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Author Topic: Specialist 1/2 Question Thread!  (Read 119941 times)  Share 

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keltingmeith

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Specialist 1/2 Question Thread!
« on: February 11, 2016, 10:16:41 pm »
+2
Given just how different specialist 1/2 and specialist 3/4 are, we've now made a second question thread for all your 1/2 needs!! This will help guarantee that your question doesn't get lost amongst all the other questions being asked, just in case anyone answering knows a lot about 3/4, but not 1/2.

qazser

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Re: Specialist 1/2 Question Thread!
« Reply #1 on: February 11, 2016, 10:50:04 pm »
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Ty Dickens for making this  ;D
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Re: Specialist 1/2 Question Thread!
« Reply #2 on: February 16, 2016, 10:17:17 pm »
0
My teacher goes thru the content quite fast in class, and I feel I need more practice as I am struggling. I did the qs in the textbook, but I found it hard and would like more practice.
What are some free resources that I can utilise?
Thnx
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MightyBeh

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Re: Specialist 1/2 Question Thread!
« Reply #3 on: February 17, 2016, 05:41:59 am »
+3
My teacher goes thru the content quite fast in class, and I feel I need more practice as I am struggling. I did the qs in the textbook, but I found it hard and would like more practice.
What are some free resources that I can utilise?
Thnx

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If you're running out of fresh questions to do, maybe try a methods textbook? Obviously not all the content will be the same but there's plenty of relevant stuff in there. You should ask your teacher too, they might have more resources. Personally, I learned the most during practice (and actual ::) ) SACs. This might also be helpful :)
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Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #4 on: February 23, 2016, 06:45:40 pm »
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I need some help on a couple questions http://m.imgur.com/a/rEeRc

For Q5b) what's wrong with my equation/method to working this out? I'm lost on how to solve this problem even though it's almost exactly the same to the others.

For 8b) the answer says 72 minutes, but I got 90. x=50 is correct as well.

For 8a and 10, I did these with a scientific calculator because the square root looked pretty messy. Are these questions intended to be solved by hand?

Thanks

tysh

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Re: Specialist 1/2 Question Thread!
« Reply #5 on: February 23, 2016, 08:24:50 pm »
+1
I need some help on a couple questions http://m.imgur.com/a/rEeRc

For Q5b) what's wrong with my equation/method to working this out? I'm lost on how to solve this problem even though it's almost exactly the same to the others.

For 8b) the answer says 72 minutes, but I got 90. x=50 is correct as well.

For 8a and 10, I did these with a scientific calculator because the square root looked pretty messy. Are these questions intended to be solved by hand?

Thanks

Hey Adequace, they were just minor errors:
For 5b, it was 600/x = 600/(x-5.5) - 220, instead of 600/x = 600/(x-5.5) + 220: remember that the plane's speed is faster, thus we substract the 220 from the plane.
For 8b, instead of 75/x (which is the usual time), they are looking for 75/(x+12.5), as this is the actual time taken for that particular journey. The question was probably a bit vague anyway though.
Questions like 10 and 8a are definitely doable by hand (although they ARE a bit tedious!). Sometimes, to speed up the process (such as in question 10) you can use km/min or another unit to avoid fractions during working out, then convert it back to km/h at the end.
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Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #6 on: February 23, 2016, 09:03:17 pm »
0
Hey Adequace, they were just minor errors:
For 5b, it was 600/x = 600/(x-5.5) - 220, instead of 600/x = 600/(x-5.5) + 220: remember that the plane's speed is faster, thus we substract the 220 from the plane.
For 8b, instead of 75/x (which is the usual time), they are looking for 75/(x+12.5), as this is the actual time taken for that particular journey. The question was probably a bit vague anyway though.
Questions like 10 and 8a are definitely doable by hand (although they ARE a bit tedious!). Sometimes, to speed up the process (such as in question 10) you can use km/min or another unit to avoid fractions during working out, then convert it back to km/h at the end.
Thanks man, appreciate it a lot!

Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #7 on: February 26, 2016, 08:01:33 pm »
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Another question.

http://imgur.com/a/vZeP5

For Q16) the answer just has p=10 but I can't get to answer. I've attached my working as well.

For Q12) I struggle greatly with worded problems. I don't think the denominators in my pipe equations are correct. Is there anything I did wrong? Haven't really tried solving this question due to my lack of confidence..

Thanks

StupidProdigy

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Re: Specialist 1/2 Question Thread!
« Reply #8 on: February 26, 2016, 09:20:50 pm »
+1
Another question.

http://imgur.com/a/vZeP5

For Q16) the answer just has p=10 but I can't get to answer. I've attached my working as well.

For Q12) I struggle greatly with worded problems. I don't think the denominators in my pipe equations are correct. Is there anything I did wrong? Haven't really tried solving this question due to my lack of confidence..

Thanks
16. You're working is good, you're reading of the question not so much haha. p has to be a positive integer. 2sqrt(21) is roughly 9.16 therefore p must =10 since that is the closest integer that is both positive and larger than 9.16
12..can't be bothered sorry
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Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #9 on: February 26, 2016, 10:06:41 pm »
0
16. You're working is good, you're reading of the question not so much haha. p has to be a positive integer. 2sqrt(21) is roughly 9.16 therefore p must =10 since that is the closest integer that is both positive and larger than 9.16
12..can't be bothered sorry
Thanks regardless, but yeah the worded one is a pain.

nerdgasm

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Re: Specialist 1/2 Question Thread!
« Reply #10 on: February 27, 2016, 12:05:35 am »
+1
I have a solution for Q12, but it is by no means elegant, and I think there's probably a better way to do it that I've missed:

Spoiler
Let C = the capacity of the tank. Let r1 and r2 be the initial flow rates through the first and second pipes respectively. Let r1_new and r2_new be the new flow rates. Let t1 and t2 be the initial time it takes to fill the tank using the initial flow rates of the first and second pipes respectively.

