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Percentages, Interest and More
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03 Nov 2015, 11:28
Percentages, Interest and MoreA Simple Approach to Percentages Let's take up a relatively simple topic – Percentages. It is extremely relevant for GMAT and your everyday life. For the critics amongst you, let me give an example: What does a 20% sale with an additional 25% off on the $85 sweater that you have your eye on mean to you? Rather than flipping open your HP12C, blink your eyes and the answer will swim in front of you… Uhh… I mean, after I tell you what you have to do in that blink (There is always a catch!). Let me begin by saying that a percentage is a fraction. A fraction where the denominator is always 100, but just a fraction nevertheless. 50% means 50 per ‘cent’ (cent being 100) or 50/100 or 50 out of every 100. So, \(50\% = \frac{50}{100} =\frac{1}{2}\) If I ask you what the 50% of 240 is, what will you do? \((\frac{50}{100})*240\) or \((\frac{1}{2})*240\)? I don’t think I need to explain why the second representation is easier to handle. A fraction in its lowest form usually helps us to calculate faster. If this makes sense, think about 12.5% i.e. 12.5/100. Let’s say, I tell you that 12.5/100 is equal to 1/8. Now, if you had to find 12.5% of 64, what would be easier to do: \((\frac{12.5}{100}) * 64\) or \((\frac{1}{8}) * 64\)? For this future ease, can you learn up some percent fraction equivalents today? In the interest of furtherance of this post, let’s assume you answered with a resounding yes! So now, if someone asks you, what is 60% of 350, all you should do is \((\frac{3}{5}) * 350 = 210\). What if instead, the question is: If you increase 105 by 20%, what do you get? You might be tempted to find the 20% of 105 and add it to 105. Something like this: \(105 + (\frac{20}{100}) * 105 = 105 + 21 = 126\). Or if you have been paying attention, then you might do it like this: \(105 + (\frac{1}{5}) * 105 = 105 + 21 = 126\). It is still not optimum! You did the calculation in two steps: In step 1, you found the 1/5th of 105. In step 2, you added the 1/5th to 105. Can we instead club it in a single step? \(105 + (\frac{1}{5}) * 105 = 105 (1 + \frac{1}{5}) = 105 * (\frac{6}{5})\) So when we have to increase a number by 20%, we just multiply it by \(\frac{6}{5}\). When we have to decrease a number by 20%, we just multiply it by \(\frac{4}{5}\) (because \(1 – \frac{1}{5} = \frac{4}{5}\)). You might feel that these are little things, not likely to help you save much time. But once we build up on these little things, they work wonders. Let’s try to decrease 120 by 33.33%. 33.33% is 1/3. When you try to decrease a number by 1/3, you will need to multiply it by \((1 – \frac{1}{3}) = \frac{2}{3}\). So to decrease 120 by 33.33%, we just need to multiply it by 2/3. We get \(120 * (\frac{2}{3}) = 80\). Try to do the following orally: What is 40% of 320? What do you get when you increase 352 by 37.5%? What do you get when you decrease 819 by 11.11%? The most important application of this method is successive percentage changes. Let’s take an example: Example: In 2008, the membership of People’s Society was 90,000. In 2009, it increased by 22.22%. In 2010, it decreased by 9.09%. What was the membership at the end of 2010?Solution: \(11.11\% = (\frac{1}{9})\), therefore 22.22% must be 2/9 (multiplying both sides of the equation by 2). To increase a number by 22.22%, we must multiply it by 11/9 (because 1 + 2/9 = 11/9). Also, \(9.09\% = \frac{1}{11}\). To decrease a number by 9.09%, we must multiply it by 10/11 (because \(1 – \frac{1}{11} = \frac{10}{11}\)) Membership at the end of \(2009 = 90,000*(\frac{11}{9})\) Membership at the end of \(2010 = 90,000*(\frac{11}{9})*(\frac{10}{11}) = 100,000\) The membership at the end of 2009 is the membership at the beginning of 2010 (i.e. \(90,000*(\frac{11}{9})\)) which decreases by 9.09%. Once you are comfortable with this method, you will directly jump to the step in Bold above. And as for the $85 sweater question, I will take it up in the next post on percentages. Attachment:
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Re: Percentages, Interest and More
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03 Nov 2015, 11:39
