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\(p\): \(\frac{2000x}{x-y}\)
\(x\): \(\frac{py}{p-2000}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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At the beginning of a given year, Sarah and Olivia each invested in two different mutual funds at simple annual interest rates of \(x\%\) and \(y\%\), respectively. Olivia invested \(p\) pounds, while Sarah invested 2,000 pounds less. By the end of the year, both earned the same total interest from their respective investments.
Select for p an expression for Olivia’s investment \(p\) in terms of x and y, and select for x an expression for Sarah’s interest rate \(x\) in terms of \(p\) and \(y\). Make exactly two selections, one from each column.
Select for p an expression for Olivia’s investment \(p\) in terms of x and y, and select for x an expression for Sarah’s interest rate \(x\) in terms of \(p\) and \(y\). Make exactly two selections, one from each column.
| \(p\) | \(x\) | |
| \(\frac{2000x}{x-y}\) | ||
| \(\frac{2000y}{x-y}\) | ||
| \(\frac{200x}{x-y}\) | ||
| \(\frac{py}{p-2000}\) | ||
| \(\frac{py}{p+2000}\) | ||
| \(\frac{p+2000}{py}\) |
ShowHide Answer
Official Answer
\(p\): \(\frac{2000x}{x-y}\)
\(x\): \(\frac{py}{p-2000}\)
09 Dec 2024, 02:40
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Bunuel
Official Solution:
Sarah's investment of \(p-2,000\) pounds in a year at at simple annual interest rates of \(x\%\) would earn \((p-2,000)*\frac{x}{100}\) pounds
Olivia's investment of \(p\) pounds in a year at at simple annual interest rates of \(y\%\) would earn \(p*\frac{y}{100}\) pounds
Given that those earnings are equal, we'd have
\((p-2,000)*\frac{x}{100} =p*\frac{y}{100}\)
\((p-2,000)x = py\)
\((p-2,000)x = py\)
From this, for the second column, we get \(x = \frac{py}{p-2,000}\).
For the first column we get
\((p-2,000)x = py\)
\(px-2,000x = py\)
\(px-py=2,000x\)
\(p(x-y)=2,000x\)
\(p=\frac{2,000x}{x-y}\)
\(px-2,000x = py\)
\(px-py=2,000x\)
\(p(x-y)=2,000x\)
\(p=\frac{2,000x}{x-y}\)
Correct answer:
\(p\) "\(\frac{2,000x}{x-y}\)"
\(x\) "\(\frac{py}{p-2,000}\)"
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09 Dec 2024, 11:30
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1. Olivia invested p pounds at an interest rate of y%, while Sarah invested p - 2000 pounds at an interest rate of x%.
2. Since both of them earned the same total interest, we can write this as: \(p * y\% = (p - 2000) * x\%\).
3. Let's derive p from this equation. \(p * y\% = (p - 2000) * x\% \rightarrow \frac{y\%}{x\%} = \frac{p - 2000}{p} \rightarrow \frac{y}{x} = 1 - \frac{2000}{p} \rightarrow \frac{2000}{p} = \frac{x - y}{x} \rightarrow p = \frac{2000x}{x - y}\).
4. Now let's also derive x too. \(p * y\% = (p - 2000) * x\% \rightarrow \frac{p * y\%}{p - 2000} = x\% \rightarrow x = \frac{py}{p - 2000}\).
5. Our answer will be: p - \(\frac{2000x}{x - y}\) and x - \(\frac{py}{p - 2000}\).
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There is another solution by analyzing the answer choices. The first three options are equations that don't include p but have x - that means they can only be equal to p. The other three options are equations that don't include x but have p - that means they can only be equal to x.
We can assume for each of the first three options that they equal p and try to derive x and see if the new equation is an answer choice:
- \(\frac{2000x}{x - y} = p \rightarrow px - py = 2000x \rightarrow x = \frac{py}{p - 2000}\), which is in the options.
- \(\frac{2000y}{x - y} = p \rightarrow px - py = 2000y \rightarrow x = \frac{(p + 2000)y}{p}\), which isn't in the options.
- \(\frac{200x}{x - y} = p \rightarrow px - py = 200x \rightarrow x = \frac{py}{p - 200}\), which isn't in the options.
So, our answer would be: p - \(\frac{2000x}{x - y}\) and x - \(\frac{py}{p - 2000}\).
2. Since both of them earned the same total interest, we can write this as: \(p * y\% = (p - 2000) * x\%\).
3. Let's derive p from this equation. \(p * y\% = (p - 2000) * x\% \rightarrow \frac{y\%}{x\%} = \frac{p - 2000}{p} \rightarrow \frac{y}{x} = 1 - \frac{2000}{p} \rightarrow \frac{2000}{p} = \frac{x - y}{x} \rightarrow p = \frac{2000x}{x - y}\).
4. Now let's also derive x too. \(p * y\% = (p - 2000) * x\% \rightarrow \frac{p * y\%}{p - 2000} = x\% \rightarrow x = \frac{py}{p - 2000}\).
5. Our answer will be: p - \(\frac{2000x}{x - y}\) and x - \(\frac{py}{p - 2000}\).
-------------------------------------------------
There is another solution by analyzing the answer choices. The first three options are equations that don't include p but have x - that means they can only be equal to p. The other three options are equations that don't include x but have p - that means they can only be equal to x.
We can assume for each of the first three options that they equal p and try to derive x and see if the new equation is an answer choice:
- \(\frac{2000x}{x - y} = p \rightarrow px - py = 2000x \rightarrow x = \frac{py}{p - 2000}\), which is in the options.
- \(\frac{2000y}{x - y} = p \rightarrow px - py = 2000y \rightarrow x = \frac{(p + 2000)y}{p}\), which isn't in the options.
- \(\frac{200x}{x - y} = p \rightarrow px - py = 200x \rightarrow x = \frac{py}{p - 200}\), which isn't in the options.
So, our answer would be: p - \(\frac{2000x}{x - y}\) and x - \(\frac{py}{p - 2000}\).
