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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
Archit3110 wrote:
length of wood
from mixed fraction to fractions would be ; 21/4, 28/15, 7/2 & 49/10
take the GCF of Nr ; ( 21,28,7,49) ; i.e 7
and LCM of Dr ; ( 4,28,2,10) ; i.e 60

LCM / GCF = 60/7
60/7 * ( 315+112+210+294)/60
we get ; 937/7 ; 133
IMO C


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



Are You Up For the Challenge: 700 Level Questions

can you explain why you did the highlighted step please?
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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
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Explanation:

Method Used: HCF & LCM from Number System Theory

= HCF (21/4, 28/15, 7/2, 49/10)
= HCF(21,28,7,49)/LCM(4,15,2,10)
=7/60

Total number of pieces: 21/4 * 60/7 + 28/15 * 60/7 + 7/2 * 60/7 + 49/10 * 60/7
= 45 + 16 + 30 + 42 = 133

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Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
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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 \(= 45+16+30+42=133\)

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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
Hi,

I got it that we need to take GCD of numerator and LCM of denominator to solve the question, but what's the logic behind doing that? Can someone please explain as I am not able to understand the logic of this solution?

Thanks !
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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
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GaurSaini wrote:
Hi,

I got it that we need to take GCD of numerator and LCM of denominator to solve the question, but what's the logic behind doing that? Can someone please explain as I am not able to understand the logic of this solution?

Thanks !


we need to cut these four logs in equal length that makes minimum number of pieces.
And GCD divides the given lengths in maximum equal parts.
suppose I have three pieces of length 4,8 and 16.
I can divide these of equal length of 2 to produce (2 + 4 + 8) 14 pieces
but when I divide these by 4, GCD in this case, it produces (1+ 2 + 4) 7 pieces. this is minimum number of pieces of equal length that we can produce from these logs.
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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
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Re: Four logs of woods of lengths 5 1/4 m, 1 13/15 m, 3 1/2 m and 4 9/10 m [#permalink]
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