9+0.99+0.999+….n terms]
S=5/9[(1-0.1)+(1-0.01)+(1-0.001)+….n terms]
S=5/9[(1+1+1+1+…n terms)-(0.1+0.01+0.001+….n terms)]—-(1)
Now find sum of each bracket. Separately
i)1+1+1+…n terms =n—-(2)
ii)0.1+0.01+0.001+….n terms
Here
First term =a = 0.1=1/10
Common ratio= r= a2/a1=0.01/0.1 =0.1=1/10
r<1
Therefore the series is in G P.
Sum =[a(1-rn)]/(1-r)

=[1/10(1-  1/10n)]/(1- 1/10)
=[1/10(1-1/10n]/(9/10)
=(1 –  1/10n)/9—-(3)
Put (2) and (3) in (1)
S=5/9[n -(1-     1/10n)/9]
S=[(45n – 5)/81]  [1-   (1/10n)]

 

The above questions deal with the difficult aspects of this topic. Hope this helped. Share your feedback in the comment section below. In case you missed out, check out our latest blog  on  Special series made easy 

Happy learning, mathemagicians!

Linear Inequations

Linear_inequalities

Before we dive into the sums, it is necessary t understand all the terms related to this topic. For instance, many misunderstand the question because they fail to figure out the difference between an inequality and an inequation.

What is an Inequality?

The open sentence which involves >, ≥, <, ≤ sign are called an inequality.

What is an Equation?

An expression that equates two expressions or values is called an equation. For e.g. 6x + 19y = 8.

Difference between the two

In contrast, if an expression relates two expressions or values with a ‘<’ (less than) sign, ‘>’ (greater than) sign, ‘≤’ (less than or equal) sign or ‘≥’ (greater than or equal) sign, then it is called as an Inequality.

What is an Inequation?

A statement indicating that value of one quantity or algebraic expression which is not equal to another are called an inequation.

For example,

(i) 8x ≥ 7
(ii) x < 6

(iii) x > 2
(iv) 3x – 18 ≤ 2

Therefore, each of the above statements is an inequation.
The statements of any of the forms ax + b > 0, ax + b ≥ 0, ax + b < 0, ax + b ≤ 0 are linear inequations in variable x, where a, b are real numbers and a ≠ 0.

1. Consider an inequation x < 2. Let the replacement be the set of Natural numbers (N).

We know that N = {1, 2, 3, …}.

Thus the solution set to the inequation is {1}.

2. If we take the example of 20x ≤ 200
This has two parts
20x < 130 which is an inequality
20x= 130 which is an equation
• For x = 0, L.H.S. = 20 (0) = 0. ∴ LHS < RHS (0 < 130) is true.
• For x = 1, L.H.S. = 20 (1) = 20 ∴ LHS < RHS (20 < 130) is true.
• For x = 2, L.H.S. = 20 (2) = 40 ∴ LHS < RHS (40 < 130) is true.
• For x = 3, L.H.S. = 20 (3) = 60 ∴ LHS < RHS (60 < 130) is true.
• For x = 4, L.H.S. = 20 (4) = 80 ∴ LHS < RHS (80 < 130) is true.
• For x = 5, L.H.S. = 20 (5) = 100 ∴ LHS < RHS (100 < 130) is true.
• For x = 6, L.H.S. = 20 (6) = 120 ∴ LHS < RHS (120 < 130) is true.
• For x = 7, L.H.S. = 20 (7) = 140 ∴ LHS < RHS (140 < 130) is false.
x = 0, 1, 2, 3, 4, 5, and 6.
{0, 1, 2, 3, 4, 5, 6} is called the solution set.x2

3.  x2 + 6x – 7 ≥ 0

Solving it further, ( x + 7) ( x- 1) ≥ 0

So, values of x lie outside the range of -7 to 1

Hence x = (- ∞, -7]  union  [1, ∞ )

 

Therefore, we conclude this topic. Comment your feedback and to support your knowledge of this topic, do check out, Linear equations in one variable.

Happy Learning! Adios.

 

 

 

Representation of linear inequalities

linear inequalities

Linear inequalities refer to a relation that holds between two values when they are different. Before moving ahead let us clear some basic concepts:-

  • If the notation a<b means that a is less than b.
  • The notation a>b means that a is greater than b.
  •  Notation a\n\neq b  means that a is not equal to b .
  • The notation a\leq b  means that a is less than or equal to b .
  • The notation a\geq b means that a is greater than or equal to b.

Types of Inequalities

  1. Linear inequalities in one variable :- Inequality which involves  a linear function in one variable .Example – Y<8
  2. Linear inequalities in two variables:- Inequality which involves a linear function in two variables. Example – 5x+2y<9

Rules of inequality

  • Equal numbers should be added or subtracted from both sides of an inequality without affecting the sign of inequality.

Example :-  x<8 is equal to  x+2 <8+2

  • Both sides should be multiplied or divided with the same positive number without affecting the sign of inequality.

