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Two years ago, Sam put $1,000 into a savings account.

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Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post Updated on: 28 Jan 2013, 09:14
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Two years ago, Sam put $1,000 into a savings account. At the end of the first year, his account had accrued $100 in interest bringing his total balance to $1,100. The next year, his account balance increased by 10%. At the end of the two years, by what percent has Sam's account balance increased from his initial deposit of $1,000 ?

A. 19%
B. 20%
C. 21%
D. 22%
E. 25%

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Originally posted by Rock750 on 19 Jan 2013, 08:05.
Last edited by Rock750 on 28 Jan 2013, 09:14, edited 1 time in total.
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Re: Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post 19 Jan 2013, 10:01
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Rock750 wrote:
Two years ago, Sam put $1,000 into a savings account. At the end of the first year, his account had accrued $100 in interest bringing his total balance to $1,100. The next year, his account balance increased by 10%. At the end of the two years, by what percent has Sam's account balance increased from his initial deposit of $1,000 ?

A- 19%
B- 20%
C- 21%
D- 22%
E- 25%

Explanations would be appreciated


investment 1000 dollars
1 st year total gained = 100
total amount end of first year = 1100
second year account increased by 10 % = 1100*0.1 = 110
therefore total amount by second year end = 1210

so total percentage increase in money = (1210-1000)*100/1000 = 21 %
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Re: Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post 29 Nov 2013, 04:06
%-change = change in value / original value

original value here: 1000 $

change value: we need to calculate the account after year 2. we take our 1100 $ after 1 year and calculate 10 % which is 110 $. we get 1210 $ after 2 years. hence there was a change in value of 210 $.

plug into formula : 210 $ / 1000 $ = 21 / 100 =21 %

Thus C.

Hope it helps.
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Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post 19 Jul 2017, 14:34
Rock750 wrote:
Two years ago, Sam put $1,000 into a savings account. At the end of the first year, his account had accrued $100 in interest bringing his total balance to $1,100. The next year, his account balance increased by 10%. At the end of the two years, by what percent has Sam's account balance increased from his initial deposit of $1,000 ?

A. 19%
B. 20%
C. 21%
D. 22%
E. 25%

Year 1 interest: $100
Year 2 interest, 10% of 1,100 = $110

Total interest = 100 + 110 = $210 (which equals the change in value)

\(\frac{change}{original}\) x 100 = percent change

\(\frac{210}{1000}\) = .21 x 100 = 21%

Answer C
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Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post 05 Apr 2018, 18:51
generis

Quote:
Two years ago, Sam put $1,000 into a savings account. At the end of the first year, his account had accrued $100 in interest bringing his total balance to $1,100. The next year, his account balance increased by 10%. At the end of the two years, by what percent has Sam's account balance increased from his initial deposit of $1,000 ?

A. 19%
B. 20%
C. 21%
D. 22%
E. 25%



Quote:
Year 1 interest: $100
Year 2 interest, 10% of 1,100 = $110


Total interest = 100 + 110 = $210 (which equals the change in value)

\(\frac{change}{original}\) x 100 = percent change

\(\frac{210}{1000}\) = .21 x 100 = 21%

Answer C


We are not given rate of interest (r) directly. Did you calculate the same by knowing principal amount (Rs.1000),
tenure (1 year), interest (Rs. 100) and using equation: Interest = PrT/100 for the first year and then using the same for the second year?

niks18
Is the highlighted part in question required to be given?
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Two years ago, Sam put $1,000 into a savings account. [#permalink]

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New post 06 Apr 2018, 13:30
adkikani wrote:
generis
Quote:
Two years ago, Sam put $1,000 into a savings account. At the end of the first year, his account had accrued $100 in interest bringing his total balance to $1,100. The next year, his account balance increased by 10%. At the end of the two years, by what percent has Sam's account balance increased from his initial deposit of $1,000 ?

A. 19%
B. 20%
C. 21%
D. 22%
E. 25%

Quote:
Year 1 interest: $100
Year 2 interest, 10% of 1,100 = $110

Total interest = 100 + 110 = $210 (which equals the change in value)

\(\frac{change}{original}\) x 100 = percent change

\(\frac{210}{1000}\) = .21 x 100 = 21%

Answer C

We are not given rate of interest (r) directly. Did you calculate the same by knowing principal amount (Rs.1000),
tenure (1 year), interest (Rs. 100) and using equation: Interest = PrT/100 for the first year and then using the same for the second year?

niks18
Is the highlighted part in question required to be given?

adkikani , I think you ask two questions.

1) did I use simple interest rate equation?
2) why does it look as if I did?

No, I did not use interest rate
I used net change in money amount for each year

Year 1's change amount is given: + $100
Year 2's change rate is given:
10% increase on $1,100 = + $110

Then I calculated percent change from net amount change,
see original post. (Change/Original * 100)

Your question about the highlighted portion
is interesting.

If Sam left the earned interest in the bank;
and if Sam left the account alone;
of course we can calculate the amount in the account.
He put in $1000. He earned $100. His total = $1,100.

But we don't know what Sam did with the account.
The highlighted portion tells us that he left it alone.

• I suspect it appears that I used interest rates
because for any given first year,
if simple interest rate = annual compound interest rate,
amount yielded is identical.

Use interest rate? Yes, but . . .
If I were to calculate percent increase
using interest rates, I would:

1) not use strict SI (it's inaccurate)*
2) use multipliers or
3) use compound annual interest


For #2 and #3, I would omit principal amount. Not needed.

Percent change using
multipliers= compound interest rate


Multipliers - TOTAL factor increase

Multiplier, Year 1? Deduce from base + interest
\(1,000 + 100 = 1,100\)
Multiplier: \(\frac{1,100}{1,000}= 1.1\)

Multiplier for Year 2? Given.
10% increase on extant amount = \(1.1\)

Multipliers: TOTAL increase factor?
(Year 1 multiplier * Year 2 multiplier) = total increase factor
Total increase factor: \((1.1 * 1.1) = 1.21\)
Original base? \(1\)

Compound annual interest: TOTAL factor increase

\(A_{final}=P(1+.10)^{nt}\)
\(A_{final}=P(1.1)^{1*2}\)
\(A_{final}=P(1.1)^2\)
\(A_{final}=1.21P\)

\(A_{original}= P\)


Percent increase:

\(\frac{New-Old}{Old}*100\)

\((\frac{1.21-1}{1}*100)=(\frac{.21}{1}*100)\)
\(= .21*100=21\)
percent

OR \((\frac{1.21P-1P}{1P}*100)=(\frac{.21P}{1P}*100)\)
\(=.21*100=21\)
percent

Hope that answers your question.

*SI amount for both years?
INCORRECT if years are taken together
Run this formula for SI:
\(A_{final} = P(1 + rt)\)
Total after two years is $1,200. Not correct.

If you separate Year 1 and Year 2;
change \(P\) from $1,000 to $1,100;
and change \(t\) from 2 to 1;
SI formula will work.

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that within me there lay an invincible summer.

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Two years ago, Sam put $1,000 into a savings account.   [#permalink] 06 Apr 2018, 13:30
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