1.2: Q7(b) Amended the y-axis to begin from −1 as required by the question.

Access it here if you’re having trouble accessing it on Facebook using your state-of-the-art smartphone 🙁

N.B. As usual, please, please leave a comment should you spot any mistake in the solution. After doing the paper at such a speed (admittedly, with some help along the way), the High Priestess is going to faint soon!

What would happen if my paper 2 graph question had a point plotted wrongly, and the subsequent parts had the wrong answers because the graph had the wrong shape. Will i get the whole graph question wrong? or can i still get some marks because of error carried forward? Thankyou!

@Erick: Granted the Q7(c)(d)(e) all carry 2 marks each, you should get some marks if you have clearly demonstrated your understanding of how to solve for x in (b), finding the gradient in (c) etc.

Hi.,. made a lot of careless mistakes T.T How much do you think this year A1 and A2? T.T For gradient of tangent, since scale is smaller, is it true that the range of accepted answer will be bigger than usual? My answer is 2.04. Then can explain to me about the probability qns last part? Thanks T.T

As long as you've demonstrated that you've obtained the gradient (within a tolerable range) at the point via calculating the gradient of your suitably-drawn tangent based on selected coordinates from your plot, you'd be fine 😉

What's the probability of picking A, then a B (without replacement of the A)? (calculating only possible outcome: AB)

Ans: (11/16)(10/15) = 11/24

What's the probabilty of picking 2 people out of the lot, and having an end result of A and B in your selection, regardless of sequence? (calculating possible outcomes: AB, BA)

Ans: 11/24 + 11/24 = 11/12

Basically:

Probability of getting AB + Probability of getting BA = 11/12

@Kester Tay: Q10(b)(ii)(b) requires us to find the probability that neither of two members selected at random is a right-handed female.

Do you first agree that both members selected die die cannot be right-handed females? This means that not even one of them can be a right-handed female.

So this translates simply into:

The first member you select from the group die die cannot be a right-handed female ⇒ probability = 11/16

AND

The second member you select from those remaining in the group die die cannot be a right-handed female ⇒ probability = 10/15

∴ P(neither is a right-handed female) = (11/16)(10/15) = 11/24

If you still don't believe Miss Loi, you may work things out the long-winded way by considering all possibilities of not selecting any right-handed female as presented in the probability tree diagram below (note: we can't have even one right-handed female selected at any point):

*This reply was typed out by a left-handed female*

P.S. If Miss Loi understands the cases you've presented, you are labelling a person of "type A or B" but in actual fact for Q10(b)(ii) there should be only one "type" i.e. "not a right-handed female".

I did the 2010 TYS question, they had a similar question to the probability one (it's Q10c in the 2010 P2) a night before the paper.

The answer told me 7/295 even though I thought it was 7/590. So I trusted the TYS answer, and x2 in the exam, and lost marks.

My working:

10)b)ii)a) 2(9/16 x 8/15) = 3/5 (when answer shld be 3/10)

10)b)ii)a) 2(11/16 x 10/15) = 11/12 (when answer shld be 11/24)

I multiplied by two for both the last part questions, will I still get any method marks out of 4m? Because I understand it's already logic error when I x2, so may not get it.

@Miss Loi: I'd post a picture, but I don't know how to :/

It's the topical TYS, Nov 02 ~ Nov 10, published by Dyna, (C) 2010. No workings in the answer key, only the answer. Didn't cross-reference with other TYS answers at the time because I thought it was correct (and also bolui keep buying TYS T_T), so in the exam I just trusted the TYS answer (7/295), and thought that AB wasn't the 'same type', oh well... Maximum minus 4m ><

@Miss Loi: Miss Lois's left-handed assistant highlighted incorrectly part of the probability tree diagram. The branch for RHF and the following branches should be all be hightlighted as RED.

@Dot Diagram: The left-handed female (who happens to be Miss Loi) initially highlighted the probabilities in red (for outcomes that resulted in selecting a RHF) in order to make it easy on her eyes when forming that long-winded expression at 2.30am in the morning *rubs eyes*

But yeah you're right - all branches following the first selection of the RHF should be highlighted as well to cover all possible outcomes that contain a RHF.

