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Aling Maria has a candy stall in the market.
One candy cost 5 cents each, the second cost 20 cents each and the 3rd cost 30 cents each.
In her absence due to personal necessity an honest buyer picked a total of
20 pcs. of candies costing 5 cents, 20 cents and 30 cents in the stall and leave 2 pesos for their exact cost.
How many each candy of 5 cents, 20 cents, and 30 cents the honest buyer picked?
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Let x be the number of 5-cent candies;
y be the number of 20-cent candies,
and z be the number of 30-cent candies.
As you read the problem, you can write these two equations
x + y + z = 20 (pieces of candies)
5x + 20y + 30z = 200 (cents, which is 2 pesos)
To simplify, divide all the terms in the second equation by 5. You will get these two equations
x + y + z = 20 (1)
x + 4y + 6z = 40 (2)
Now subtract equation (1) from equation (2). You will get
3y + 5z = 20.
You should find the solutions to this equation in integer positive numbers, according to the problem's meaning.
Using this additional restrain, you have ONLY ONE unique solution z = 1, y = 5 (simply by "trial and error" method
which is VERY EASY in this case).
Then from equation (1), you get x = 20 - 1 - 5 = 14.
ANSWER. 14 5-cent candies; 5 20-cent candies and 1 30-cent candy.
CHECK. 14*5 + 5*20 + 1*30 = 70 + 100 + 30 = 200 cents = 2 pesos, the total cost. ! Correct !
Solved.
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Usually (and as a rule), three unknowns require three equations, in order for the problem had a unique solution.
But this problem demonstrates an exclusive case.
The problem gives only 2 equations for 3 unknowns, but the additional restrain
that the solution should be in integer positive numbers provides the UNIQUE answer (!)
So, this restrain works as a replacement to third equation (which absents (!) )