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Lesson Posted on 08 Mar CBSE/Class 10/Mathematics CBSE/Class 12/Science/Mathematics

Honesty and Dedication helps in education and success in life.

Jagdish S.

Mathematics teaching for CBSE ICSE,PUC state CET,JEE IIT and Engineering maths ,CAT,MAT,International...

Honesty and Dedication helps in education and success in life. As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel... read more

Honesty and Dedication helps in education and success in life.

As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel happiness. Get less mark but with honesty that will make you feel happiness.

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Answered on 24 Feb CBSE/Class 12/Science/Mathematics

Srinivas Rao

Math Guru

Super
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Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics

What Is The Value Of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)? Answer: Method: f(x) = f(x – 2) + x(x – 2) f(1) + f(6) = 0 => f(6)... read more

Question:

A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?

Answer:

Method:

f(x) = f(x – 2) + x(x – 2)

f(1) + f(6) = 0 => f(6) = -f(1)

f(2) = f(0)

f(3) = f(1) + 3

f(4) = f(2) + 8

f(5) = f(1) + 18

f(6) = f(4) + 24 = f(2) + 32

f(2) = -f1 - 32

f(4) = -f1 -24

f(1) + ... f(6) = -f1 -32 -f1 - 24 + f1 + 3 + f1 + 18

= -56 + 21 = -35.

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Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics

I have Rs. x And y Paise. After Spending Rs. 7.20, I Am Left With Rs 3y And 3x Paise. Find The Value Of (x + y)

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Answer: Rs. paise x y- 7 20(x-1-7) (100+y-20) x-8 y+80x-8 = 3y y+80 = 3x x = 29 y = 7 x+y = 36 I have actually made two columns one for Rupees one for Paise! Here messed up that format! Anyway, I divided it into Rupees and Paise separately instead of converting the entire thing into paise! That... read more

Answer:

Rs. paise
x y
-
7 20

(x-1-7) (100+y-20)

x-8 y+80

x-8 = 3y

y+80 = 3x

x = 29

y = 7

x+y = 36

I have actually made two columns one for Rupees one for Paise! Here messed up that format! Anyway, I divided it into Rupees and Paise separately instead of converting the entire thing into paise! That way we are getting two variable but only one eqation.

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Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics

How Many Pairs Of Integers (x, y) Exist Such That x^2+ 4*y^2< = 100

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: How many pairs of integers (x, y) exist such that x^2+ 4*y^2<= 100? Solution: Method 1: Case and Fundamental counting: y = 0, 1, 2 =>19 ways y = 3 =>17ways y = 4 =>13 ways y = 5 => 1 way Total = 19*3+17+13+1=88 Similarly for y negtve total ways =19*2+17+13+1=69 Again... read more

Question:

How many pairs of integers (x, y) exist such that x^2+ 4*y^2<= 100?

Solution:

Method 1:

Case and Fundamental counting:

y = 0, 1, 2 =>19 ways

y = 3 =>17ways

y = 4 =>13 ways

y = 5 => 1 way

Total = 19*3+17+13+1=88

Similarly for y negtve total ways =19*2+17+13+1=69

Again for y=10 x=0 and y=-10 x=0

So total ways = 88+69+2=159

Method 2:

Ellipse: Geometrical approach:

x^2/10^2 + y^2/5^2 = 1 => Ellipse

Case 1: 4 Boundary points cutting x and y axis.

Case 2: x =+/- 9 => y = +/-2 and 0 => 19*5 = 95 points

Case 3: y = +/- 3 => x^2 <= 64 => 2 * 17 = 34 points

Case 4: y = +/- 4 => x^2 <= 36 => 2 * 13 = 26 points

Total: 4 + 95 + 60 = 159 points

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Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics

The Equation 3x+11y = K Where K Is A 2 Digit Number Has Exactly Three Solutions In Which Both X And Y...

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: The equation 3x+11y=k where k is a 2 digit number has exactly three solutions in which both x and y are positive integers. Which of the following is not a value of x? 5,17,19,25 Answer: In equation 3x+11y = k, x increases by 11 and y decreases by 3 as k is constant. Now take the value... read more

Question:

The equation 3x+11y=k where k is a 2 digit number has exactly three solutions in which both x and y are positive integers.
Which of the following is not a value of x? 5,17,19,25

 Answer:

In equation 3x+11y = k, x increases by 11 and y decreases by 3 as k is constant.

Now take the value of x= 5 and put y =1 we get k= 26(true) Next value of x = 5 + 11 = 16, k = 59(true).

Next value of x = 27, k = 92(true)

So solution set = {5, 16, 27}

Similarly check for x = 17.

For x = 19, we have k = 68 (true),

Next x = 30 then k = 108(false)

Even you take x = 19 - 11 = 8, k = 35 (true) but on x = 30 is false which is the third possible value of a solution set. and
 
At last x = 25, k = 86 (true)

Next value of x = 3, 14 having k = 20, 53 respectively.

So solution set {3, 14, 25}.

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Answered on 05/06/2017 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC)

Ca Prashanth Reddy

Tutor

S chand publications
Answers 107 Comments
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Answered on 23/05/2017 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC)

Ca Prashanth Reddy

Tutor

You can start at 12th, but you must understand that this subject will require much more attention than the other subjects. Only understanding concept will not help, you will have to do lot of practice questions. You will need precise help and very good reference books. You also need to be ready to refer... read more
You can start at 12th, but you must understand that this subject will require much more attention than the other subjects. Only understanding concept will not help, you will have to do lot of practice questions. You will need precise help and very good reference books. You also need to be ready to refer back the earlier class maths book, if required. read less
Answers 160 Comments
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Answered on 25 Feb CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC)

Salman Ahmad Siddiqui

Maths Tutor

7th term =40=a + 6d sum of 13 term =(13/2)(2a+12d)=13(a+6d)=13*40=520
Answers 16 Comments
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