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2019 AMC 10A Problems/Problem 3: Difference between revisions

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The answer is <math>16-4 = 12 \implies \boxed{\textbf{(D)}}.</math>
The answer is <math>16-4 = 12 \implies \boxed{\textbf{(D)}}.</math>
==Solution 2 (Guess and Check)==
Simple guess and check works. Start with all the square numbers - <math>1</math>, <math>4</math>, <math>9</math>, <math>16</math>, <math>25</math>, <math>36</math>, etc. (probably stop at around <math>100</math> since at that point it wouldn't make sense). If Ana is <math>9</math>, then Bonita is <math>3</math>, so in the previous year, Ana's age was <math>4</math> times greater than Bonita's. If Ana is <math>16</math>, then Bonita is <math>4</math>, and Ana's age was <math>5</math> times greater than Bonita's in the previous year, as required. The difference in the ages is <math>16 - 4 = 12</math>.


==See Also==
==See Also==

Revision as of 12:16, 28 July 2019

Problem

Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

Solution

Solution 1

Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,

\[A-1 = 5(B-1)\] and \[A = B^2.\]

Substituting the second equation into the first gives us

\[B^2-1 = 5(B-1).\]

By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$

The answer is $16-4 = 12 \implies \boxed{\textbf{(D)}}.$

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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