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Eight years ago, the ratio of Alex's age to Brian's age was 1:2. In four years, the ratio of Brian's age to Charlie's age will be 2:3. If the average (arithmetic mean) of the current ages of Alex and Charlie is 27 years, what is the current age of Charlie?
A. 16
B. 24
C. 30
D. 38
E. 42
A. 16
B. 24
C. 30
D. 38
E. 42
02 Feb 2024, 07:41
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Official Solution:
Eight years ago, the ratio of Alex's age to Brian's age was 1:2. In four years, the ratio of Brian's age to Charlie's age will be 2:3. If the average (arithmetic mean) of the current ages of Alex and Charlie is 27 years, what is the current age of Charlie?
A. 16
B. 24
C. 30
D. 38
E. 42
Assuming the current ages of Alex, Brian, and Charlie are \(a\), \(b\), and \(c\), respectively, we are given:
\(\frac{a-8}{b-8} = \frac{1}{2}\), which simplifies to \(b = 2 a - 8\).
\(\frac{b+4}{c+4} = \frac{2}{3}\), which simplifies to \(2c = 3b + 4 \).
\(a + c = 2*27 = 54\)
Substituting the value of \(b\) from the first equation into the second equation gives:
\(2c = 3(2a - 8) + 4 \), which simplifies to \(c = 3a - 10\)
Now, substitute the value of \(a\) from the third equation into the above:
\(c = 3(54 - c) - 10\)
\(c =38\).
Answer: D
Eight years ago, the ratio of Alex's age to Brian's age was 1:2. In four years, the ratio of Brian's age to Charlie's age will be 2:3. If the average (arithmetic mean) of the current ages of Alex and Charlie is 27 years, what is the current age of Charlie?
A. 16
B. 24
C. 30
D. 38
E. 42
Assuming the current ages of Alex, Brian, and Charlie are \(a\), \(b\), and \(c\), respectively, we are given:
\(\frac{a-8}{b-8} = \frac{1}{2}\), which simplifies to \(b = 2 a - 8\).
\(\frac{b+4}{c+4} = \frac{2}{3}\), which simplifies to \(2c = 3b + 4 \).
\(a + c = 2*27 = 54\)
Substituting the value of \(b\) from the first equation into the second equation gives:
\(2c = 3(2a - 8) + 4 \), which simplifies to \(c = 3a - 10\)
Now, substitute the value of \(a\) from the third equation into the above:
\(c = 3(54 - c) - 10\)
\(c =38\).
Answer: D
