[gabriel 41]

Hello guys,


There's a question I got here, and the second part of the question is giving me problems. Here's the question:

A manufacturing company has to replace a piece of equipment in three years’ time,
and it is assumed that the replacement machine will cost £45,000.
(i) If the company already has the £45,000, calculate how much of this must be
invested now, to buy this piece of equipment in three years’ time, if the interest
rate which can be earned on investments is 8%. (Interest earned each year will be
re-invested and no money will be withdrawn.) Give your answer to the nearest £.
(ii) Instead of investing all the money now, the company decides to invest £15,000
at the beginning of each of the three years. Interest at 8% will be earned on
each year’s investment and re-invested as in part (i). Calculate the value of this
investment at the end of the third year, giving your answer to the nearest £.
(15 marks)


So part (i) of the question is very okay for me; however, part (ii) is the problem. I'm not getting the correct answer in it. I use the simple interest formula and in year one I get interest of 1200 then in the second year I use the simple interest formula but this time around I work it out on the principle of 16200 and get an interest of 1296. Then in year 3 I use the formula again on the principle of 17496 and get and interest of 1400, to the nearest pound. so then the value of the investment at the end of 3 years should be 1400+ 17496 = 18896. However, this isn't the answer and the answer is 52,592. I don't understand how we reach this answer because in the mark scheme they show the principles of 3 years being added up together. Why? Because we are supposed to find the value of the investment at the end of three years and not add the principles of three years together, so why isn't my method working? Please help me understand this or show me an alternative way of solving the problem. Thanks.

[msmith2512]

What happens at the beginning of year 2?

Interest is added to the investment and a further investment is made. I can see you dealing with the interest but not the additional investment.

[gabriel 41]

Please answer my question for equations of straight lines too. See below.

[SpiggyTopes]

(Original post by gabriel 41)
Hello guys,


There's a question I got here, and the second part of the question is giving me problems. Here's the question:

A manufacturing company has to replace a piece of equipment in three years’ time,
and it is assumed that the replacement machine will cost £45,000.
(i) If the company already has the £45,000, calculate how much of this must be
invested now, to buy this piece of equipment in three years’ time, if the interest
rate which can be earned on investments is 8%. (Interest earned each year will be
re-invested and no money will be withdrawn.) Give your answer to the nearest £.
(ii) Instead of investing all the money now, the company decides to invest £15,000
at the beginning of each of the three years. Interest at 8% will be earned on
each year’s investment and re-invested as in part (i). Calculate the value of this
investment at the end of the third year, giving your answer to the nearest £.
(15 marks)


So part (i) of the question is very okay for me; however, part (ii) is the problem. I'm not getting the correct answer in it. I use the simple interest formula and in year one I get interest of 1200 then in the second year I use the simple interest formula but this time around I work it out on the principle of 16200 and get an interest of 1296. Then in year 3 I use the formula again on the principle of 17496 and get and interest of 1400, to the nearest pound. so then the value of the investment at the end of 3 years should be 1400+ 17496 = 18896. However, this isn't the answer and the answer is 52,592. I don't understand how we reach this answer because in the mark scheme they show the principles of 3 years being added up together. Why? Because we are supposed to find the value of the investment at the end of three years and not add the principles of three years together, so why isn't my method working? Please help me understand this or show me an alternative way of solving the problem. Thanks.

For part two, treat the money put in each year separately. i.e. work out 3 years interest on £15000, two years interest and finally one years interest and then add all of these values together.

[gabriel 41]

(Original post by SpiggyTopes)
For part two, treat the money put in each year separately. i.e. work out 3 years interest on £15000, two years interest and finally one years interest and then add all of these values together.

Okay your method is absolutely perfect but please could you explain it to me. Why do we do like that? Give me the understanding behind it please.

