2019 AMC 10 A
Complete problem set with solutions and individual problem pages
Ana and Bonita were born on the same date in different years, years apart. Last year Ana was times as old as Bonita. This year Ana's age is the square of Bonita's age. What is ? (2019 AMC 10A Problem, Question#3)
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Solution
Let be the age of Ana and be the age of Bonita. Then,
and .
Substituting the second equation into the first gives us .
By using difference of squares and dividing, .Moreover, .
The answer is .
Solution (Guess and Check)
Simple guess and check works. Start with all the square numbers , , , , , , etc. (probably stop at around since at that point it wouldn't make sense). If Ana is , then Bonita is , so in the previous year, Ana's age was times greater than Bonita's. If Ana is , then Bonita is and Ana's age was times greater than Bonita's in the previous year, as required. The difference in the ages is .