Successive Percentage Changes Here I will take a topic I briefly introduced in the previous post on percentages. Let me start with the question I posted there. What does a 20% sale with an additional 25% off on the $85 sweater that you have your eye on mean to you?It means a big rebate. Let’s see how much: If you reduce 85 by 20%, it becomes \(85 * \frac{4}{5} = $68\). Now, you reduce it again by 25% and it becomes \(68 *(\frac{3}{4}) = $51\) Notice that a 20% discount and then a 25% discount is not equal to a 45% discount (\(85 * \frac{55}{100} = $46.75\)). It is less than 45% but to the imperceptive, oblivious customer, it registers as 45% (Now you know why the retailers use the strategy of marking down by 20% and then giving an ‘additional’ 25% later!). The difference arises because the 20% discount was given on $85 but the 25% discount was given on $68. 25% of 68 is definitely less than 25% of 85 and hence the overall percentage decrease is less than 45%. This is called successive percentage change – a number is changed by some percentage and then the new number is changed by another percentage. Both the percentage changes are not applied to the same original number. The most popular example of successive percentage change is population change. Let us look at an example to understand this. Example 1:A city’s population was 10,000 at the end of 2008. In 2009, it increased by 10% and in 2010, it decreased by 18.18%. What was the city’s population at the end of 2010?Solution:Population at the end of 2008 = 10,000 Population at the end of 2009 \(= 10,000 * (\frac{11}{10}) = 11,000\) Population at the end of 2010 \(= 11,000 * (\frac{9}{11}) = 9000\) Simply put, population at the end of 2010 \(= 10,000 * (\frac{11}{10}) * (\frac{9}{11}) = 9000\) It is best to do the calculations in a single step because you do not need to calculate the intermediate population values. Besides, there is a good possibility that factors will get canceled out and hence, you will need to do fewer calculations. Obviously, there is no limit to the number of successive percentage changes that can be made to a number. The approach remains unchanged in any case. Let me elaborate with another example: Example 2:Six months back, the cost of an air ticket from Detroit to San Francisco was $400. Four months back, the fares increased by 12.5%. Last month, the fares increased by 25% and yesterday, the airlines again increased the fares by 11.11%. What is the price of a Detroit to San Francisco ticket today?Solution:Price of a ticket today \(= 400 * (\frac{9}{8}) * (\frac{5}{4}) * (\frac{10}{9}) = $625\) This is much faster than finding the ticket price at every price change which would need the following steps: Price of a ticket four months back \(= 400 * (\frac{9}{8}) = $450\) Price of a ticket last month back \(= 450 * (\frac{5}{4}) = $562.5\) and finally, price of a ticket today \(= 562.5 * (\frac{10}{9}) = $625\) So, in case you do not need the intermediate values, do not calculate them. When there are only two successive percentage changes, we can derive a formula. In some cases, the formula makes the solution very simple. When a number, N, changes by x% and then changes again by y%, we do the following to find the new number: New number \(= N * (1 + \frac{x}{100}) * (1 + \frac{y}{100})\) Now, \((1 + \frac{x}{100}) * (1 + \frac{y}{100}) = 1 + \frac{x}{100} +\frac{y}{100} + \frac{xy}{10000}\) If we say that \(x + y + \frac{xy}{100} = z\), then \((1 + \frac{x}{100}) * (1 + \frac{y}{100}) = 1 + \frac{z}{100}\) Here, z is the effective percentage change when a number is changed successively by two percentage changes. Let’s take another example to see the formula in action: Example 3:A city’s population was 10,000 at the end of 2008. In 2009, it increased by 20% and in 2010, it decreased by 10%. What was the city’s population at the end of 2010?Solution:x% = 20% y% = – 10% (Notice the negative sign here because this is a decrease) Effective percentage change \(= x + y + \frac{xy}{100} = 20 + (– 10) + \frac{20*(10)}{100} = 8\%\) Population at the end of 2010 \(= 10,000 * (\frac{108}{100}) = 10800\) Note: When the percentage is a decrease, a negative sign is used as shown above. This formula is used only when there are two successive percentage changes and the percentages are easy to work with e.g. 15% and 25%, 10% and – 30% etc. With more than two successive percentage changes or trickier percentage values e.g. 11.11% and 18.18%, 9.09% and 6.25% etc, stick to the method shown above. A major application of successive percentage changes in GMAT is the MarkUpDiscountProfit questions. We will take that topic next week but I will leave you with a question to ponder upon: If a retailer marks up his goods by 40% and then offers a discount of 10%, what is his profit%?