Example :- x+y<9 is equal to (x+y)\times3 < 7\times3

  •  You should reverse the sign of inequality in case you divide or multiply both sides by a negative number.

Example :- x+y <9 is equal to (x+y)\times(-3) >7\times(-3)

Representation of linear inequalities in one variable on number line and graphical solution in two variables 

 

Class 11 Maths Chapter 6 Linear Inequalities

 

 

Trigonometry Functions

Trigonometry

Trigonometry studies the relationships between the angles and sides of a triangle.Furthermore, it is used in many spheres of life from describing the state of an atom to analysing a musical note. So trigonometric functions are sine, cosine and tangent. Given below are the trigonometric ratios at a glance and their relation with the functions.

 

Here,   θ is any given angle, but let’s look at some standard angles.

POINTS TO REMEMBER

1. The ‘co’ in the functions cosine, cotangent and cosecant stands for complementary

cos(θ) = sin(90 – θ)

cosec(θ) = sec(90 – θ)

cot(θ) = tan(90 – θ)

2. Relation between the functions

cosec(θ) = 1/sin(θ)

Similarly, sec(θ) = 1/cos(θ)

tan(θ) = 1/cot(θ)

*So, cot and tan functions are complementary and their ratios of reciprocals of each other. This will help us solve some tricky sums.

EXAMPLE

Find  tan10 . tan80 . tan45

=  tan10. tan80. 1                                Reason: tan45 = 1

= tan10 .    1/cot80                              Reason: tan(θ) = 1/cot(θ)

=   tan10/cot80

=   tan10/tan( 90 – 80) which is equal to tan10 / tan10

Therefore, ans is 1.

3. Signs

If A + B = 180° then:

  • sin(A) = sin(B)
  • cos(A) = -cos(B)
  • tan(A) = -tan(B)
  • cot(A) = -cot(B)

Similarly, if A + B = 90° then:

  • sin(A) = cos(B)
  • cos(A) = sin(B)
  • tan(A) = cot(B)
  • cot(A) = tan(B)

4. cos2(A) + sin2(A) = 1

cot2(A) + tan2(A) = 1

sec2(A) + cosec2(A) = 1

5. Quadrants

Most noteworthy, depending on the quadrant in which the angle is situated, a sign is assigned to the value.

Certainly, this can be remembered by a SHORTCUT. For instance,

The bold letters stand for ASTC

All   Silver  Tea  Cups

Also, if you’re a gujju, A SThay  Che (What is happening ), a question most students find themselves asking during Math class. But don’t worry, we’re here to help.

5. Radian

Radian is another unit to measure angles.

In terms of radian measure, a complete rotation (360 degrees) is 2π radians. As a result, half of a rotation, or 180 degrees, must therefore be π radians, and 90 degrees must be one-half pi.

So from the above statement, let’s figure out the values of standard angles:

90°=  

45°= 

30°= 

 

Consequently, we will be using the radian unit of measurement.

SOME SOLVED PROBLEMS:

 sin2(39°) + cos2(39°)=?

Since, cos2(A) + sin2(A) = 1.

Hence,  Ans) 1

Another question, find cosθ,   if sinθ = 5/ 13  , when π​/2 < θπ

Because θ belongs to the second quadrant, cosθ will be negative.

co=− Square root of (1−sinθ ) 

=−Square root of (1− 25/169 )

Hence,  Ans) − 12/ 13

 

So, this topic now concludes. Do comment and share your feedback. Hope this helped! In case you missed it,check out further application of Trigo Functions in  our blog on Heights and Distances. Finally, Happy Learning!

Bayes’ Theorem

Bayes

Before beginning with Bayes’ Theorem let’s skim through the formula for Conditional Probability. The conditional probability of an event B is the probability that the event will occur given the knowledge that another event A has already occurred. This expression for probability is as P(B|A), notation for the probability of B given A.

If events A and B are not independent, then the probability of the intersection of A and B (the probability that both events occur) is defined by
P(A and B) = P(A)P(B|A).

From this definition, the conditional probability P(B|A) is easily obtained by dividing by P(A):

Conditional Probability

           Conditional Probability

Bayes’ Theorem is also known as the Inverse Probability Theorem. The theorem simply suggests that if {E1, E2,…, En} are mutually exclusive and exhaustive events associated with a sample space, and A is any event of non-zero probability, then

Baye's Theorem

Consider A to be any event in a sample space of say, s. The occurrence of A could be the result of various reasons [1,2,3,4…….n]. Of all the possible causes that made A occur, suppose we are to find the probability that it has occurred due to a cause say ‘i’. Thus, for that we will have to calculate P(Ai/A). It is for this calculation that Bayes’ Theorem comes handy.

Question: A company has two plants which manufacture scooters. Plant I manufactures 80% of the scooters while Plant II manufactures 20% of the scooters. At Plant I,  85 out of 100 scooters are rated as being of standard quality, while at Plant II only 65 out of 100 scooters are rated as being of standard quality. If a scooter is of standard quality , what is the probability that it come from Plant I.

