Archive for April, 2016

DSE Math 2016 Paper 2 ANS

Posted: April 12, 2016 in Mathematics

Cristiano Ronaldo’s HAT TRICK miracle saved the day tonight! So happy although I am not a Madrid fan. But I do like Ronaldo, Modric, and Sergio Ramos. Can’t sleep and a few more hours to get to work so I decided to post the solution to the exam earlier from this morning. This year’s DSE core math exam paper 2 again isn’t that difficult, although I had a little problem with the very last question (It’s a stat problem, I am not very good at stat and I didn’t do any review or studying before it). Anyhow I managed to finish the whole thing (45 questions) in about 70 minutes.

So let’s begin1

Q1) A

8^{222}\cdot5^{666}=2^{666}\cdot5^{666}=10^{666}

Q2) A

Multiple both sides by xy yields

ay+bx=3xy \Rightarrow x=\dfrac{ay}{3y-b}

Q3) D

16-(2x-3y)^2 = (4-2x+3y)(4+2x-3y)

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Q4) C

Obviously!

Q5) A

Solve the system of

4\alpha+\beta=5 \quad \text{and} \quad 7\alpha+3\beta=5

Multiple the first equation by 3 and then by the method of elimination,

\alpha=2 \Rightarrow \beta=-3

Q6) B

f(-\frac12)=4(-\frac12)^3+k(-\frac12)+3=0

\therefore k=5

\therefore f(-1)=4(-1)^3-5+3=-6

Q7) A

x<-7 \text{ and } x<3 \Rightarrow x<-7

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Q8) C

\Delta = k^2-4(8k+36) = k^2 - 32k - 4\cdot36 = 0

(k-36)(k+4)=0

\therefore k=36\text{ or } k=-4

Q9) D

Let’s say a=-\frac12, then the equation becomes

(\frac{x}{2}-1)^2-\frac12

So it’s a “happy face” or “open up” parabola with vertex (2, -\frac12). Then let x=0, we can find the y-int to be \frac12 which is positive.

Q10) C

D=1.25P \text{ and } P=0.75T

\therefore T=\dfrac{D}{1.25\cdot 0.75} = \dfrac{33360}{\frac54\cdot\frac34} = \dfrac{33360\cdot16}{15} =35584

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Q11) D

\dfrac{3y-4x}{2x+y}=\dfrac{5}{6}

18y-24x=10x+5y

13y=34x

\dfrac{x}{y}=\dfrac{13}{34}

Q12) D

z=\dfrac{k\sqrt{x}}{y}\quad \text{for some constant } k

\dfrac{k\sqrt{0.64x}}{1.6y}=\dfrac{0.8k\sqrt{x}}{1.6y}=\dfrac{1}{2}z

Q13) A

3(42)+2y=36(5) \Rightarrow y=27

Q14) C

t_7=9+6\cdot5=39

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Q15) B

Add a parallel line in the middle then it is obvious that

180-c+a+b=360 \Rightarrow a+b-c=180

and obviously I and III are wrong!

Q16) D

By inspection, notice that

AB=24=3\times8

BD=32=4\times8

AD=40=5\times8

which form a 3-4-5 right triangle and hence \angle DBC is a right angle as well. By Pythagoras, we have

BC=\sqrt{68^2-32^2}=4\sqrt{17^2-8^2}=4\sqrt{289-64}=60

Q17) A

\angle CBE = \angle BCE = 180-114=66

\therefore \angle ABE = 114 - 66 = 48

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Q18) C

It’s obvious that there is a 5-12-13 right triangle, hence the volume is

\dfrac{(4+4+12)5}{2}\cdot6 = 300

Q19) A

Let \angle AOB = \theta

(39^2\pi-33^2\pi)\dfrac{\theta}{360}=72\pi

\dfrac{9(13^2-11^2)\theta}{360}=72

\theta = \dfrac{72\cdot40}{169-121} = 60

So, I is right, now we can ignore II and check III. If you are smart, you don’t need to calculate the perimeter to know III must be wrong, because the arc length must be an integer over 6 times \pi and the two radii add to an integer. The perimeter must be in the form a\pi+b where a\in\mathbb{Q}, b\in\mathbb{Z}.

Q20) C

By mid-point theorem, we can divide those two region as follow and the answer comes out clearly!

tmp1

7

Q21) B

Join BE such that point E is on AD and line BE is parallel to the line CD. then

AD = AE+ED = AB\cos a + BC \sin c

Q22) D

Since rhombus is a parallelogram, \angle C = 180-118 = 62. And then by inscribed angle is half the central angle, \angle E = 31. Therefore

DFE = 180-31-(180-118) =87

Q23) A

Obviously!

8

Q24) B

180(n-2)=3240 \Rightarrow n=20, so A cannot be right. 360\div20=18, hence B is right!

Q25) D

The two lines are perpendicular hence

\dfrac{h}{k}=-\dfrac{3}{4} \Rightarrow h=-\dfrac{3k}{4}

Intersection lies on x-axis means

4x=5 \Rightarrow x=\dfrac54

\therefore -\dfrac{3k}{4}\cdot\dfrac{5}{4}+15=0

\therefore k=16

Q26) B

A(9,-2), B(-1,8), C(x,y), \quad x=2y

(x-9)^2+(y+2)^2 = (x+1)^2+(y-8)^2

x^2-18x+81+y^2+4y+4=x^2+2x+1+y^2-16y+64

20y+20=20x

2y+2=2x

x+2=2x

x=2

Q27) C

By completing the square,

3(x^2-4x+4-4)+3(y^2+10y+25-25)+65=0

3(x-2)^2-12 + 3(y+5)^2-75 + 65 = 0

(x-2)^2 + (y+5)^2 = \dfrac{22}{3}

r=\sqrt{\dfrac{22}{3}} \approx 2.7

Hence C is a circle with radius about 2.7 centered at (2,-5).