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Re: Percentages, Interest and More
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03 Nov 2015, 11:47
Mark Up, Discount, and Profit Mark Up, Discount and Profit questions confuse a lot of people. But, actually, most of them are absolute sitters — very easy to solve — a free ride! How? We will just see. Let me begin with the previous post’s question. Question: If a retailer marks up an article by 40% and then offers a discount of 10%, what is his percentage profit?Let us say the retailer buys the article for $100 ($100 is his cost price of the item). He marks it up by 40% i.e. increases his cost price by 40% (100 * 140/100) and puts a tag of $140 on the article. Now, the article remains unsold and he puts it on sale – 10% off everything. So the article marked at $140, gets $14 off and sells at $126 (because 140 * 9/10 = 126). This $126 is the selling price of the article. To recap, we obtained this selling price in the following way: \(Cost \ Price * (1 + Mark \ Up\%) * (1 – Discount\%) = 100 * (1 + \frac{40}{100}) * (1 – \frac{10}{100}) = 126 = Selling \ Price\) The profit made on the item is $26 (obtained by subtracting 100, the retaile’s cost price, from 126, the retailer’s selling price). He got a profit % of \((\frac{26}{100}) * 100 = 26\%\) (Profit/Cost Price x 100) Or we can say that \(Cost \ Price * (1 + Profit\%) = 100 * (1 + \frac{26}{100}) = 126 = Selling \ Price\) The italicized parts above show the two ways in which you can reach the selling price: using markup and discount or using profit. The same thing is depicted in the diagram below: Therefore, \(Cost \ Price * (1 + Mark \ Up\%) * (1 – Discount\%)= Cost \ Price * (1 + Profit\%)\) Or \((1 + Mark \ Up\%) * (1 – Discount\%)= (1 + Profit\%)\) Look at the numbers here: Mark Up: 40%, Discount: 10%, Profit: 26% (Not 30% that we might expect because 40% – 10% = 30%) Why? Because the discount offered was on $140, not on $100. So a bigger amount of $14 was reduced from the price. Hence the profit decreased. This leads us to an extremely important question in percentages – What is the base? 100 was increased by 40% but the new number 140 was decreased by 10%. So in the two cases, the bases were different. Hence, you cannot simple subtract 10 from 40 and hope to get the Profit %. Also, mind you, almost certainly, 30% will be one of the answer choices, albeit incorrect. (The GMAT doesn’t forego even the smallest opportunity of tricking you into making a mistake!) Let’s see this concept in action on a tricky third party question: A dealer offers a cash discount of 20%. Further, a customer bargains and receives 20 articles for the price of 15 articles. The dealer still makes a profit of 20%. How much percent above the cost price were his articles marked?a) 100% b) 80% c) 75% d) 66+2/3% e) 50% This question involves two discounts: 1. the straight 20% off 2. discount offered by selling 20 articles for the price of 15 – a discount of cost price of 5 articles on 20 articles i.e. a discount of 5/20 = 25% Using the formula given above: \((1 + \frac{m}{100})(1 – \frac{20}{100})(1 – \frac{25}{100}) = (1 + \frac{20}{100})\) m = 100 Therefore, the mark up was 100%. Answer (A) This question is discussed HERE. Note: The two discounts are successive percentage discounts. Another application of successive percentage changes is the concept of compounding. But more on that, in the next post. Attachment:
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Re: Percentages, Interest and More
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03 Nov 2015, 11:54
Simple Interest and the NotSoSimple One I am sure you have heard of the phenomenal “power of compounding.” Elders love to preach about the wisdom of starting a savings account at the age of 25 (when you don’t have any money left from your pay check after your notsosensible Vuitton/Gucci/Chanel escapades!) rather than at the age of 40 (when you have 2 mortgages, 2 kids and a high maintenance Cadillac). Let’s crunch some numbers to see if they are right. Starting at age 25, if you put $200 every month for 40 years at 10% per annum, you will have more than 1.25 million at the age of 65. Starting at age 40, if you put $200 every month for 25 years at 10% per annum, you will have only $265,000 at the age of 65. You might have reservations about the fact that in the first case, you are investing more, after all! It’s not just about the longer time period. So look