THE TOTAL COST OF A CERTAIN LENGTH OF A PIECE OF CLOTH IS RS 200.IF THE PIECE WAS 5 METRE LONGER AND EACH METRE OF CLOTH COSTS RS 2 LESS .THE COST OF THE PIECE WOULD HAVE REMAINED UNCHANGED .HOW LONG IS THE PIECE AND WHAT IS ITS ORIGINAL RATE PER METRE.
Answers
Now, its price = Rs.200
Price per m = 200/x --Eq.1
Now, if the cloth was 5 m longer, its length = x+5
and
Price per m would be 200/x+5 --Eq.2
(As total price is unchanged)
Or
Price per m would be Rs.2 less than the original one i.e.
200/x -2 -- Eq.3
Equating Eq.2 and Eq.3,
200/x+5 = 200/x - 2
200/x+5 = 200-2x/x
200x/x+5 = 200 -2x
200x = (x+5)(200-2x)
200x = 200x -2x² +1000 -10x
Cut 200x from both sides
-2x² +190x +1000 = 0
Solve the quadratic equation, you'll get the answer .
Substituting values in the formula.
Root 1 = 10 +√[10² - 4(-2)(1000)]/ -4
= 10+90/-4
= 100/-4
= -25
Root 2 = 10 -√[10² - 4(-2)(1000)]/ -4
= 10- 90/-4
= -80/-4
= 20
Now, the length of the original piece cannot be negative.
So, the length of original piece is 20 m.(Root 2)
And original rate = 200/x = 200/20 = Rs.10
Solutions :-
Let the length of a piece of cloth be x
And the original rate per metre be y
According to the question,
x × y = 200
=> xy = 200 _______(i)
(x + 5) × (y - 2) = 200
=> xy - 2x + 5y - 10 = 200 ________(ii)
Putting the value of xy in equation (ii) we get,
=> 200 - 2x + 5y - 10 = 200
=> - 2x + 5y = 10
=> y = (2x + 10)/5 ______(iii)
Putting the value of y in equation (i) we get,
=> x (2x + 10)/5 = 200
=> 2x² + 10x = 200 × 5
=> 2x² + 10x - 100 = 0
=> 2(x² + 5x - 500) = 0
=> x² + 5x - 500 = 0/2
=> x² + 25x - 20x - 500 = 0
=> x(x + 25) - 20(x + 25) = 0
=> (x + 25) (x - 20) = 0
=> x = - 25 or x = 20
Length of cloth be taken positively.
Putting the value of x in equation (iii) we get,
=> y = (2 × 20 + 10)/5
=> y = 50/5 = 10
Hence,
The length of a piece of cloth = 20 m
And The original rate per metre = Rs 10 per metre