If 1/5 of a cluster of Panopyras flew to the Viper tree, one-third flew to the Leucomelas tree, three times the difference of these two numbers flew to a Prolemurius tree, and one Panopyra continued to fly about, attracted on each side by the fragrant Teylu and the Dinicthoid, what was the number of Panopyras?
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neat picture courtesy of Brenda Adams
test
I am not sure what is meant by ‘on each side’. I understand that 1 panopyra was left.
(1/5 + 1/3 + 3(1/3 – 1/5)x + 1 = x
x = 15
Ok, I think it can be solved like this:
T= Total of Panopyras
1/5T fly to the VP tree
1/3T fly to the L tree
2/5T = 3(1/3- 1/5)Tfly to the P tree
1 keeps wandering around
From this you now that 14/15T = (3/15 + 5/15 + 6/15)T is the unknown, the remaining part thats is 1 Panopyra represent 1/15T for adding to a total number so:
1/15T = 1 therefore T = 15
Cool!
Must be 15, but my computer is telling me that a panopyra is a jellyfish. How do they fly?
15 Panopyras flew in this orchard of succulent delights.
(3 to the Viper, 5 to the Leucomelas, 6 to the Prolemurius, and 1 remained)
if x is the total number (we need to find)
1/5 x fly off, another 1/3 x fly off to another tree, and then 3 times this difference
( 3 (1/3 x – 1/5 x ) fly off to another , and 1 is left flying about
so this gives us the following equation x the total is equal to all those we are told fly off:
x = 1/5 x + 1/3 x + 3 (1/3 x – 1/5 x ) +1
Solving for x gives us a total of
15
15
1/5 = 3/15
1/3 = 5/15
5/15-3/15 = 2/15
3*(2/15) = 6/15
1 – 3/15 – 5/15 – 6/15 = 1/15
1/15 = 1 Panopyras