The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Problem solving

A man runs to a telephone and back in 900 seconds. On the way to the telephone his speed is 5m/s and on his way back, his speed is 4m/s. find the distance to the telephone
Call the distance x.

Write an equation in x, then solve it to get the distance.
No idea but i would know if it was 200 times easier. Am i right or am I right
Original post by Greatthinker
No idea but i would know if it was 200 times easier. Am i right or am I right

I think you're trying to hint at the answer, which is not allowed, however your answer is wrong.
I've pm you instead morgan
Miatbailey did you get the question out in the end?
Reply 6
Avatar for Arieisit
Arieisit
3
Don't mean to hijack the thread but since it's problem solving too


A cancer patient is injected with 250ml of a chemotherapy drug. Every 6hrs 32% of the drug passes out of her blood system. To compensate, a further 55ml dose is given every 6 hrs.

(1)find the recurrence relation for the amount of drugs in her blood stream.
(2) remaining drugs in her system after 24hrs.

I get trouble forming the recurrence relation.

Posted from TSR Mobile
Original post by Arieisit
Don't mean to hijack the thread but since it's problem solving too


A cancer patient is injected with 250ml of a chemotherapy drug. Every 6hrs 32% of the drug passes out of her blood system. To compensate, a further 55ml dose is given every 6 hrs.

(1)find the recurrence relation for the amount of drugs in her blood stream.
(2) remaining drugs in her system after 24hrs.

I get trouble forming the recurrence relation.

Posted from TSR Mobile

xn+1=f(xn)x_{n+1} = f(x_{n}) where f(x) is a function of x

Take each 6 hours as a step and think about how the volume of dosage changes each time.
Reply 8
Avatar for Arieisit
Arieisit
3
Original post by morgan8002
xn+1=f(xn)x_{n+1} = f(x_{n}) where f(x) is a function of x

Take each 6 hours as a step and think about how the volume of dosage changes each time.


Yeah, I know what recurrence relations look like but I just get everything to add up when I try it so I know thats wrong so I'm asking for help.

I had some thing like U6=U1−73U_6 = U_1 - 73
Original post by Arieisit
Yeah, I know what recurrence relations look like but I just get everything to add up when I try it so I know thats wrong so I'm asking for help.

I had some thing like U6=U1−73U_6 = U_1 - 73

Yeah I didn't know where you got stuck.
The first part is that the drug passes out of her system. For this you need to multiply by something as 32% is lost each 6 hrs.

The second part is that you add 55 ml every 6 hours.

You can put these two effects into one relation.
Original post by Arieisit
Yeah, I know what recurrence relations look like but I just get everything to add up when I try it so I know thats wrong so I'm asking for help.

I had some thing like U6=U1−73U_6 = U_1 - 73


U0 = 250
U1 = 250*0.68 + 55

Etc
Original post by TenOfThem
...
TenOfThem I was trying to get them to work that out themselves.
(edited 9 years ago)
Original post by TenOfThem
U0 = 250
U1 = 250*0.68 + 55

Etc


Why multiply by .68 and not .32?

Posted from TSR Mobile
Original post by Arieisit
Why multiply by .68 and not .32?

Posted from TSR Mobile


U1 is how much you have at that point

If you lose 32% how much do you have
Original post by TenOfThem
U1 is how much you have at that point

If you lose 32% how much do you have


Ok so I'm at

U1=250[br][br]U2=250×0.68+55[br][br]U3=250×0.36+55[br][br]Un=250(?)+55U_1 = 250[br][br]U_2 = 250 \times 0.68 + 55[br][br]U_3 = 250 \times 0.36 + 55[br][br]U_n = 250( ? ) + 55



Posted from TSR Mobile
Original post by Arieisit
Ok so I'm at

U1=250[br][br]U2=250×0.68+55[br][br]U3=250×0.36+55[br][br]Un=250(?)+55U_1 = 250[br][br]U_2 = 250 \times 0.68 + 55[br][br]U_3 = 250 \times 0.36 + 55[br][br]U_n = 250( ? ) + 55



Posted from TSR Mobile


The u3 is nonsense

U3 has 68% of the amount that was in at hour 6 plus 55
Original post by Arieisit
Ok so I'm at

U1=250[br][br]U2=250×0.68+55[br][br]U3=250×0.36+55[br][br]Un=250(?)+55U_1 = 250[br][br]U_2 = 250 \times 0.68 + 55[br][br]U_3 = 250 \times 0.36 + 55[br][br]U_n = 250( ? ) + 55



Posted from TSR Mobile

[br]U1=250[br]U2=(0.68)U1+55[br]U3=(0.68)U2+55[br][br]U_1 = 250[br]U_2 = (0.68)U_1 + 55[br]U_3 = (0.68)U_2 + 55[br]
and so on because the same relation occurs each time.
Original post by morgan8002
[br]U1=250[br]U2=(0.68)U1+55[br]U3=(0.68)U2+55[br][br]U_1 = 250[br]U_2 = (0.68)U_1 + 55[br]U_3 = (0.68)U_2 + 55[br]
and so on because the same relation occurs each time.


Thank you for your help Morgan. I was actually helping a friend out but thanks for reminding me that I can be clueless to some A level questions even while at uni. :tongue:

Posted from TSR Mobile
Original post by Arieisit
Thank you for your help Morgan. I was actually helping a friend out but thanks for reminding me that I can be clueless to some A level questions even while at uni. :tongue:

Posted from TSR Mobile

I thought this was GCSE? Which module do people do this stuff in?
Original post by morgan8002
I thought this was GCSE? Which module do people do this stuff in?


GCSE? Nooo. What module? I'm not sure, at the moment.

Posted from TSR Mobile

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you