How to divide 33 photos into 2 groups so the ratio is 4 to 7
1 answer:
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Problem 1
<h3>Answer: 26/32</h3>-----------------------
Work Shown:
There are is 1 way to get 0 tails (aka 5 heads)
There are 5 ways to flip the coins so that exactly one coin shows tails and the rest are heads.
We have 1+5 = 6 ways of getting at most 1 tail. The phrasing "at most 1" means "1 is the largest number we can get".
There are 2^5 = 32 ways to flip five coins, so there are 32-6 = 26 ways of getting at least two tails. Note the events "getting at most 1 tail" and "getting at least 2 tails" are complementary events. One or the other must happen.
The probability of getting two or more tails is therefore 26/32.
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Problem 2
<h3>Answer: 56 total villages</h3>-----------------------
Work Shown:
We'll work our way backwards starting with the last (7th) son
x = number of villages left before the 7th son gets his share
x/8 + 7 = number of villages the 7th son gets
x - (x/8+7) = 7x/8 - 7 = number of villages left after the 7th son gets his share
This amount should be 0, so,
7x/8 - 7 = 0
7x/8 = 7
7x = 7*8
7x = 56
x = 56/7
x = 8
There are 8 villages left before the 7th son gets his share; this means he gets x/8+7 = 8/8+7 = 1+7 = 8 villages. Or you can note that if there are 8 villages left, and he's the last in line, then he gets the remaining 8.
Plugging x = 8 into 7x/8 - 7 leads to 0.
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y = amount of villages left before the 6th son gets his share
y/8 + 6 = amount of villages the 6th son gets
y - (y/8+6) = 7y/8 - 6 = amount of villages left after 6th son gets his share
We match up the x = 8 found earlier with this new expression and we solve for y. This is because x represents the number of villages leftover after the 6th son gets his share. After the 6th son gets his share, that leftover x is then handled for the 7th son as discussed above.
7y/8 - 6 = x
7y/8 - 6 = 8
7y/8 = 14
7y = 14*8
7y = 112
y = 112/7
y = 16
There are 16 villages leftover before the 6th son gets his share.
This means the 6th son gets y/8 + 6 = 16/8 + 6 = 2+6 = 8 villages.
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z = amount of villages left before the 5th son gets his share
z/8 + 5 = amount of villages the 5th son gets
z - (z/8+5) = 7z/8 - 5 = amount of villages left after 5th son gets his share
You can probably spot the pattern by now. We'll follow the same idea as before. Set 7z/8-5 equal to the 16 villages and we get
7z/8-5 = y
7z/8-5 = 16
7z/8 = 16+5
7z/8 = 21
7z = 21*8
7z = 168
z = 168/7
z = 24
There are 24 villages before the 5th son gets his share.
He gets z/8 + 5 = 24/8 + 5 = 3 + 5 = 8 villages.
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I'll do one more iteration
w = amount of villages left before the 4th son gets his share
w/8 + 4 = amount of villages the 4th son gets
w - (w/8+4) = 7w/8 - 4 = amount of villages left after 4th son gets his share
Set that equal to the previous value 24 and solve for w
7w/8-4 = z
7w/8-4 = 24
7w/8 = 24+4
7w/8 = 28
7w = 28*8
7w = 224
w = 224/7
w = 32
There are 32 villages before the 4th son gets his share.
He gets w/8 + 4 = 32/8 + 4 = 4 + 4 = 8 villages.
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The rest of the steps to find the other village counts will follow this template. You should find that each son gets 8 villages. There are 7 sons, so that means there are 7*8 = 56 villages total.
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It makes sense that the answer is some multiple of 8 because when taking 1/8 of the value leads to an integer. In other words, (1/8)*x is an integer.
Furthermore, it also makes sense that each son gets a multiple of 8 villages because if let's say the first son gets 9 villages (some non-multiple of 8), then 56-9 = 47 is not a multiple of 8. Plugging that into x/8+2 for the second son will lead to some non-integer result. So it's a bit nice that everything is a multiple of 8.
A stronger verification is shown in the table below. The table keeps track of how many leftover villages there are for any given son, whether its before or after they get their share. At the top we start with 56 and then slowly make our way down to 0 villages.
