My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Thanks Tree2Thanks
  • 1 Post By Denis
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
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?
Ganesh Ujwal is offline  
 
April 25th, 2018, 07:24 PM   #2
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,797
Thanks: 715

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Ganesh Ujwal View Post
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.

Does this answer your question? I'm not sure.

-Dan
topsquark is offline  
April 25th, 2018, 10:11 PM   #3
Senior Member
 
Joined: Aug 2014
From: India

Posts: 238
Thanks: 1

Quote:
Originally Posted by topsquark View Post
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.
Ganesh Ujwal is offline  
April 26th, 2018, 01:29 AM   #4
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,165
Thanks: 867

Quote:
Originally Posted by Ganesh Ujwal View Post
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:
$P_2 = 32$
Country Boy is offline  
April 26th, 2018, 01:41 AM   #5
Senior Member
 
Joined: Aug 2014
From: India

Posts: 238
Thanks: 1

Quote:
Originally Posted by Country Boy View Post
O......
Your reply is confusing, I am satisfied with my answer.
Ganesh Ujwal is offline  
April 26th, 2018, 05:57 AM   #6
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 12,421
Thanks: 832

GOOD! Thread closed!
Thanks from jonah
Denis is offline  
April 26th, 2018, 10:19 AM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 18,962
Thanks: 1606

Quote:
Originally Posted by Ganesh Ujwal View Post
$\dfrac{PR}{Q^2} = R$ (R is constant)
That doesn't hold for $(P,\, Q,\, R) = (24,\, 6,\, 3)$.

In contrast, $\dfrac{PR}{Q^2} = 2$ does work (unless $Q = 0$).
Thanks from topsquark
skipjack is offline  
April 26th, 2018, 12:13 PM   #8
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,797
Thanks: 715

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Ganesh Ujwal View Post
$\dfrac{PR}{Q^2} = R$ (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.
topsquark is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
problem, ratio, solve



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
A Ratio Problem magicmetal03 Elementary Math 2 July 5th, 2017 01:29 AM
Ratio Problem marcus2015 Elementary Math 1 April 23rd, 2015 10:37 AM
Ratio problem jon619 Algebra 2 October 1st, 2013 01:12 PM
Ratio Problem. Please Help. homeschooling Elementary Math 2 June 5th, 2013 03:34 PM
Solve the problem in ratio ngm Algebra 2 January 21st, 2010 03:05 AM





Copyright © 2018 My Math Forum. All rights reserved.