If I have an inaccurate "P" value for the graph question, however my tangent and x- coordinates are correct, do I still qualify for the marks for part C) onwards?

@John: Well it seems like your inaccurate value of p wasn't "inaccurate" enough to affect your answers from Q7(c) onwards so it shouldn't matter as far as accuracy marks are concerned.

1)For 9(c), i mistakenly wrote OM as 4cm instead of my calculated OM value. So my ans for part c is wrong, so will there be method marks given for part c, d and e and will there be 'error carry forward?' 2) for 4 ci) i did divide the 3.97x10^13 by 60k and then 12, will i get method mark for that?

@RY: 1) Apparently quite a number have used the wrong value of OM in Q9(c). You may still retain some marks in (d) & (e) if you've subbed the remaining values correctly into the appropriate formulae i.e. volume of pyramid, volumes of similar figures.

2) You mean 4(c)(ii)? Yes you should get a mark or two for the bulk of your workings if you missed out on only the year conversion or number of significant figures.

For question 10a)i)a), can it be 37 instead of 37.4? Also, 10a)iii) can it be saying that the chem students had a lower median and higher upper quartile?

@Denton: IMHO, 37 is a bit off since the median point on the curve lies clearly between 37 and 38.

For Q10(iii) you should describe how each statistical parameter is used to compare the students' performances (i.e. a higher Physics median mark indicates that the students performed better in Physics) instead of simply stating them.

Hey wanted to ask whether the graphquestions 7'c) , 7d) and 7e) has a range of answers? If so, is answers to 7c) x= 2.4 ad x=3.3 , 7d) gradient = 1.51 and 7e) min pt. (2.8, -0.3) acceptable? Thank you for the solutions. Really appreciate it. 🙂

@Navin: Yes they'll always be a tolerable range of answers when it comes to graph plotting questions. As mentioned many times, don't worry too much about your graphical answers if you have done this. 😀

@Mary: By multiplying the prices i.e. $3 × 1.05 = $3.15 & $2.50 × 1.05 = $2.625, you should still arrive at the same answer albeit in a slightly longer way:

New P =

Total fees taken in a week after the increase = $525 + $708.75 = $1233.75 ≈ $1234 (nearest dollar)

There must be a careless mistake somewhere within your workings 🙁

For the graph can I start with 1 on the x-axis and -1 on the y-axis.For part c can 2.35 and 3.55 be accepted and the gradient 1.73 be accepted. Can the min point be (3,-0.35)

@Ng Guan Zhi: For Q7(b), yes you should start with −1 on the y-axis as stipulated in the question - we've actually uploaded v1.2 of the solutions for this little amendment.

However, Miss Loi is wondering how you managed to begin with 1 on the x-axis, since the y-axis has to intersect the x-axis at x=0, and the next unit along the x-axis is 1 (so no 'zig-zag line' required)?

@Miss Loi: I mean when the two axes intersect we usually put 0 in the middle so the coordinates of the starting point will be (0,0) but now I put 1 on x-axis and -1 on the y-axis so now the coordinates of the starting point is (1,-1) instead of (0,0).By the way for part c can 2.35 and 3.35 be accepted and the gradient 1.73 be accepted

@Miss Loi: I mean the coordinates where the axes intersect is usually (0,0) but mine is (1,-1) is this accepted.By the way for part c will 2.35 and 3.55 be accepted and the gradient 1.73 be accepted.will the min point (3,-0.35) be accepted

@Ng Guan Zhi: So you mean to say that your origin is at (1, −1)???!!! NOOOOOOOO ...

Even in cases where you draw "zigzag lines" due to the starting values on the x- & y-axes being far from the origin, the axes still have to meet at (0, 0) beyond the zigzag lines.

@Miss Loi: In this case will I still get marks for the graph if I plotted my points correctly even though my origin is wrong and will I still get marks for the other parts of the question

@Ng Guan Zhi: Miss Loi hopes/prays/wishes that moving your axes around will the affect your marks for other parts, since your values are close enough.

But part (c) may be a problem since you presumably moved your x-axis down to −1? If this is the case, how did you find the x-intercepts? You drew another line y = 0?