at it another way. If you invest $25,000 at age 25 at 10% per annum, you will have $1.13 million at age 65. But if you invest $25,000 at age 40 at 10% per annum, you will have $271,000 at age 65. In the first case, you invest for less than double the time (for 40 years as compared to 25 years in the second case) but you get four times the return (1.31 million as compared to 271,000)… It seems that elders might have been right about this after all. (Now I wish I had listened to my dad!) Since compounding has a powerful influence on finances, it is something you will come across often during your MBA studies. So GMAT likes to test you on it too. Let’s get crackin’ then. As I mentioned in my previous post, compound Interest is an important application of successive percentage changes. GMAT tests you on simple and compound interest and sometimes, may even test you on the relation between the two. So let’s look at both of them one by one. When you say that the rate of simple interest is 10% per annum, it means that you earn 10% on your original principal every year. Say I deposit $1000 for 4 years at 10% simple interest per annum. Amount at the end of one year is simply 1000*(11/10) = $1100 i.e. I earn $100 in a year. Since it is simple interest, every year I will earn the same amount i.e. $100. So total simple interest earned will be $100*4 = $400. If you observe carefully, we have calculated total simple interest using the following concept: \(Simple \ Interest = \frac{Interest \ Rate}{100} *Principal * No. \ of \ years\) which is exactly what the formula for calculation of simple interest gives us. Now let’s go on to compound interest. Compounding means successive percentage changes. It means that a sum of money increases by a certain percentage in a year. At the end of the year, the interest earned is combined with the principal and next year, interest is earned on this combined amount. Say I deposit $1000 for 4 years at 10% compound interest per annum. Amount at the end of one year is simply 1000*(11/10) = $1100 i.e. I earn $100 in a year. Till now it is just like simple interest. But from next year on, we will earn on this extra $100 that we earned this year too. Amount at the end of 2nd year = 1100*(11/10) = 1210. Amount at the end of third year will be 1210*(11/10) and so on… As you can see, this is just 1000*(11/10)(11/10)(11/10)(11/10) or \(A = P(1 + r)^i\) (“i” is the number of time periods and r is the percentage rate of interest) which is nothing but the ‘amount in case of compound interest’ formula How does knowing this help us? GMAT tests your ingenuity and conceptual understanding. I will give below two examples where you will see how knowing this helps. Example 1:A bank launched a new financial instrument called VDeposit. A VDeposit offers you variable rate of compound interest in accordance with the current market rate. Ethan deposited $8000 in a VDeposit. If he gets interest rates of 10% in the first two years and 12.5% in the third year, what is the total amount at the end of 3 years?As you can see, solving it using the standard formula is slightly cumbersome since we would have to use it twice. I would rather view it as: \(8000*(\frac{11}{10})(\frac{11}{10})(\frac{9}{8}) = 1000*(\frac{11}{10})*(\frac{11}{10})*(9)\) Notice that even though 12.5% compound interest was offered in the third year, we can still cancel off the 8 of 8000 with 8 of 12.5% increase when we view the calculation this way. Amount = $10,890 Example 2:Mark deposited $D in a scheme offering 5% simple interest per annum. Tetha deposited $D in a scheme offering 5% compound interest per annum. At the end of second year, Tetha had earned a total of $2.50 more than Mark. What is the value of D?Till the end of first year, simple interest and compound interest cases are exactly the same. The difference comes in at the end of second year when compound interest offers interest on previous year’s interest too. $2.50 is 5% interest earned in the second year on first year’s interest. \(2.5 = (\frac{5}{100}) * I\) I = $50 So interest earned in the first year is $50, which is 5% of the deposited amount D \(50 = (\frac{5}{100})*D\) D = $1000 In these and many more case, it pays to understand the concept of simple and compound interest. Now, I will leave you with a question: In the case of yearly compound interest, the ratio of amounts at the end of the 20th year to the amount at the end of the 22nd year is 0.81. What is the rate of interest?
