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###### Biology paper 3 2019 A level question

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1 year ago

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

(edited 1 year ago)

Reply 1

1 year ago

Woah, you're already started doing past papers (paper THREE even!)

Out of curiousity have you teachers finished all the content yet?

Out of curiousity have you teachers finished all the content yet?

Reply 2

1 year ago

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%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

Reply 3

1 year ago

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%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

Reply 4

1 year ago

Original post by Sid NandulaA 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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

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 patientās tumour before

treatment with Trexall.

The formula for the volume of a sphere is 4

3

Ļr3 where Ļ = 3.14

You can see the graph that refers to this question in Q 6.5 in the link below.

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-3%2FQP%2FJune%25202019%2520QP.pdf

Take it step-by-step. First, determine the size of the tumour after treatment in mm

(edited 1 year ago)

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

- 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 ā 32016 mm^3

hope this helps

(edited 1 year ago)

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