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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

Math Assignment Class X Ch-4 | Quadratic Equations

 Math Assignment / Class X / Chapter 4 /  Quadratic Equations

Extra questions of chapter 4 class 10 Quadratic Equations with answer and  hints to the difficult questions. Important and useful math assignment for the students of class 10

For better results

  • Students should learn all the basic points of  Quadratic Equations 
  • Student should revise N C E R T book thoroughly with examples.
  • Now revise this assignment. This assignment integrate the knowledge of the students.

ASSIGNMENT FOR 10 STANDARD QUADRATIC EQUATIONS

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Question 28. 

Out of a number of Saras birds, (1/4)th of the number are moving about in Lotus plants, (1/9)th coupled with (1/4)th,  as well as 7 times the square root of the number moving on a hill, 56 birds remain on trees.   What is the total number of words.

Solution

Let total birds = x

No. of birds moving in lotus plant = (1/4)x

No. of birds coupled with = (1/9 +1/4)x = (13/36)x

No of birds moving on the hill = 7x

No. of birds on the tree = 56

Acoording to the question

\[ \frac{1}{4}x+\frac{13}{36}x+7\sqrt{x}+56=x\]

\[ \frac{22}{36}x+7\sqrt{x}+56=x\]

\[ x-\frac{22}{36}x-7\sqrt{x}-56=0\]

\[ \frac{14}{36}x-7\sqrt{x}-56=0\]

\[\frac{2}{36}x-\sqrt{x}-8=0\]

\[ \frac{1}{18}x-\sqrt{x}-8=0\]

\[ x-18\sqrt{x}-144=0\]

\[ Putting \: \: \: \: \sqrt{x}=y \: \: \: \Rightarrow\: \: x=y^{2}\]

\[ y^{2}-18y-144=0\]

\[ y^{2}-24y+6y-144=0\]

\[ y(y-24)+6(y-24)=0\]

\[ (y-24)(y+6)=0\]

\[ \Rightarrow y=24, -6\]

\[ y=-6 \: \: is \: \: rejected\]

\[ y=\sqrt{x}=24\Rightarrow x=(24)^{2}=576\]

Hence Total Number of birds = 576

Question : 38

7 years ago, age of Varun was five times the square of the age of Swati. After 3 years, age of Swati will be 2/5 of the age of Varun. Find their present ages.

Solution :

Let 7 years ago the age of Swati = x years

Let 7 years ago the age of Varun = y years

ATQ       y = 5x2 …………………(1)

Present age of Swati = x + 7

Present age of Varun = y + 5

After 3 years age of Swati = x + 10

After 3 years age of Varun= y + 10

Again ATQ :  Age of Swati = (2/5) age of Varun

x + 10 = (2/5)[y + 10]

5(x + 10) = 2y + 20

5x + 50 = 2y + 20

Putting  y = 5x2  we get

5x + 50 = 2(5x2 ) + 20

10x2  - 5x + 20 - 50 = 0

10x2 - 5x - 30 = 0

2x2 - x - 6 = 0    (Dividing by 5)

(2x + 3)(x - 2) = 0

x = - 3/2 (Rejected)     and   x = 2

Putting x = 2 in equation (1) we get

y = 5 (2)2    ⇒     y = 20

Present age of Swati = x + 7  = 2 + 7 = 9

Present age of Varun = y + 7 =  20 + 7 = 27


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