Hi, I would like to ask, is there some mistake in qn 6aiii ? Shouldnt the answer be only 5,7,9,10,11,13,14,15. Why is there no. 3,6,12 ? Can you explain? Thank you. (:

Hi! If my Venn diagram already had a mistake from start, hence affecting all of my answers in the later parts of te Venn diagram question, will I get any working marks?

For Question 5, if one of the angles is proven wrongly does it mean that all my subsequent answers will be incorrect even if the methods used or angle is correct?

@Jim Li: If any, probably only in part (ii)(b), and only if it's clear that it was simply due to the wrong value of ∠ADE/BAD in your concise working that led directly to the wrong answer.

@Jim Li: If you're referring to part (i) which carries 4 marks, you should get 1 mark for the correct answer for ∠AEB. With two other marks lost through your wrong answers for ∠ADE/BAD, Miss Loi believes the remaining mark may be allocated towards your reasons/intermediate step(s) required for each answer (which you will only get if all the relevant properties are stated correctly).

@Jim Li: Please note that the mark scheme isn't set by Miss Loi so no way she can be 100% sure about marks allocation.

But having say that, Q5(a)(ii)(a) seems to be a do-or-die "correct answer or zero" 1 mark case, while Q5(a)(ii)(b) might still give you a mark for correct method with wrong values/ECF etc.

@Jim Li: Hard to say, depends if the answer mark is tied to having the correct working leading to it or whether a benefit of doubt is given. But Miss Loi thought you said your ∠ADE was wrong?!

@Raymond: 37 is unlikely to be accepted by an examiner (with no crossed-eye tendencies) as a good estimate since the line clearly doesn't touch the 37 mark in the chart 🙁

@Navin: It appears that a significant number of students are asking this with regards to the format of the y-intercept answer in Q3(b)(i)(b).

Strictly speaking, the y-intercept should be the y-value (i.e. −2) of the coordinates of the intercept point.

Leaving the answer in coordinate form (0, −2) will really depend on whether this is condoned in the mark scheme (something which an eternally optimistic Miss Loi thinks they will 🙂 )

@Miss Loi: One last question, for the graph in part (b)where we are told to draw the graph, it is 3 marks. How are the 3 marks allocated. My graph wasn't drawn neatly . Will they penalize marks? Though my answers in parts (c) (d) and (e) were correct.

@Navin: Please note that the mark scheme isn't set by Miss Loi so no way she can be 100% sure about marks allocation.

For Q7(b), however, an educated guess could be 2 marks for all points plotted correctly and joined by a smooth curve + 1 mark for correctly drawing and labeling the axes to the required scale.

So your "wasn't drawn neatly" could fall under the "smooth curve" part but it's kinda hard for Miss Loi to objectively gauge how badly drawn your graph is.

Im very worried about qn 2d. I got my a) b) and c) right. Just that i somehow got 2d wrong. I used cosine rule for triangle ECD... So i may have gotten some values wrong. Ended up getting DE = 12.9 or something. So would i get 0/3m for this? Or would i get method mark for using cosine rule?

8e) I showed my working : 2000/161.8, somehow i evaluated it wrongly cause i may have typed it in cal wrongly. Would i be given 1m for this?

10a)i)a) Median i put as 37.5?

10b)ii)b) Probability of getting neither right handed female, if i got my ans as 11/12, would i get 0/2m? Or is there any chance of 1m?

@Jsn: For Q2(d), if you read through how method marks are awarded, they are awarded based on the correct application of the formula rather simply stating the use of the formula.

From your description, you appear to have used the cosine rule on the wrong triangle so, strictly speaking, no method mark will be awarded if this is the case 🙁 (Miss Loi hopes she's wrong here for your sake)

For Q8(e), it depends on the mark scheme but since there are 2 marks in this part, there's a chance to for a 1 mark redemption here.

For Q10(a)(i)(a), 37.5 should be fine ^^

For Q10(b)(ii)(b), again it ultimately depends on the mark scheme but for probability questions like this one where you need to get the concept right from the start, it's often a case of 0 or full marks 🙁

@Tom: Q4(c)(iii): Miss Loi have come across some arguing that 1 year should be 365.25 days, but by assuming 1 month = 31 days ⇒ 1 year = 31 × 12 = 372 days you are committing a seditious act of Monthistism in discriminating against the lesser months of Feb/April/June/Sep/Nov!!! 😮

Q7(c): Would think the 2 marks here goes towards the correct values for x obtain directly from the graph, rather than anything else.

