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Bunuel
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A savings account accrues 5% interest at the end of every four months. 29 Jun 2018, 02:23
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55% (hard)

Question Stats:

66% (02:21) correct 34% (02:35) wrong based on 125 sessions

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A savings account accrues 5% interest at the end of every four months. Given an initial savings deposit of s, if an extra 10 dollars was deposited into the account immediately after the first year, which of the following choices represents the ending balance of the savings account after y years, where y > 1?


A. \((s)^{3y} - 3(1.05) + (10)^3y(1.05)\)

B. \((s)^{3y}(1.05) + (10)3^{y-3}(1.05)\)

C. \((s)(1.05)^{3y} + (10)3^{y - 3}\)

D. \((s)(1.05)^{3y - 3} + (10)3^y\)

E. \((s)(1.05)^{3y} + (10)(1.05)^{3y - 3}\)
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A savings account accrues 5% interest at the end of every four months. 29 Jun 2018, 03:10
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Bunuel
A savings account accrues 5% interest at the end of every four months. Given an initial savings deposit of s, if an extra 10 dollars was deposited into the account immediately after the first year, which of the following choices represents the ending balance of the savings account after y years, where y > 1?


A. \((s)^3y - 3(1.05) + (10)^3y(1.05)\)

B. \((s)^3y(1.05) + (10)3y-3(1.05)\)

C. \((s)(1.05)^3y + (10)3y - 3\)

D. \((s)(1.05)^3y - 3 + (10)3y\)

E. \((s)(1.05)^3y + (10)(1.05)^3y - 3\)
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We'll first translate our words into equations so we don't get confused.
This is a Precise approach.

5% interest per 4 months
initial value: s
extra $10 added after one year

so - after four months we had \(s*1.05\), after 8 \(s*1.05^2\) and after one year \(s*1.05^3\)
then we added 10 giving \(s*1.05^3 + 10\)
so after another four months we had \((s*1.05^3 + 10)*1.05\), another 8 \((s*1.05^3 + 10)*1.05^2\) and another year \((s*1.05^3 + 10)*1.05^3\)
so for y = 2 the answer is \((s*1.05^3 + 10)*1.05^3\)
Since we multiply by another \(1.05^3\) for every year,
y = 3 gives \((s*1.05^3 + 10)*1.05^6\)
y = 4 gives \((s*1.05^3 + 10)*1.05^9\)
and in general
\((s*1.05^3 + 10)*1.05^{3(y-1)} = s*1.05^3*1.05^{3(y-1)} + 10*1.05^{3(y-1)} = s*1.05^{3y} + 10*1.05^{3(y-1)}\)

This is answer (E).

Bunuel if I'm not mistaken there is an error in the formatting of the powers in the answers.
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A savings account accrues 5% interest at the end of every four months. 29 Jun 2018, 03:34
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DavidTutorexamPAL
Bunuel
A savings account accrues 5% interest at the end of every four months. Given an initial savings deposit of s, if an extra 10 dollars was deposited into the account immediately after the first year, which of the following choices represents the ending balance of the savings account after y years, where y > 1?


A. \((s)^3y - 3(1.05) + (10)^3y(1.05)\)

B. \((s)^3y(1.05) + (10)3y-3(1.05)\)

C. \((s)(1.05)^3y + (10)3y - 3\)

D. \((s)(1.05)^3y - 3 + (10)3y\)

E. \((s)(1.05)^3y + (10)(1.05)^3y - 3\)

We'll first translate our words into equations so we don't get confused.
This is a Precise approach.

5% interest per 4 months
initial value: s
extra $10 added after one year

so - after four months we had \(s*1.05\), after 8 \(s*1.05^2\) and after one year \(s*1.05^3\)
then we added 10 giving \(s*1.05^3 + 10\)
so after another four months we had \((s*1.05^3 + 10)*1.05\), another 8 \((s*1.05^3 + 10)*1.05^2\) and another year \((s*1.05^3 + 10)*1.05^3\)
so for y = 2 the answer is \((s*1.05^3 + 10)*1.05^3\)
Since we multiply by another \(1.05^3\) for every year,
y = 3 gives \((s*1.05^3 + 10)*1.05^6\)
y = 4 gives \((s*1.05^3 + 10)*1.05^9\)
and in general
\((s*1.05^3 + 10)*1.05^{3(y-1)} = s*1.05^3*1.05^{3(y-1)} + 10*1.05^{3(y-1)} = s*1.05^{3y} + 10*1.05^{3(y-1)}\)

This is answer (E).

Bunuel if I'm not mistaken there is an error in the formatting of the powers in the answers.
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Yes, you are right. Edited. Thank you.
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A savings account accrues 5% interest at the end of every four months. 29 Jun 2018, 04:23
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Bunuel
A savings account accrues 5% interest at the end of every four months. Given an initial savings deposit of s, if an extra 10 dollars was deposited into the account immediately after the first year, which of the following choices represents the ending balance of the savings account after y years, where y > 1?


A. \((s)^{3y} - 3(1.05) + (10)^3y(1.05)\)

B. \((s)^{3y}(1.05) + (10)3^{y-3}(1.05)\)

C. \((s)(1.05)^{3y} + (10)3^{y - 3}\)

D. \((s)(1.05)^{3y - 3} + (10)3^y\)

E. \((s)(1.05)^{3y} + (10)(1.05)^{3y - 3}\)
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Initial investment is "\(s\)" dollars, invested at 5% interest rate for 4 months

After 4 months the total amount becomes = \(s(1+0.05) = 1.05s\)

After 8 months the total amount becomes = \(1.05s(1+0.05) = (1.05)^{2}*s\)
&
After 12 months the total becomes = \((1.05)^{2}*s*(1+0.05) = (1.05)^{3}*s\)

So Total balance after 1 year = \((1.05)^{3}*s\)

To this balance $10 are added, hence new principal amount is \((1.05)^{3}*s + 10\)

Now we are asked to calculate the balance after total \(y\) years, so for year 1 of the \(y\) years we get \((1.05)^{3}*s\) & $10 dollars are added.

This amount = \((1.05)^{3}*s + 10\) is then invested for the remaining \((y-1)\) years at 5% interest accrued every 4 months

We get,

At the end of year 1, the balance amount = \([(1.05)^{3}*s + 10] * (1.05)^{3*1}\)

At the end of year 2, the balance amount = \([(1.05)^{3}*s + 10] * (1.05)^3 * (1.05)^3\) = \([(1.05)^{3}*s + 10] * (1.05)^{3*2}\)

Hence at the end of \((y-1)\) years, the balance amount = \([(1.05)^{3}*s + 10] * (1.05)^{3(y-1)}\)

Simplified as,

\([(1.05)^{3}*s + 10] * (1.05)^{3(y-1)}\) = \((s)(1.05)^{3y} + (10)(1.05)^{3y - 3}\)


Answer E.


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GyM
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A savings account accrues 5% interest at the end of every four months. 01 Jul 2023, 11:29
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