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Re: Percentages, Interest and More
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03 Nov 2015, 12:12
Profit on One, Loss on Another I am no fan of formulas, especially the unintuitive ones but the one we are going to discuss today has proved quite useful. It is for a concept tested on GMAT Prep so it might be worth your while to remember this little formula. When two items are sold at the same selling price, one at a profit of x% and the other at a loss of x%, there is an overall loss. The loss% = (x^2/100)% We will see how this formula is derived but the algebra involved is tedious. You can skip it if you wish. Say two items are sold at $S each. On one, a profit of x% is made and on the other a loss of x% is made. Say, cost price of the article on which profit was made = Ct \(Ct (1 + \frac{x}{100}) = S\) \(Ct = \frac{S}{(1 + \frac{x}{100})}\) Cost Price of the article on which loss was made = Cs \(Cs (1 – \frac{x}{100}) = S\) \(Cs = \frac{S}{(1 – \frac{x}{100})}\) Total Cost Price of both articles together \(= Ct + Cs = \frac{S}{(1 + \frac{x}{100})} + \frac{S}{(1 – \frac{x}{100})}\) \(Ct + Cs = S(\frac{1}{(1 + \frac{x}{100})} + \frac{1}{(1 – \frac{x}{100})})\) \(Ct + Cs = \frac{2S}{(1 – (\frac{x}{100})^2)}\) Total Selling Price of both articles together = 2S Overall Profit/Loss = \(2S – (Ct + Cs)\) Overall Profit/Loss % = \(\frac{2S – (Ct + Cs)}{Ct + Cs} * 100\) \(= [\frac{2S}{(Ct + Cs)} – 1] * 100\) = [2S/[2S/(1 – (x/100)^2)] – 1] * 100 \(= (\frac{x}{100})^2 * 100\) \(=\frac{x^2}{100}\) Overall there is a loss of \((\frac{x^2}{100})\%\). Let’ see how this formula works on a GMAT Prep question. Question: John bought 2 shares and sold them for $96 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had(A) a profit of $10 (B) a profit of $8 (C) a loss of $8 (D) a loss of $10 (E) neither a profit nor a loss Solution:Note that the question would have been straight forward had the COST price been the same, say $100. A 20% profit would mean a gain of $20 and a 20% loss would mean a loss of $20. Overall, there would have been no profit no loss. Here the two shares are sold at the same SALE price. One at a profit of 20% on cost price which must be lower than the sale price (to get a profit) and the other at a loss of 20% on cost price which must be higher than the sale price (to get a loss). 20% of a lower amount will be less in dollar terms and hence overall, there will be a loss. The loss % \(= \frac{(20)^2}{100} \% = 4\%\). But we need the amount of loss, not the percentage of loss. Total Sale price of the two shares = \(2*96 = \$192\) Since there is a loss of 4%, the 96% of the total cost price must be the total sale price \((\frac{96}{100})*Cost \ Price = Sale \ Price\) \(Cost \ Price = \$200\) \(Loss = \$200 – \$192 = \$8\) Answer (C) This question is discussed HERE.
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Re: Percentages, Interest and More
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Re: Percentages, Interest and More
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04 Jul 2017, 11:57
How to Solve Advanced Compound Interest Questions on the GMAT We have discussed simple and compound interest in a previous post. We saw that simple and compound interest (compounded annually) in the first year is the same. In the second year, the only difference is that in compound interest, you earn interest on previous year’s interest too. Hence, the total two year interest in compound interest exceeds the two year interest in case of simple interest by an amount which is interest on year 1 interest. So a question such as this one is very simple to solve: Question 1: Bob invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 2 years at the same rate of interest and received $605 as interest. What was the annual rate of interest?(A) 5% (B) 10% (C) 12% (D) 15% (E) 20% Solution:Simple Interest for two years = $550 So simple interest per year = 550/2 = 275 But in case of compound interest, you earn an extra 605 – 550 = $55 This $55 is interest earned on year 1 interest i.e. if rate of interest is R, it is 55 = R% of 275 R = 20 Answer (E) This question is discussed HERE. The question is – what happens in case you have 3 years here, instead of 2? How do you solve it then? Here is a small table of the difference between simple and compound interest to help you. Say the Principal is P and the rate of interest if R It gets a bit more complicated though not very hard to solve. All you need to do is solve a quadratic, which, if the values are well thought out, is fairly simple to solve. Let’s look at the same question adjusted for three years. Question 2: Bob invested one half of his savings in a bond that paid simple interest for 3 years and received $825 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 3 years at the same rate of interest and received $1001 as interest. What was the annual rate of interest?(A) 5% (B) 10% (C) 12% (D) 15% (E) 20% Simple Interest for three years = $825 So simple interest per year = 825/3 = $275 But in case of compound interest, you earn an extra $1001 – $825 = $176 What all is included in this extra $176? This is the extra interest earned by compounding. This is R% of interest of Year1 + R% of total interest accumulated in Year2This is R% of 275 + R% of (275 + 275 + R% of 275) = 176 \((\frac{R}{100}) *[825 + (\frac{R}{100})*275] = 176\) Assuming \(\frac{R}{100}= x\) to make the equation easier, \(275x^2 + 825x – 176 = 0\) \(25x^2 + 75x – 16 = 0\) \(25x^2 + 80x – 5x – 16 = 0\) \(5x(5x + 16) – 1(5x + 16) = 0\) \(x = \frac{1}{5}\) or \(\frac{16}{5}\) Ignore the negative value to get \(\frac{R}{100} = \frac{1}{5}\) or \(R = 20\). Attachment:
CompoundInterest.jpg [ 27.85 KiB  Viewed 24626 times ]
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Re: Percentages, Interest and More
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11 Jun 2018, 18:14
Bunuel wrote: How to Solve Advanced Compound Interest Questions on the GMAT We have discussed simple and compound interest in a previous post. We saw that simple and compound interest (compounded annually) in the first year is the same. In the second year, the only difference is that in compound interest, you earn interest on previous year’s interest too. Hence, the total two year interest in compound interest exceeds the two year interest in case of simple interest by an amount which is interest on year 1 interest. So a question such as this one is very simple to solve: Question 1: Bob invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 2 years at the same rate of interest and received $605 as interest. What was the annual rate of interest?(A) 5% (B) 10% (C) 12% (D) 15% (E) 20% Solution:Simple Interest for two years = $550 So simple interest per year = 550/2 = 275 But in case of compound interest, you earn an extra 605 – 550 = $55 This $55 is interest earned on year 1 interest i.e. if rate of interest is R, it is 55 = R% of 275 R = 20 Answer (E) This question is discussed HERE. The question is – what happens in case you have 3 years here, instead of 2? How do you solve it then? Here is a small table of the difference between simple and compound interest to help you. Say the Principal is P and the rate of interest if R It gets a bit more complicated though not very hard to solve. All you need to do is solve a quadratic, which, if the values are well thought out, is fairly simple to solve. Let’s look at the same question adjusted for three years. Question 2: Bob invested one half of his savings in a bond that paid simple interest for 3 years and received $825 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 3 years at the same rate of interest and received $1001 as interest. What was the annual rate of interest?(A) 5% (B) 10% (C) 12% (D) 15% (E) 20% Simple Interest for three years = $825 So simple interest per year = 825/3 = $275 But in case of compound interest, you earn an extra $1001 – $825 = $176 What all is included in this extra $176? This is the extra interest earned by compounding. This is R% of interest of Year1 + R% of total interest accumulated in Year2This is R% of 275 + R% of (275 + 275 + R% of 275) = 176 \((\frac{R}{100}) *[825 + (\frac{R}{100})*275] = 176\) Assuming \(\frac{R}{100}= x\) to make the equation easier, \(275x^2 + 825x – 176 = 0\) \(25x^2 + 75x – 16 = 0\) \(25x^2 + 80x – 5x – 16 = 0\) \(5x(5x + 16) – 1(5x + 16) = 0\) \(x = \frac{1}{5}\) or \(\frac{16}{5}\) Ignore the negative value to get \(\frac{R}{100} = \frac{1}{5}\) or \(R = 20\). Attachment: CompoundInterest.jpg is this This is R% of principal + R% of total interest accumulated in Year2This is 275 + R% of (275 + 275 + R% of 275) = 176 because as per the formula, I + R%of(I+I+R%of I)



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Percentages, Interest and More
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25 Jul 2018, 09:04
Dear Bunuel, I also do not understand why we need R % of 275 for the first year and not just 275 as supposed according to the formula? Plus, I also think that 176 must include the interests compounded from the first, second, and third years like it is done in the formula, and if they are actually included then I do not understand why it is mentioned like "R% of total interest accumulated in Year2" and not like "R% of total interest accumulated in Year3" . Could you please elaborate on this. Thanks!



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Re: Percentages, Interest and More
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03 Sep 2018, 07:53
avinash R1 Gayluk its R% of 275(275 is interest earned at the end of year 1) + R% of interest accumulated in year 2{which is 275(interest earned in year 1) + 275(interest yearned in year 2 on principle) + R% of 275 ( here R% of 275 is interest earned on the year 1 interest since its compounded)}



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Re: Percentages, Interest and More
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04 Sep 2018, 00:11
GaylukI was also confused by the equation so I decided to make an overview of the interest earned each year with the compound interest rate 1 Year > 275 2 Year > 550 + 275*X 3 Year > 825 + 275*X + (550+275*X)*X We know that 1001 should be total of compound interest earned, so: 825 + 275X + (550+275X)*X = 1101 This simplifies to: > 275X^2 + 825X  176 = 0
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Re: Percentages, Interest and More &nbs
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04 Sep 2018, 00:11