For last probability qn, i thought it meant it cannot be both right handed female. The qn is phrased such that it is actually trying to convey that cannot have AT LEAST ONE right handed female....... 🙁

hey if i draw my venn diagram wrong..like if it was written D={factors of 6} and i wrote the multiples and drew it and there were in total 5 parts each carrying 1.5 marks expect one which was carrying 1.... i knew exactly what to do... but silly mistakes T.T

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

Gratitudes

I can see that she is more confident now, especially when she has been previously suffering from lack of confidence and kept on telling me that she “can’t do it”. Now she says “Yes! I can do it!”. … [read more]

## 93 Comments

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What would happen if my paper 2 graph question had a point plotted wrongly, and the subsequent parts had the wrong answers because the graph had the wrong shape. Will i get the whole graph question wrong? or can i still get some marks because of error carried forward?

Thankyou!

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@Erick: Granted the Q7(c)(d)(e) all carry 2 marks each, you should get some marks if you have clearly demonstrated your understanding of how to solve for

xin (b), finding the gradient in (c) etc.曜

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My paper 2 was not tied in the qns order for the graph

qns, which is qns7. Instead, I tied it to the last page. Is it alright?? :/

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@Rachel: o.O Don't worry the sky isn't falling!

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@Miss Loi: Thanks Miss Loi!! ^_^

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Hi.,. made a lot of careless mistakes T.T

How much do you think this year A1 and A2? T.T

For gradient of tangent, since scale is smaller, is it true that the range of accepted answer will be bigger than usual? My answer is 2.04. Then can explain to me about the probability qns last part?

Thanks T.T

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@wotagei: Every year, Miss Loi has to say this:

Regarding Q10(b)(ii)(b), please see here.

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Q10c)

Let the two people you pick be labeled A and B.

What's the probability of picking A, then a B (without replacement of the A)? (calculating only possible outcome: AB)

Ans: (11/16)(10/15) = 11/24

What's the probabilty of picking 2 people out of the lot, and having an end result of A and B in your selection, regardless of sequence? (calculating possible outcomes: AB, BA)

Ans: 11/24 + 11/24 = 11/12

Basically:

Probability of getting AB + Probability of getting BA = 11/12

Shouldn't this be the case, Ms Loi? :0

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@Kester Tay: Q10(b)(ii)(b) requires us to find the probability that

neitherof two members selected at random is a right-handed female.Do you first agree that

bothmembers selecteddie diecannot be right-handed females? This means that not even one of them can be a right-handed female.So this translates simply into:

The first member you select from the group

die diecannot be a right-handed female ⇒ probability = 11/16AND

The second member you select from those remaining in the group

die diecannot be a right-handed female ⇒ probability = 10/15∴ P(neither is a right-handed female)

= (11/16)(10/15) = 11/24

If you still don't believe Miss Loi, you may work things out the long-winded way by considering all possibilities of

notselecting any right-handed female as presented in the probability tree diagram below (note: we can't have even one right-handed female selected at any point):So P(neither is a right-handed female)

= (6/16)(5/15 + 3/15 + 2/15) + (3/16)(6/15 + 2/15 + 2/15) + (2/16)(6/15 + 3/15 + 1/15)

= (6/16)(10/15) + (3/16)(10/15) + (2/16)(10/15)

= (11/16)(10/15)

= 11/24

*This reply was typed out by a left-handed female*

P.S. If Miss Loi understands the cases you've presented, you are labelling a person of "type A or B" but in actual fact for Q10(b)(ii) there should be only one "type" i.e. "not a right-handed female".

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@Miss Loi: しまった！T_T

I did the 2010 TYS question, they had a similar question to the probability one (it's Q10c in the 2010 P2) a night before the paper.

The answer told me 7/295 even though I thought it was 7/590. So I trusted the TYS answer, and x2 in the exam, and lost marks.

