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Argument questions 23/03

As.1997

Could someone please explain the question(s) below - I will add some more later.
Please refer to the attachment(s).

For Q2
1) Why are options B and E wrong?
2) Why is option C correct? As far as I have understood the passage, I thought it was telling us that altruism exists and the example of birds tells us that animals show some form of altruism. However, the passage claims the traditional evolutionary theory states animals are never altruistic (selfless). Therefore, does it not make sense that reciprocal altruism is "not" compatible with traditional evolutionary theory?

(edited 4 years ago)

s1, 08, Q2 .png159.7KB

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Reply 1

nerak99

Original post by As.1997

you need to move this to the correct forum

Reply 2

mqb2766

Original post by As.1997

The second half of the paragraph fully supports C. They would only be altruistic if there was something in it for them - reciprocal.
B - unrelated to the text
E - no real mention of survival of the fittest.

Reply 3

As.1997

Original post by mqb2766

For option C, doesn't the fact that the birds give warning calls and put themselves at personal risk indicate an act of selflessness i.e. altruism?

Reply 4

mqb2766

Original post by As.1997

For option C, doesn't the fact that the birds give warning calls and put themselves at personal risk indicate an act of selflessness i.e. altruism?

Yes, but it's defined as reciprocal altruism in the text. They're safer overall.

Reply 5

As.1997

Original post by mqb2766

Yes, but it's defined as reciprocal altruism in the text. They're safer overall.

Fair enough. So I believe it is the idea that their act of altruism is not true altruism since it is carried out in the hope that they will get protection back.

(edited 4 years ago)

Reply 6

As.1997

These are some more questions

For Q11
The answer is D. My question is what is option D even saying to us and how did they get this conclusion?
What I do understand is the table shows a higher risk for left-handed people for breast cancer in premenopausal women (2.41), which reduces postmenopause (1.12). It also shows cancer risk increases after having children (1.58).

For Q13
The answer is E which means statements 2 and 3 are correct. My question is why is option 3 correct?
According to the passage, 47% thought air travel should be limited but only 15% were willing to fly less often. Therefore, the 15% who are not willing to fly less often may not fall into the 47% who think air travel should be limited. Therefore, shouldn't option 3 be wrong???

For Q17
The answer is A. My question is what is statement 2 on about?

For Q18
The answer is 2. Not sure how you get this answer without finding the minimum droplets needed. (Note the question asks for the maximum no. of droplets needed).

For Q21
The answer is C. I got A which is wrong and I'm not sure why. Also, not sure why C is correct.

For Q22
The answer is D. Why is E wrong? Is it because E is stated in the passage already?

For Q26
The answer is D. Not sure what statement 1 is saying.

(edited 4 years ago)

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s1, 08, Q13.png220.9KB

s1, 08, Q17.png227.7KB

s1, 08, Q18.png75.7KB

S1, 08, Q21.png97.9KB

S1, 08, Q22.png76.8KB

S1, 08, Q26.png193.8KB

Extract for Q11, s1, 08 .png177.0KB

Extract for Q21-22, s1, 08 .png317.0KB

Reply 7

mqb2766

Can you do one at a time pls? Start with 11. This is a classic correlation/causal argument. You'd not expect left handedness to cause cancer, but there is a correlation. For D, the question tells you to assume the fact is true, so it says there is some hormonal change which causes both left handedness and cancer. This would cause the correlation mentioned in the question, so D is good.

(edited 4 years ago)

Reply 8

As.1997

Original post by mqb2766

That makes sense :smile:

(One at a time would be perfect)

Reply 9

mqb2766

Q13) 15℅ were willing to fly less often, so 85℅ were not willing. There must be an overlap with the 47% who thought it should be limited?

Reply 10

As.1997

Original post by mqb2766

Q13) 15℅ were willing to fly less often, so 85℅ were not willing. There must be an overlap with the 47% who thought it should be limited?

That makes sense too.
Btw I just thought of it as a worst-case scenario -- all 15% fell in the 47% category. This still leaves 32% which say flights should be limited but are not willing to fly less.

(edited 4 years ago)

Reply 11

mqb2766

Original post by As.1997

That makes sense too.
Btw I just thought of it as a worst-case scenario -- all 15% fell in the 47% category. This still leaves 32% which would said flights should be limited but are not willing to fly less.

Exactly.

Reply 12

mqb2766

Q18) you have to think about min/max volumes first, then the maximum number of droplets to give less than 1℅.
*) The volume of the water is 5 -15 cm^3. The concentration must be less than 1℅. What chemical volume should we use?
*) The volume of one droplet is 0.015-0.025 cm^3. What volume should we use to have an upper bound on the number of droplets?

(edited 4 years ago)

Reply 13

As.1997

Original post by mqb2766

1) We want the max volume of water --> this is 15cm^3.
2) We want the droplet size to be the smallest possible --> 0.015cm^3
The resulting volume would be 15cm^3 of water + xcm^3 of chemical

(edited 4 years ago)

Reply 14

mqb2766

Original post by As.1997

1) We want the max volume of water --> this is 15cm^3.
2) We want the droplet size to be the smallest possible --> 0.015cm^3
The resulting volume would be 15cm^3 of water + xcm^3 of chemical

The number of droplets = chemical volume / droplet volume
The question has the potential to be misleading as you want the maximum number to always ensure less than 1℅ concentration. The 1℅ concentration is the key.
You've maximised the numerator. If this was 5cm^3 of water, the concentration would be 3℅, hence invalid.

Reply 15

As.1997

Original post by mqb2766

Isn't the equation supposed to be something like this:

0.015x/(15+0.015) <1/100
0.015x/(15.015) <0.01
If it was this, then x would have to be less than 2 for this equation to work -- so this must be wrong.
But I don't know where I'm going wrong with this.

(edited 4 years ago)

Reply 16

mqb2766

Original post by As.1997

Isn't the equation supposed to be something like this:

0.015x/(15+0.015) <1/100
0.015x/(15.015) <0.01

You've assumed the wrong volumes. To write it like that it would be something like
n*v/(V+n*v) <= 0.01
Or
n <= V/(99v)
What volumes v,V would we use to ensure
n = V/(99v)
Has the smallest value? That is the only way to guarantee 1℅ concentration.

The maximum part comes from saying we want equality or as close to it as possible.

(edited 4 years ago)

Reply 17

As.1997

Original post by mqb2766

If I do 0.025x/5 <0.01
x=2 but aren't we finding the minimum number of droplets?
(I'm lost :l)

Reply 18

mqb2766

Original post by As.1997

If I do 0.025x/5 <0.01
x=2 but aren't we finding the minimum number of droplets?

No, it's the maximum number of droplets that has at most 1℅ concentration. The minimum is 0 droplets. Any more than that will exceed the 1℅ concentration in some cases.

So minimum V and maximum v.

(edited 4 years ago)

Reply 19

As.1997

I think I get what the question is asking. It is asking what is the safest maximum number of droplets that we can put to ensure we are not going above 1%. Therefore, to find this safe maximum that would be what we usually refer to the minimum number of droplets whereby we would use the upper bound for the droplet and lower bound for the water.
So that equates to:

0.025x/(5+0.025x) <0.01 (I guess we can simplify 5+0.025x to 5 because in the grand scheme it is negligible)
Therefore x=2 by trial and error.

(technically saying minimum is wrong as the minimum safest is 0)