Peter, Paul and you decided to divide some marbles among yourselves........
nop, i am not gonna share my marbles
Peter, paul and you decided to divide some marbles among yourselves. Suppose you got half of the marbles and one more. If Peter took one more then half of those left, and Paul had 8 remaining marbles, how many marbles were there originally ?
Please help me set this up......my mind just isn't working very good right now...lol
\(\large m - \frac{(\frac{1}{2} m + 1)}{2} + 1 = 8\)
do I multiply everything by 2 ?
wait ive just put something, lets see if it fits the given data or not lol
say, total number of marbles = m
I got half of marbles and 1 more :- \(\frac{1}{2}m + 1\)
ok......"you' has got 1/2m + 1....
and Paul has 8
after ive taken half +1, remaining marbles = \(m - (\frac{1}{2}m+1)\)
from this remaining, peter took half + 1 more, so Peter has : \(\frac{m-(\frac{1}{2}m+1)}{2} + 1\)
oh....I see
whats left is 8, so :-
\(m - ME - PETER = 8\)
m - (1/2m + 1)/(2 ) + 1 = 8.....
do I multiply everything by 2 to get rid of the fraction ?
\(m - (\frac{1}{2}m + 1) - (\frac{m-(\frac{1}{2}m+1)}{2}+1) = 8\)
solve m, yes multiply everything with 2 and see
2m - (m + 1) - 2m - m + 1 = 16 amI doing this right ?
\(m - (\frac{1}{2}m + 1) - (\frac{m-(\frac{1}{2}m+1)}{2}+1) = 8 \) \(2m - (m+2) - (m-(\frac{1}{2}m+1)+2) = 16 \)
simplify a bit
\(m - (\frac{1}{2}m + 1) - (\frac{m-(\frac{1}{2}m+1)}{2}+1) = 8 \) \(2m - (m+2) - (m-(\frac{1}{2}m+1)+2) = 16 \) \(2m - m-2 - m+(\frac{1}{2}m+1)-2 = 16 \) \(-4 + (\frac{1}{2}m+1) = 16 \)
-4 + 1/2m + 1 = 16 1/2m = 16 + 4 - 1 1/2m = 19 m = 8 Is that correct ?
you mean m = 38 ?
oh...oops...your right..... so " you " then has: 1/2m + 1 = 1/2(38) + 1 = 19 + 1 = 20 Peter has : (m - 1/2m + 1)/(2) + 1 = (38 - 1/2(38) + 1)/2 (38 - 19 + 1)/2 = 20/2 = 10 peter has 8 check.... 20 + 10 + 8 = 38 OMG...your good
nope, you're really good :) good work !!
oops paul has 8 not peter ......thank you so much for your help
you're wlcme :D
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