Bunuel wrote:
Four logs of woods of lengths \(5 \frac{1}{4}\) m, \(1 \frac{13}{15}\) m, \(3 \frac{1}{2}\) m and \(4 \frac{9}{10}\) m are to be cut into smaller pieces of equal length. What is the minimum number of the small pieces possible?
A. 74
B. 105
C. 133
D. 166
E. 266
Let:
\(a=5\frac{1}{4}\)
\(b=1\frac{13}{15}\)
\(c=3\frac{1}{2}\)
\(d=4\frac{9}{10}\)
The LCM of the four denominators is 60.
Multiplying each equation by 60, we get:
\(60a=60(5+\frac{1}{4})\)
\(60a=300+15\)
\(60a=315\)
\(60a=3*3*5*7\)\(60b=60(1+\frac{13}{15})\)
\(60b=60+52\)
\(60b=112\)
\(60b=2*2*2*2*7\)\(60c=60(3+\frac{1}{2})\)
\(60c=180+30\)
\(60c=210\)
\(60c=2*3*5*7\)\(60d=60(4+\frac{9}{10})\)
\(60d=240+54\)
\(60d=294\)
\(60d=2*3*7*7\)In the resulting blue equations, the GCF on the right side is 7.
Divide each equation by 60, applying 60 on the right side to the GCF of 7:
\(60a=3*3*5*7\)\(a = (3*3*5)(\frac{7}{60}) = 45*\frac{7}{60}\)
Implication:
\(a\) can be split into 45 smaller logs, each of length \(\frac{7}{60}\)
\(60b=2*2*2*2*7\)\(b = (2*2*2*2)(\frac{7}{60}) = 16*\frac{7}{60}\)
Implication:
\(b\) can be split into 16 smaller logs, each of length \(\frac{7}{60}\)
\(60c=2*3*5*7\)\(c = (2*3*5)(\frac{7}{60}) = 30*\frac{7}{60}\)
Implication:
\(c\) can be split into 30 smaller logs, each of length \(\frac{7}{60}\)
\(60d=2*3*7*7\)\(d = (2*3*7)(\frac{7}{60}) =42*\frac{7}{60}\)
Implication:
\(d\) can be split into 42 smaller logs, each of length \(\frac{7}{60}\)
Total number of smaller logs \(= 45+16+30+42=133\)