My working:

10)b)ii)a) 2(9/16 x 8/15) = 3/5 (when answer shld be 3/10)

10)b)ii)a) 2(11/16 x 10/15) = 11/12 (when answer shld be 11/24)

I multiplied by two for both the last part questions, will I still get any method marks out of 4m? Because I understand it's already logic error when I x2, so may not get it.

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@Kester Tay: Miss Loi just took this photo from a TYS at The Temple of the worked solutions for the question you mentioned:

... and it's answer & method is consistent with Miss Loi's

leh. Which TYS are you using?曜

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@Miss Loi: I'd post a picture, but I don't know how to :/

It's the topical TYS, Nov 02 ~ Nov 10, published by Dyna, (C) 2010. No workings in the answer key, only the answer. Didn't cross-reference with other TYS answers at the time because I thought it was correct (and also bolui keep buying TYS T_T), so in the exam I just trusted the TYS answer (7/295), and thought that AB wasn't the 'same type', oh well... Maximum minus 4m ><

Anyway, 本当にありがとうございます 🙂

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@Miss Loi: Miss Lois's left-handed assistant highlighted incorrectly part of the probability tree diagram. The branch for RHF and the following branches should be all be hightlighted as RED.

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@Dot Diagram: The left-handed female (who happens to be Miss Loi) initially highlighted the probabilities in red (for outcomes that resulted in selecting a RHF) in order to make it easy on her eyes when forming that long-winded expression at 2.30am in the morning *rubs eyes*

But yeah you're right - all branches following the first selection of the RHF should be highlighted as well to cover all possible outcomes that contain a RHF.

*image amended*

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If I have an inaccurate "P" value for the graph question, however my tangent and x- coordinates are correct, do I still qualify for the marks for part C) onwards?

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@John: Well it seems like your inaccurate value of

pwasn't "inaccurate" enough to affect your answers from Q7(c) onwards so it shouldn't matter as far as accuracy marks are concerned.曜

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1)For 9(c), i mistakenly wrote OM as 4cm instead of my calculated OM value. So my ans for part c is wrong, so will there be method marks given for part c, d and e and will there be 'error carry forward?'

2) for 4 ci) i did divide the 3.97x10^13 by 60k and then 12, will i get method mark for that?

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@RY: 1) Apparently quite a number have used the wrong value of

OMin Q9(c). Youmaystill retain some marks in (d) & (e) if you've subbed the remaining values correctly into the appropriate formulae i.e. volume of pyramid, volumes of similar figures.2) You mean 4(c)(ii)? Yes you should get a mark or two for the bulk of your workings if you missed out on only the year conversion or number of significant figures.

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For question 10a)i)a), can it be 37 instead of 37.4? Also, 10a)iii) can it be saying that the chem students had a lower median and higher upper quartile?

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@Denton: IMHO, 37 is a bit off since the median point on the curve lies clearly between 37 and 38.

For Q10(iii) you should describe how each statistical parameter is used to compare the students' performances (i.e. a higher Physics median mark indicates that the students performed better in Physics) instead of simply stating them.

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10biib answer is it wrong? Cos for most of my friends and i get 11/24 for the answer

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@---: 11/24 it is! See explanation.

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Hey wanted to ask whether the graphquestions 7'c) , 7d) and 7e) has a range of answers? If so, is answers to 7c) x= 2.4 ad x=3.3 , 7d) gradient = 1.51 and 7e) min pt. (2.8, -0.3) acceptable? Thank you for the solutions. Really appreciate it. 🙂

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@Navin: Yes they'll always be a tolerable range of answers when it comes to graph plotting questions. As mentioned many times, don't worry too much about your graphical answers if you have done this. 😀

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Dear miss Loi , for qn 9d, if I got the height wrong and used it for other parts will I get carried forward marks ?

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@Ling: *wave wave*

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@Miss Loi: meaning? dont have?

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@Ling: The *wave wave* is to direct you to this standard reply to those in the same predicament, which basically says all may not be lost 🙂

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Hi is question 1d (ii) confirmed correct? Cause just now I saw some other ans key something diff 🙁 thanks!

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@Ron: For Q1(d)(ii) you are referring to

others' vs Miss Loi's which essentially are the same?曜

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Miss Loi, can you predict what might be the A1 range?

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For 4(a)(i), can the answer be left as $420?

