Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Could someone please help with the question posted on top of this page. Cheers.

InteGrand · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33305

Dont know how to answer part d)

Trying to show there is one pair of parallel lines i.e. DC parallel to AB

Answers says cointerior angles are supplementary.

appleibeats said:
Could someone please help with the question posted on top of this page. Cheers.

$\noindent Here's one way to do it. Join $AC$ and $BD$ and let their point of intersection be $X$. It follows from part (b) that $AX = BX$ (equal sides in isosceles triangle $ABX$). Let $\angle XAB = x = \angle XBA$. Let $\angle BXA = \alpha$, so $\alpha + 2x = 180^\circ$ (angle sum of a triangle). Now, $\angle CXD = \alpha$ (vertically opposite to angle $BXA$). Also, $CA = DB$ (corresponding sides of congruent triangles from part (a)). Since $AX=BX$, we must then have that $XC = XD$. Thus in $\triangle XCD$, the equal base angles $D$ and $C$ must each be $x$, as the angle sum is $180^\circ$ and the other angle is $\alpha$, and $\alpha + 2x = 180^\circ$. So we have $\angle CDB = x = \angle DBA$. So $AB\parallel CD$ (alternate angles $CDB$ and $DBA$ are equal), and this proves that $ABCD$ is a trapezium.$

appleibeats · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thank you

Screen Shot 2016-07-01 at 1.06.49 PM.png

I don't know how to do part a). Answers say AAS

and the first part of part b), the show bit.

Thanks.

InteGrand · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Thank you

View attachment 33309

I don't know how to do part a). Answers say AAS

and the first part of part b), the show bit.

Thanks.

$\noindent a) Note that $\angle AGD = \angle HGB$ (vertically opposite angles). Also, $\angle DAG = \angle BHG = 90^\circ$ (angles of a rectangle). Finally, $AD = HB$ (corresponding sides of congruent rectangles are equal). Hence $\triangle ADG \equiv \triangle HBG$ (AAS).$

$\noindent b) (First part) Note that by Pythagoras' Theorem in $\triangle ADG$, we have $AG^2 = DG^2 - AD^2$. But $DG=BG$ (corresponding sides of the congruent triangles in part a)). So $DG = BG = AB - AG$. So $AG^2 = \left(AB-AG\right)^2 - AD^2\Rightarrow AG^2 = AB^2 - 2AB\cdot AG + AG^2 - AD^2$ (expanding the square). Cancelling $AG^2$, we have $AB^2 - 2AB \cdot AG -AD^2 = 0$. Rearranging this yields $AG = \frac{AB^2 - AD^2}{2AB}$, as required.$

leehuan · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Thank you

View attachment 33309

I don't know how to do part a). Answers say AAS

and the first part of part b), the show bit.

Thanks.

You have vertically opposite angles, and two angles that both equal to 90deg because they belong in a rectangle
Given that they are congruent rectangles, you also have AD=BC=DF=HB
That's enough to prove congruency

AG^2 + AD^2 = DG^2 taking the hint
What is DG

DG = GB from congruent triangles
AG^2 + AD^2 = GB^2

RHS = GB^2 = (AB-AG)^2 = AB^2 - - 2AB.AG + AG^2

So we are saying
AG^2 + AD^2 = AB^2 - 2AB.AG + AG^2
Therefore
2AB.AG = AB^2 - AD^2 and divide through by 2AB.

Always be prepared to use difference of sides.

Oops, IG already did it.

appleibeats · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-07-01 at 7.04.14 PM.png

Did part b) however got a different answer to the answers in the book.

Could someone see if they get my answer of 4root91 cm

The answers in the book says 4root11 cm .

Thanks

InteGrand · Jul 1, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33310

Did part b) however got a different answer to the answers in the book.

Could someone see if they get my answer of 4root91 cm

The answers in the book says 4root11 cm .

Thanks

I think it should be what you got (4*sqrt(91) cm).

I.e. 2*sqrt(20^2 – 6^2), which follows from Pythagoras's Theorem and the fact that diagonals of a rhombus bisect each other at right angles.

appleibeats · Jul 3, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

If you have an parabola that is partly under the x axis and the shaded area is the part under the x axis and you were told the Area was 20. Is that area positive 20 or negative 20 . I thought it was positive 20 but the apparently since its below the axis it is -20.