Using the formula Capacity = Rate * Time (and its rearrangements), we can form the following six equations:
Equation 1: r1 + r2 = 3C/20   
Equations 2 and 3: r1 = C/t1 ; r2 = C/t2
Equations 4 and 5: r1_new = C/(t1 - 1) ; r2_new = C/(t2 + 2)
Equation 6: r1_new + r2_new = C/7.

I now substitute Eqs. 2 and 3 into 1, and Eqs. 4 and 5 into 6, to get the following:

Now I factor out C from both equations (C > 0, or else the question makes no sense). We arrive at:
Our goal is therefore to work out the value of t1 and t2, so we can work out (t1-1) and (t2+2), which is the ultimate aim.

Algebraically manipulating the fractions, we get the following:;. I now treat this as a system of (non-standard) simultaneous equations, and try to solve. We get:
. But notice that we previously had Therefore, we may subtract these two equations to get We can substitute this back into 5t1-4t2=27 to get This quadratic can be factorised to give . If you substitute these values back into 5t1-4t2=27, you will get the corresponding solutions Of course, we can't have a negative time value, so it follows that . Those are the initial times to fill the tank for each pipe. To get the new times, we simply evaluate t1-1 and t2+2, to get new times of 14 minutes for both pipes! Therefore, under the new flow rates, each pipe will take 14 minutes to fill the tank!

In summary: Initially, the first pipe took 15 minutes to fill the tank, and the second pipe took 12 minutes. Hence the first pipe fills 1/15 of the tank per minute, and the second pipe fills 1/12 of the tank per minute. If you combine the two pipes, they fill 1/15 + 1/12 = 3/20 of the tank per minute, and hence it takes 20/3 minutes to fill the tank. After the flow rates were adjusted, the two pipes both take 14 minutes to fill the tank individually, and hence both fill 1/14 of the tank per minute. So when you combine the two pipes, they fill 1/14 + 1/14 = 1/7 of the tank per minute, and hence it takes 7 minutes to fill the tank.

There has seriously got to be a better way to do this :-\.


Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #11 on: February 27, 2016, 11:40:35 am »
0
I have a solution for Q12, but it is by no means elegant, and I think there's probably a better way to do it that I've missed:

Spoiler
Let C = the capacity of the tank. Let r1 and r2 be the initial flow rates through the first and second pipes respectively. Let r1_new and r2_new be the new flow rates. Let t1 and t2 be the initial time it takes to fill the tank using the initial flow rates of the first and second pipes respectively.

Using the formula Capacity = Rate * Time (and its rearrangements), we can form the following six equations:
Equation 1: r1 + r2 = 3C/20   
Equations 2 and 3: r1 = C/t1 ; r2 = C/t2
Equations 4 and 5: r1_new = C/(t1 - 1) ; r2_new = C/(t2 + 2)
Equation 6: r1_new + r2_new = C/7.

I now substitute Eqs. 2 and 3 into 1, and Eqs. 4 and 5 into 6, to get the following:

Now I factor out C from both equations (C > 0, or else the question makes no sense). We arrive at:
Our goal is therefore to work out the value of t1 and t2, so we can work out (t1-1) and (t2+2), which is the ultimate aim.

Algebraically manipulating the fractions, we get the following:;. I now treat this as a system of (non-standard) simultaneous equations, and try to solve. We get:
. But notice that we previously had Therefore, we may subtract these two equations to get We can substitute this back into 5t1-4t2=27 to get This quadratic can be factorised to give . If you substitute these values back into 5t1-4t2=27, you will get the corresponding solutions Of course, we can't have a negative time value, so it follows that . Those are the initial times to fill the tank for each pipe. To get the new times, we simply evaluate t1-1 and t2+2, to get new times of 14 minutes for both pipes! Therefore, under the new flow rates, each pipe will take 14 minutes to fill the tank!

In summary: Initially, the first pipe took 15 minutes to fill the tank, and the second pipe took 12 minutes. Hence the first pipe fills 1/15 of the tank per minute, and the second pipe fills 1/12 of the tank per minute. If you combine the two pipes, they fill 1/15 + 1/12 = 3/20 of the tank per minute, and hence it takes 20/3 minutes to fill the tank. After the flow rates were adjusted, the two pipes both take 14 minutes to fill the tank individually, and hence both fill 1/14 of the tank per minute. So when you combine the two pipes, they fill 1/14 + 1/14 = 1/7 of the tank per minute, and hence it takes 7 minutes to fill the tank.

There has seriously got to be a better way to do this :-\.
Thanks a lot man! Also, when you factorised 5t^2 -87t +180 =0. Is this expected to be done by hand?

keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #12 on: February 27, 2016, 12:59:25 pm »
+1
Thanks a lot man! Also, when you factorised 5t^2 -87t +180 =0. Is this expected to be done by hand?

Yes - you are expected to be able to solve all quadratics by hand. Don't forget about techniques like completing the square and the quadratic equation!

Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #13 on: February 27, 2016, 09:44:54 pm »
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Yes - you are expected to be able to solve all quadratics by hand. Don't forget about techniques like completing the square and the quadratic equation!
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)

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Re: Specialist 1/2 Question Thread!
« Reply #14 on: February 27, 2016, 09:52:32 pm »
+1
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)
For part f, x^2 -x isn't an irreducible quadratic factor because it can be written as x(x-1) so you would write A/x + B/x-1

I can't help with h at the moment as I don't have pen and paper with me sorry
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