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Is there a simpler way to solve this question

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by [email protected] » Wed Apr 17, 2013 6:24 pm
225. 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
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Source: — Problem Solving |

by srcc25anu » Wed Apr 17, 2013 6:55 pm
At 1/3rds, the markings would be at 1/3, 2/3 and 1
At 1/4ths. the markings would be at 1/4, 1/2, 3/4 and 1

lets take the LCM of 3 and 4 that will be 12
Now denoting all the above lengths in terms of denominator as 12:

4/12, 8/12, 12/12 for one-third markings
3/12, 6/12, 9/12 and 12/12 for one-forth markings

different lengths could now be easily calculated as 1/12, 2/12 = 1/6 and 3/12 = 1/4
hence Ans D
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by Anju@Gurome » Wed Apr 17, 2013 7:30 pm
[email protected] wrote:225. 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?
Image
Refer to the figure above,
B, D and F are the marks at the fourths.
C and E are the marks at the thirds.

Now, we will only have three different lengths. They are,
  • 1. AB = FG = 1/4
    2. CD = DE = (1/3)/2 = 1/6
    3. BC = EF = (1/3 - 1/4) = 1/12
The correct answer is D.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
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