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

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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>.
  
 
===Solution 2 (Guess and Check)===
 
===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 = 12. \boxed{D}</math>
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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
 +
 
 +
~Education, The Study of Everything
 +
 
 +
==Video Solution 2==
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https://youtu.be/rbKRwUubUWo
 +
 
 +
~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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