mang kanor has some 20 and 10 coins. The total amount of these coins is at most 800 pesos.Suppose there are 40 10 coins. What is the possible number of 20 coins that Mang Kanor has?
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Solving problems involving Linear inequalities in two variables po
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Answer:
Let x be the number of 20 coins that Mang Kanor has. Then, the number of 10 coins he has is 40, since it is given in the problem.
The total value of the coins can be expressed as:
20x + 10(40) ≤ 800
Simplifying and solving for x:
20x + 400 ≤ 800
20x ≤ 400
x ≤ 20
Therefore, the possible number of 20 coins that Mang Kanor has is any integer from 0 to 20, inclusive.