Author Topic: 3U Maths Question Thread (Read 1237954 times) Tweet Share

shreya_ajoshi · « **Reply #1515 on:** February 18, 2017, 07:30:37 pm »

Quote from: RuiAce on February 18, 2017, 07:23:41 pm

$\text{The equation factorises to }y=3x(x-1)(x+1)\\ \text{so we know where our }x\text{-intercepts are, making our cubic easy to graph}\\ \text{The equation also expands to }y=3x^3-3x^2$
(Image removed from quote.)
$\text{Clearly, there is a split at }x=1\\ \text{Hence the area is given by}\\ \left|\int_0^1 (3x^3-3x^2)\,dx\right|+\int_1^3 (3x^3-3x^2)\,dx$

ooooh yes thank you so much!!!

K9810 · « **Reply #1516 on:** February 18, 2017, 09:58:20 pm »

Hey, need help with these

12. Twenty students, ten male and ten female, are to travel from school to the sports ground. Eight of them go in a minibus, six of them in cars, four of them on bikes and two walk. (b) In how many ways can they be distributed if none of the boys walk?

22. Nine players are to be divided into two teams of four and one umpire. (a) In how many ways can the teams be formed? (b) If two particular people cannot be on the same team, how many diﬀerent combinations are possible?

5. The letters A, E, I, P, Q and R are arranged in a circle. Find the probability that: (a) the vowels are together

8. Find how many arrangements of n people around a circle are possible if:(d) three particular people sit together.

15. A tank contains 20 tagged ﬁsh and 80 untagged ﬁsh. On each day, four ﬁsh are selected at random, and after noting whether they are tagged or untagged, they are returned to the tank. Answer the following questions, correct to three signiﬁcant ﬁgures. (a) What is the probability of selecting no tagged ﬁsh on a given day? (b) What is the probability of selecting at least one tagged ﬁsh on a given day? (c) Calculate the probability of selecting no tagged ﬁsh on every day for a week. (d) What is the probability of selecting no tagged ﬁsh on exactly three of the seven days during the week?

Ty

kiwiberry · « **Reply #1517 on:** February 18, 2017, 10:34:11 pm »

Quote from: K9810 on February 18, 2017, 09:58:20 pm

5. The letters A, E, I, P, Q and R are arranged in a circle. Find the probability that: (a) the vowels are together

a) Total arrangements = 5! (because it's a circle not a line)
Now, treating the vowels as one group AEI - this can be arranged in 3! ways
The groups AEI, P, Q, R can be arranged in 3! ways in a circle
So arrangements with vowels together = 3! x 3!
∴ Probability = (3! x 3!)/5!

Quote from: K9810 on February 18, 2017, 09:58:20 pm

8. Find how many arrangements of n people around a circle are possible if:(d) three particular people sit together.

if 3 people sit together, they can be arranged in 3! ways
The remaining n-3 people can be arranged in (n-3)! ways
∴ 3!(n-3)!

RuiAce · « **Reply #1518 on:** February 18, 2017, 10:45:33 pm »

Quote from: K9810 on February 18, 2017, 09:58:20 pm

Hey, need help with these

12. Twenty students, ten male and ten female, are to travel from school to the sports ground. Eight of them go in a minibus, six of them in cars, four of them on bikes and two walk. (b) In how many ways can they be distributed if none of the boys walk?

22. Nine players are to be divided into two teams of four and one umpire. (a) In how many ways can the teams be formed? (b) If two particular people cannot be on the same team, how many diﬀerent combinations are possible?

5. The letters A, E, I, P, Q and R are arranged in a circle. Find the probability that: (a) the vowels are together

8. Find how many arrangements of n people around a circle are possible if:(d) three particular people sit together.

15. A tank contains 20 tagged ﬁsh and 80 untagged ﬁsh. On each day, four ﬁsh are selected at random, and after noting whether they are tagged or untagged, they are returned to the tank. Answer the following questions, correct to three signiﬁcant ﬁgures. (a) What is the probability of selecting no tagged ﬁsh on a given day? (b) What is the probability of selecting at least one tagged ﬁsh on a given day? (c) Calculate the probability of selecting no tagged ﬁsh on every day for a week. (d) What is the probability of selecting no tagged ﬁsh on exactly three of the seven days during the week?