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@Cat: No problem at all ^^

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Miss Loi , for qn 9d paper 2, if my height is wrong and I used it for other parts , is there error carried forward marks?

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@Ling: Moved your comment here since you're talking about a question from Paper 2.

In any case, see Miss Loi's standard reply to the legions of students who have unfortunately used the wrong height in Q9(c).

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For 6.(b)(v) I multiplied 1.05 to the original prices ($2.50/$3), then I found a new P and got a different total new fee. Is this method acceptable?

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@Mary: By multiplying the prices i.e. $3 × 1.05 = $3.15 & $2.50 × 1.05 = $2.625, you should still arrive at the same answer albeit in a slightly longer way:

New

P=Total fees taken in a week after the increase

= $525 + $708.75 = $1233.75 ≈ $1234 (nearest dollar)

There must be a careless mistake somewhere within your workings 🙁

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For the graph can I start with 1 on the x-axis and -1 on the y-axis.For part c can 2.35 and 3.55 be accepted and the gradient 1.73 be accepted. Can the min point be (3,-0.35)

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@Ng Guan Zhi: For Q7(b), yes you should start with −1 on the

y-axis as stipulated in the question - we've actually uploaded v1.2 of the solutions for this little amendment.However, Miss Loi is wondering how you managed to begin with 1 on the

x-axis, since they-axis has to intersect thex-axis atx=0, and the next unit along thex-axis is 1 (so no 'zig-zag line' required)?曜

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@Miss Loi: I mean when the two axes intersect we usually put 0 in the middle so the coordinates of the starting point will be (0,0) but now I put 1 on x-axis and -1 on the y-axis so now the coordinates of the starting point is (1,-1) instead of (0,0).By the way for part c can 2.35 and 3.35 be accepted and the gradient 1.73 be accepted

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@Miss Loi: I mean the coordinates where the axes intersect is usually (0,0) but mine is (1,-1) is this accepted.By the way for part c will 2.35 and 3.55 be accepted and the gradient 1.73 be accepted.will the min point (3,-0.35) be accepted

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@Ng Guan Zhi: So you mean to say that your

originis at (1, −1)???!!! NOOOOOOOO ...Even in cases where you draw "zigzag lines" due to the starting values on the

x- &y-axes being far from the origin, the axes still have to meet at (0, 0) beyond the zigzag lines.曜

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@Miss Loi: In this case will I still get marks for the graph if I plotted my points correctly even though my origin is wrong and will I still get marks for the other parts of the question

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@Ng Guan Zhi: Miss Loi hopes/prays/wishes that moving your axes around will the affect your marks for other parts, since your values are close enough.

But part (c) may be a problem since you presumably moved your

x-axis down to −1? If this is the case, how did you find thex-intercepts? You drew another liney= 0?曜

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Hi, just a question.

For example after i put P(Neither Blahblah)= 3/24

I rewrite the sentence : The probability of Neither Blah Blah is 3/24.

Accepted? Haha

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@Rayner: P( rewriting the sentence: "The probability of Neither Blah Blah is 3/24" is accepted ) = 1. ^^

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What is the minimum score to get a C6?

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Hi, I would like to ask, is there some mistake in qn 6aiii ?

Shouldnt the answer be only 5,7,9,10,11,13,14,15.

Why is there no. 3,6,12 ?

Can you explain? Thank you. (:

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@Alicia: Since

A= { 1, 2, 3, 4, 6, 8, 12 }A′ = { 5, 7, 9, 10, 11, 13, 14, 15 }B= { 3, 6, 9, 12, 15 }Q6(a)(iii) asks for the union of sets

A′ andB.By definition,

A′ ∪Bcontains elements present inA′ or inBor in bothA′ &B→ basically putting all elements ofA′ andBtogether.So since 3, 6, 12 belongs to Set

Bthey are contained inA′ ∪Bas shown in the diagram below:曜

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Hello. Just wanted to check that whether for Q3bib) the y intercept I can leave the answer in coordinate form like (0,-2)

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Hi, if I didn't put any curly wordy line for the equal sign but I stated (3sf) or (nearest dollars) etc. will I still be penalise?

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@Junwei: Not using that curly wordy approximation sign (≈) is fine as long as you have stated that it's an approximation to 3sf, 1 d.p. etc.

