
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
August 28th, 2017, 05:31 AM  #1 
Newbie Joined: Aug 2017 From: uk Posts: 3 Thanks: 0  Simple percentage problem on how much I can eat :)
Right. Mayonaise bottle states... it now contains 40% less fat than in the original recipe. Therefore in my mind I can now consume more and not consume any additional fat over than I would of done with the original. In my mind I can now eat an additional 2/3. Such that, if I ate 100gms before I can now eat 166gms and take in the same amount of fat. Is this correct and if so how did I arrive at this answer? (old mayo was 50% fat, new mayo is 30% fat) Thanks Paul 
August 28th, 2017, 05:55 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 823 Thanks: 335 
That is a very good approximation. An exact answer is possible only with grams of mayo evenly divisible by three. Of course, the claim on the label is undoubtedly not exact either. 100 gms old mayo * 50% fat per gm. = 50 gm. fat. 166 gms new mayo * 30% fat per gm = 49.8 gm. fat. EDIT: My tractor uses gasoline, which is far cheaper per gallon than the mayo I buy, regardless of fat content. Give it a try. Last edited by JeffM1; August 28th, 2017 at 05:58 AM. 
August 28th, 2017, 09:48 AM  #3 
Newbie Joined: Aug 2017 From: uk Posts: 3 Thanks: 0 
Any idea how to work it out?

August 28th, 2017, 09:50 AM  #4 
Newbie Joined: Aug 2017 From: uk Posts: 3 Thanks: 0 
So am I right in saying for a 40% reduction in fat I can have an extra 2/3 of it

August 28th, 2017, 12:15 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 823 Thanks: 335 
$\text {Let } p = \text { percentage of old mixture that was ingredient X.}$ $\text {Let } r = \text { percentage reduction of X in new mixture.}$ $\text {Let } q = \text { percentage of new mixture that is ingredient X.}$ $\therefore q = \dfrac{100  r}{100} * p \implies \dfrac{p}{q}= \dfrac{100}{100  r.}$ $\text {Let } f = \text { quantity of old mixture that formerly contained desired quantity of X.}$ $\text {Let } c = \text { quantity of new mixture that currently contains desired quantity of X.}$ $\therefore fp = \text { quantity desired } = cq \implies \dfrac{c}{f} = \dfrac{p}{q} = \dfrac{100}{100  r} \implies$ $\text {Percentage new required relative to old } = \dfrac{10000}{100 r}.$ And low and behold $\dfrac{10000}{100  40} = \dfrac{10000}{60} = \dfrac{1000}{6} \approx 166.66.$ That answer is a percent. I think you can see why people who have to deal with numbers a lot use ratios rather than percentages, which are a nuisance. 
August 29th, 2017, 02:18 AM  #6 
Senior Member Joined: Apr 2014 From: UK Posts: 780 Thanks: 292 
Some low fat mayo has more sugar in it, which can be far worse than the fat for your health. I'm starting to think Lard is actually the healthy option these days...


Tags 
eat, percentage, problem, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Percentage problem  which is probably really simple!!  Tubbyfej  Elementary Math  3  March 23rd, 2017 08:49 AM 
Simple percentage question  MBI  Elementary Math  6  November 9th, 2016 08:19 AM 
Simple percentage problem  AKofTroy  Elementary Math  2  July 15th, 2014 07:28 PM 
Simple Percentage Problem?  hs1103  Algebra  2  October 26th, 2011 06:53 PM 
Simple percentage problem  noon  Algebra  3  November 27th, 2007 02:56 PM 