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Just take x = 200%
You can solve it in a minute!
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email2vm
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

====looking for the answer choices

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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Got this wrong on an official practice test. Resolving here for retention:

Method 1 - Plug-in values for x (quickest method)

x= 20

Compounded
New balance after compounding is equal to 1000(1+20/200)^2
= 1000(11/10)^2
= 1000(121/100)
= 1210

Simple interest
New balance with interest 1000(1+20/100)^1
= 1000(6/5)
= 1200

'How much more interest' = 1210 - 1200 = 10 more

Plug-in 20 to each answer choice to get 10
A - 5(20) = 200 --> Wrong
B - 10(20) = 200 --> wrong
C - (20)^2/ 20 = 20 --> wrong
D - (20)^2/40 = 10 --> correct
E - (10(2) + (20)^2/40) = 30 --> wrong


Method 2 - Algebraic, but since we have variables, breakdown the compounding periods instead of calculating it all in one go)


Compounding:
1000(1+x/100/2)^1
= 1000 (1 + x/200)^1
= 1000 + 5x
New balance after first compounding period = 1000 + 5x

Second compounding period
Principle *(Interest rate/#compound periods + 1)^compounding period
(1000 + 5x)(1 + x/200)^1
= 1000 + 5x + 5x + (5x^2)/200
= 1000 + 10x + x^2/40
Now we need to eliminate the principle to get the interest portion
= 1000 + 10x + x^2/40 - 1000
= 10x + x^2/40

Simple interest amount
1000(1+ x/100)^1
= 1000 + 10x
Interest portion = 1000+10x - 1000
= 10x

How much more interest?
= 10x + x^2/40 - 10x
= x^2/40
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chetan2u
email2vm
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

====looking for the answer choices

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.
Can someone please help me here.
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chetan2u

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


Hi chetan2u,

Thanks for the solution!

It might be a silly question, but here goes - in your breakdown of the second solution ( 2) when semi annually
\(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)+r/100\) ), where does the additional r/100 in the end come from?

It seems like \(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)\) is the full result, as \(P*(1+\frac{r}{c*100})^(c*y)\), where C is the compounding period and Y is the years compounding.

What am I missing?
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BeepBoop
chetan2u

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


Hi chetan2u,

Thanks for the solution!

It might be a silly question, but here goes - in your breakdown of the second solution ( 2) when semi annually
\(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)+r/100\) ), where does the additional r/100 in the end come from?

It seems like \(1000( 1+\frac{r}{2*100})^2 = 1000(1+\frac{r}{200}^2)\) is the full result, as \(P*(1+\frac{r}{c*100})^(c*y)\), where C is the compounding period and Y is the years compounding.

What am I missing?

Hi
The bracket is put wrongly.
Otherwise it is correct as it is.
Just expansion of (1+r/200)^2=1+(r^2/100^2)+2*1*r/200)
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Daaaaaamn that makes perfect sense. Thanks a lot!
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Asked: 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?

The interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually = 1000(1+x%/2)^2 - 1000
The interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually = 1000(1+x%) - 1000

The difference = (1000(1+x%/2)^2 - 1000) - (1000(1+x%) - 1000) = 1000 {(1+x%/2)^2 - (1+x%)} = 1000 (1 + x^2/40000 + x/100 - 1 - x/100) = 1000*x^2/40000 = x^2/40

IMO D
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How much more interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semi-annually, than if she invests 1000$ for one year at x percent annual interest, compounded annually?

This question tests your understanding of interest as an application of percentage increase.

Case 1: she invests 1000$ for one year at x percent annual interest, compounded annually ==>The interest is calculated annually and at the end of the year, 1000 $ will be increased by x% . She will get an interest of x% of 1000 $.

Case 2: she invests 1000$ for one year at x % annual interest, compounded semiannually===>Here the interest is calculated semi-annually ( every half year ). Since x% is the annual interest rate, for the half-year, the rate of interest would be x/2%. So we can say that 1000$ is increased by x/2 % at the end of every half-year ( 6 months ) for 1 year. i.e 1000 $ is successively increased by x/2 % for 2 times.

If you are the clear with above statements, then this question would be easy for you. We are asked to find how much more interest she got in case 2 compared with case 1.
There are multiple ways to approach this question

#1: Using the successive increase formula

When a value is successively increased by a % then by b%, the overall effective % increase would be a + b + ab/100.

