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compound interest question help

ayumi

I have tried doing this but its confusing so I wanted to know exactly the right way to do it.
its ord maths mock 2008

5000 was invested at compoud interest.
rate of interest for the first two yrs was 2.5% per annum.
tax is deducted from the interest earned each year at a rate of 20%

i)calculate the value of the investment at the end of the second yr
then the next part asked:

at the beginning of the third year a sum of money was withdrawn.
rate of interest for third yr was 2.75% per annum.
tax rate remained unchanged.
at the end of the third yr the total investment amounted to 4088.

II) calculate the sum of money withdrawn at the beginning of the third yr

I wish that anyone can help me.thank you

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eoins23456

5000x0.025=125 for interest
deduct 20 percent 125/5=25x4=100
Amount after first year=5100

Do the same process again for the second year
5100x0.025=127.50/5=25.50x4=102
5202 amount after second year

eoins23456

think the answer to the second one is 1202.now try and work towards that answer

ayumi

this mock paper is hard,
I have beening getting - answers all over the place!
one q asked to express z in terms of x & y where x=3y-2z allover z+5 hence evalute z when x=1/2 and y=1/3 ,answe form a/b
and find the local minimum and maximum points,
x cubed -5x squared+3x-2
and w=3-i express 1/w +w in form x+yi
hence find real nos p & q such that
p(1/w +w)+3qi=33

i wish i have been not asking alot,I have tried doing these qs and going to the book but its no use!

MathsManiac

Just think about the third year: if the government is going to take a fifth of the 2.75%, then the person is left with four fifths. So this is just the same as if the interest rate was 2.2% without any tax. (2.75 X 0.8 = 2.2)

So to get from the end of the third year back to the start of the third year:

??? X 102.2% = 4088
=> ??? = 4088 / 1.022 = 4000

The difference between that and your answer to the earlier part is the answer.

MathsManiac

ayumi wrote: »

one q asked to express z in terms of x & y where x=3y-2z allover z+5 hence evalute z when x=1/2 and y=1/3 ,answe form a/b

Multiply both sides by z+5 to clear the fraction.
simplify, and then bring the terms with z in them to the left, and other terms to the right.
Then z can be factored out on the left.
Then you can divide across by the other factor.

Once you have z in terms x and y, the last part can be done on a calculator (carefully!)

MathsManiac

ayumi wrote: »

this mock paper is hard,
Find the local minimum and maximum points,
x cubed -5x squared+3x-2

The max and min points will be at the values of x for which the derivative is zero.

Differentiate y to get dy/dx = 3x^2 - 10x + 3.

Set this equal to 0 and solve using factors or the formula to get x=3 or x=(1/3).

Substitute each of these into the original equation to get the corresponding y-values.

For a cubic function, the max. point has to be higher than the min. point, so whichever of the two points has the highest y-value is the max.

MathsManiac

ayumi wrote: »

w=3-i express 1/w +w in form x+yi
hence find real nos p & q such that
p(1/w +w)+3qi=33

To find 1/w, which is 1/(3-i), you multiply above and below by the conjugate: (3+i). When you simplify above and below, the bottom is just the real number 10, so you can divide it in. Then add on w (which is 3-i) to get the answer for 1/w+w.

Once that's done, then to do the next part, substitute in your answer for w+1/w and then multiply out the bracket.

Then equate the real terms on both sides the equation (the terms without an i) and equate the imaginary terms on both sides. From this, you should be able to find p first and then find q.

Post back after trying these, and you'll get more help if you need it.

ayumi

i thank yous all for the help,Ill be doing these later today and I write back

ayumi

MathsManiac wrote: »

To find 1/w, which is 1/(3-i), you multiply above and below by the conjugate: (3+i). When you simplify above and below, the bottom is just the real number 10, so you can divide it in. Then add on w (which is 3-i) to get the answer for 1/w+w.

Once that's done, then to do the next part, substitute in your answer for w+1/w and then multiply out the bracket.

Then equate the real terms on both sides the equation (the terms without an i) and equate the imaginary terms on both sides. From this, you should be able to find p first and then find q.

