Ximena L.

asked • 03/09/16

how many cups of tea did each have?

A café sold tea at 30 cents a cup and cakes at 50 cents each. Everyone in a group had the same number of cakes and the same number of cups of tea. (Note: This is not to say that the number of cakes is the same as the number of teas.) The bill came to $13.30. How many cups of tea did each have?

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David W. answered • 03/09/16

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Let:
  t = total number of cups of tea
  T = number of cups of tea per person (same for everyone)
  c = total number of cakes
  C = number of cakes per person (same for everyone)
  n = number of people in the group
 
Translate:
  "tea at 30 cents a cup and cakes at 50 cents each"  "bill came to $13.30"    means
            $0.30t + $0.50c = $13.30
  "Everyone in a group had the same number of cakes and the same number of cups of tea"   means
          cost of tea     +      cost of cake      =    $13.30
           $0.30nT       +     $0.50nC            =   $13.30
            n ($0.30T    +    $0.50C)          = $13.30
Math:
  Just like a positional number system, we may let n go from 1 to 44 (note: if 44 people got only tea for $0.30, that would be $13.20), let T go from 0 to 44 (1 person with 44 cups of tea), and C go from 0 to 26 (note: 1 person with 26 cakes is $13.00).  We then find which sets of (n,C,T) = $13.30.
 
Alas, that gives us only:
   n       C        T
  ----   ----    -----
   1       1       26
   1       6       23
   1      11      20
   1      16      17
   1      21      14
   1      26      11
   1      31        8
   1      36        5
   1      41        2
   7        3        2
 
Now, we could have set the range for n to be 2 to 44 (since "group" implies more than one person), but it is o.k. to get these answers just to check that the cost is $13.30.
 
A group of 7 people bought 3 cups of tea each and 2 cakes each for a total of $13.30.   There is no other possibility.
       7 * 3 * $0.30  +  7 * 2 * $0.50
           $6.30  +  $7.00
               $13.30
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Alan G.

I have read your second comment and you are being a little harsh and abrasive in your explanation. Please keep in mind your target audience and that you are trying to be helpful and not assert your ego over the student and other tutors.
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03/10/16

David W.

Very frequently, people assign motives and baggage to my very factual comments:
 
  -  there is only one answer  --  fact; clearly demonstrated in the answer
  -  a mundane repeat of the standard mediocre problems on this forum -- fact; it is very easy to create a frequency distribution of problem types (note: any non-math problems really stand out); I have collected questions and answers to prepare a data base for an interactive LMS to teach math (starting with Algebra I);  anyone would get the same frequency results
  -  "this is an excellent problem"  -- my opinion, based on the previous observation  (partially because this problem is unusual and causes students to think)
 
Early in my career, I read Gerald Weinberg's excellent book, The Psychology of Computer Programming, and adopted the "egoless programming" commandments;  see --
   http:  //  blog.  codinghorror.  com  /the-ten-commandments-of-egoless-programming/
 
I would not write, "Now, could you solve this with the information you gave in the problem? You need to know something more about t and c."   The factual answer is "yes" and this statement is demeaning.  Note:  the person is not the process is not the product;  with good people, a better process generally produces a better product.  It is very, very different to write, "Your second and third sentences seem to be of little help here" as opposed to "The second and third sentences seem to be of little help here."
 
THX for you observations; I'm working to make it clearer when I presenting verifiable facts; I never want to be charged with displaying "ego" and using inappropriate tone; I desire to take the role of tutor (mentor, coach), rather than teacher (who must grade answers).
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03/11/16

Alan G. answered • 03/09/16

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Let c be the number of cakes and t the number of cups of tea.
 
The cost of the food is
 
30t + 50c = 1330.
 
Now, could you solve this with the information you gave in the problem? You need to know something more about t and c. Your second and third sentences seem to be of little help here. Could you be more specific about what these say?
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Ximena L.

That is all the problem provides of information.
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03/09/16

Alan G.

I am sorry, but my reply and the previous gentleman's reply (David W.) show that your question is missing some information or may be wrong. You should check the source of the problem and be sure that everything you have given here is ALL there is. If there isn't something missing, the source of the problem is a bad one.
 
I apologize for the confusion this has cause for you. Keep trying.
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03/10/16

David W.

 

My thought was that this is an excellent problem (not just a mundane repeat of the standard mediocre problems on this forum).

There is only one answer (7,3,2).  So, why can’t people solve this with the information given?

 The multiples of 30 are:
   0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360, 390, 420, 450, 480, 510, 540, 570, 600, 630, 660, 690, 720, 750, 780, 810, 840, 870, 900, 930, 960, 990, 1020, 1050, 1080, 1110, 1140, 1170, 1200, 1230, 1260, 1290, 1320    --- One of these MUST be the total amount paid for tea.
 
The multiples of 50 are:
   0, 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600, 650, 750, 800, 850, 900, 1000, 1050, 1100, 1150, 1200, 1300   -- One of these MUST be the total amount paid for cakes.
 
The total amount the group paid for cakes + total amount paid for tea is 1330.  Let’s consider amounts for cakes first and determine whether the remaining amount is in the list for tea.
 
t-cost          c-cost       #t=t/50    #c=c/30
0                   1330            0                  
50                 1280            1                  
100              1230            2                   41
150              1180            3                  
200              1130            4                  
250              1080            5                   36
300              1030            6                  
350              980              7                  
400              930              8                   31
450              880              9                  
500              830              10                
550              780              11                 26
600              730              12                
650              680              13                
700              630              14                 21   (14=2*7 and 21=3*7; common factor of 7)
750              580              15                
800              530              16                
850              480              17                 16
900              430              18                
950              380              19                
1000            330              20                 11
1050            280              21                
1100            230              22                
1150            180              23                 6
1200            130              24                
1250            80                 25                
1300            30                 26                 1
 
The only solution is:
 700             630              14                 21   (14=2*7 and 21=3*7)
 
which means 7 people buying 2 cakes each and 3 tea each.
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03/10/16

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