A \(1\)-meter tree limb was divided and labeled by fifths and sixths. The limb was then cut at each label. If pieces

[Gmat_mission]

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

Source: Princeton Review

[deloitte247]

$$Pieces\ to\ be\ kept\ between\ \frac{1}{9\ }\ and\ \frac{1}{20}$$
$$L.C.M\ \ of\ \frac{1}{9\ }and\ \frac{1}{20}=180$$
If thetree limb is cut at 5ths;
cut intervals = 180/5 = 36
Hence, the cut intervals at 6ths = 30, 60, 90, 120, 150, 180
pieces to be kept must be between
$$\frac{1}{9}\cdot180=20$$ $$=9$$
Combining values from 5th to 6ths cut intervals
30, 6, 24, 12, 18, 18, 12, 24, 6, 30

For the pieces to be kept unique, pieces to be kept are between 9 and 20. So, they are 12, 18, 24, 30 making a total of 4 pieces.
$$Answer\ is\ Option\ C$$

[Scott@TargetTestPrep]

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