Ty

In the future, please avoid immediately posting a whole bundle of questions all at once.
$\textit{Deal with the restriction first}\\ \text{The boys only have 18 forms of transport to choose from: }^{18}P_{10}\\ \text{Then the girls can choose any of the remaining 10: }10!\\ \text{Answer: }(18\times 17\times 16\times \dots \times 9)\times 10!$
__________
$\text{Choose the umpire: }\binom91\\ \text{Then choose 4 out of 8 for the first team, and 4 out of 4 for the second team:}\binom84\binom44\\ \text{However, the teams are of equal size. We have repetition.}\\ \text{This is because we can just swap the teams around to get the same outcome}\\ \text{Hence, we must cancel out the repetition factor: }2!\\ \text{Answer: }\frac{\binom91 \binom84\binom44}{2!}$
$\textit{Two possible cases for part b):}\\ \text{Case 1: One of them is the umpire}\\ \text{Then there is no way for the two to be on the same team.}\\ \text{Similar to part a), this yields }\frac{\binom84\binom44}{2!}$
$\text{Case 2: Neither is the umpire}\\ \textit{Deal with the restriction first}$
$\text{Place the first player in one of two teams: }2\\ \text{Then the other player is forced into the other team: }1\\ \text{Of the remaining players, take 1 to be the umpire: }\binom71\\ \text{Then, choose 3 out of 6 players to be on team 1:} \binom63\\ \text{Then, choose 3 out of 3 players to be on team 2:}\binom33\\ \text{Since the teams are equal sized, we still have to cancel out the repetition factor of }2!\\ \text{So this case yields }\frac{2\times 1\times \binom63\times \binom 33}{2!}$
$\text{Combining all cases for the final answer: }\frac12\binom84 + \binom63$
Note: This one may be off. I kept doubting myself here.
__________
$\text{Group the vowels as one letter: }\boxed{AEI}\\ \text{Because the ordering of the vowel is interchangeable, we must arrange the vowels}\\ \text{among each other:} 3!\\ \text{Then, fix the vowels for the sake of a circular arrangement.}$
$\text{Then, we are arranging }\boxed{AEI}, P, Q\text{ and }R\text{ in a circle: }(4-1)!\\ \text{i.e. }3!\\ \therefore \text{ Favourable outcomes: }3!\times 3!$
$\text{The total possible outcomes is just}\\ \text{arranging 6 letters in a circle: }5!\\ \therefore \text{answer: }\frac{3!\times 3!}{5!}$
__________
$\text{Arrange the 3 together in some order: }3!\\ \text{And then group them}\\ \text{Then, treating the group as one person there's }(n-2)\text{ people}\\ \text{to be arranged in a CIRCLE.}\\ \text{Answer: }3!(n-3)!$
__________

This question is somewhat famous. It is also in the Cambridge 3U textbook.
$\text{For any one day, the total possible outcomes is }\binom{100}{4}$
$\text{a) Only untagged fish may be selected.}\\ \therefore \text{We have }\binom{80}4\text{ favourable outcomes.}\\ \text{Answer: }\frac{\binom{80}4}{\binom{100}4}\approx 0.403$
$\text{b) is the complementary event of a)}\\ \text{Answer: }1-\frac{\binom{80}4}{\binom{100}4}$
$\text{c) is recursively getting the same thing in a)}\\ \text{7 times in a row.}\\ \text{Answer: }\left(\frac{\binom{80}4}{\binom{100}4}\right)^7$
$\text{The problem translates over into a binomial probability in d)}\\ \text{In 7 days, we require exactly 3 days of which no tagged fish are used.}\\ \text{For sake of neatness, I'm going to use the rounded answers of }0.403\text{ and }0.597\text{ here}$
$\text{Define the success to be where no tagged fish are drawn}\\ \text{and failure to be where at least one tagged fish is drawn}\\ \text{Since we require 3 successes out of 7 trials}\\ \text{Answer: }\binom 73 (0.403)^3 (0.597)^4$

shreya_ajoshi · « **Reply #1519 on:** February 19, 2017, 04:30:17 pm »

Hello!
How would you do these?

Write down the most concise general solution for:
a) 2sinx=cosx
b)3 + 2cosx = 15cosx
c) 12sin2x = 5sin36degrees

THANK YOU

jamonwindeyer · « **Reply #1520 on:** February 19, 2017, 05:23:03 pm »

Quote from: shreya_ajoshi on February 19, 2017, 04:30:17 pm

Hello!
How would you do these?

Write down the most concise general solution for:
a) 2sinx=cosx
b)3 + 2cosx = 15cosx
c) 12sin2x = 5sin36degrees

THANK YOU

Hey! So the general solution of a trig equation is basically just asking you to write, in the neatest way possible, an expression which can evaluate to all possible answers. This is in contrast to normal problems which just ask for a few (normally the ones between $0$ and $2\pi$). The formulae for these expressions is on your reference sheet (3U section, bottom left).

