SomeOneO
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#1
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#1
Hello, I need help with a few questions:

1) Brom travels to work at 3km/h and back home at 6km/h. if he takes one hour to and from work, how many km is work from home?

My attempt: V=d/t, D=V.T, avg speed = 4.5km/h, therefore 4.5km = 2D, D = ...2.25km, which is wrong. im missing something trivial here.

2) ratio by volume b:a:p is 2:50:100. Liquid is altered such that ratio b to a is doubled whereas ratio of b to p is halved. if new liquid has has 100cc^m of a, how much of P is there?

My attempt: 1:25:50, so new liquid is 1:50:25? hence there is 25cc, which is wrong, again my im confused here

Thanks.
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ghostwalker
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#2
Report 3 years ago
#2
(Original post by SomeOneO)
Hello, I need help with a few questions:

1) Brom travels to work at 3km/h and back home at 6km/h. if he takes one hour to and from work, how many km is work from home?

My attempt: V=d/t, D=V.T, avg speed = 4.5km/h, therefore 4.5km = 2D, D = ...2.25km, which is wrong. im missing something trivial here.
Your problem is the "average speed" - it's not 4.5 km/h

If I travelled at 3km/h for an hour say, and 6km/h for another hour, then the average speed would be 4.5km/h. But here, since the distance is the same - out and then back - the times at each speed will be different.

If we let D be the distance in km to work from home, then:

1) How long does it take (in terms of D) going at 3km/h?

2) And at 6km/h?

3) And we know the sum of those two is 1. So....

2) ratio by volume b:a:p is 2:50:100. Liquid is altered such that ratio b to a is doubled whereas ratio of b to p is halved. if new liquid has has 100cc^m of a, how much of P is there?

My attempt: 1:25:50, so new liquid is 1:50:25? hence there is 25cc, which is wrong, again my im confused here

Thanks.
Since "b" is common to your two ratio changes, I would leave that unchanged and instead change the "a" and "p". Any change in "b" would effect both b:a and b : p simultaneously, and confuse things.

Since b:a is doubled, we would usually double the value of b, but since we're leaving b unchanged, instead we halve the value of a. This has the same effect on the ratio, without upsetting the b : p ratio.

Perform a similar process on b : p.

In your original answer using 1:50:25, we have 50cc of a to 25 cc of p. So, 100cc of a would have given what volume of p? Not 25.

Repeat that with the correct ratios obtained from above.
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Foreverton
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For part 1 the time there can be expressed as D/3 and the time back as D/6. Adding these expressions together gives 3D/6 which we know is equal to 1. As D/2 = 1, D = 2km. As for part 2 I believe that doubling b:a would make 4:50 from 2:50 and halving b would make 50:200 from 50:100 resulting in a ratio of 4:50:200 and there would therefore be 400 cm3 of p. Less sure about that one though
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SomeOneO
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#4
Report Thread starter 3 years ago
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(Original post by ghostwalker)
Your problem is the "average speed" - it's not 4.5 km/h

If I travelled at 3km/h for an hour say, and 6km/h for another hour, then the average speed would be 4.5km/h. But here, since the distance is the same - out and then back - the times at each speed will be different.

If we let D be the distance in km to work from home, then:

1) How long does it take (in terms of D) going at 3km/h?

2) And at 6km/h?

3) And we know the sum of those two is 1. So....



Since "b" is common to your two ratio changes, I would leave that unchanged and instead change the "a" and "p". Any change in "b" would effect both b:a and b : p simultaneously, and confuse things.

Since b:a is doubled, we would usually double the value of b, but since we're leaving b unchanged, instead we halve the value of a. This has the same effect on the ratio, without upsetting the b : p ratio.

Perform a similar process on b : p.

In your original answer using 1:50:25, we have 50cc of a to 25 cc of p. So, 100cc of a would have given what volume of p? Not 25.

Repeat that with the correct ratios obtained from above.
(Original post by Foreverton)
For part 1 the time there can be expressed as D/3 and the time back as D/6. Adding these expressions together gives 3D/6 which we know is equal to 1. As D/2 = 1, D = 2km. As for part 2 I believe that doubling b:a would make 4:50 from 2:50 and halving b would make 50:200 from 50:100 resulting in a ratio of 4:50:200 and there would therefore be 400 cm3 of p. Less sure about that one though
Thanks, so;

1) Easily seen now, t1+t2=1hr, v=d/t, sub in, ans=2km.

2) Using Ghosts method, which makes sense, I get final ratio of 2:25:200, by keeping ratio B the same to keep a and p ratios consistent. Since a = 100cc, then p = 800cc.

Thanks all, if above is still incorrect please let me know!

SomeOneO
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ghostwalker
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#5
Report 3 years ago
#5
(Original post by SomeOneO)
Thanks, so;

1) Easily seen now, t1+t2=1hr, v=d/t, sub in, ans=2km.

2) Using Ghosts method, which makes sense, I get final ratio of 2:25:200, by keeping ratio B the same to keep a and p ratios consistent. Since a = 100cc, then p = 800cc.

Thanks all, if above is still incorrect please let me know!

SomeOneO
Looks good to me.
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