[gabriel 41]

Okay guys there's another problem I have in maths. So here's the question:

A company wishes to hire a photocopying machine on a monthly basis. The company has
received the following hire tariffs from three different office equipment suppliers:
Supplier A: a fixed rate charge of £100 per month, up to a maximum of 10,000 copies.
Supplier B: a fixed rate charge of £40 per month, plus a 1 pence charge per copy.
Supplier C: a charge of 2 pence per copy, with no fixed charge.
(Note that £1 = 100 pence (p))
(a) For each supplier, represent their tariff as a linear equation in the form y = mx + c,
where y is the cost of photocopying and x is the number of copies. (6 marks)
(b) Plot all three linear equations on one graph, with y having a maximum value of
10,000 copies. (8 marks)

Now I'm totally blank in this question and have no idea of how to answer part (a) or part (b). I understand the equation of a straight line, y=mx+c, but I don't get how to apply it in this situation. So I need help. Can someone please fully explain to me how to put these tariffs in the form of y=mx+c? For example, if you take (a) then for supplier A the equation should read something like: 100=10,000m + c ; however for supplier A the answer is simply y=100. So I don't really understand how we reach y=100. Can someone please help me and make me understand this question. Also please help for part (B) and the other suppliers i.e. supplier B and C.

Thank you.

[SpiggyTopes]

(Original post by gabriel 41)
Okay guys there's another problem I have in maths. So here's the question:

A company wishes to hire a photocopying machine on a monthly basis. The company has
received the following hire tariffs from three different office equipment suppliers:
Supplier A: a fixed rate charge of £100 per month, up to a maximum of 10,000 copies.
Supplier B: a fixed rate charge of £40 per month, plus a 1 pence charge per copy.
Supplier C: a charge of 2 pence per copy, with no fixed charge.
(Note that £1 = 100 pence (p))
(a) For each supplier, represent their tariff as a linear equation in the form y = mx + c,
where y is the cost of photocopying and x is the number of copies. (6 marks)
(b) Plot all three linear equations on one graph, with y having a maximum value of
10,000 copies. (8 marks)

Now I'm totally blank in this question and have no idea of how to answer part (a) or part (b). I understand the equation of a straight line, y=mx+c, but I don't get how to apply it in this situation. So I need help. Can someone please fully explain to me how to put these tariffs in the form of y=mx+c? For example, if you take (a) then for supplier A the equation should read something like: 100=10,000m + c ; however for supplier A the answer is simply y=100. So I don't really understand how we reach y=100. Can someone please help me and make me understand this question. Also please help for part (B) and the other suppliers i.e. supplier B and C.

Thank you.

y=mx+c where y is the total cost, m is the cost per sheet, x is the number of sheets photocopied and c is the fixed rate charge.

For part a)....

Supplier A: y=100
Supplier B: y= x+40
Supplier C: y=2x

For part b) put some values of x(number of sheets) into each equation to get values of y and then plot a graph of the points.

Supplier A will give a horizontal line, Supplier B will give a line pointing NW and starting at 40 and Supplier C will give a steep(ish) line.



I think your method of working out the interest didn't work because you didn't account for interest gained from the money invested at later stages.

[gabriel 41]

(Original post by SpiggyTopes)
y=mx+c where y is the total cost, m is the cost per sheet, x is the number of sheets photocopied and c is the fixed rate charge.

For part a)....

Supplier A: y=100
Supplier B: y= x+40
Supplier C: y=2x

For part b) put some values of x(number of sheets) into each equation to get values of y and then plot a graph of the points.

Supplier A will give a horizontal line, Supplier B will give a line pointing NW and starting at 40 and Supplier C will give a steep(ish) line.



I think your method of working out the interest didn't work because you didn't account for interest gained from the money invested at later stages.

Okay. Thank you very much for your help. The answers are absolutely correct; however you said " where y is the total cost, m is the cost per sheet, x is the number of sheets photocopied and c is the fixed rate charge." How is y the total cost? They said in the question y is the cost of photocopying. Furthermore, how did you know what m and c were? They didn't give us the values of m and c in the question, so how did you figure out these? Please reply. Or, is there an alternate easier way of doing this? Please let me know if there is and explain the alternate way to me, if you have time.