Solution : –

Let A and B denotes the scooters manufactured in Plant I and Plant II respectively and Z represent the event that the scooter manufactured is of standard quality.

Then,

  1. Probability   P(A) = 80/100 = 0.8 &  P(B) = 20/100 = 0.2
  2. P(Z/A) = 85/100 = 0.85
  • P(Z/B) = 65/100 = 0.65 .

Applying Bayes’ Theorem , the probability of that the selected scooter is of standard quality produced by Plant I ,

P(A/Z) = [P(A) P(Z/A)]/[P(A)P(Z/A) + P(B)P(Z/B)]

= [0.8×0.85]/[(0.8 ×0.85) + (0.2×0.65)]

Therefore, 0.68/(0.68 + 0.13) = 0.68/0.81 = 68/81 = 0.84

So quick and simple, just practice the steps given above and never lose a mark for Bayes’ Theorem!

Also, read our post on Trigonometry Functions!

CBSE announces “Know Your Aptitude” tests for Class 9, 10 students

Aptitude

CBSE has recently introduced a Know Your Aptitude Test. To learn what it is and how it will be helpful, read the post!

Know Your Aptitude is an initiative by Central Board of Secondary Education. The exam primarily aims at allowing the students to asses the correct field for future-studies. It measures the child’s inclination towards a particular stream [Science, Commerce or Arts] or a particular field.

The exam-paper is curated by the National Council of Educational Research and Training (NCERT) and will be available online to all schools willing to conduct the test. The test is voluntary and can be conducted by schools, willing to do so. Schools have to download the test guides, manual, test booklet and answer sheets, which will be available on CBSE website from January 29.

Know what you are best at

Know what you are best at

Often post standard 10th, students are perplexed with the choice to make amongst the three different streams. It becomes difficult to lay their fingers on any one field and conclusively affirm that they want to pursue this for the rest of their lives. This is a good opportunity to enable the students to measure their skills, tendencies and inclinations. Whether there lives a smart banker in your head or is it a defensive lawyer; know what comes to you naturally by knowing your aptitude. Don’t just choose what is popular, choose the shoe that fits you best! As Albert Einstien correctly said;

“Everybody is a genius. But if you judge a fish by its ability to climb a tree, it will live its whole life believing that it is stupid.”

Apparently, A similar initiative was taken by the State Governments in the name of Kal Chachini. The Kal Chachani project is working towards realizing this outcome by equipping Grade 10 students with the right tools to make the right career choices at the right time. Till date, the Kal Chachani Interest Test and the Maha Career Mitra portal have helped numerous adolescents find the right direction in life; and with your support, we can help many more youngsters choose a better future for themselves.

So, gear up to do what you are best at. It is better to ace the field of your interest than just be failing in a career that is popular. Use the test and use it to your best!

Triangles [Types and Properties]

Types of Triangles

The plane figure bounded by three lines, joining three non collinear points, is called a triangle. As per the number of sides and the measure of the angles of triangle, there can be various types of triangles as given in the figure below.

Figure 1: Types of Triangles

Figure 1: Types of Triangles

Important Properties of a Triangle

  1. The sum of the three interior angles of a triangle is always 180 degrees.
  2. The sum of the three exterior angles of a triangle is always 360 degrees.
  3. Sum of two sides of a triangle is always more than the length of the 3rd side.
  4. Difference of lengths of any 2 sides is always smaller than the length of the 3rd side.
  5. The shortest side is always opposite to the smallest interior angle. Similarly, the longest side is always opposite to the largest interior angle.
  6. The orthocenter, the centroid, and the circumcenter of a triangle line on a straight line. This line is  the Euler Line (Black line in the figure below).

    Figure 2: Orthocentre, Centroid and Median

                 Figure 2: Circumcenter, Orthocentre, Centroid and Median

  7. The centroid divides each median in the ratio of 2:1
  8. The line segment, on the Euler line, between the circum-centre & orthocentre is divide in the ratio of 1:2 by the centroid.

Different ways to calculate the area of a triangle:

  1. Conventional Formula of ½ x base x height. This formula is for cases when we have the base and the height of the triangle.
  2. Area of Equilateral Triangle = (√3/4)a2(where a = side of the equilateral triangle)- Substituting a=b=c in the Heron’s formula, we get the above.
  3. Area of Triangle = (1/2)(ab)sin(C)– This formula is for when we know, two sides and the angle in between them. It can be obtained from the basic formula, Area of Triangle = 0.5 × base × height. The height with respect to side ‘a’ can be written in as b×sin(C), where C is the angle between a and b.
  4. Area of Triangle = √(s × (s-a) × (s-b) × (s-c)) (where s = (a + b + c)/2) – This is the Heron’s Formula. It is useful when we know the three sides of the triangle.

    Figure 3: Monster Munch

    Figure 3: Monster Munch

Now look at the figure given below and play the monster munch game as per the instructions to practice different types of triangle.

Also read out post on Area of Parallelograms and Triangle, for step-by-step solutions!