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Alejandro and his children went into a grocery store and will buy apples and mangos. Each apple costs 
$2 and each mango costs 
$1. Alejandro has a total of 
$20 to spend on apples and mangos. Write an inequality that would represent the possible values for the number of apples purchased, 
a, and the number of mangos purchased, 
m.
Answer:

Alejandro and his children went into a grocery store and will buy apples and mangos. Each apple costs $2 \$ 2 and each mango costs $1 \$ 1 . Alejandro has a total of $20 \$ 20 to spend on apples and mangos. Write an inequality that would represent the possible values for the number of apples purchased, a a , and the number of mangos purchased, m m .\newlineAnswer:

Full solution

Q. Alejandro and his children went into a grocery store and will buy apples and mangos. Each apple costs $2 \$ 2 and each mango costs $1 \$ 1 . Alejandro has a total of $20 \$ 20 to spend on apples and mangos. Write an inequality that would represent the possible values for the number of apples purchased, a a , and the number of mangos purchased, m m .\newlineAnswer:
  1. Denote Variables: Let's denote the number of apples Alejandro buys as aa and the number of mangos as mm. Each apple costs $2\$2, so the total cost for apples is 2a2a. Each mango costs $1\$1, so the total cost for mangos is mm. Alejandro has $20\$20 to spend, so the combined cost of apples and mangos should be less than or equal to $20\$20.
  2. Write Inequality: We can write the inequality that represents the total spending on apples and mangos as:\newline2a+m202a + m \leq 20\newlineThis inequality shows that the amount Alejandro spends on 22 dollars per apple plus 11 dollar per mango should not exceed 2020 dollars.
  3. Check Math Errors: To check for any math errors, let's consider a simple case. If Alejandro buys no mangos m=0m = 0, he could buy up to 1010 apples a=10a = 10 since 2×10=202 \times 10 = 20, which satisfies the inequality 2a+m202a + m \leq 20. Similarly, if he buys no apples a=0a = 0, he could buy up to 2020 mangos m=20m = 20 since 1×20=201 \times 20 = 20, which also satisfies the inequality. This confirms that our inequality is set up correctly.

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