[SpiggyTopes]

(Original post by gabriel 41)
Okay. Thank you very much for your help. The answers are absolutely correct; however you said " where y is the total cost, m is the cost per sheet, x is the number of sheets photocopied and c is the fixed rate charge." How is y the total cost? They said in the question y is the cost of photocopying. Furthermore, how did you know what m and c were? They didn't give us the values of m and c in the question, so how did you figure out these? Please reply. Or, is there an alternate easier way of doing this? Please let me know if there is and explain the alternate way to me, if you have time.

Cost of photocopying=total cost=y <-- three sides of the same coin

I know what c is because c is always a constant value and so doesn't change no matter what everything else in the equation is.

I know what m is because this is the gradient, the gradient is the rate of change. i.e. if an extra photocopy is done, how much y will increase by. In other words, this is the cost of 1 photocopy.

I know my explanations aren't great, I just don't know how to word it, sorry!

[gabriel 41]

(Original post by SpiggyTopes)
Cost of photocopying=total cost=y <-- three sides of the same coin

I know what c is because c is always a constant value and so doesn't change no matter what everything else in the equation is.

I know what m is because this is the gradient, the gradient is the rate of change. i.e. if an extra photocopy is done, how much y will increase by. In other words, this is the cost of 1 photocopy.

I know my explanations aren't great, I just don't know how to word it, sorry!

Okay there's no problem. I get you well. You've been so nice helping me around. I hope I'm not too annoying but I have another question that I would like you to help me in. This question is on probability and here's the question:

Q8 (a) An inspection carried out on 300 electronic items manufactured by two companies
(Company A and Company B) found that 30 were faulty. Of the 150 electronic
items inspected that were manufactured by Company A, 10 were found to be faulty.
In contrast, 20 were found to be faulty out of the 150 inspected electronic items
manufactured by Company B.
If one electronic item is randomly selected from these 300 electronic items, fi nd the
probability that this electronic item is:
(i) Manufactured by Company B; (3 marks)
(ii) Faulty; (3 marks)
(iii) Not faulty and is manufactured by Company B; (4 marks)
(iv) Manufactured by Company A, given that it is faulty. (4 marks)

So (i) and (ii) are very easy and I'm okay there but (iii) and (iv) is giving me the problem. I like solving probability questions using venn diagrams, and I've attached my venn diagram, see below. This venn diagram is my answer to (iii) and (iv). From my venn diagram you can see that (P) of faulty and manufactured by company B (answer to (iii)) is 20/300, so (P) of not faulty and manufactured by company B will be 1-(20/300) which is 14/15. However, this answer seems to be wrong and the mark schemes shows (P) of this question as 13/30. Isn't my method correct. What's the problem. And then for (iv) (P) of manufactured by A and faulty, from my venn diagram is 10/300 which is 1/30 but in the mark scheme the (P) of this question is (1/3). I think for (iv) the mark scheme is wrong. What do you think? Please help me. Thank you for your time and co-operation.

[SpiggyTopes]

(Original post by gabriel 41)
Okay there's no problem. I get you well. You've been so nice helping me around. I hope I'm not too annoying but I have another question that I would like you to help me in. This question is on probability and here's the question:

Q8 (a) An inspection carried out on 300 electronic items manufactured by two companies
(Company A and Company B) found that 30 were faulty. Of the 150 electronic
items inspected that were manufactured by Company A, 10 were found to be faulty.
In contrast, 20 were found to be faulty out of the 150 inspected electronic items
manufactured by Company B.
If one electronic item is randomly selected from these 300 electronic items, fi nd the
probability that this electronic item is:
(i) Manufactured by Company B; (3 marks)
(ii) Faulty; (3 marks)
(iii) Not faulty and is manufactured by Company B; (4 marks)
(iv) Manufactured by Company A, given that it is faulty. (4 marks)

So (i) and (ii) are very easy and I'm okay there but (iii) and (iv) is giving me the problem. I like solving probability questions using venn diagrams, and I've attached my venn diagram, see below. This venn diagram is my answer to (iii) and (iv). From my venn diagram you can see that (P) of faulty and manufactured by company B (answer to (iii)) is 20/300, so (P) of not faulty and manufactured by company B will be 1-(20/300) which is 14/15. However, this answer seems to be wrong and the mark schemes shows (P) of this question as 13/30. Isn't my method correct. What's the problem. And then for (iv) (P) of manufactured by A and faulty, from my venn diagram is 10/300 which is 1/30 but in the mark scheme the (P) of this question is (1/3). I think for (iv) the mark scheme is wrong. What do you think? Please help me. Thank you for your time and co-operation.

