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Are Compound Interest problems really worth the time? [#permalink]

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28 Jun 2013, 08:15

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General observation from GMAT prep software some of these questions can be really time consuming. Its solvable but time consuming, so what should one do? _________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Re: Are Compound Interest problems really worth the time? [#permalink]

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01 Jul 2013, 08:49

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In what way do you consider them time consuming? The steps you should be taking with these problems are:

1. Determine number of compounding periods. 2. Determine interest rate as a total percentage, and not a relative percentage (i.e. convert "a gain of 20% per period" into "120% per period = 1.20" 3. Raise the total percentage to the number of periods (1.20^3 for three periods). 4. Multiply step 3 by the principle.

Steps 1 and 2 are simple reasoning and math that you should be able to do in your head. Step 3 and 4 are the time consuming parts. Step 3 there is a trick to make the math easier, IMO, which is using a trick to multiply numbers together fast.

The trick is that when you multiply two numbers together to use a reference number. So for example, let's say the problem has a growth rate of 30% per period. First, convert to total percentage - 130%. Then you'll need to raise it to the number of periods. A period of 1 is easy obviously. A period of 2 is 130 squared. If you do this the traditional way it can be time consuming.

However, the trick is to use 100 as a reference number and use the fast method of multiplication. So here are the steps: 1. Rewrite 130^2 as 130*130 2. Use 100 as a reference number. Since 130 is 30 units above 100, our reference number, add 30 to it to get 160. Then multiply by our reference number to get 16000. 3. Take the distance of both numbers from our reference number - in this case, they're both 30 - and multiply those together to get 900. Add that to step 2 to get the answer: 16000 + 900 = 16900. 130^2 is 16900. Because this is a percentage, you'd add the decimals back in to get 1.69. You then can multiply the principle by 1.69 (depending on the value of the principle you might be able to do that calculation in your head, too).

Using this method, you can easily do this problem entirely in your head.

Now, there is a rule that I didn't explain here, which is that if the number is above your reference number, you add the difference (like in step 2). If it's beneath the reference number, you subtract it. For example, if you have a problem where the value is decreasing by 20% per period, you'd first make that a total percentage per period (20% decline is 80% total) then:

80^2 = 80*80 Use 100 as our reference number. Since 80 is 20 units under our reference, we subtract 20: 80-20 = 60 Multiply by our reference number 60*100 = 6000 Multiply the distances of both numbers from the reference number: 20*20 = 400 Add 6000 + 400 = 6400 Convert back to decimal if needed (add 4 decimal places): 0.64

Using 100 as a reference number, and considering that most growth rates are around 100, make the tough calculation easy to do in your head.

For problems involving more than two periods, you can use this method to decrease calculation time. For example, if you have 4 periods of 20% growth per period, instead of calculating (1.20)(1.20)(1.20)(1.20) you can use the above method to get 1.69, then square 1.69, and even in that case do another iteration of the above method, just with 169*169!

If you have an odd number of periods, you can use the squares and then multiply again. For example, for 5 periods of 20% growth, in your head you can get to 1.69 for two periods quickly, then square that again using the above method to get to 2.8561, and then hand calculate the last period (2.8561*1.69). You might even be able to round 2.8561 to 2.86 and do the above method for 2.86*1.69 to get your final answer.

This all sounds complicated but with a little bit of practice it becomes pretty easy. It also makes you seem like a genius when you can multiply in your head this quickly haha.

General observation from GMAT prep software some of these questions can be really time consuming. Its solvable but time consuming, so what should one do?

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number. _________________

Re: Are Compound Interest problems really worth the time? [#permalink]

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09 Jul 2013, 01:47

VeritasPrepKarishma wrote:

fozzzy wrote:

General observation from GMAT prep software some of these questions can be really time consuming. Its solvable but time consuming, so what should one do?

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?

This question is from GMAT prep _________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?

