Percentage problem

[elouise55]

Hi all

I'm stuck on working out one particular problem involving percentages, and would really appreciate an explanation as to how to solve it!

Example: company x wants to reach sales of £40,000. In March total sales were £600, and sales were increasing by 2% each month. What month will it be when the company reaches its target sales?

I'm totally stumped on this one... the only way I know is to work out 2% of the previous number every time, but obviously that takes a lifetime!

Thanks

 

[dsk2]

Hi all

I'm stuck on working out one particular problem involving percentages, and would really appreciate an explanation as to how to solve it!

Example: company x wants to reach sales of £40,000. In March total sales were £600, and sales were increasing by 2% each month. What month will it be when the company reaches its target sales?

I'm totally stumped on this one... the only way I know is to work out 2% of the previous number every time, but obviously that takes a lifetime!

Thanks

elouse55,

Assuming the sales started in the month of March.

Month 1 sales (March)=600
2 months sale=600+600*1.02 (since the sales went up by 2 percent in the second month)
3 months sale=600+600*1.02+ (600*1.02)*1.02
=600(1+1.02+1.02^2)
...
n months sale=600(1+1.02+1.02^2+1.02^3+.....1.02^(n-1))
=600*(1-1.02^n)/(1-1.02) (See formula for "geometric series": Sum of n terms in geometric series=1+r+r^2+..+r^(n-1)=(1-r^n)/(1-r) )

Hopefully at some point, the sales would have reached 40,000 pounds.

40000=600*(1-1.02^n)/(-0.02) => 1.3333=1.02^n-1=>2.3333=1.02^n =>n=log(2.333)/log(1.02)=42.77 months.

If n=1 is march, what is the 43rd month? 43 months=3 years and 7 months, or just 7th month starting March =>September.

In September (which is 42.77 months later), the total sales will exceed 40,000 pounds.

Cheers,
Sai.

P.S: I made some edits to the grammar of my earlier post 12/31

 

[elouise55]

elouse55,

Assuming the sales started in the month of March.

Month 1 sales (March)=600
2nd month sale=600+600*1.02 (since the sales went up by 2 percent in the second month)
3rd month sale=600+600*1.02+ (600*1.02)*1.02
=600(1+1.02+1.02^2)
...
nth month sale=600(1+1.02+1.02^2+1.02^3+.....1.02^(n-1))
=600*(1-1.02^n)/(1-1.02) (See formulat for geometric series: 1+r+r^2+..+r^(n-1)=(1-r^n)/(1-r) )

Hopefully at some point, the sales would have reached 40,000 pounds.

40000=600*(1-1.02^n)/(-0.02) => 1.3333=1.02^n-1=>2.3333=1.02^n =>n=log(2.333)/log(1.02)=42.77 months.

If n=1 is march, what is the 43rd month? 43 months=3 years and 7 months, or just 7th month starting March =>September.

In September (which is 42.77 months later), the total sales will exceed 40,000 pounds.

Cheers,
Sai.

Thanks so much for your speedy response.

Sorry to be a pain, but I'm just trying to get my head around this (I'm teaching myself maths - I haven't studied it since I was at school and I'm still really intimidated by it!)

My understanding of this problem beforehand was if I wanted to increase sales 2% each month, I'd have to repeatedly divide the next figure up by 100 and multiply it by 102, and do this constantly until I reach 40,000. But is that not correct? - Using this method, my answer is much larger than 43 months.

I'm not quite sure what this means - 40000=600*(1-1.02^n)/(-0.02) => 1.3333=1.02^n-1=>2.3333=1.02^n =>n=log(2.333)/log(1.02)=42.77 months. - How would I put this through on a calculator, for example?

I've asked this particular question because I've seen similar ones crop up on numerical reasoning tests and I never know how to answer them- I just don't know how people manage to answer questions like this in under a minute!!

 

[HallsofIvy]

"Divide by 100 and multiply by 102" is the same thing as multiplying by $\displaystyle \frac{102}{100}= 1.02$.
So if you start at month "0" with 600, you have, after month 1, 600(1.02), after month 2, $\displaystyle [600(1.02)](1.02)= 600(1.02)^2$, after month 3, $\displaystyle [600(1.02)^2](1.02)= 600(1.02)^3$, etc. It should be easy to see that after month "n" you will have $\displaystyle 600(1.02)^n$ so you want to solve $\displaystyle 600(1.02)^n= 40000$.

And obvious first step is to divide both sides by 600: $\displaystyle (1.02)^n= \frac{40000}{600}= 66$ and 2/3. To finish that, because the unknown value, n, is an exponent, you will have to use a logarithm. Do you know how to do that?

 

[elouise55]

"Divide by 100 and multiply by 102" is the same thing as multiplying by $\displaystyle \frac{102}{100}= 1.02$.
So if you start at month "0" with 600, you have, after month 1, 600(1.02), after month 2, $\displaystyle [600(1.02)](1.02)= 600(1.02)^2$, after month 3, $\displaystyle [600(1.02)^2](1.02)= 600(1.02)^3$, etc. It should be easy to see that after month "n" you will have $\displaystyle 600(1.02)^n$ so you want to solve $\displaystyle 600(1.02)^n= 40000$.

And obvious first step is to divide both sides by 600: $\displaystyle (1.02)^n= \frac{40000}{600}= 66$ and 2/3. To finish that, because the unknown value, n, is an exponent, you will have to use a logarithm. Do you know how to do that?

Not quite... how would I use a logarithm?

 

[elouise55]

Do you not have a "Ln" key on your calculator?

I don't!

thanks very much for that table, it makes things a little clearer...

 

[JeffM]

I don't!

thanks very much for that table, it makes things a little clearer...

elouise

If you return, a couple of thoughts.