I'm on my phone at the moment so I can're read the question and respond very well.I'll be back at my laptop in an hour or two so I'll have a look then.

Don't worry,I love helping people with maths!

[SpiggyTopes]

(Original post by gabriel 41)
Okay there's no problem. I get you well. You've been so nice helping me around. I hope I'm not too annoying but I have another question that I would like you to help me in. This question is on probability and here's the question:

Q8 (a) An inspection carried out on 300 electronic items manufactured by two companies
(Company A and Company B) found that 30 were faulty. Of the 150 electronic
items inspected that were manufactured by Company A, 10 were found to be faulty.
In contrast, 20 were found to be faulty out of the 150 inspected electronic items
manufactured by Company B.
If one electronic item is randomly selected from these 300 electronic items, fi nd the
probability that this electronic item is:
(i) Manufactured by Company B; (3 marks)
(ii) Faulty; (3 marks)
(iii) Not faulty and is manufactured by Company B; (4 marks)
(iv) Manufactured by Company A, given that it is faulty. (4 marks)

So (i) and (ii) are very easy and I'm okay there but (iii) and (iv) is giving me the problem. I like solving probability questions using venn diagrams, and I've attached my venn diagram, see below. This venn diagram is my answer to (iii) and (iv). From my venn diagram you can see that (P) of faulty and manufactured by company B (answer to (iii)) is 20/300, so (P) of not faulty and manufactured by company B will be 1-(20/300) which is 14/15. However, this answer seems to be wrong and the mark schemes shows (P) of this question as 13/30. Isn't my method correct. What's the problem. And then for (iv) (P) of manufactured by A and faulty, from my venn diagram is 10/300 which is 1/30 but in the mark scheme the (P) of this question is (1/3). I think for (iv) the mark scheme is wrong. What do you think? Please help me. Thank you for your time and co-operation.

I personally don't like using venn diagrams- I think they complicate things.

I think you are making part iii) difficult for yourself. There are 130 items which aren't faulty and made by company B so the probability is just 130/300, or 13/30.

For part iv) the item removed is definitely faulty, so you are only considering the 30 faulty items, not all 300 items. Because 10 of these 30 items are made by A, the chance is 10/30, or 1/3.


Once again, I love helping people with maths! Ask me as many questions as you like- I need something to do in my summer holidays!

[SpiggyTopes]

(Original post by gabriel 41)
Okay there's no problem. I get you well. You've been so nice helping me around. I hope I'm not too annoying but I have another question that I would like you to help me in. This question is on probability and here's the question:

Q8 (a) An inspection carried out on 300 electronic items manufactured by two companies
(Company A and Company B) found that 30 were faulty. Of the 150 electronic
items inspected that were manufactured by Company A, 10 were found to be faulty.
In contrast, 20 were found to be faulty out of the 150 inspected electronic items
manufactured by Company B.
If one electronic item is randomly selected from these 300 electronic items, fi nd the
probability that this electronic item is:
(i) Manufactured by Company B; (3 marks)
(ii) Faulty; (3 marks)
(iii) Not faulty and is manufactured by Company B; (4 marks)
(iv) Manufactured by Company A, given that it is faulty. (4 marks)

So (i) and (ii) are very easy and I'm okay there but (iii) and (iv) is giving me the problem. I like solving probability questions using venn diagrams, and I've attached my venn diagram, see below. This venn diagram is my answer to (iii) and (iv). From my venn diagram you can see that (P) of faulty and manufactured by company B (answer to (iii)) is 20/300, so (P) of not faulty and manufactured by company B will be 1-(20/300) which is 14/15. However, this answer seems to be wrong and the mark schemes shows (P) of this question as 13/30. Isn't my method correct. What's the problem. And then for (iv) (P) of manufactured by A and faulty, from my venn diagram is 10/300 which is 1/30 but in the mark scheme the (P) of this question is (1/3). I think for (iv) the mark scheme is wrong. What do you think? Please help me. Thank you for your time and co-operation.