This question is from GMAT prep

Here, you don't really need to calculate \((1.01)^5\). Look, 1% is a very small percentage. Usually you get 5%/8%/10% etc. Then why do you have 1% here? Recall the concept of compound interest vs simple interest. CI basically gives you extra interest on interest. When we have only 1% interest, a 1% on top will have hardly any effect. So my best bet would be an option a little bit greater than 26,250 (5% more than 25000, 1% for every year). You haven't given the options - I would assume that there will not be two options satisfying this criteria. _________________

In that case, I will go with 26000 since a little greater than 26250 will be closer to 26000 rather than to 27000. If you use a calculator, you will see that 25000(1.01)^5 = 26275

If you want to be further certain, think this way: 1% of 25000 every year will be 250. So you will get 250*5 = 1250 interest Second year onwards, you will also earn 1% of 250 (first year interest) which is 2.5. You will get this for 4 yrs so 2.5*4 = 10 Third year onwards, you will also earn 1% on 250 (second year interest) and 2.5... Do you see this going anywhere? The additional amounts are very small. _________________

Re: Are Compound Interest problems really worth the time? [#permalink]

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21 Jul 2013, 21:46

fozzzy wrote:

VeritasPrepKarishma wrote:

fozzzy wrote:

General observation from GMAT prep software some of these questions can be really time consuming. Its solvable but time consuming, so what should one do?

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?

This question is from GMAT prep

1% growth makes this really easy. To get 1% of a value, you just move the decimal place over two places, so you can do this with simple arithmetic.

Exact solution: Year 0 = 25,000 Year 1 = 25,000 + 250 = 25,250 Year 2 = 25,250 +252.5 = 25,502.5 Year 3 = 25,502.5 + 255.025 = 25,757.525 Year 4 = 25,757.525 + 257.57525 = 26,015.10025 Year 5 = 26, 015.10025 + 260.1510025 = 26,275.25

Rounding to Ones Digit: Year 0 = 25,000 Year 1 = 25,000 + 250 = 25,250 Year 2 = 25,250 + 253 = 25,503 Year 3 = 25,503 + 255 = 25,758 Year 4 = 25,758 + 258 = 26,016 Year 5 = 26,016 + 260 = 26,276

Rounding to Tens Digit: Year 0 = 25,000 Year 1 = 25,000 + 250 = 25,250 Year 2 = 25,250 + 250 = 25,500 Year 3 = 25,500 + 260 = 25,760 Year 4 = 25,760 + 260 = 26,020 Year 5 = 26,020 + 260 = 26,280

Alternatively, use 25,000 + 5*250 = 26,250 as a lower bound and 25,000 + 5*300 = 26,500 as an upper bound. Because the interest amount is so miniscule, you know that it's not going to hit 300, the minimum amount required to round up to 27,000, proving that it's 26,000.

Re: Are Compound Interest problems really worth the time? [#permalink]

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31 Jul 2013, 23:43

Quote:

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

the problem is : $1200 is invested at a given interest rate for two years. The difference between the simple 2-year non-compounded return at the end of the two [b]years and an annually compounded return is $132. What is the interest rate ?options are : 10% 11% 12% 13% 14%[/b]

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

the problem is : $1200 is invested at a given interest rate for two years. The difference between the simple 2-year non-compounded return at the end of the two [b]years and an annually compounded return is $132. What is the interest rate ?options are : 10% 11% 12% 13% 14%[/b]

The interest earned with compounding is more than simple in the second year because one earns interest on previous year's interest too. In the first year, interest earned in the two cases is exactly the same.

Here there is some error in the given numbers:

132 = r% of I I = r% of 1200

132 = (r/100)*(r/100)*1200 You need to get the value of r. With the correct numbers, you can easily find r. _________________

Re: Are Compound Interest problems really worth the time? [#permalink]

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10 Aug 2013, 13:04

VeritasPrepKarishma wrote:

fozzzy wrote:

General observation from GMAT prep software some of these questions can be really time consuming. Its solvable but time consuming, so what should one do?

I don't recall any official problem which is actually time consuming. Tricky - yes, cumbersome - no. Post the problems; perhaps you haven't come across the easier, more intuitive solutions. I will be surprised if you need to multiply anything more than a 2 digit number by a 2 digit number.

Thanks for this great information _________________

Re: Are Compound Interest problems really worth the time? [#permalink]

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21 Mar 2015, 06:15

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Re: Are Compound Interest problems really worth the time? [#permalink]

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29 Jul 2016, 17:53

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