Let me show you the first one as a guide. We have:

$2\sin{x}=\cos{x}$

We need it in terms of a simple trig ratio. We divide:

$2\sin{x}=\cos{x}\\2\tan{x}=1\\\tan{x}=\frac{1}{2}$

That is now in the form required on our reference sheet! So the general solution is:

$\tan{x}=\frac{1}{2}\implies x=n\pi+\tan^{-1}{\frac{1}{2}} \left(n\text{ is an integer}\right)$

This expression will give you ANY answer to the equation you like, just by substituting different values of $n$. The other two questions should be much the same, rearrange then apply the formula, just remember to use the right expression on your reference sheet!

Does that make sense?

shreya_ajoshi · « **Reply #1521 on:** February 19, 2017, 07:04:42 pm »

Quote from: jamonwindeyer on February 19, 2017, 05:23:03 pm

Hey! So the general solution of a trig equation is basically just asking you to write, in the neatest way possible, an expression which can evaluate to all possible answers. This is in contrast to normal problems which just ask for a few (normally the ones between $0$ and $2\pi$). The formulae for these expressions is on your reference sheet (3U section, bottom left).

Let me show you the first one as a guide. We have:

$2\sin{x}=\cos{x}$

We need it in terms of a simple trig ratio. We divide:

$2\sin{x}=\cos{x}\\2\tan{x}=1\\\tan{x}=\frac{1}{2}$

That is now in the form required on our reference sheet! So the general solution is:

$\tan{x}=\frac{1}{2}\implies x=n\pi+\tan^{-1}{\frac{1}{2}} \left(n\text{ is an integer}\right)$

This expression will give you ANY answer to the equation you like, just by substituting different values of $n$. The other two questions should be much the same, rearrange then apply the formula, just remember to use the right expression on your reference sheet!

Does that make sense?

Hi! yes that makes sense. I wasn't too sure whether i needed to work out what x is from tan x = 1/2.

For the other questions would this be correct:
b) x = 2

n plus/minus cos^-1 3/13

c) I'm not too sure about how to do this one

shreya_ajoshi · « **Reply #1522 on:** February 19, 2017, 07:36:42 pm »

Hey!
I need some help with drawing 3D trig diagrams. Was wondering whether if someone could give tips on drawing them
Could someone also pretty please help me drawing this diagram:
The Eiffel Tower stands 80m high at point A, and the Arc de Triomphe stands 90m high at B, which is due south of A. A man stands at the point C on the ground due east of A. The top, D, of the Eiffel Tower at A has an angle of elevation of 55degrees from C. The top, E, of the Arc de Triopmphe at B has an angle of elevation of 35degrees from A.

jamonwindeyer · « **Reply #1523 on:** February 19, 2017, 08:11:59 pm »

Quote from: shreya_ajoshi on February 19, 2017, 07:04:42 pm

Hi! yes that makes sense. I wasn't too sure whether i needed to work out what x is from tan x = 1/2.

For the other questions would this be correct:
b) x = 2n plus/minus cos^-1 3/13

c) I'm not too sure about how to do this one

Hey great work! Second one is correct!

the third one looks tricky, but remember two things:

- $\sin{36^\circ}$ is just a number, you could even pop it in your calculator and replace it with a decimal if you wanted to!
- The formula gives you the solution to anything in the form $\sin{\theta}=a$, that includes $a=2x$, you'll just divide everything in the formula by 2 to get an answer in terms of $x$!

Let me know if that helps you get the answer out

jamonwindeyer · « **Reply #1524 on:** February 19, 2017, 08:33:07 pm »

Quote from: shreya_ajoshi on February 19, 2017, 07:36:42 pm

Hey!
I need some help with drawing 3D trig diagrams. Was wondering whether if someone could give tips on drawing them
Could someone also pretty please help me drawing this diagram:
The Eiffel Tower stands 80m high at point A, and the Arc de Triomphe stands 90m high at B, which is due south of A. A man stands at the point C on the ground due east of A. The top, D, of the Eiffel Tower at A has an angle of elevation of 55degrees from C. The top, E, of the Arc de Triopmphe at B has an angle of elevation of 35degrees from A.

Hey! Two tips:

1 - Cross out bits of information AS YOU USE THEM, to make sure you don't forget anything
2 - Lots of practice

I'll draw this one for you!

gummyybear · « **Reply #1525 on:** February 20, 2017, 08:29:05 pm »

Hi all, hope it's okay to post two (well twoish) questions.

