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

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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15</math>
 
<math>\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15</math>
  
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==Solution 1==
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==Solution==
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===Solution 1===
 
Let <math>A</math> be the age of Ana and <math>B</math> be the age of Bonita. Then,
 
Let <math>A</math> be the age of Ana and <math>B</math> be the age of Bonita. Then,
  
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By using difference of squares and dividing, <math>B=4.</math> Moreover, <math>A=B^2=16.</math>
 
By using difference of squares and dividing, <math>B=4.</math> Moreover, <math>A=B^2=16.</math>
  
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The answer is <math>16-4 = 12 \implies \boxed{\textbf{(D)}}.</math>
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The answer is <math>16-4 = \boxed{\textbf{(D) }12}</math>.
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===Solution 2 (Guess and Check)===
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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 = \boxed{\textbf{(D) }12}</math>.
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===Solution 3 (Answer Choices)===
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The second sentence of the problem says that Ana's age was once <math>5</math> times Bonita's age. Therefore, the difference of the ages <math>n</math> must be divisible by <math>4.</math> The only answer choice which is divisible by <math>4</math> is <math>\boxed{\textbf{(D) }12}</math>.
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~awesome_weisur
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==Video Solution 1==
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https://youtu.be/vNRY85ORir4
  
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Solution by Baolan
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~Education, The Study of Everything
  
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==Solution 2 (Guess and Check)==
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==Video Solution 2==
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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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https://youtu.be/rbKRwUubUWo
  
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iron
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 12:02, 16 July 2024

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 $16-4 = \boxed{\textbf{(D) }12}$.

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 $16 - 4 = \boxed{\textbf{(D) }12}$.

Solution 3 (Answer Choices)

The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $\boxed{\textbf{(D) }12}$.

~awesome_weisur

Video Solution 1

https://youtu.be/vNRY85ORir4

~Education, The Study of Everything

Video Solution 2

https://youtu.be/rbKRwUubUWo

~savannahsolver

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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