Maths question from 2011 hsc Mathmatics.

[girlworld_club]

I am a prelim student, trying to do some hsc work so excuse the unfamiliarity.

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.
(i)
What will be the cost of building the 25th floor?
2
(ii)
What will be the cost of building all 110 floors of the skyscraper?


To find out my answer i re-arranged to make this formula:
3,000,000 + (f x 500000) - 500000
where f = the number of floors

is this the right approach to answering the question, or does it produce the wrong answer? any other ways of doing it?

[bleakarcher]

Cost of the nth floor C(n)=3+(n-1)0.5=3+0.5n-0.5=0.5n+2.5
i) When n=25, => C(25)=0.5*25+2.5=$15 million
Now you can do part ii)

[girlworld_club]

Cost of the nth floor C(n)=3+(n-1)0.5=3+0.5n-0.5=0.5n+2.5
i) When n=25, => C(25)=0.5*25+2.5=$15 million
Now you can do part ii)

How did you figure this our? and also i tried using no decimal points to determine cost of second floor and got this

(c)2 = 5000000x25+25000000 = a number which is completely off 3,500,000

[girlworld_club]

I also have another question:
The number of members of a new social networking site doubles every day. On Day 1 there were 27 members and on Day 2 there were 54 members.
(i)
How many members were there on Day 12?
1
(ii)
On which day was the number of members first greater than 10 million?
2
(iii)
The site earns 0.5 cents per member per day. How much money did the site earn in the first 12 days? Give your answer to the nearest dollar.

[RealiseNothing]

I am a prelim student, trying to do some hsc work so excuse the unfamiliarity.

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.
(i)
What will be the cost of building the 25th floor?
2
(ii)
What will be the cost of building all 110 floors of the skyscraper?


To find out my answer i re-arranged to make this formula:
3,000,000 + (f x 500000) - 500000
where f = the number of floors

is this the right approach to answering the question, or does it produce the wrong answer? any other ways of doing it?

For the first one, the first floor cost 3mil, then add 0.5mil for each floor after. So when we get to the 25th floor, we should of added 12mil onto the original 3mil. Hence answer is 15mil.

The second one, we find the sum of all the floors, which will be:

3+3.5+4+4.5+5+5.5......... (left out the mil to make it easier to type)

Since there is 110 floors to be built, this will go up to 57.5mil.

Let's split it up:

3+4+5+...+57 which is (55/2)(3+57) = 22.5x60 = 1350

Now we have the 3.5 + 4.5 + ... + 57.5

Notice this is the same as the previous series just with .5's on the end? Let's make it that then:

3 + 0.5 + 4 + 0.5 .... + 57 + 0.5

So we know from the last series that it equals 1350, so add this on to get 2700.

Now we just add on the amount of 0.5's we have, that is 0.5x55=27.5

So the answer is 2700 + 27.5 = 2727.5 million or 2.7275 billion

[girlworld_club]

thanks but it seems extremely tedious to add one by one, isn't there a formula?

[Alkanes]

Dw if you can't do it. This is a year 12 topic, you won't learn it until next year.

[SpiralFlex]

thanks but it seems extremely tedious to add one by one, isn't there a formula?

No. No. No. No. No. No. No. Bad human.

[girlworld_club]

But these sort of questions don't really require you to "learn" anything- such as formulas and what not, these questions are more on the basis of your understanding.

[girlworld_club]

No. No. No. No. No. No. No. Bad human.

?

[bleakarcher]

For the first one, the first floor cost 3mil, then add 0.5mil for each floor after. So when we get to the 25th floor, we should of added 12mil onto the original 3mil. Hence answer is 15mil.

The second one, we find the sum of all the floors, which will be:

3+3.5+4+4.5+5+5.5......... (left out the mil to make it easier to type)

Since there is 110 floors to be built, this will go up to 57.5mil.

Let's split it up:

3+4+5+...+57 which is (55/2)(3+57) = 22.5x60 = 1350

Now we have the 3.5 + 4.5 + ... + 57.5

Notice this is the same as the previous series just with .5's on the end? Let's make it that then:

3 + 0.5 + 4 + 0.5 .... + 57 + 0.5

So we know from the last series that it equals 1350, so add this on to get 2700.

Now we just add on the amount of 0.5's we have, that is 0.5x55=27.5

So the answer is 2700 + 27.5 = 2727.5 million or 2.7275 billion

Dude how did your answer end up in the billions?

[bleakarcher]

the answer to ii should be 57.5 million dollars.

[girlworld_club]

the answer to ii should be 57.5 million dollars.

and how did you get that?

[RealiseNothing]

the answer to ii should be 57.5 million dollars.

It's the cost of all floors, not just the last one.

[IamBread]

the answer to ii should be 57.5 million dollars.

Its building ALL the floors. It's 57.5 million to build the 110th floor, but we need to add on the cost of building all the previous ones.

[bleakarcher]

I am a prelim student, trying to do some hsc work so excuse the unfamiliarity.

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.
(i)
What will be the cost of building the 25th floor?
2
(ii)
What will be the cost of building all 110 floors of the skyscraper?


To find out my answer i re-arranged to make this formula:
3,000,000 + (f x 500000) - 500000
where f = the number of floors

is this the right approach to answering the question, or does it produce the wrong answer? any other ways of doing it?

You will notice that for the second floor the cost would have 3+(2-1)0.5 (million), the third floor 3+(3-1)0.5 (million), fourth floor would have 3+(4-1)0.5 so it follows that for the nth floor the cost would be 3+(n-1)0.5 million.

[bleakarcher]

Ah shit, disregard me didnt read the question properly.

[girlworld_club]

ya! so anyone know how to answer it without adding all the floors up one by one.

[Gigacube]

I think this is the formula:



n is the number of terms
a is the first term
l is the last term

This is a question from the topic Series.

[Genero]

It's an arithmetic series so you know Tn = a (n-1)d where a is the first term, d is the difference and n is the number of terms. To get the sum of the first n terms, use Sn = n/2(a + l). Those are the formulas since you were looking for one. It's a topic called Series and Sequences if you're interested.

[girlworld_club]

I think this is the formula:



n is the number of terms
a is the first term
l is the last term

This is a question from the topic Series.

Do you learn the topic series in year 12 or ll?

[Gigacube]

Do you learn the topic series in year 12 or ll?

I learnt it in year 12. Which textbook does your school use?

[girlworld_club]

I learnt it in year 12. Which textbook does your school use?

maths in focus. I just decided to do some questions from hsc 2011 ( it looked very easy at first, like that sort of sc question that you make up your own formula to solve a problem). But everything i tried it didn't work.

[Gigacube]

maths in focus. I just decided to do some questions from hsc 2011 ( it looked very easy at first, like that sort of sc question that you make up your own formula to solve a problem). But everything i tried it didn't work.

If you use the old MiF then you'll learn part in year 11 & part in year 12. If you use the newer edition of MiF then you'll learn it in year 12.

I found Series one of the easier topics & it is very useful. It shouldn't be too hard to learnt it by yourself. Just do some practice questions on series & you'll get the hang of it. I think you should be able to do most of the topic with the year 11 knowledge so far. You may need to use logs for a few questions.

[girlworld_club]

If you use the old MiF then you'll learn part in year 11 & part in year 12. If you use the newer edition of MiF then you'll learn it in year 12.

I found Series one of the easier topics & it is very useful. It shouldn't be too hard to learnt it by yourself. Just do some practice questions on series & you'll get the hang of it. I think you should be able to do most of the topic with the year 11 knowledge so far. You may need to use logs for a few questions.

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