18. Twelve points are arranged in order around a circle.
a) How many triangles can be drawn with these points as vertices? (I did this)
b) How many pairs of such triangles can be drawn if the vertices of the two triangles are distinct? (it's just the last step for this that I can't figure out, I figured you'd have to do 12C3 X 9C3 to get the two triangles but then you have divide by 2 to get the right answer, I'm not sure why division by 2 is the right way to go but it has something to do with cancelling out repetition right?)

23. By considering its prime factorisation, find the number of positive divisors of 315000.

RuiAce · « **Reply #1526 on:** February 20, 2017, 08:43:02 pm »

Quote from: gummyybear on February 20, 2017, 08:29:05 pm

Hi all, hope it's okay to post two (well twoish) questions.

18. Twelve points are arranged in order around a circle.
a) How many triangles can be drawn with these points as vertices? (I did this)
b) How many pairs of such triangles can be drawn if the vertices of the two triangles are distinct? (it's just the last step for this that I can't figure out, I figured you'd have to do 12C3 X 9C3 to get the two triangles but then you have divide by 2 to get the right answer, I'm not sure why division by 2 is the right way to go but it has something to do with cancelling out repetition right?)

23. By considering its prime factorisation, find the number of positive divisors of 315000.

Yeah. It's repetition.

In general, repetition occurs when "things" have the same "size". Here, the "things" are the points you are choosing, and the size is always 3 because you're making triangles. The repetition occurs as a consequence of the fact if you swap the order the triangles were chosen from, you get back to where you started. This creates a double-counting scenario which you need to eliminate. Since there's only two triangles here, you divide by 2!, which is equal to 2.
____________________________________
$\text{Note that }315000 = 315 \times 1000$
$\text{The prime factorisation of }1000\text{ is easy}\\ \begin{align*}1000&=10^3\\ &=5^3\times 2^3\end{align*}$
$\text{Clearly, 5 is a factor of 315}\\ \text{From the calculator, }\frac{315}{5}=63\\ \text{And then we can find the prime factorisation of 315}$
$\begin{align*}315&=5\times 63\\ &= 3^2\times 5\times 7\end{align*}$
$\text{Hence, the prime factorisation of 315000 is}\\ \boxed{315000=2^3\times 3^2\times 5^4\times 7}$
$\text{Now, since the prime factors of 315000 are }2,3,5,7\\ \text{If }x\text{ is a factor of 315000}\\ \text{Then notice how }x\text{ will be in the form:}\\ x=2^a \times 3^b \times 5^c\times 7^d\\ \text{Where }a,b,c,d\text{ are integers.}$
$\text{Question: What are the restrictions on }a,b,c,d?\\ \text{Well, clearly powers can be at least 0.}\\ \text{But the powers are capped off because of the prime factorisation. Note how }2^4\text{ won't work.}$
$\text{So we note that:}\\ 0\le a \le 3\\ 0\le b \le 2\\ 0\le c\le 4\\ 0\le d\le1$
$\text{Therefore, we have }4\text{ possible choices for the value of }a:\, 0,1,2\text{ or 3}\\ \text{3 possible choices for }b:\, 0,1\text{ or 2} \\\text{ and etc.}$
$\text{Final answer: }4\times 3\times 5\times 2 = \boxed{120}$

gummyybear · « **Reply #1527 on:** February 21, 2017, 08:15:33 pm »

ahh it finally makes sense, thank you!!

legorgo18 · « **Reply #1528 on:** February 22, 2017, 11:08:02 pm »

Hey guys got an interesting series question here

Sn = 0.2 + 0.22 + 0.222 + ... + 0.222...2 (n 2s). Simplify Sn in terms of n.

Pref not solved and clues instead, ruins the satisfaction

RuiAce · « **Reply #1529 on:** February 22, 2017, 11:12:18 pm »

Quote from: legorgo18 on February 22, 2017, 11:08:02 pm

Hey guys got an interesting series question here

Sn = 0.2 + 0.22 + 0.222 + ... + 0.222...2 (n 2s). Simplify Sn in terms of n.

Pref not solved and clues instead, ruins the satisfaction

This is hard to give a clue for.

Try to split 0.22 into 0.2 + 0.02
0.222 into 0.2 + 0.02 + 0.002

Then do some clever rearranging.

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Author Topic: 3U Maths Question Thread (Read 1237954 times) Tweet Share

shreya_ajoshi

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K9810

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