SOLUTION: Hi. I am facing problem in this word problem. It is quadratic equation. In may 1994 Mr Chauhan changed 1140 Indian Rupees into German Marks when the rate of exchange was x rupee

Question 1006861: Hi. I am facing problem in this word problem. It is quadratic equation.
In may 1994 Mr Chauhan changed 1140 Indian Rupees into German Marks when the rate of exchange was x rupees =1 Mark.
(a) write down an expression in terms of x , for the number of Marks he received
In july, Mr Chuhan again changed 1140 rupees into Marks. The rate of exchange was then (x+1) rupees=1 Mark
(b) write down an expression in term of x, for the number of Marks he received this time,
(c)Given that he received 3 Marks less in July than he did in May, form an equasion in x and show that it reduces to x^2 +x-380=0
(d) Solve this equation to find the rate of exchange in May 1994.
Thank you for reading my question. Please reply.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
here goes.

in may the ratio of rupees to marks is x to 1 = x/1.

in july the ratio of rupees to marks is (x+1) to 1 = (x+1)/1.

he exchanged 1140 rupees to marks in both months.

in may, the ratio becomes 1140/m = x/1.

in july, the ratio becomes 1140/m = (x+1)/1.

solve each of these equations for m to get:

in may, m = 1140/x.

in july, m = 1140/(x+1).

to distinguish the marks between may and july, we'll call m in may m1 and we'll call m in july m2.

you get:

m1 = 1140/x.

m2 = 1140/(x+1)

you are given that the number of marks in july is equal to the number of marks in may less 3.

that means that m2 = m1 - 3.

since m2 = 1140/(x+1) and m1 = 1140/x, you get:

m2 = m1 - 3 becomes:

1140/(x+1) = 1140/x - 3.

multiply both sides of this equation by (x+1)*(x) and you get:

1140/(x+1)*(x)*(x+1) = 1140/x*(x)*(x+1) - 3*(x)*(x+1) which results in:

1140*x = 1140*(x+1) - 3*(x)*(x+1) which can be further simplified to:

1140x = 1140x + 1140 - 3x^2 - 3x.

if you subtract 1140x from both sides of this equation, you get:

0 = 1140 - 3x^2 - 3x

multiply both sides of this equation by -1 and you get:

0 = -1140 + 3x^2 + 3x

divide both sides of this equation by 3 and you get:

0 = -380 + x^2 + x

reorder the terms of the expression on the right in descending order of degree to get:

0 = x^2 + x - 380

by the law of commutivity of equations (if a = b, then b = a), this is the same as:

x^2 + x - 380 = 0.

that matches the equation that you showed in statement c.

factor this equation to get:

(x+20)*(x-19) = 0

solve for x to get:

x = 20 or x = 19.

x can't be negative, so x has to be 19.

replace x in the original equation with 19 to get:

m1 = 1140/19 = 60

m2 = 1140/20 = 57

m2 - m1 = -3 which is correct since there are 3 less marks in july than in may.

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