Inversely proportional question

Original post by Fizzle-A

\displaystyle V=\frac{1}{P} \cdot k

That is inversely proportional.

Reply 2

"Inversely proportional" means that as one thing gets twice as big, the other thing gets twice as small, or if one thing gets ten times as big, the other thing gets ten times as small, etc.

Mathematically, if our things are x and y, then x is proportional to 1/y, or x = k (1/y), where k is some constant number (called the constant of proportionality).

In your case, the things are pressure and volume, so P = k (1/V), where k is some number. You don't need to know k to finish the question off!

Reply 3

Original post by RDKGames

\displaystyle V=\frac{1}{P} \cdot k

That is inversely proportional.

How do I now solve the question? Do I work out K first?

1.25V = K/P ????

Reply 4

Original post by Fizzle-A

How do I now solve the question? Do I work out K first?

1.25V = K/P ????

You don't need to know

k

. You simply need to observe what happens to P as V increases by 25%.

You should rewrite it as

\displaystyle 1.25V=\frac{1.25k}{P}

and then see what you can do the fraction on the right to get it into the form of

\displaystyle 1.25V=\frac{k}{lP}

where l will tell you the change in P.

(edited 7 years ago)

Reply 5

You don't need to work out k.

Rearrange the equation so that it says k = ...

You should now be able to write one equation for P and V before the change in volume and P and V after the change in volume.

Remember that k is constant, so it will be the same before and after the change in volume. Now solve your equations simultaneously.

Reply 6

Original post by Pangol

Thank you for explaining it.

V = K (1/P)
V increases by 25% to become 1.25V
1.25V = K (1.25/P) ??

Is this the answer, P also increases by 1.25 ?

I don't feel that that is correct because from your explanation earlier you said 'Inversely proportional' means "that as one thing gets twice as big, the other thing gets twice as small" ... so should P decrease by 25%?

Reply 7

Original post by RDKGames

You don't need to know

k

. You simply need to observe what happens to P as V increases by 25%.

You should rewrite it as

\displaystyle 1.25V=\frac{1.25k}{P}

and then see what you can do the fraction on the right to get it into the form of

\displaystyle 1.25V=\frac{k}{lP}

where l will tell you the change in P.

How did you get it into the form 1.25V = K/lP ??

Reply 8

Original post by Fizzle-A

How did you get it into the form 1.25V = K/lP ??

Are you aware of the simple technique where you can divide the numerator and the denominator by the same number?

So

\displaystyle \frac{a}{b}=\frac{\frac{a}{c}}{ \frac{b}{c} }

Apply the same idea here by dividing top and bottom by 1.25 and expressing the denominator as a decimal which should allow you to see the decrease in P.

Reply 9

1.25V = 1.25K/P

I am aware that you can divide the numerator and denominator by the same number :u:

1.25V = 1.25K/P
If I divide the fraction on the right by 1.25 then I get:

1.25V = K/1.25P ??

Reply 10

Original post by Fizzle-A

You are correct when you say that this is not correct! But you have the right idea.

If V has got 1.25 times larger, then P has got 1.25 times smaller.

What is P divided by 1,25?

Reply 11

Original post by Fizzle-A

1.25V = 1.25K/P

I am aware that you can divide the numerator and denominator by the same number :u:

1.25V = 1.25K/P
If I divide the fraction on the right by 1.25 then I get:

1.25V = K/1.25P ??

That's wrong because you dividing the top but multiplied the bottom by 1.25

Reply 12

Original post by Pangol

You are correct when you say that this is not correct! But you have the right idea.

If V has got 1.25 times larger, then P has got 1.25 times smaller.

What is P divided by 1,25?

= P/1.25 ?

... So if I use V = k(1/P)
1.25V = 1.25k/P
1.25V = k/P/1.25...

k = 1.5625V / P - does this have any relevance?

(edited 7 years ago)

Reply 13

Original post by Fizzle-A

= P/1.25 ?

... So if I use V = k(1/P)
1.25V = 1.25k(1/P)
1.25V = k(1/P/1.25)...

Yes now turn P/1.25 into P multiplied by some decimal. So in other words, what is 1/1.25 as a decimal?

Reply 14

Original post by RDKGames

Yes now turn P/1.25 into P multiplied by some decimal. So in other words, what is 1/1.25 as a decimal?

0.8 ... therefore P will decrease by 20%?

Reply 15

Original post by Fizzle-A

0.8 ... therefore P will decrease by 20%?

Precisely. :smile:

Reply 16

Original post by Fizzle-A

= P/1.25 ?

... So if I use V = k(1/P)
1.25V = 1.25k/P
1.25V = k/P/1.25...

k = 1.5625V / P - does this have any relevance?

Not exactly wrong, I think, but hugely and unecessarily overcomplicated. If you multiply V by 1.25, you have to divide P by 1.25. So the question is saying, what is the new pressure, in terms of the old pressure P, when this happens. The answer is P/1.25. But there is a slightly nicer way to write P/1.25, as P times something, rather than P divided by something,

Reply 17

Original post by RDKGames

Precisely. :smile:

Yes! Thank you for helping me out :biggrin:

Reply 18

Original post by Fizzle-A

0.8 ... therefore P will decrease by 20%?

I think you posted this while I was writing my last reply... This is it!

Reply 19