Gmat_mission wrote: ↑Fri Nov 20, 2020 1:47 am
A \(1\)-meter tree limb was divided and labeled by fifths and sixths. The limb was then cut at each label. If pieces between \(\dfrac19\) and \(\dfrac1{20}\) of a meter are kept, how many pieces are kept?
a. 0
b. 2
c. 4
d. 6
e. 30
Answer:
C
Solution:
The markings on the tree limb by fifths are 1/5, 2/5, 3/5, and 4/5. Similarly, the markings by sixths are 1/6, 2/6 = 1/3, 3/6 = 1/2, 4/6 = 2/3, and 5/6. Including 0 and 1 (the two ends of the limb) and arranging these numbers in ascending order, we have:
0, 1/6, 1/5, 1/3, 2/5, 1/2, 3/5, 2/3, 4/5, 5/6, 1
Now, let’s find the difference of every pair of consecutive numbers (i.e., 1/6 - 0, 1/5 - 1/6, 1/3 - 1/5, etc.):
1/6, 1/30, 2/15, 1/15, 1/10, 1/10, 1/15, 2/15, 1/30, 1/6
The numbers in the above list are the lengths of the pieces being cut into. Of these numbers, 1/6 and 2/15 are greater than 1/9, and 1/30 is less than 1/20. Thus, they are not between 1/9 and 1/20. The remaining 4 numbers (two 1/15’s and two 1/10’s) are. So 4 pieces are kept.
Answer: C 