I think the easiest way I can explain this is by rewording the question...

A faulty item is chosen, what is the probability it is made by company A?

Now, you are only considering faulty items, not all 300 items.

So, there are 30 faulty items and 10 of them are made by company A. Therefore the probability a faulty item is made by Company A is 10/30, or 1/3.

[gabriel 41]

(Original post by SpiggyTopes)
I think the easiest way I can explain this is by rewording the question...

A faulty item is chosen, what is the probability it is made by company A?

Now, you are only considering faulty items, not all 300 items.

So, there are 30 faulty items and 10 of them are made by company A. Therefore the probability a faulty item is made by Company A is 10/30, or 1/3.

Okay so here's the another disturbing probability question I need help in. In this question every sub question is a problem and I scored 0/10 marks in it. Here's the question:

(b) Three men and four women are available to tackle a specific task which requires two
people.
If two people are chosen at random for the task, find the probability that there will be:
(i) no men,
(ii) one man,
(iii) two men. (10 marks)

So for part (i) this is what I did. No men means 1-P(men). So P(men) = 3/7, so 1-(3/7) = 4/7 but the answer says 2/7. How? Please comprehensively explain to me how to reach 2/7. Then for part (ii) the question ask for P(of one men) so that's 3/7 but the answer says 4/7. How and why? Please fully explain this two me. For (iii) the question ask for too men which is the same as 1-P(women) so P(women) is 4/7, so 1-(4/7) which gives 3/7. Again my answer is wrong and the correct answer is 1/7. Why? Please fully explain this to me. As you can see, I did very poorly in this question so please explain everything in a way that I can really understand. Thank you.

[gabriel 41]

(Original post by SpiggyTopes)
I think the easiest way I can explain this is by rewording the question...

A faulty item is chosen, what is the probability it is made by company A?

Now, you are only considering faulty items, not all 300 items.

So, there are 30 faulty items and 10 of them are made by company A. Therefore the probability a faulty item is made by Company A is 10/30, or 1/3.

There's yet another probability question that I'm totally blank in and I need you to take me through each step for this question because I'm totally blank in it. I'm so lost in this question that I didn't even attempt it. So I really need a comprehensive explanation from you for this question. I need help in all parts of the question because I'm very lost. Please find the question attached. Please give me a detailed explanation for each part of the question and take me through the question. Please find the answer attached too.

[gabriel 41]

So there's another small question I have. Please find the question attached as "probs question 7 2005". In this question (i), (ii) (iii) and (v) are very clear to me; however, (iv) is giving me a problem. The question says" (iv) is manufactured by company A or is defective so this is a or question, so I decided to add up the two probabilities together. I.e. P(A) = 150/300 AND P(DEFECTIVE) = 30/300. When I add the two, I get an answer of 0.6. But, this answer appears to be false and the correct answer is 0.57. Why is this so? Could you please explain this to me. Thank you very much for your help

[SpiggyTopes]

For the last question, there are 10 batteries which are made by company AND are faulty, you have counted these twice when fining your total number of batterries which are made by company A and are faulty.

I hope they help and please ask me as many as you like!

You do a lot of probability don't you!!

[gabriel 41]

(Original post by SpiggyTopes)
For the last question, there are 10 batteries which are made by company AND are faulty, you have counted these twice when fining your total number of batterries which are made by company A and are faulty.

I hope they help and please ask me as many as you like!

You do a lot of probability don't you!!

Thank you very very much! May you get all you desire and may God bless a good person like you.

[SpiggyTopes]

(Original post by gabriel 41)
Thank you very very much! May you get all you desire and may God bless a good person like you.

HaHa, thank you! Do they make sense? Any more questions?

[gabriel 41]

(Original post by SpiggyTopes)
HaHa, thank you! Do they make sense? Any more questions?

Yes I've got so many more questions ; however, I don't want to bug you regularly, so in those questions I try to apply what you teach me and try to figure out how the theories you gave me will work.


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