Even some TYS do this 😛

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Hi. My friends told me thour ans script MUST be tied in order. But I tied my graph qns, which is qn 7 at the last page. Is is alright?

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@Rachel: You are the same Rachel right? 😛

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@Miss Loi: Heehe yeah. Sorry, my phone gt prob and i dunno why it posted twice. But thanks again, Miss Loi! 😀

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Hi!

If my Venn diagram already had a mistake from start, hence affecting all of my answers in the later parts of te Venn diagram question, will I get any working marks?

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@Xan: For Q6(a), it's a bit hard to allocate any method mark when (ii) and (iii) only carry a mark each 🙁

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For Question 5, if one of the angles is proven wrongly does it mean that all my subsequent answers will be incorrect even if the methods used or angle is correct?

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@Jim Li: Not sure which angle let you down (∠

ADE?) but parts (i) & (ii)(b) appear to contain a method mark or two to salvage something.曜

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@Miss Loi: Hi, yup for angle ADE and BAD if they are proven wrongly, would there be any marks for the subsequent angles?

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@Jim Li: If any, probably only in part (ii)(b), and only if it's clear that it was simply due to the wrong value of ∠

ADE/BADin your concise working that led directly to the wrong answer.曜

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@Miss Loi: Hi, you mean that even if the answer and method is correct for finding angle AEB , i will not receive any mark?

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@Jim Li: If you're referring to part (i) which carries 4 marks, you should get 1 mark for the correct answer for ∠

AEB. With two other marks lost through your wrong answers for∠ADE/BAD, Miss Loi believes the remaining mark may be allocated towards your reasons/intermediate step(s) required for each answer (which you will only get ifallthe relevant properties are stated correctly).曜

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@Miss Loi: Hi, sorry for continuously troubling you but may i inquire about the mark allocation for 5a ii?

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@Jim Li: Please note that the mark scheme isn't set by Miss Loi so no way she can be 100% sure about marks allocation.

But having say that, Q5(a)(ii)(a) seems to be a do-or-die "correct answer or zero" 1 mark case, while Q5(a)(ii)(b) might still give you a mark for correct method with wrong values/ECF etc.

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@Miss Loi: Also, if the answer for angle ADE is correct with wrong reasoning, will i receive an answer mark?

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@Jim Li: Hard to say, depends if the answer mark is tied to having the correct working leading to it or whether a benefit of doubt is given. But Miss Loi thought you said your ∠

ADEwas wrong?!曜

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Hi miss loi. If I put units inside the matrices for qn 5aii and 5aiii, will I be given the marks?

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@Raymond: You mean Q6(b)(ii) & 6(b)(iii)?

Aiyoh ... a matrix is strictly an array of

numbersso don't think that is condoned 🙁曜

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@Miss Loi: How likely will they condoned the mistake?

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@Raymond: P (putting units inside matrices is acceptable in the O Level mark scheme) ≈ 0

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Oh and I thought for the median marks should be only to accuracy of 0.5? cos this is the case in actual...

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@Raymond: For the median in Q10(a)(i)(a), a crossed-eye-free estimate somewhere between 37 and 37.5 (inclusive) will do 😉

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@Miss Loi: So is 37 acceptable?

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@Raymond: 37 is unlikely to be accepted by an examiner (with no crossed-eye tendencies) as a good estimate since the line clearly doesn't touch the 37 mark in the chart 🙁

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Sorry Ms Loi >.< For question 3), is it okay to put (0,-2) as the y-intercept instead of -2?

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Hi just wanna ask, whether Q3bib) the y interct I could leave the answer in coordinate form like (0,-2). Thnks!

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@Navin: It appears that a significant number of students are asking this with regards to the format of the

y-intercept answer in Q3(b)(i)(b).Strictlyspeaking, they-intercept should be they-value (i.e. −2) of the coordinates of the intercept point.Leaving the answer in coordinate form (0, −2) will really depend on whether this is condoned in the mark scheme (something which an eternally optimistic Miss Loi thinks they will 🙂 )

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@Miss Loi: One last question, for the graph in part (b)where we are told to draw the graph, it is 3 marks. How are the 3 marks allocated. My graph wasn't drawn neatly . Will they penalize marks? Though my answers in parts (c) (d) and (e) were correct.

