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April 25th, 2018, 07:14 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 238 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  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Does this answer your question? I'm not sure. Dan  
April 25th, 2018, 10:11 PM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 238 Thanks: 1  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  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867  Quote:
Quote:
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:
Quote:
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:
 
April 26th, 2018, 01:41 AM  #5 
Senior Member Joined: Aug 2014 From: India Posts: 238 Thanks: 1  
April 26th, 2018, 05:57 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,421 Thanks: 832 
GOOD! Thread closed!

April 26th, 2018, 10:19 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606  
April 26th, 2018, 12:13 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \frac{P}{Q^2} = \text{(constant)} R$ 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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