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27 Jan 2018, 12:48
1
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Hi All,

This question requires knowledge of the formulas for Simple Interest and Compound Interest:

Simple Interest = Principal x (1 + RT)
Compound Interest = Principal x (1 + R)^T

Where R is the yearly interest rate and T is the number of years. When compounding MORE than once a year, you must divide R by the number of periods and multiply T by the same number of periods.

We're asked for the DIFFERENCE in the amount of interest that is generated under two different situations. We can TEST VALUES to answer this question. IF... X = 20...

$1,000 at 20% compounded semi-annually (meaning twice a year) =$1,000 x (1.1)^2 =
$1,000 x (1.21) =$1210

$1,000 at 10% compounded annually =$1,000 x (1.2)^1 =
$1200 The difference is$10, so we're looking for an answer that equals 10 when we plug X=20 into it. There's only one answer that matches....

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Re: How much more interest will maria receive if she invests 1000$for one [#permalink] ### Show Tags 28 Jan 2018, 19:50 [quote="FightToSurvive"]How much more interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually, than if she invest 1000$for one year at x percent annual interest, compounded annually? a> 5x b> 10x c> x^2/20 d> x^2/40 e>10x+ (x^2/40) I dont from what source these questions are from. See from simplification sake, take x =10% (i) investing 1000$ for one year at x % annual interest, compounded semiannually
1000*1.05*1.05=1102.5
Interest amount = 1102.5-1000=102.5
(ii)investing 1000$for one year at x percent annual interest, compounded annually 1000*1.1=1100 Interest amount = 1100-1000=100 Difference in two cases = 102.5-100 = 2.5 Putting x=10 in the above option , D gives you the answer Intern Joined: 01 Jun 2017 Posts: 19 How much more interest will maria receive if she invests 1000$ for one  [#permalink]

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20 Aug 2018, 06:09
can we use the concept that CI will be slightly larger than SI. hence the difference should also be slightly larger.

in this case, the difference in SI is 0 hence the difference in CI should be slightly larger than 0. Hence x^2/40 is the smallest number out of all and should be the answer.

Bunuel chetan2u
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03 Jan 2019, 08:28
email2vm wrote:
How much more interest will maria receive if she invests 1000$for one year at x % annual interest, compounded semianually, than if she invest 1000$ for one year at x percent annual interest, compounded annually?

A. 5x

B. 10x

C. $$\frac{x^2}{20}$$

D. $$\frac{x^2}{40}$$

E. $$(10x+\frac{x^2}{40})$$

Here is how i solved only to get it wrong:

let x=100%

annually :

SI = 1000*100/100*1 = 1000
total =1000+1000=2000

semiannually:

Si (for 1st 6 months) = 1000
total after 6 months =2000

Si(for last 6 months) = 2000
total after 6 months= 4000

more interest maria will receive = 2000

100^2/40 != 2000!!!

where am I wrong? Today it seems I am back to square 1

It is easier to test numbers in this question. Try with x = 20%. One can solve this question in less than 2 minutes.
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21 Jun 2020, 00:55
chetan2u wrote:
email2vm wrote:
How much more interest will maria receive if she invests 1000$for one year at x % annual interest, compounded semianually, than if she invest 1000$ for one year at x percent annual interest, compounded annually?

a> 5x
b> 10x
c> x^2/20
d> x^2/40
e>10x+ (x^2/40)

Here is how i solved only to get it wrong:

let x=100%

annually :

SI = 1000*100/100*1 = 1000
total =1000+1000=2000

semiannually:

Si (for 1st 6 months) = 1000
total after 6 months =2000

Si(for last 6 months) = 2000
total after 6 months= 4000

more interest maria will receive = 2000

100^2/40 != 2000!!!

where am I wrong? Today it seems I am back to square 1

Ravi

Hi,

the correct way is--

whenever amount is compounded other than annually, reduce the interest that many time and increase the time period that many times..

lets see in this Q..

it is semi annually that is twice in a year...
so our time becomes 2n and rate of interest = r/2..

1) when annually
Amount =$$1000( 1+\frac{r}{100})^1 = 1000(1+\frac{r}{100})$$

2) when semi annually
$$1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)+r/100$$

subtract 1 from 2..
$$1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2+\frac{r}{100} - 1000( 1+\frac{r}{100})^1$$..
$$1000(1+\frac{r}{200}^2+\frac{r}{100}-(1+\frac{r}{100})$$..
$$1000(\frac{r}{200}^2)$$ = $$r^2/40$$
D

Hello Team,
I am stuck at the calculation part.

For Compounded Annually I got,
1000(1+(x/100))^1

For compounded Semi Annually, I got,
1000(1+(x/200))^2.

I have to subtract 1 from 2 but I am not able to get the correct answer.