So II and III are right!

9.jpg

Q28) C

Obviously!

Q29) B

0.1\times90 + 0.3\times 20 + 0.6\times 10 = 9+6+6=21

Q30) B

Since mode is 68 and there are two 98 already,

so the data are 32 , 68 , 68 , 68 , 79 , 86 , 88 , 98 , 98 , unknown.

Since the mean is 77 therefore the unknown is 770-685 = 85

Therefore the median is average of 79 and 85 which is 82.

10.jpg

Q31) C

Obviously

Q32) D

\log_9 y = \dfrac12 x - 2

y=9^{\frac{x}{2}-2} = \dfrac{3^x}{81}

\therefore b=3

Q33) A

11\times16^{12}+12\times16^{11}+13\times16^7+14\times16^6

= (11\cdot16+12)\times16^{11} + (13\cdot16+14)\times16^6

= 188\times16^{11} + 222\times16^6

11.jpg

Q34) B

uv = \dfrac{49}{a^2+1}

Since a can be any real number such as \pi. \dfrac{49}{a^2+1} may not be rational. II doesn’t need to be checked, just check III. III is obviouly wrong. So answer is B.

Q35) D

Let the value be

t=7y-5x+3 \Rightarrow y=\dfrac{5}{7}x + \dfrac{t-3}{7}

Draw a family of lines with slope \dfrac{5}{7} and the line that touches the feasible set with the maximum value of t is at S obviously.

Q36) B

r^4 = \dfrac{189}{21} = 9 \Rightarrow r=\pm\sqrt3

I is wrong. So the answer is either B or D, so we can ignore II and check III.

If r=\sqrt3, then

S_{99} = \dfrac{7((\sqrt3)^{99}-1)}{\sqrt3-1}\approx 3.96 \times 10^{24}

However if r=-\sqrt3, then

S_{99} = \dfrac{7((-\sqrt3)^{99}-1)}{-\sqrt3-1}\approx 1.06 \times 10^{24}

Therefore only II must be true.

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Q37) A

\cos 2x means cosine curve is compressed horizontally by a half, so the period is 180 degrees and therefore b=90. Amplitude is obviously 2 and the graph is flipped vertically, so a must be negative and hence a=-2.

Q38) B

(5 \sin\theta - 4)(\sin \theta + 1)=0

Two solution for the first bracket and one solution for the second bracket.

Q39) A

By 3-4-5 right triangle,

AC=20 \Rightarrow PQ=\sqrt{181}

By 3-4-5 right triangle,

FQ=25

By 5-12-13 right triangle,

FP=26

By cosine law,

PQ^2 = FQ^2 + FP^2 - 2 FQ \cdot FP \cos \angle PFQ

\cos \angle PFQ = \dfrac{25^2+26^2-181}{2\cdot25\cdot26} = \dfrac{56}{65}

\therefore \sin\angle PFQ = \sqrt{1-\left(\dfrac{56}{65}\right)^2} = \dfrac{33}{65}

13.jpg

Q40) D

By inscribed angle of a semi-circle,

\angle ABC = 90^\circ

Let the center of the circle be O, then in quadrilateral OBPD,

\angle BOD = 360^\circ - 90^\circ - 90^\circ - 68^\circ = 112^\circ

By inscribed angle is half the central angle,

\angle BAD = 56^\circ

By angle sum of a triangle,

\therefore \angle AQB = 180^\circ - 90^\circ - 56^\circ = 34^\circ

Q41) C

Plug y=2x-6 into x^2+y^2-8y-14=0 gives

x^2 - 8x + 14=0

x=4\pm\sqrt2

So the midpoint is 4, then plug x=4 into 2x-y-6=0 gives

y=2

Q42) A

\displaystyle \frac{\binom{3}{2} \binom{9}{2}+\binom{3}{3}\binom{9}{1}}{\binom{12}{4}} = \frac{3\cdot\frac{9\cdot8}{2}+9}{\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}} = \frac{13}{55}

14

Q43) D

\displaystyle \binom{15}{0} \binom{20}{6} + \binom{15}{1}\binom{20}{5} + \binom{15}{2}\binom{20}{4} = 780045

Q44) B

I is obviously wrong, 55 is lower quartile. Only need to check III. Use a calculator to find the standard deviation is about 11.58. So II only.

Q45) C

Because I forgot \mathrm{Var}(aX+b) = a^2 \; \mathrm{Var}(X) (later I read the book). I still remember what standard deviation is. I spent like 10 minutes to do this. Notice that

\mathrm{Var}(4X+9) = \sigma^2 = \dfrac{\sum(4 x_i+9-(4\mu+9))^2}{n}

= 16\dfrac{\sum(x_i-\mu)^2}{n} = 16 \; \mathrm{Var}(X)

Therefore variance of the new set of number is 49\times16=784.

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