Farmer Frank's pumpkin patch had 360 pumpkins and he was searching for perfect pumpkins to sell at his farm stand. As he walked through his pumpkin patch, he noted the following:
Every third pumpkin had no stem.
Every sixth pumpkin was too small.
Every fourth pumpkin was too big.
Every fifth pumpkin was not perfectly round.
How many perfect pumpkins were in Farmer Frank's pumpkin patch?
Only algebraic solutions will be accepted.
Bad Pumpkins = (1/3 + 1/6 + 1/4 + 1/5)360 = ((20 + 10 + 15 + 12) / 60)360 = (57/60)360 =(19/20)360 = 342.
Perfect P umpkins = 360 - 342 = 18.
That can't be right. At least all the prime-numbered pumpkins are perfect, and there are than 70 primes less than 360 (excluding 3 and 5).
I think you just didn't carry through the procedure far enough. If you scratch out all the 1/3 and 1/4, you have counted 1/12 twice, so you need to add it back in. Same for other pairs of factors.
360
-360(1/3 + 1/4 + 1/5 + 1/6)
+360(1/12 + 1/15 + 1/18 + 1/20 + 1/24 + 1/30)
-360(1/60 + 1/72 + 1/120)
+360(1/360)
= 124
But that doesn't count #1, which isn't one of the multiples. So, I think
125 is the final count.
144. 2 ways to solve:
1st response:
Step 1: take out all the 3s, 4s and 5s
(360 - (360 / 3) - (360 / 4) - (360 / 5)) = 78
Step 2: add back in all the overlaps of 3 and 4, 3 and 5, and 4 and 5
(78 + (360 / (3 * 4)) + (360 / (3 * 5)) + (360 / (4 * 5))) = 150
Step 3: take out the overlaps of 3, 4 and 5 (which were just taken out)
(150 - (360 / (3 * 4 * 5))) = 144
2nd method:
Count all the remaining numbers in the set of 1-30:
1, 2, 7, 11, 13, 14, 17, 19, 22, 23, 26, 29 -- 12
Multiply that times the number of sets of 1-30 in 360 (i.e., 12)
12 * 12 = 144
To solve this problem algebraically, we can set up a system of equations. Let's represent the number of pumpkins in each category with variables:
Let P be the number of perfect pumpkins.
Let N be the total number of pumpkins.
Based on the given information, we can write the following equations:
1. Every third pumpkin had no stem: (N/3) pumpkins had no stem.
2. Every sixth pumpkin was too small: (N/6) pumpkins were too small.
3. Every fourth pumpkin was too big: (N/4) pumpkins were too big.
4. Every fifth pumpkin was not perfectly round: (N/5) pumpkins were not perfectly round.
We know that the sum of all these categories should be equal to the total number of pumpkins:
(N/3) + (N/6) + (N/4) + (N/5) + P = N
Now, we can simplify and solve for P:
Multiply both sides of the equation by 60 to eliminate the denominators:
20N + 10N + 15N + 12N + 60P= 60N
Combine like terms:
57N + 60P = 60N
Move the terms containing N to one side:
60P = 60N - 57N
Simplify further:
60P = 3N
Divide both sides by 3:
20P = N
So, we have found that the number of perfect pumpkins (P) is equal to one-twentieth (1/20) of the total number of pumpkins (N).
Since the problem tells us that Farmer Frank's pumpkin patch had 360 pumpkins, we can substitute this value into the equation and solve for P:
20P = 360
P = 360/20
P = 18
Therefore, there were 18 perfect pumpkins in Farmer Frank's pumpkin patch.