Walkabout wrote:
A straight pipe 1 yard in length was marked off in fourths and also in thirds. If the pipe was then cut into separate pieces at each of these markings, which of the following gives all the different lengths of the pieces, in fractions of a yard?
(A) 1/6 and 1/4 only
(B) 1/4 and 1/3 only
(C) 1/6, 1/4, and 1/3
(D) 1/12, 1/6 and 1/4
(E) 1/12, 1/6, and 1/3
This problem is best solved by setting up a diagram to represent the markings. We are given that we are marking the pipe into 3rds and 4ths. So we will have:
1/3, 2/3, 3/3, and ¼, 2/4, ¾, 4/4
However, for these “markings” to be more meaningful, we should create a common denominator. Because the denominators are 3 and 4, we know that the common denominator is 12. Converting each fraction, we have:
4/12, 8/12, 12/12 = 1, and 3/12, 6/12, 9/12, 12/12 = 1
Putting these in order we have:
3/12, 4/12, 6/12, 8/12, 9/12, 1
We now label this on a diagram:
We are then asked, if the pipe was then cut into separate pieces at each of these markings, which of the following gives all the different lengths of the pieces, in fractions of a yard?
To determine this, we need to calculate the difference between each consecutive pair of markings. This is represented in another diagram below.
1) 3/12 – 0 = 3/12 = ¼
2) 4/12 – 3/12 = 1/12
3) 6/12 – 4/12 = 2/12 = 1/6
4) 8/12 – 6/12 = 2/12 = 1/6
5) 9/12 – 8/12 = 1/12
6) 1 – 9/12 = 3/12 = ¼
Thus, the different lengths are 1/12, 1/6, and ¼.
The answer is D.
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