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September 19th, 2018, 12:18 AM  #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, 04:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,988 Thanks: 1855 
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, 10:52 AM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,632 Thanks: 954  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, 01:58 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 497  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 02:13 PM.  
September 19th, 2018, 11:47 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 20th, 2018, 12:14 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,988 Thanks: 1855 
It's correct.


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