Difference between revisions of "2019 AMC 10A Problems/Problem 3"

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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 by Baolan
 
  
 
==Solution 2 (Guess and Check)==
 
==Solution 2 (Guess and Check)==
Simple guess and check works. Start with all the square numbers - 1, 4, 9, 16, 25, 36, etc. (probably cap off at like 100 since at that point it wouldn't make sense) If Ana is 9, then Bonita is 3, so their ages were 4x apart the year prior. If Ana is 16, then Bonita is 4, and thus their ages were 5x apart the year prior, which is what we are trying to find. The difference in the ages is 16 - 4 = 12.
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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>.
 
 
iron
 
  
 
==See Also==
 
==See Also==

Revision as of 20:59, 17 February 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 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 $16-4 = 12 \implies \boxed{\textbf{(D)}}.$

Solution 2 (Guess and Check)

Simple guess and check works. Start with all the square numbers - $1$, $4$, $9$, $16$, $25$, $36$, etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$, then Bonita is $3$, so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$, then Bonita is $4$, and Ana's age was $5$ times greater than Bonita's in the previous year, as required. The difference in the ages is $16 - 4 = 12$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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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