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@Navin: Please note that the mark scheme isn't set by Miss Loi so no way she can be 100% sure about marks allocation.

For Q7(b), however, an educated guess could be 2 marks for all points plotted correctly and joined by a smooth curve + 1 mark for correctly drawing and labeling the axes to the required scale.

So your "wasn't drawn neatly" could fall under the "smooth curve" part but it's kinda hard for Miss Loi to objectively gauge how badly drawn your graph is.

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Hi,

Im very worried about qn 2d. I got my a) b) and c) right. Just that i somehow got 2d wrong. I used cosine rule for triangle ECD... So i may have gotten some values wrong. Ended up getting DE = 12.9 or something. So would i get 0/3m for this? Or would i get method mark for using cosine rule?

8e) I showed my working : 2000/161.8, somehow i evaluated it wrongly cause i may have typed it in cal wrongly. Would i be given 1m for this?

10a)i)a) Median i put as 37.5?

10b)ii)b) Probability of getting neither right handed female, if i got my ans as 11/12, would i get 0/2m? Or is there any chance of 1m?

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@Jsn: For Q2(d), if you read through how method marks are awarded, they are awarded based on the correct

applicationof the formula rather simply stating the use of the formula.From your description, you appear to have used the cosine rule on the wrong triangle so, strictly speaking, no method mark will be awarded if this is the case 🙁 (Miss Loi hopes she's wrong here for your sake)

For Q8(e), it depends on the mark scheme but since there are 2 marks in this part, there's a chance to for a 1 mark redemption here.

For Q10(a)(i)(a), 37.5 should be fine ^^

For Q10(b)(ii)(b), again it ultimately depends on the mark scheme but for probability questions like this one where you need to get the concept right from the start, it's often a case of 0 or full marks 🙁

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Hi Ms Loi ,

For qn 4c i) I wrote my ans as 8.14X10^13 instead of 8.1356x10^13. minus marks for that ?

Qn 10b) i counted the total group of students as 17 instead of 16 , thus i got all the answers wrong for 10b , will i get any marks for that qn ?

I did not put the curly sign or (3.sf) for most of the qns . Will i still get penalised for that ?

Thanks 😀

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@Hari: Q4(c)(i): At most 1 mark for accuracy.

Q10(b): This

reallywill depend on the benevolence of the mark scheme & your examiner *sweats*For curly

~~fries~~signs *sorry, feeling hungry replying comments in the middle of the night*, read this ^^曜

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Hi Ms Loi, for question 4ciii) is it okay to write "Assume one month has 31 days" and divide the value by (31x12) instead of 365?

And for question 7c), will marks be deducted for not showing "Hence y=0"?

Lastly, what do you think this year A1 mark will be?

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@Tom: Q4(c)(iii): Miss Loi have come across some arguing that 1 year should be 365.25 days, but by assuming 1 month = 31 days ⇒ 1 year = 31 × 12 = 372 days you are committing a seditious act of Monthistism in discriminating against the lesser months of Feb/April/June/Sep/Nov!!! 😮

Q7(c): Would think the 2 marks here goes towards the correct values for

xobtain directly from the graph, rather than anything else.曜

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Will there be error carried forward marks? For example, for the venn diagram, if I drew it wrongly.

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@Cindy: Generally, ECFs are a possibility whenever a part carries >1 mark.

But Q6(a)(ii) and (iii) contain only 1 mark each so it's doubtful if you can get anything here due to your Venn Diagram error 🙁

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For last probability qn, i thought it meant it cannot be both right handed female. The qn is phrased such that it is actually trying to convey that cannot have AT LEAST ONE right handed female....... 🙁

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wow thanks miss loi for the answers, are the qn paper available to be posted too for easy x ref thanks for the great work

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@weng: See this 🙁

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Hey, just wondering if I did not add units for most of my answers, will they minus 1 mark for all? What if for those "prove" questions?

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hey if i draw my venn diagram wrong..like if it was written D={factors of 6} and i wrote the multiples and drew it and there were in total 5 parts each carrying 1.5 marks expect one which was carrying 1.... i knew exactly what to do... but silly mistakes T.T