The exchange rate for the Australian dollar in January 2000 was AU$100 = S$x. In June 2000, the exchange rate had become AU$100 = S$(x-5). Karen found out that she could get an extra AU$32 for every S$672 that she exchanged in June compares to January. Form an equation in x and solve it.
January:
100 AU = x S
1 S = (100/x) AU ---> 672 S = (67200/x) AU
June:
100 AU = (x-5) S
1 S = (100/(x-5)) AU --- > 672 S = (67200/(x-5)) AU
67200/(x-5) AU - 67200/x AU = 32 AU
divide by AU, multiply by x(x-5)
67200x - 67200x + 336000 = 32x^2 - 160x
32x^2 - 160x - 336000 = 0
x^2 - 5x - 10500 = 0
(x-105)(x+100) = 0
x = 105 or x = -100 , but x > 0
so x = 105
so in Jan: AU$ 100 = S$ 105
in June : AU$ 100 = S$ 100
67200/(x-5) -
67200/ x AU=32AU
To solve this problem, let's break it down into steps:
Step 1: Determine the exchange rate in January and June.
In January 2000, the exchange rate was AU$100 = S$x.
In June 2000, the exchange rate became AU$100 = S$(x-5).
Step 2: Determine the difference in exchange rates.
The difference in exchange rates between January and June is 5 Singapore dollars.
Step 3: Set up an equation using the given information.
Karen found out that she could get an extra AU$32 for every S$672 exchanged in June compared to January.
We can set up the equation: (AU$100/ S$x) * (S$(x-5))- (AU$100/ S$x) * S$x = AU$32.
Step 4: Solve the equation.
Simplifying the equation: (100/x)(x-5) - 100 = 32.
To solve this equation, we'll first multiply out the brackets: 100 - (500/x) - 100 = 32.
Simplifying further, we get: 500/x = 32.
To isolate x, we can cross-multiply: 500 = 32x.
Dividing both sides by 32, we find: x = 15.625.
Therefore, the value of x is approximately 15.625.
Note: This solution assumed that the exchange rate remains constant throughout the entire year.