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hi Jenny.Jenny

Welcome to the forum.

This looks like a compu_hi question. They get a bit upset when their questions are posted on-line for 'copyright reasons'. It has been asked before, although it was a goat last time.

You will find an analysis of how to do the calculation here:

http://www.mathisfunforum.com/viewtopic.php?id=18391

It's a long thread and starts with other questions. Your question starts at post 48. I haven't checked the measurements … it is possible the distances have been changed along with the animal so you'll need to adapt your answer to fit.

Bob

It depends on the question. Sometimes it is wrong to give too many figures. In the UK GCSE exam, candidates are penalised if they fail to round off sensibly!

In your question population figures are being compared. How accurately could you count a population? The figure could go up or down during the count and the people doing the counting might miss somebody. The question wants to demonstrate that the growth is (approximately) exponential, so two decimal places is quite sufficient to show that.

As a practice, consider how many figures are appropriate in each of these cases:

1. The number of supporters at a football match is counted as 34,567.

2. You want to order some 'ready-mix' concrete to make a good surface to park your car. You calculate the area of the space, multiply by the required depth of concrete and get 12.3678 cubic yards.

3. Do trees for timber grow better if the are closely planted or if they are spread out more? To test this you grow some trees on two plots. For one the trees are planted close together; on the other bigger spaces are left between the trees. Then you measure the height of each tree and the circumference of the trunk. Typical measures are 30.36 feet and 81.5 inches.

Bob

hi

I used Microsoft Excel to do a few years' calculations. I have made an image of these and alongside the formulas I used to do the calculations.

You should be able to see that the 'Principle' is gradually going down but not as fast as you thought.

Hope that answers all your questions.

Bob

10% + 5

and

5 + 10%

The sum is badly stated as it is not clear whether it means "add another 5% to 10%' or 'find 10% of something and then add another $5 to that'

It's interesting that google gives two different answers. It seems that it treats 5 + 10% as (5+10)%. Which is correct depends on what the question is. I have made up two questions here; one answer is correct for one question; and the other answer is correct for the second question.

question 1. A man is told he will get a 10% increase in wages. But then the boss decides to be more generous and make it 5% more than he originally said. What is the percentage increase now?

question 2. A money lender offers to lend money using the following rule to work out how much extra you have to pay when you pay back the loan: 10% of the amount borrowed plus $5. How much does the loan cost?

Bob

hi

To work out a percentage increase you have to do this:

[amount of increase] / [original amount] times 100%

eg. The price of something goes up from $20 to $25. What is the percentage increase?

increase = 25 - 20 = 5

% increase = 5/20 x 100 = 25%

In your question the 'old' population is 12.853 M and the 'new' is 13.290 M

So the increase is 13.290 M - 12.853 M = 0.437 M

[note: the 'Millions' = M cancel out here]

% increase = 0.437 / 12.853 x 100 = 3.4%

The question didn't do this calculation; rather

13.290 M / 12.853 M = 1.034

You are wondering how they got the correct percentage from this.

So an easy way to get the % increase from is just to take off the 1 (= 12.853/12.853)

In situations like these taking off the 1 will always work as 1 is the same as 100% which is the original amount before the increase.

The law that you have discovered is the one you should be using … it's the same as the one I used above.

The question calculates the increase for each year compared with the year before.

So every time you should be doing

[old population] / [population one year later] x 100%

The answers I get are

3.399984

3.438676

3.477122

3.486819

3.484818

3.43311

The 'book' has rounded off the answers. They are not exactly the same but near enough to say the growth is a constant percentage.

Bob

Let's try an algebraic approach.

Let the initial amount in the account be P; the monthly interest rate be r; and the amount taken out each month be M.

At the end of each month

new amount in the account = old amount plus interest less the monthly amount out

P_new = P_old + (P_old x r/100) - M

If P_new is to be the same as P_old then

P = P + P x r/100 - M

So M = P x r/100

If I choose P = 10000 and r = 0.5 then

M = 10000 x 0.5/100 = 100 x 0.5 = 50

So the algorithm will not end if M = 50

You could vary a different value.

If r = 0.5 and M = 500 then

500 = P x 0.5/100 = P/200 so P = 100,000

So if you have a much bigger amount saved then the interest earned can exactly cover the monthly amount out.

In the above I have kept the P constant from month to month. You could also let the amount in the account grow. So P = 10000, r = 0.5, M = 40 means that P will grow, month by month.

Bob

hi Monox D. I-Fly

Wow! Am I glad I made this post.

Best wishes!

Bob

hi Faith,

That'll do nicely.

Bob

If you buy a maths book its title will tell you what topics are covered. eg. Algebra for beginners.

The MIF site is for anybody, at any level, and tries to cover most mathematical topics. You can dip into any topic freely. So you'll have to decide where to start. eg. If your topic is algebra, click the algebra index link and then have a look at what is covered. If you know some basics, then skip forward to topics you don't yet know. At the bottom of each page there are a set of test yourself questions so you can find out if you have really understood something.

The site has very good cross linking … so if you read something that is not covered on that page you will be able to click across to the page that explains what you need.

Bob

The interest after the first month is $50 not $500.

10000 x 0.5/100 = 50

Bob

**bob bundy**- Replies: 3

I have just investigated a post that contained a link claiming to tell members about stalkerware, a form of malware. But my security software identified the link itself as dangerous as it led to malware! I checked further and found this warning to be correct. I have deleted the post and banned the poster.

MIF does allow members to post links (in the post, the signature and in the member panel to the left of the post). Be very careful of these as there have been a number of attempts recently to use these links to trick members. In windows, if you hover your mouse pointer over a link, you can see at the bottom left of the page where the link is really taking you. If in doubt, don't click such links.