Could someone explain why?

InteGrand · Jul 3, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
If you have an parabola that is partly under the x axis and the shaded area is the part under the x axis and you were told the Area was 20. Is that area positive 20 or negative 20 . I thought it was positive 20 but the apparently since its below the axis it is -20.

Could someone explain why?

The area is positive. The signed area (which is the integral over that interval of the function) is negative.

What is the question's wording?

appleibeats · Jul 3, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

There was a sketch of a parabola and it says the parabola has this equation y = kx^2 .... where k is a positive constant.

The shaded region has area 15 u^2
Find the value of k

I get it now as since were finding an integral it is the signed area were looking for.

appleibeats · Jul 4, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-07-04 at 1.38.03 PM.png

Need help with part b). Not sure how to use the similar triangles to find x

InteGrand · Jul 4, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33317

Need help with part b). Not sure how to use the similar triangles to find x

$\noindent Note $AD = AB = x$ (sides of isosceles triangle $BDA$, which is isosceles as it is similar to isosceles triangle $ABC$). You should be able to show from easy angle-chasing that $\triangle ADC$ is isosceles with $AD=CD$ (show that the relevant angles in this triangle are equal to show that it is isosceles). So $CD = AD = x \Rightarrow BD = CB - CD=1-x$. From similar triangles, we have $\frac{AB}{AC}=\frac{BD}{BA}$, i.e. $\frac{x}{1}=\frac{BD}{x}\Rightarrow BD = x^2$. Since $BD = 1-x$, we have $1-x=x^2 \Rightarrow x^2 + x -1 = 0$. So $x=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$. Since $x>0$, $x=\frac{-1+\sqrt{5}}{2}$.$

marxman · Jul 4, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Chapter 10E Q14 c) Cambridge Year 12 Ext 1:

If repetitions are not allowed, how many EVEN four-digit numbers can be formed from the
digits 1, 2, . . . 8, 9?

leehuan · Jul 4, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

marxman said:
Chapter 10E Q14 c) Cambridge Year 12 Ext 1:

If repetitions are not allowed, how many EVEN four-digit numbers can be formed from the
digits 1, 2, . . . 8, 9?

I'm assuming 0 was intentionally excluded.

$\text{Fix the last digit first. The last digit has to be either }2, 4, 6\text{ or }8.\\ \text{There are 4 possible numbers we can choose from.}$

$\text{But then the other three digits can be any of the remaining eight numbers, without repetition. Clearly, there will be }8\times 7\times 6 = ^8 P_3\text{ choices.}$

$4\times8\times7\times6 = 1344$

marxman · Jul 4, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thanks!

appleibeats · Jul 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-07-05 at 10.30.11 AM.png

Did this question and got the monthy repayments being $360 to the nearest dollar

The answers says $345

Not sure if the answer are correct or I am.

leehuan · Jul 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I am getting $345

$A_n=15000(1.01125)^{n} - M\left(1+1.01125+...+1.01125^{n-1}\right)$

$A_{60}=0$

$15000(1.01125)^{n}-M\left(\frac{1.01125^{60}-1}{0.01125}\right)$

InteGrand · Jul 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33319

Did this question and got the monthy repayments being $360 to the nearest dollar

The answers says $345

Not sure if the answer are correct or I am.

What was your working?

appleibeats · Jul 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i had let Tn be the amount owed after n years and M be the montly repayments

so T1 = 15000 (1.135) - 12M
T2 = T1(1.135) - 12M
= 15000(1.135)^2 - 12M(1.135) - 12M
.....
T5 = 15000(1.135)^5 - 12M(1.135^4 + 1.135^3 + ... + 1)

and then i made T5 = 0

and found M to be 360

Why is this incorrect ?

leehuan · Jul 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
i had let Tn be the amount owed after n years and M be the montly repayments

so T1 = 15000 (1.135) - 12M
T2 = T1(1.135) - 12M
= 15000(1.135)^2 - 12M(1.135) - 12M
.....
T5 = 15000(1.135)^5 - 12M(1.135^4 + 1.135^3 + ... + 1)

and then i made T5 = 0

and found M to be 360

Why is this incorrect ?

Months. Not years.

You did not cater for the fact that interest is being compounded monthly in doing 1.135

Edit: I will give you a bit of the benefit of the doubt though - the question didn't explicitly say interest was compounded monthly either.

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