can anyone please answer and explain this
A little monkey had 75 peaches.
Each day, he kept a fraction of his peaches, gave the rest away, and then ate one.
These are the fractions he decided to keep:
1/2,1/4,3/4,3/5,5/6,11/15
In which order did he use the fractions so that he was left with just one peach at the end?
75 * 11/15 = 55 - 1 = 54
54 * 5/6 = 45 - 1 = 44
44 * 3/4 = 33 - 1 = 32
32 * 1/2 = 16 - 1 = 15
15 * 3/5 = 9 - 1 = 8
8 * 1/4 = 2 - 1 = 1
my name
thanks john
i dont really get it (sad life)!!!
I can’t figure out the last question to this it dosent make sense
help meh
To solve this problem and determine the order of fractions used by the little monkey, we need to work backwards from the end result of having only one peach left.
Let's start by considering the last fraction used, which leaves the monkey with one peach. The monkey ate one peach after the last fraction, so before that, he should have had two peaches.
Now, let's look at the fractions available. We need to find a fraction that, when the monkey keeps it, results in him having two peaches. Looking at the given fractions, 11/15 is the only fraction that, when kept, would yield two peaches.
Next, we consider the peaches the monkey had before using the fraction 11/15. The monkey ate one peach after using this fraction, so before that, he should have had three peaches.
Again, we go through the remaining fractions to find the one that, when kept, results in three peaches. Among the available fractions, 5/6 is the only one that fulfills this condition.
Continuing this process, we determine the order of fractions that will give the monkey the desired number of peaches at each step:
11/15 (leaves 2 peaches)
5/6 (leaves 3 peaches)
3/5 (leaves 4 peaches)
3/4 (leaves 5 peaches)
1/4 (leaves 6 peaches)
1/2 (leaves 7 peaches)
So, the order in which the monkey used the fractions, starting from 75 peaches, is:
1/2, 1/4, 3/4, 3/5, 5/6, 11/15.
This sequence allows the little monkey to end up with just one peach.