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Скачать или смотреть class 10 maths , Chapter 3 Exercise 3.1 Questions 1 - 3

  • Mathsdiscussion
  • 2019-10-30
  • 116
class 10 maths , Chapter 3 Exercise 3.1 Questions 1 - 3
class 10 mathsLinear EquationFormation of equationformation of linear equationClass 10 NCERT exercise 3.1solutions of linear equationgraphs of linear equationmathsdiscussionmathsdiscussion.comNTSE
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Описание к видео class 10 maths , Chapter 3 Exercise 3.1 Questions 1 - 3

How to form the linear equation in different types of word problem is made easy in this video.
Video has well explained formation of linear equation in two variables from
EXERCISE 3.1
QUESTION 1.
Aftab tells his daughter , "Seven years ago , I was seven times as old as you were then. Also , three years from now , I shall be three times as old as you will be.

In this question we can see we are talking about ages in three different years . In the video it is explained nicely that if we formulate table for different years and then wright the relations between ages for that particular year the formation of equation becomes very easy.
like Years 7 yrs ago present age age 3 yrs hence.
(x-7) Aftab x (x+3)
(y-7) Daughter y (y+3)
----------------------------------------------- ---------------------------------------------
(x-7)=7(y-7) (x+3)=3(y+3)
Hence x+42=7y ---------------- (1) x = 3y + 6 ----------------------------(2)
These are two linear equation which can be formed easily using table with thee columns. QUESTION 2
The coach of a cricket team buys 3 bats and 6 balls for Rs 3600. Later , she buys another bat and 3 more balls of same kind for Rs 1300 . Represent this situation algebrically and geometrically.
In these types of questions where we talk about purchase of certain articles or sell articles one of the easiest way to form linear equation is develop simple bill for purchase/sell.
Bill 01.
Item Qty Rate Amount
Bat 3 Rs x /- 3x
Balls 6 Rs y /- 6y
----------------------------------------------------------------------
Total Cost = 3x + 6y = 3900 (given)
Bill 02.
Bat 1 Rs x/- x
Balls 3 Rs y/- 3y
------------------------------------------------------------------
Total cost x + 3y = 1300 (given )
Hence two linear equations are 3x + 6y = 3900 and x + 3y = 1300
QUESTION 3
The cost of 2 Kg apples and 1 Kg grappes on a day was found to be Rs 160 /- . After a month, the cost of 4 Kg of apples and 2 Kg of grappes is Rs 300/- . Represent the situation algebrically and geometrically.
Once again it is a case of sell we go by formation of bill.
Bill 01 Date 26/09/2019
Item QTY Rate Amount
Apples 2 Rs x/- 2x
Grappes 1 Rs y /- y
------------------------------------------------------------
Total Cost 2x + y = 160 (Given)
Bill 02 Date 26/10/2019
Apples 4 Rs x 4x
Grappes 2 Rs y 2y

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