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 April 25th, 2018, 07:14 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 295 Thanks: 1 How to solve this ratio problem? P varies directly with the square of Q when R is constant and inversely with R when Q is constant when Q = 6 and R = 3, then P = 24, Find P when Q = 8 and R = 4. Solution: P varies directly with $Q^2$ (when R is constant): $\dfrac{P}{Q^2} = c$ for some constant c. P varies inversely with R when Q is constant: PR = d (for some other constant d). Put together, $\dfrac{PR}{Q^2} = k$ for some constant. Or, $P = k\dfrac{Q^2}{R}$. How "c" is taken as constant instead of R in first step?
April 25th, 2018, 07:24 PM   #2
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Quote:
 Originally Posted by Ganesh Ujwal P varies directly with the square of Q when R is constant and inversely with R when Q is constant when Q = 6 and R = 3, then P = 24, Find P when Q = 8 and R = 4. Solution: P varies directly with $Q^2$ (when R is constant): $\dfrac{P}{Q^2} = c$ for some constant c. P varies inversely with R when Q is constant: PR = d (for some other constant d). Put together, $\dfrac{PR}{Q^2} = k$ for some constant. Or, $P = k\dfrac{Q^2}{R}$. How "c" is taken as constant instead of R in first step?
For the first part we have $\displaystyle \frac{PR}{Q^2} = a \implies \frac{P}{Q^2} = \frac{a}{R}$ when R is constant. But we can't use this in the second part where R is not constant. So instead we just use $\displaystyle \frac{P}{Q^2} = c$ and let the R's figure themselves out in the end.

-Dan

April 25th, 2018, 10:11 PM   #3
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Quote:
 Originally Posted by topsquark For t.....sure.
I know you are confused, but right solution is this:

$P\, \alpha\, Q^2$ (P varies directly with the square of Q)

$P \,\alpha\, \dfrac{1}{R}$ ( inversely with R)

$P \alpha \dfrac{Q^2}{R}$ (combining them)

$\dfrac{PR}{Q^2} = R$ (R is constant)

$\dfrac{P_1 R_1}{Q_1^2} = \dfrac{P_2 R_2}{Q_2^2}$

$\dfrac {24(3)}{6^2} = \dfrac {P_2}{8^2}$

$P_2 = 32$

Last edited by Ganesh Ujwal; April 25th, 2018 at 10:15 PM.

April 26th, 2018, 01:29 AM   #4
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Quote:
 Originally Posted by Ganesh Ujwal I know you are confused, but right solution is this: $P\, \alpha\, Q^2$ (P varies directly with the square of Q)
Or $P= aQ^2$ where a might involve variables other than P and Q (like R).

Quote:
 $P \,\alpha\, \dfrac{1}{R}$ ( inversely with R)
Or $P= \frac{b}{R}$ where b might involve variables other than P and R (like Q).

Putting those together $P= \frac{kQ^2}{R}$ where k might involve variables other than P, Q or R. But since P, Q, and R are the only variables in this problem, we can treat k as a constant.

Quote:
 $P \alpha \dfrac{Q^2}{R}$ (combining them) $\dfrac{PR}{Q^2} = R$ (R is constant)
Since you already have "R" as a variable, you should not call the separate constant "R"!

Quote:
 $\dfrac{P_1 R_1}{Q_1^2} = \dfrac{P_2 R_2}{Q_2^2}$
This is true but I dislike writing proportions like this.

Instead, going back to $P= \frac{kQ^2}{R}$, since we are told that "when Q = 6 and R = 3, then P = 24" we have $24= \frac{k(6^2)}{3}= 12k$ so k= 24/12= 2. $P= \frac{2Q^2}{R}$. When Q= 8 and R= 4, $P= \frac{2(8^2)}{4}= \frac{128}{4}= 32$.

Quote:
 $P_2 = 32$

April 26th, 2018, 01:41 AM   #5
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Quote:
 Originally Posted by Country Boy O......

 April 26th, 2018, 05:57 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,892 Thanks: 882 GOOD! Thread closed! Thanks from jonah
April 26th, 2018, 10:19 AM   #7
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Quote:
 Originally Posted by Ganesh Ujwal $\dfrac{PR}{Q^2} = R$ (R is constant)
That doesn't hold for $(P,\, Q,\, R) = (24,\, 6,\, 3)$.

In contrast, $\dfrac{PR}{Q^2} = 2$ does work (unless $Q = 0$).

April 26th, 2018, 12:13 PM   #8
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Quote:
 Originally Posted by Ganesh Ujwal $\dfrac{PR}{Q^2} = R$ (R is constant) $\dfrac{P_1 R_1}{Q_1^2} = \dfrac{P_2 R_2}{Q_2^2}$
You are incorrect about the top line. The only reason you ended up with the correct answer is that the correct equation is
$\displaystyle \frac{P}{Q^2} = \text{(constant)} R$ where R is constant and that the constant cancels out in the ratio in the second line. Again this only works for the first step. Your equation top equation is not valid for the second step where R is not constant. The ratio approach hides this fact and as skipjack shows this does not always give the correct answer. You got the correct answer in this case but with the wrong reasoning.

-Dan

Last edited by topsquark; April 26th, 2018 at 12:19 PM.

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