Question

Exercise 8.4
Solve the following
2. How much salt must be added to 60 litres of a 20% solution of salt to increase it to a 40% solution of salt?
7. The length of a rectangle exceeds it's width by 3m.If the width is increased by 4m and the length is decreased by 6m, the area is decreased by 22 sq.m.Find the dimensions of the rectangle.
Please give me soon.

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Answer 1

Answer:

xy=2467

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Answer 2

Answer:

2) Ans : 12liters

7) Ans :

Step-by-step explanation:

2)

Given solution = 60 liters

There 20% of salt in 60 liters of solution

=> ( 20/ 100 )( 60 ) = 12 liters of Salt

Let's find out how much Salt we get when 40%of salt exists in 60 liters of Solution.

=> ( 40/100 )( 60 ) = 24 liters of Salt

So we need to add 24 - 12 = 12 liters of Salt to be added

7)

Given that

For a Rectangle ,Area = xy

Length exceeds it's width by 3 m ------(condition)

let's length = x and width/breadth = y say

from ( condition ) we get

=> x = y + 3 ---------(1)

If the width is increased by 4m and the length is decreased by 6m,the area is decreased by 22 sq.m

so

=> ( y + 4 )( x - 6 ) = xy - 22

=> ( y + 3 )( y + 3 - 6 ) = ( y + 3 )y - 22

=> ( y + 3 )( y - 3 ) = ( y + 3 )y - 22

=> (y)^2 - 9 = (y)^2 + 3y - 22

=> 3y = 13

=> y = 13/3

from eqn (1)

=> x = y + 3

=> x = 13/3 + 3

=> x = ( 13 + 9 ) / 3

=> x = 22/3

Therefore, Length = 22/3 m and Breadth = 13/3 m

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