
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 18th, 2018, 11:18 PM  #1 
Member Joined: Sep 2018 From: Japan Posts: 35 Thanks: 2  Number of muffins in a bakery
In a bakery, (1/4) of the muffins for sale are banana muffins, (2/5) are walnut muffins and the rest are chocolate muffins. There are 18 more chocolate muffins than banana muffins. How many muffins are there for sale altogether? My work: Number of chocolate = C Number of walnut = W Number of banana = B [1] banana muffins are (1/4) of the total muffins [2]Walnut muffins are (2/5) of the total muffins [3]There are 18 more cocolate muffins than banana muffins. so, C = B + 18 from [1], 1  (1/4) = 3/4 muffins remaining. from [2], I'm not sure why they did (3/4)  (2/5) = (7/20). Unless the wording is suppose to be this way. "In a bakery, (1/4) of the muffins for sale are banana muffins, and the remaining (2/5) are walnut muffins and the rest are chocolate muffins." Then I got stuck from [2] by not getting to one of the correct paths. 
September 19th, 2018, 03:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2219 
The stated problem implies B = 45, W = 72, and C = 63, so there are 180 muffins for sale in total. If that isn't the intended solution, the wording of the problem needs amendment. 
September 19th, 2018, 09:52 AM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
b = t/4 w = 2t/5 c = b+18 c = t/4 + 18 [1] also: c = t  b  w c = t  t/4  2t/5 [2] [1] = [2] t/4 + 18 = t  t/4  2t/5 Solve for t  
September 19th, 2018, 12:58 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
With four unknowns you need four independent and consistent equations. $t = b + c + w.$ Notice that you were unable to translate the facts that you labelled 1 and 2 into equations. That was a clue that you had not associated each unknown with a unique symbol. $b = \dfrac{1}{4} * t.$ $w = \dfrac{2}{5} * t.$ $c = b + 18.$ Now the step that you did not understand becomes $b + c + w = t \implies c + w = t  b = t  \dfrac{1}{4} * t = \dfrac{3}{4} * t \implies$ $c = \dfrac{3}{4} * t  w = \dfrac{3}{4} * t  \dfrac{2}{5} * t = \dfrac{15  8}{20} * t = \dfrac{7}{20} * t.$ I am guessing that they then did $c = b + 18 \implies \dfrac{7}{20} * t = \dfrac{1}{4} * t + 18 \implies \dfrac{7}{20} * t  \dfrac{1}{4} * t = 18 \implies$ $\dfrac{2}{20} * t = 18 \implies t = 180.$ EDIT: I must admit I find that a very ugly way to solve this system. I would do it as follows. $b + c + w = t \implies 20b + 20c + 20w = 20t.$ $b = \dfrac{1}{4} * t \implies 20b = 5t.$ $c = b + 18 \implies 20c = 20b + 360 = 5t + 360.$ $w = \dfrac{2}{5} * t \implies 20w = 8t.$ $\therefore 20t = 5t + 5t + 360 + 8t = 18t + 360 \implies 2t = 360 \implies t = 180.$ BUT UNTIL YOU TRANSLATE THE PROBLEM INTO EQUATIONS, you do not have the luxury of figuring out a clean way to solve it. And you cannot translate into equations UNTIL YOU HAVE ASSIGNED A UNIQUE SYMBOL TO EACH UNKNOWN. Last edited by JeffM1; September 19th, 2018 at 01:13 PM.  
September 19th, 2018, 10:47 PM  #5 
Member Joined: Sep 2018 From: Japan Posts: 35 Thanks: 2  Follwing what Denis wrote. This is what I did. Number of chocolate = C Number of walnut = W Number of banana = B Number of Total = T B + W + C = T (T/4) + (2T/5) + T/4 + 18 = T T = 180. If that's wrong, please let me know. Thank you to everyone who helped 
September 19th, 2018, 11:14 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2219 
It's correct.


Tags 
bakery, muffins, number 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
The relationships between Prime number and Fibonacci number (Part 2)  thinhnghiem  Math  0  May 15th, 2018 08:07 AM 
Prove a large number of samplings converge to a set of small number of samples  thanhvinh0906  Advanced Statistics  3  August 30th, 2017 04:27 PM 
natural number multiple of another number if its digit sum equal to that number  Shen  Elementary Math  2  June 5th, 2014 07:50 AM 
Number of Necklace/Bracelets With Fixed Number of Beads  UnreasonableSin  Number Theory  2  June 13th, 2010 12:03 AM 
find unique n number combination in total n number  jsonliu  Algebra  3  May 18th, 2010 05:01 PM 