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08 May 2024, 01:30
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A
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Difficulty:
95%
(hard)
Question Stats:
48% (02:18) correct
52%
(02:43)
wrong
based on 230
sessions
History
Date
Time
Result
Not Attempted Yet
If \(6a=3b=7c\) and \(5b=8a+4c\), what is the value of \(a+b+c\)?
A. 0
B. 16
C. 17
D. 42
E. Cannot be determined from the given information
A. 0
B. 16
C. 17
D. 42
E. Cannot be determined from the given information
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Official Solution:
If \(6a=3b=7c\) and \(5b=8a+4c\), what is the value of \(a+b+c\)?
A. 0
B. 16
C. 17
D. 42
E. Cannot be determined from the given information
Start with \(6a = 3b = 7c\). The least common multiple of 6, 3, and 7 is 42, so we can write: \(6a = 3b = 7c = 42x\), for some number \(x\). Thus, \(a = 7x\), \(b = 14x\), and \(c = 6x\).
Substituting these values into \(5b=8a+4c\) we get:
\(5(14x)=8(7x)+4(6x)\);
\(70x=80x\);
\(10x=0\);
\(x=0\)
As \(x = 0\), we find that \(a = b = c = 0\) and \(a + b + c = 0\).
Answer: A
If \(6a=3b=7c\) and \(5b=8a+4c\), what is the value of \(a+b+c\)?
A. 0
B. 16
C. 17
D. 42
E. Cannot be determined from the given information
Start with \(6a = 3b = 7c\). The least common multiple of 6, 3, and 7 is 42, so we can write: \(6a = 3b = 7c = 42x\), for some number \(x\). Thus, \(a = 7x\), \(b = 14x\), and \(c = 6x\).
Substituting these values into \(5b=8a+4c\) we get:
\(5(14x)=8(7x)+4(6x)\);
\(70x=80x\);
\(10x=0\);
\(x=0\)
As \(x = 0\), we find that \(a = b = c = 0\) and \(a + b + c = 0\).
Answer: A
General Discussion
08 May 2024, 01:46
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Bunuel
\(6a = 3b\)
\(a = \frac{1}{2 }b\)
\(3b = 7c\)
\(c = \frac{3}{7}b\)
\(5b=8a+4c\)
\(5b=8*(\frac{1}{2 }b)+4*(\frac{3}{7}b)\)
\(5b=4b+\frac{12}{7 }b\)
\(b = \frac{12}{7 }b\)
This is only possible when \(b = 0\)
Hence, \(a = b = c = 0\)
\(a + b + c = 0\)
Option A
