Biology paper 3 2019 A level question

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Sid Nandula
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#1
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#1
A lung cancer patient received 15 mg of Trexall per week. After treatment, the
diameter of his lung tumour was 35.8mm
Assuming the tumour was spherical, use the mean percentage change in tumour
volume shown in Figure 10 to calculate the volume of the patient’s tumour before
treatment with Trexall.
The formula for the volume of a sphere is 4
3
πr3 where π = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.
https://www.physicsandmathstutor.com...019%2520QP.pdf
Last edited by Sid Nandula; 3 months ago
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Bookworm_88
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#2
Woah, you're already started doing past papers (paper THREE even!)
Out of curiousity have you teachers finished all the content yet?
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Jess_Lomas
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#3
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#3
(Original post by Sid Nandula)
A lung cancer patient received 15 mg of Trexall per week. After treatment, the
diameter of his lung tumour was 35.8mm
Assuming the tumour was spherical, use the mean percentage change in tumour
volume shown in Figure 10 to calculate the volume of the patient’s tumour before
treatment with Trexall.
The formula for the volume of a sphere is 4
3
πr3 where π = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.
https://www.physicsandmathstutor.com...019%2520QP.pdf
Personally I would use the 35.8 to find the volume afterwards then use the graph to find the change and apply that to the value you get for the volume
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Bookworm_88
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#4
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(Original post by Sid Nandula)
A lung cancer patient received 15 mg of Trexall per week. After treatment, the
diameter of his lung tumour was 35.8mm
Assuming the tumour was spherical, use the mean percentage change in tumour
volume shown in Figure 10 to calculate the volume of the patient’s tumour before
treatment with Trexall.
The formula for the volume of a sphere is 4
3
πr3 where π = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.
https://www.physicsandmathstutor.com...019%2520QP.pdf
In terms of the question:
- Calculate the end size of the tumour after treatment (thy give you the diameter of tumour post-treatment and you need radius!)
- From there, I've checked the markscheme and Idk how they got that answer

(Original post by Jess_Lomas)
Personally I would use the 35.8 to find the volume afterwards then use the graph to find the change and apply that to the value you get for the volume
that's what I did
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BlueChicken
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#5
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#5
(Original post by Sid Nandula)
A lung cancer patient received 15 mg of Trexall per week. After treatment, the
diameter of his lung tumour was 35.8mm
Assuming the tumour was spherical, use the mean percentage change in tumour
volume shown in Figure 10 to calculate the volume of the patient’s tumour before
treatment with Trexall.
The formula for the volume of a sphere is 4
3
πr3 where π = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.
https://www.physicsandmathstutor.com...019%2520QP.pdf
Take it step-by-step. First, determine the size of the tumour after treatment in mm3 (as that's what answer units are) using the formula and information given. Second, use dosage information from question to read mean percentage change in tumour volume from graph. Third, apply this % change to the volume you determined originally to give the starting volume (which is what the question wants you to find).
Last edited by BlueChicken; 3 months ago
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hello_m
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(Original post by Bookworm_88)
In terms of the question:
- Calculate the end size of the tumour after treatment (thy give you the diameter of tumour post-treatment and you need radius!)
- From there, I've checked the markscheme and Idk how they got that answer


that's what I did
at first i was very confused because i was multiplying V2 by 1.25 but after talking this out with someone, i understand where i made my mistake so i'll try and explain ( i know how frustrating it can be when the MS doesn't explain the answer )

for reference
V1 - volume before treatment
V2 - volume after treatment

they've given us diameter and formula for a sphere so we can calculate V2 = 24011.95 mm^3

the graph indicates a 25% decrease in volume after treatment i.e.
V2 = V1 x 0.75
rearrange to get
V1 = V2 / 0.75

V1 = 24011.95 / 075
= 32015.93 ≈ 32016 mm^3

hope this helps
Last edited by hello_m; 4 weeks ago
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