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 November 28th, 2010, 02:42 AM #1 Member   Joined: Oct 2010 Posts: 43 Thanks: 0 Word problem After the examination session was clear that group of very good grades of students passed the session is less than 4/33 the number of students in the group, but higher than 5/42 the total number of group. What is the minimum number of students could be in group?
November 28th, 2010, 12:36 PM   #2
Math Team

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Re: Word problem

Hello, MIDI!

Quote:
 $\text{The number of students who passed the exam is less than }\frac{4}{33}$ $\text{ the number of students,}$ [color=beige]. . [/color]$\text{but greater than }\frac{5}{42}$ $\text{ the number of students.}$ $\text{What is the minimum number of students that could be in group?}$

$\text{Let }\,N \,=\,\text{ number of students in the group.}$

$\text{Then: }\;\frac{5}{42}N \;<\;\text{Passed} \;<\;\frac{4}{33}N$

$\text{And we see that }N\text{ is a multiple of 42 and a multiple of 33.}$

$\text{The smallest such number is: }\:N \:=\:\text{LCM}(42,\,33) \:=\:462$

$\text{Therefore, the smallest group is }462\text{ students.}$

 December 4th, 2010, 02:36 PM #3 Member   Joined: Nov 2010 Posts: 34 Thanks: 0 Re: Word problem i hate lcm
 December 16th, 2010, 08:06 AM #4 Member   Joined: Oct 2010 Posts: 43 Thanks: 0 Re: Word problem But answer in book is 25.
 December 16th, 2010, 06:28 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Word problem Like soroban did, use the inequality: $\frac{5}{42}N<\text{Passed}<\frac{4}{33}N$ write in approximate decimal form: $0.119N<\text{Passed}<0.121N$ Now, we need a decimal in between the two decimals with the smallest denominator... Let $\text{Passed}=.12N=\frac{3}{25}N$ thus, 25 is the smallest group that could have the required fractional portion satisfying the given conditions.
December 19th, 2010, 08:02 AM   #6
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Re: Word problem

Quote:
 Originally Posted by MarkFL Now, we need a decimal in between the two decimals with the smallest denominator... Let $\text{Passed}=.12N=\frac{3}{25}N$ thus, 25 is the smallest group that could have the required fractional portion satisfying the given conditions.
how u get from 12N=3/25 ?

 December 19th, 2010, 08:20 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Word problem $.12=\frac{12}{100}=\frac{3\cdot4}{4\cdot25}=\frac{ 3}{25}$

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