Post back after trying these, and you'll get more help if you need it.

i got 3-i/10 soo what do i do next?
i was confused by simplify above and below

ayumi

MathsManiac wrote: »

The max and min points will be at the values of x for which the derivative is zero.

Differentiate y to get dy/dx = 3x^2 - 10x + 3.

Set this equal to 0 and solve using factors or the formula to get x=3 or x=(1/3).

Substitute each of these into the original equation to get the corresponding y-values.

For a cubic function, the max. point has to be higher than the min. point, so whichever of the two points has the highest y-value is the max.

i got (3,-11) and (1/3,-41/27)
and then it asks to write the range of values of x for which f(x) is decreasing and explain using diagram or otherwise why f(x)=0 has one solution.
thanx mathsmaniac for putting your time and helping me

ayumi

eoins23456 wrote: »

think the answer to the second one is 1202.now try and work towards that answer

i didnt get that

but why did you divide by 5 and multiplied by 4?
i got 1228.444
by mutipling 5202*2.75% to get 143.055
then divide by 5 and multipling by 4 to get 114.444
then i added it to 5202 to get 5316.444
then i subtracted from 4088 to get 1228.444?

MathsManiac

ayumi wrote: »

i got 3-i/10 soo what do i do next?
i was confused by simplify above and below

When you multiply 1/(3-i) above and below by 3+i, the top of the fraction is 1(3+i) and the bottom is (3-i)(3+i). When you simplify the top, you just get 3+i and when you simplify the bottom you get 9 + 3i - 3i + i^2, which equals 10. So you have (3+i)/10. This is the same as (3/10) + (1/10) i, which means that you have written it in the right form. (That is, x is 3/10 and y is 1/10.)

For the next part, we now have:
p[(3/10) + (1/10) i] = 3qi = 33.

In an equation with complex numbers, the real parts are equal and the imaginary parts are equal. So we have:

(3/10)p = 33 and (1/10)p+3q=0.

The first of these tells you that p=110. Once you know that, you substitute it into the second one and get q=-11/3.

MathsManiac

ayumi wrote: »

i got (3,-11) and (1/3,-41/27)
and then it asks to write the range of values of x for which f(x) is decreasing and explain using diagram or otherwise why f(x)=0 has one solution.
thanx mathsmaniac for putting your time and helping me

I assume you have drawn graphs of cubic functions and are therefore familiar with their shape. Look it up in your book if you don't remeber. Plot your two points and note that the max and min points are where the curve changes from going up to going down (as you move left to right). So the curve comes from bottom left of the diagram UP to the point (1/3,-41/7), then it goes DOWN to (3,-11) and then goes UP again and disappears off to the top right.

The function is decreasing when it's going down, so it's decreasing for the bit between 1/3 and 3. Ans: 1/3 < x < 3.

If you've drawn the graph, you'll see that the max. point and the min. point are both below the x-axis. This means that the curve can only cross the x-axis once. The solutions of f(x)=0 (which is the same as y=0) are the places where the curve crosses the x-axis, so there is only one solution. (I fthe max was above the x-axis and the min was below it, then the curve would cross it three times, and there would be three solutions.)

MathsManiac

ayumi wrote: »

i didnt get that
but why did you divide by 5 and multiplied by 4?
i got 1228.444
by mutipling 5202*2.75% to get 143.055
then divide by 5 and multipling by 4 to get 114.444
then i added it to 5202 to get 5316.444
then i subtracted from 4088 to get 1228.444?

The problem is that you can't get 2.75% of 5202, because some of the money was taken out before the third year started, so you don't know how much to get 2.75% of.

My earlier post (5th post in the thread) showed you how to work back from the end of the third year to the start of the third year.

Since we now know that there was €4000 starting into the third year, but that there was €5202 at the end of the second year, the person must have taken out €1202 at the end of the second year.

seandoiler

MathsManiac wrote: »

The solutions of f(x)=0 (which is the same as y=0) are the places where the curve crosses the x-axis, so there is only one solution. (I fthe max was above the x-axis and the min was below it, then the curve would cross it three times, and there would be three solutions.)

Real solution