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P6 SPAR Worksheet5. Q3. -- Here's the proof that there is definitely more than one possible set of answers
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P6 SPAR Worksheet5

Question3

A courier company charged $18 for every large parcel and

$10 for every small parcel safely delivered.  

However, it paid paid a penalty of $20 for every lost or damaged parcel regardless of size.  

If the company delivered 145 parcels (inclusive of safely-delivered or damaged)

and collected an amount of $1030, how many of each type of parcels did it deliver safely?  

(ie. how many Large, Small and Damaged parcels were delivered)

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My Solution by Algebra

Let S be no. of small parcels delivered safely

Let L be no. of large parcels delivered safely

Let D be no. of lost or damaged parcels

18L + 10S - 20D = 1030 …. (1)

L + S + D =  145 …. (2)

From (2):

D = 145 - L - S  …. (3)

Substitute (3) into (1):

18L + 10S - 20(145 - L - S) = 1030

       18L + 10S - 2900 + 20L + 20S = 1030

   38L + 30S = 3930

Simplifying further to make S the subject,

It turns out that:

    S  =  131  -   L x 19/15 .

By observation of the above equation,

so long as L is a multiple of 15

and that the expression " L x 19/15 " does not exceed 131,

there will be a set of solution for the problem.

ie.  L must be less than or equal to 131x15/19 or 103

Which means, L can take the values of 15, 30, 45, 60, 75 and 90

… and hence there will be 6 corresponding sets of answers for S and D we have to accept.  

… End of Proof …

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