1/6 of a box of paperclips are red, 5/12 of them are blue and 2/5 of the remainder are black. If there are 10 black paperclips, how many more blue paper clips are there than red ones?
number of paperclips ---- x
reds ---- (1/6)x
blue ---- (5/12)x
sofar: x/6 + 5x/12 = (7/12)x
remainders ---- x - 7x/12 = 5x/12
so (2/5)(5x/12) = 10
10x/60 = 10
x/6 = 10
x = 60
reds = 60/6 = 10
blues= 5(60)/12 = 25
So there are 15 more blues than reds
check:
reds and blues = 35
remainders = 25
is (2/5) of 25 = 10 ??
YES, all is good
To find the number of red paperclips, blue paperclips, and the remainder, we need to convert the given fractions into a common denominator.
The common denominator for 6, 12, and 5 is 60.
So, let's calculate the number of red paperclips:
1/6 of the box = (1/6) * 60 = 10 red paperclips
Next, let's calculate the number of blue paperclips:
5/12 of the box = (5/12) * 60 = 25 blue paperclips
Now, let's find the remainder, which is the part of the box not accounted for by the red and blue paperclips:
Total - (Red + Blue) = Remainder
60 - (10 + 25) = 25 paperclips
According to the given information, 2/5 of the remainder are black paperclips. Since we know there are 10 black paperclips, we can set up an equation to find the value of the remainder:
(2/5) * Remainder = 10
To solve for the Remainder, we can multiply both sides of the equation by the reciprocal of 2/5, which is 5/2:
Remainder = (10 * (5/2)) / (2/5)
Remainder = 25
Now we have the values for red paperclips (10), blue paperclips (25), and the remainder (25).
To find the difference between the number of blue and red paperclips:
Difference = Blue - Red = 25 - 10 = 15
Therefore, there are 15 more blue paperclips than red ones.