OA is C
This is the OE:
This question asks whether the amount of compound interest earned in a special account is greater than the amount earned by the same amount of money in an account earning simple interest.
The basic account has an interest rate of b%, so for ease of notation, let the rate of interest earned be named B (where B = b%). Then the amount of interest the basic account earns in two years will be:
Tbasic = 2 × B × $100,000 = 2B(100,000).
Similarly, let the rate of interest earned by the special account be called S (where S = s%). Then the amount of interest earned in the first year is S(100,000). During the second year, the principal has increased to the original principal plus the interest earned in the first year, or (100,000 + S(100,000)) = (1 + S)(100,000). The amount of interest earned on this principal during the second year is S(1 + S)(100,000) = (S2 + S)(100,000), so the total interest for the two years is:
Tspecial = (S2 + S)(100,000) + S(100,000) = (S2 + 2S)(100,000).
We could also calculate the total interest earned by using the formula for compound interest, Pt = (1 + r)t(Pi), where t = 2 and r = S. For the new principal after 2 years, the formula gives P2 = (1 + S)2 (100,000). The difference is: total interest = P2 - P i= (1 + S)2(100,000) - 100,000 = (1 + 2S + S2 - 1)(100,000) = (2S + S2)(100,000). Note that since this expression is the same as the one we found, it can be derived as shown above without memorizing the actual formula.
We wish to know whether Tspecial > Tbasic, or:
Is (S2 + 2S)(100,000) > 2B(100,000)?
Since 100,000 is positive, we can divide both sides of the inequality by 100,000, leaving us with the question: Is S2 + 2S > 2B?
Statement 1 alone tells us that b < 1.05s. Since B = b × 100% and S = s × 100%, we also have B < 1.05S.
Thus, if S2 + 2S > 2(1.05S), then since 2(1.05S) must be greater than 2B by statement 1, S2 + 2S > 2B, making the answer "Yes." We can rearrange the first inequality to get:
S2 + 2S - 2(1.05)S > 0 → S2 + 2S - 2.1S > 0 →
S2 - 0.1S > 0 → S(S - 0.1) > 0
Since S is positive, we can divide both sides by S to get:
S - 0.1 > 0 → S > 0.1
Converting 0.1 into a percentage gives:
S > 10%
So if S > 10%, then the answer to the question is "Yes."
However, we do not know what S is. For high values of S, such as S = 40% and B = 41% (which is less than 1.05 × 40% = 42%), then the interest earned by the basic account is Tbasic = 2 × 0.41 × 100,000 = 82,000. The interest earned by the special account is Tspecial = (0.402 + 2(0.40))(100,000) = (0.16 + 0.8)(100,000) = 96,000, which is greater than the interest earned by the basic account.
But for a low value of S, the interest earned by the basic account may be greater. For instance, if S = 1% and B = 1.04% (which is less than 1.05 × 1% = 1.05%), the interest earned by the basic account is Tbasic = 2 × 0.0104 × 100,000 = $2,080. The interest earned by the special account is Tspecial = (0.012 + 2(0.01))(100,000) = $2,010, which is less than the interest earned by the basic account.
Statement 1 is not sufficient to answer the question definitively. Eliminate answer choices A and D.
Statement 2 alone tells us that the amount of interest earned from the basic account in the first year is greater than $11,000. In other words, b > 11, or B > 11%.
But this does not tell us anything about the special account. For example, if B = 12% and S = 1%, then the amount earned by the basic account is Tbasic = 2 × 0.12 × 100,000 = 24,000 dollars, while the amount earned by the special account would be 2,010 dollars, as we found above. But if B = 12% and S = 40%, then the special account earns 96,000 dollars, as we found above, which is more than the basic account.
Statement 2 alone is insufficient. Eliminate choice B.
Now let's take the two statements together. We know from Statement 1 that if S > 10%, then the special account earned more interest. We know from Statement 2 that B > 11%. We also know that b < 1.05s, or, B < 1.05S. Since B > 11%, if we plug B = 11 into the inequality, we can determine the smallest possible value of S:
B < 1.05S → 11 < 1.05S
We could divide both sides by 1.05, but that would be cumbersome arithmetic. It is enough to notice that if S = 10, the inequality would read 11 < 10.5, which is false. Thus, S must be greater than 10 to make the inequality true. Since S > 10%, the special account must have earned more interest than the basic account. The two statements together are sufficient to answer the question.
Answer choice C is correct.
Regards,
Karthik
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