Question 1207394: Sofia has 45 markers and 60 crayons that she wants to put into
boxes.
• All the boxes will have an equal number of markers.
• All the boxes will have an equal number of crayons.
What is the greatest number of boxes Sofia can make using all
the markers and crayons?
Found 4 solutions by Theo, MathTherapy, ikleyn, Edwin McCravy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 60 crayons and 45 marker.
the ratio of crayons to markers is 60 / 45 = 4/3 = 1.25
that's 1.25 crayons for every marker.
if you assume 1 marker per box, then the number of crayons has to be 1.25 per box which is not possible because integers are required.
if you assume 2 markers per box, then the number of crayons has to be 2.5 per box which is not possible becuse integers are required.
if you assume 3 markers per box, then the number of crayons has to be 3 per box which is possible because integers are reuired and that's what you have.
1.25 * 1 = 1.25
1.25 * 2 = 2.5
1.25 * 3 = 3
the ratio is 1.25 to 1
so smallest number of markers per box is 3 which requires 4 crayons per box which keeps the ratio at 4 to 3 which simplifies to 1.25 to 1.
since the largest number of boxes is when each box has the smallest number of crayons and markers, you have 4 crayons and 3 markers per box which gets you 15 boxes total because 15 * 4 = 60 crayons and 15 * 3 = 45 maekrs.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Sofia has 45 markers and 60 crayons that she wants to put into
boxes.
• All the boxes will have an equal number of markers.
• All the boxes will have an equal number of crayons.
What is the greatest number of boxes Sofia can make using all
the markers and crayons?
Greatest number of boxes she can place all markers and crayons in: GCF (Greatest
Common Factor) of 45 and 60, which is 15
That's ALL!!
Answer by ikleyn(52780)  (Show Source):
You can put this solution on YOUR website! .
I am glad that tutor @MathTherapy came with right solution.
Surely, the problems like this one require analysis using integer numbers only.
The reasons with decimals, as used by the other tutor, are not appropriate.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
Let M = the number of markers in each a box
Let C = the number of crayons in each a box
Let N = the number of boxes.
So N(M+C) must equal the total number of markers and crayons.
N(M+C) = NM + NC = 45 + 60, so N must be a factor of both 45 and 60
For N to be as large as possible, it must be the greatest common factor
of 45 and 60, which is 15.
N(M+C) = 15M + 15C and 15M=45 and 15C=60, or M=3 and C=4
So there are 15 boxes, with 3 markers and 4 crayons in each box.
Edwin
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