If you have already clicked that stalkerware link, do a full system scan to check your device is not infected.

If you have suspicions about a post please click the report link so that moderators and admin can investigate for you.

Bob

12th September 2019

hi IMustDoItTheWrongWayFirst

Welcome to the forum.

Any chance you can adopt a shorter and more optimistic name?

I have just recommended another member to look here:

I think this site is an excellent way to start learning maths. Our forum grew out of the teaching site to give posters a place where they could raise questions.

I recommend you use both.

Bob

This is just 'How much the car has been used'. It is measured by saying how many miles are 'on the clock'. Generally a second hand car will be better if its mileage is lower.

Bob

hi Hannibal lecter

You are obviously working hard to learn so don't give up. There are members who will try to help you. I worked for 37 years as a teacher of maths. I was constantly searching for ways to explain something that my students were finding difficult. Sometimes I had a great success. One that will always stick in mind was with a class of pupils all of whom were struggling with maths in preparation for the GCSE exam. I brought in a pack of sweets and used these to illustrate the particular problem. It worked. They were able to visualise what was happening and could move sweets about to demonstrate the stages of the problem. By the time we had finished the lesson they were working at 'C' level, which is a pass in that exam. Then we ate the sweets! Great lesson!

The best on-line site for learning maths is the one created by our administrator MathsIsFun. You'll find it here:

Bob

hi Hannibal lecter

So that savers can compare schemes it is usual to state the yearly (annual) rate of interest. Some schemes then calculate and add interest once per year. But some schemes add interest every month, so it is necessary to state what rate will be applied for this. I live in the UK and here there is a formula that does this. It's more complicated than what is given in your question so I won't confuse you by describing it now. The USA system for working out a monthly rate is to divide the yearly rate by 12. Thus 6% per year becomes 0.5% per month.

Each month this is what happens:

Start with the amount in the account.

Add interest from the previous period.

Subtract $500.

This gives the new amount that will apply for the next interest calculation, one month later.

Bob

Rounding is explained here: https://www.mathsisfun.com/rounding-numbers.html

In the example from the book 33.69375, how many digits are required? If one decimal place is required then the '9' causes a round up so the correct answer is 33.7

If two decimal places are required then the '3' requires a round down to 70.69

If this number is an amount in $ or £ then it may be ok to round to 33.70 if the number is wanted to the nearest tenth. 33.69375 is nearer to 33.70 than to 33.60.

So many rounding answers are possible depending what is required.

Bob

In post 2 I showed how you can make a multiplier that shortens the steps for increasing and decreasing by a given percentage.

There is another way:

(1) Calculate the amount of the increase (decrease).

(2) Add (subtract) that amount from the old amount.

eg.

Increase $300 by 20%

(1) 20/100 x 300 = 60

(2) $300 + $60 = $360

With a multiplier you only need one step:

1.20 x 300 = 360

Now you are asking about changing a percentage into a fraction. This is also a two step process.

(1) Put the percentage over 100. (Always 100 as that's what a percent means cent = 100th part)

(2) Simplify the fraction. There is not just one way to do this. It depends on the numbers. I'll give several examples.

eg1. Change 25% into a fraction.

25/100 ….divide top and bottom by 25 = 1/4

eg2. Change 87.5% into a fraction.

87.5/100 eliminate the decimal within the fraction so that both the top number and the bottom number are whole

875/1000 now simplify … divide by 25 = 35/40 divide by 5 = 7/8

eg3. Change 222.10% into a fraction.

222.10/100 As the top ends in a zero it is sufficient to times by 10 = 2221/1000 No cancelling is possible here.

eg4. Change 52.55% into a fraction.

52.55/100 times by 100 = 5255/10000 divide by 5 = 1051/2000 No further cancelling possible.

Hope that helps.

Bob

hi SDKannan

Welcome to the forum.

Bob

hi lovely77

Welcome to the forum.

Other members have posted numerous times on this topic. The quickest way you can get help is to use the search facility to look up what was previously said.

Bob

hi monto

Welcome to the forum.

It looks like there is an error in the wording of this problem. Here's why:

Let x be the number of cows and y be the number of chickens.

Assuming each animal has two eyes; the cows have four legs; and the chickens have two legs then

2x + 2y = 25

4x + 2y = 76

Subtracting

2x = 51 so x = 25.5 and y = -13

So there's an algebraic solution but not one that makes sense for these animals.

Please check the problem wording.

Bob

hi Darcykm

Welcome to the forum.

The creator of the forum first made this site:

It is well indexed and cross referenced and the teaching materials are **excellent!**

Have a look and hopefully you'll find what you need. If not, post specific questions here and someone will try to help.

Bob

You need two answers; one for surface area and one for volume. For each, just add up the three component answers to get the totals.

Bob

ps. If you post what you have worked out, I'll check them for you.

hi h3691218

Thanks. I've got the image of the cone/cylinder/hemisphere and I can show you how to calculate that. I'm not getting the other image.

Here's the formulas you need:

cone:

h is given as 8; the radius is the same for all three parts: 10/2 = 5

s is the length of the sloping edge. You'll have to use Pythagoras for that:

cylinder:

h' is what's left after you've taken off 8 and 5 from 25.

hemisphere:

If you had been asked for the surface area of each part then you'd need to include the base(s) but, as they are joined, this shouldn't be needed.

Bob

hi ashraf

Welcome to the forum.

It comes directly from the dot or scalar product.

https://www.mathsisfun.com/algebra/vect … oduct.html

so

Bob