Mr.Hash bough some plates at a a yard sale. After arriving home he found that 2/3 of the plates were chipped, 1/2 were cracked, 1/4 were both chipped and cracked. Only 2 plates were w/o chips or cracks. How many plates did he buy in all?
Suppose there were p plates. If you count both the ships and the cracks, then you count chipped&cracked twice.
p - 2p/3 - p/2 + p/4 = 2
12p - 8p - 6p + 3p = 24
p = 24
So, 16 were chipped
12 were cracked
6 both
2 undamaged
36
To determine the total number of plates Mr. Hash bought, we need to solve this problem step by step.
Let's start by representing the total number of plates as "x".
According to the problem, "2/3 of the plates were chipped". That means (2/3) * x plates were chipped.
Next, it is stated that "1/2 were cracked". Therefore, (1/2) * x plates were cracked.
Now, we are given that "1/4 were both chipped and cracked". So, (1/4) * x plates were both chipped and cracked.
Finally, it is mentioned that "only 2 plates were without chips or cracks". This means 2 plates were not chipped or cracked.
To solve this problem, we can set up an equation:
(x - (2/3)x - (1/2)x + (1/4)x) + 2 = x
Now, we'll simplify the equation:
((2/3)x + (1/2)x - (1/4)x) + 2 = x
Combining the fractions:
((8/12)x + (6/12)x - (3/12)x) + 2 = x
Adding the x terms:
(11/12)x + 2 = x
Now we'll isolate x:
(11/12)x - x = -2
Multiplying every term by 12 to eliminate the fraction:
11x - 12x = -24
Simplifying the equation:
-x = -24
Multiplying both sides of the equation by -1:
x = 24
Therefore, Mr. Hash bought 24 plates in total.