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Two years ago, Sam put $1,000 into a savings account.
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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.
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19 Jan 2013, 10:01
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 = (12101000)*100/1000 = 21 %




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Re: Two years ago, Sam put $1,000 into a savings account.
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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.
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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.
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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.
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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 rateI 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 interestFor #2 and #3, I would omit principal amount. Not needed. Percent change using multipliers= compound interest rate
• Multipliers  TOTAL factor increaseMultiplier, 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: \(\frac{NewOld}{Old}*100\)
\((\frac{1.211}{1}*100)=(\frac{.21}{1}*100)\) \(= .21*100=21\) percent OR \((\frac{1.21P1P}{1P}*100)=(\frac{.21P}{1P}*100)\) \(=.21*100=21\) 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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Re: Two years ago, Sam put $1,000 into a savings account.
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Re: Two years ago, Sam put $1,000 into a savings account.
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