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

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<cmath>A-1 = 5(B-1)</cmath>
 
<cmath>A-1 = 5(B-1)</cmath>
 
and
 
and
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<cmath>A = B^2</cmath>
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<cmath>A = B^2.</cmath>
  
 
Substituting the second equation into the first gives us
 
Substituting the second equation into the first gives us
Line 16: Line 16:
 
<cmath>B^2-1 = 5(B-1).</cmath>
 
<cmath>B^2-1 = 5(B-1).</cmath>
  
−
Solving this quadratic yields <math>B=4.</math> Moreover, <math>A=B^2=16.</math> The answer is <math>16-4 = 12 \implies \boxed{D}.</math>
+
Solving this quadratic yields <math>B=4.</math> Moreover, <math>A=B^2=16.</math>
 +
 
 +
The answer is <math>16-4 = 12 \implies \boxed{D}.</math>
  
 
Solution by Baolan
 
Solution by Baolan

Revision as of 17:11, 9 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

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

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

Solving this quadratic yields $B=4.$ Moreover, $A=B^2=16.$

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

Solution by Baolan

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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