(1) I agree with how Halls read the question. To show the second month's sales as 1212 means a growth rate of 102% a month. Pretty implausible.

(2) Either way, the problem is easy to solve using a spread sheet like excel. Unfortunately, using excel will not teach you much math. (Of course, with spread sheets around, fewer people may need much math.)

(3) If you have a Windows machine ( and probably if you have an Apple), it will have a built-in calculator that can be set to scientific mode. You will then have either an e^x or a ln key (probably both). In my version of Windows, the calculator is under Accessories, and you set it to scientific mode by clicking on View.You have to toggle between e^x and ln. If you do not understand these directions, please ask for clarification.

(4) I am going to expand a bit on Hall's mathematical answer.

In math notation

$\displaystyle a^0 = 1.$ This can also be shown as a^0 = 1.

$\displaystyle If\ n\ is\ a\ positive\ whole\ number,\ a^n = a * a^{n-1}.$ This can also be shown as a^n = a * a^(n - 1).

Let's calculate April's sales the hard way.

$\displaystyle 600 + 600 * 2\% = 600 + 600 * \dfrac{2}{100} = 600 * \left(1 + \dfrac{2}{100}\right) = 600 * \dfrac{100 + 2}{100} = 600 * 1.02 = 612.$

With me so far?

Let's calculate May's sales the hard way

$\displaystyle 612 + 612 * 2\% = 612 + 612 * \dfrac{2}{100} = 612 * \left(1 + \dfrac{2}{100}\right) = 612 * \dfrac{100 + 2}{100} = 612 * 1.02 = 624.24.$ Still with me?

But we could have done it more easily

$\displaystyle 612 = 600 * 1.02 \implies 612 * 1.02 = (600 * 1.02) * 1.02 = 600 * 1.02^2.$

Now we see that

$\displaystyle March's\ sales = 600 = 600 * 1.02^0.$

$\displaystyle April's\ sales = 612 = 600 * 1.02 = 600 * 1.02^1.$

$\displaystyle May's\ sales = 624.24 = 600 * 1.02^2.$

In other words, if we number March of this year as month 0

$\displaystyle Sales\ for\ month\ n = 600 * 1.02^n.$ Make sense?

This gives a nice easy formula, and you can solve the problem fairly quickly by trial and error witha scientific calculator.

(5) Now if you know about logarithms, the problem becomes a simple equation where x is the unknown month when sales first equal or exceed 40,000

$\displaystyle 600 * 1.02^x \approx 40,000 \implies 1.02^x \approx \dfrac{40000}{600} \implies ln\left(1.02^x\right) \approx ln\left(\dfrac{40000}{600}\right) \implies x * ln(1.02) \approx 4.1997 \implies 0.0198x \approx 4.1997 \implies$

$\displaystyle x \approx 212.1 \implies x = 213.$

Let's check:

$\displaystyle 600 * 1.02^{212} < 40000.$

$\displaystyle 600 * 1.02^{213} > 40000.$

It is to solve problems like this that you learn logarithms.

 

[dsk2]

elouise,

Quite an interesting problem there. You already may have figured out the problem by now, if not, you must just hate the intruiging methodology. But all the contributors gave excellent solutions.

Sorry for being so concise in my explanation earlier. But let me spit out a few more thoughts, if you care.

The problem may seem intimidating, but if you go back to the book where you found the problem, you might find that there is a standard formula using which you can solve similar problems very quickly. All of us here are trying to show the solution from the basics, which with patience and some back ground, you can easily understand. Ideally, it would help if you had a basic knowledge of series, in specific, geometric series, and also a basic knowledge of application of logarithms. If you calculator does not have a log or ln (typically log is to the base 10 and ln is to the base e), then your windows computer should have the calculator that you can use in the scientific mode. In fact, you can just do most math in google ( may be not during the exam though).

Before you do the problem, I think the problem statement certainly begs for clarification, since the problem can be interpreted in two ways.

Well, we know that total sales in March is $600, and we also know that the sales are increasing at a rate of 2% per month. The question is,

1. Do we want the sales in one single month to reach $40000 or
2. The cumulative sales starting March to add up to a total of $40000.

Attaining the target as in item1 is quite ambitious, and it would take 213 months, and target as in item2 would take 43 months.

You are right in your approach that all that you need to do is multiply the sales in the previous month by 1.02 to obtain the sales of the new month. And yes, it would take for ever to continue to do that. So, the smart way to do it is to find a pattern and try and use a formula to simplify your calculation. In fact, this is where 'series' help. Series, are bunch of numbers that are set in a certain order, that follow a certain perfect pattern. And to add those pattern of numbers there are certain formula. Look up 'Arithmetic Series', 'Geometric Series' in google to understand them.

Unfortunately, you still need to use logarithms, to solve the problem. For instance, how would you solve a simple question like 2^x=3. Seemingly, simple, you cannot solve for x unless you use logarithms. 2 raised to something (x) gives 3. We know 2^1=2 and 2^2=4. 2^0.5=1.414 ( 2^0.5 is also called sqrt(2) or squareroot of 2). So, you can kind of guess that x should be between 1 and 2. But you see, the only way to find out is to use the law of logarithms. Answer is to apply log, which converts exponents into products, so that you can use regular algebra. So, step1. log(2^x)=log3, step 2. x*log2=log3 (note that this is formula that you need to take for granted atleast in the beginning, log (x^y)=y*logx). Now use regular algebra, x=log3/log2 which is that same as x=ln3/ln2 ( I wont explain why though). So x is '1.5849'. What I did to find x was to type log(3)/log(2) in google search. Try it!! You see why we had to logarithms??

I just gave you an analogy as to how we apply logarithms. Now, go back to each of the solutions that you were given by all the contributors and study the solutions patiently. You will understand it then.

Good luck!!

Cheers,
Sai.