Applying above formula in Case 2: When a value is successively increased by x/2 % then again by x/2% , the overall effective % increase would be x/2 + x/2 + x/2*x/2/100 = x + \(x^2\)/400 %

How much more interest she got in case 2 compared with case 1. = (x + \(x^2\)/400 )% of 1000 - x % of 1000 = (\(x^2\)/400 ) % of 1000 = (\(x^2\)/400)/100 *1000 = \(x^2\)/40

Option D is the answer.

#2: Plugin values for x

Let's assume that x =20%

Why did I pick 20 %? Because it's easy to calculate 20 % as well as x/2 i.e 10 %. Whenever you plugin values, always choose the number wisely.

Case 1: Interest = 20 % of 1000 = 200$

Case 2: 1000 is increased by 10 % ,2 times. So the final amount = 1210 $
Interest = 1210 - 1000 = 210$

Difference in the interest = 210 -200 = 10$

Substituting x as 20 in the answer choices, only Option D will satisfy.

Option D :\( x^2 \)/40= \(20^2\)/40 = 400/40 =10

Thanks,
Clifin J Francis,
GMAT Mentor.
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Hi all,
Shouldn't the right answer be \(\frac{x^2 }{ 40,000}\) since we need to write x down as number (instead of %)? Thanks.
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1000 invested at x% compounded annually \(=1000(1+\frac{x}{100})^1\)
\(=1000+10x\)

1000 invested at x% compounded semiannually \(=Principal(1+\frac{x%}{Periods.In.a.Year})\)^(periods in a year*years)
\(=1000(1+(\frac{x}{2}))\)^2*1
\(=1000*(1+2*\frac{x}{200}+(\frac{x}{200})^2)\)
\(1000+\frac{1000x}{100}+\frac{1000x^2}{40000}\)
\(=1000+10x+\frac{x^2}{40}\)

now substract both values

\(1000+10x+\frac{x^2}{40}-(1000+10x)\)
\(=\frac{x^2}{40}\)

IMO D
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I solved it using algebra, a bit long but I did not find any other method when solving.

Translate it to maths: first interest investment-second interest investment

\(First. Investment .Interest=1000(1+\frac{r}{{100*2}})^{2*1}-1000\)
\(1000(\frac{200+x}{200})^2-1000\)
\(1000(\frac{200-^2+400x+x^2}{200^2})-1000\)
\(\frac{1000(200^2+400x+x^2)-200^2*1000}{200^2}\)
\(\frac{1000((200^2+400x+x^2)-200^2)}{4000}\)
\(\frac{400x+x^2}{40}\)

second investment interest\(1000(1+\frac{x}{100})\)
\(1000(\frac{x}{100})=10x\)

now substract 1 from 2
\(10x-\frac{400x+x^2}{40}\)
\(\frac{400x-400x+x^2}{40}\)
\(\frac{x^2}{40}\)

IMO D
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20 Second solution.

Make-up a value for X% (You want to make up a number that easily cancels with the denominator, hence X=20 works best here)

From the compound interest formula = p(1+r%/n)^rt, if we just use this part - p(r%/n)^rt, we get the "difference" between the simple and compound interest, i.e. $40 in this case. When you plug back the value of into ans. choices, you will get D.
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A very simple and effective way of solving this. Don't do too many expansions and formulas

Finding interest earned when compounded semi annually:

First let me calculate what will be the interest if I put annually: That will be 1000*x/100=10x
So by first half year: I will have 10x/2 which 5x as interest earned in first 6 months

Now I am going to calcuate for next 6 months which is (1000+5x)*(x/100)
This becomes 10x+(5x^2/100). Again note that this is interest for entire year, I now have to divide the same by half. So the second half of the year my interest is 5x+(5x^2/200)

So total interest earned in semi annual compounding is 5x+5x+(5x^2/200) which is 10x+x^2/40

Now calculating compounded annual interest which we have done also in above steps will be 1000*x/100=10x

Subtracting the two we get diff as x^2/40

Refer the video link here from GMAT Ninja for seeing this done: GMAT Ninja Percents



email2vm
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

====looking for the answer choices

100^2/40 != 2000!!!

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