Question 1006861
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.