A magician marks off on a stick of length 1 yard in thirds and fifths

Question

A magician marks off on a stick of length 1 yard in thirds and fifths and breaks the stick at the marked points. what is the maximum number of pieces which are equal in length?

(A) 2
 (B) 3
 (C) 4
 (D) 5
 (E) 6

shashankism · Accepted Answer

Since LCM of 3 and 5 is 15 
So, let the length of stick be 15 units.
So, on division in thirds, stick will be broken at 5,10 units from left position 
Again on division in fifths , stick will be broken at 3,6,9,12 units from left position.

So, lengths of parts will be 3-0, 5-3, 6-5, 9-6, 10-9, 12-10, 15-12 = 3,2,1,3,1,2,3

So the maximum number of equal parts are there each equal to 3 units.

Answer B

rohit8865 · Answer

Please follow attached fig. 
Let length be LCM of 5,30 = 150
equal lengths are marked

Ans B

elegantm · Answer

Dear Nidhi,

Though I am not a MOD, A piece of advice.....

Kindly search the forum before posting.....Your patience would be appreciated by everyone else and such approach would be handy for you.

tapabrata · Answer

Since we are marking at the thirds and fifths, we take a stick length equal to the LCM of (3 and 5) i.e. 15 for ease of calculation/depiction.

So we will have 15 equal lengths of 1 unit each. I will represent each dash as 1 unit
_ _ _||_ _|_||_ _ _||_|_ _||_ _ _

I have marked the third positions by a single vertical line and the fifths by double vertical line. Let us list down the number of sticks of different lengths(count the number of dashes before a bar and in between bars to do so)

1 unit => 2
2 units => 2
3 units => 3

So maximum number of sticks of equal length = 3. Answer B

Bunuel · Answer

ScottTargetTestPrep · Answer

Solution:
 
If the magician only marks the stick by thirds, we have (including 0 and 1): 0, 1/3, 2/3, and 1. If he only marks the stick by fifths, we have (including 0 and 1): 0, 1/5, 2/5, 3/5, 4/5, and 1. Since he does both, these fractions in ascending order (including 0 and 1, but counting them only once) are:

0, 1/5, 1/3, 2/5, 3/5, 2/3, 4/5, and 1.

The lengths of the pieces can be found by finding the difference of two consecutive numbers in the above list. That is:

Piece 1 = 1/5 - 0 = 1/5

Piece 2 = 1/3 - 1/5 = 2/15
 
Piece 3 = 2/5 - 1/3 = 1/15

Piece 4 = 3/5 - 2/5 = 1/5

Piece 5 = 2/3 - 3/5 = 1/15

PIece 6 = 4/5 - 2/3 = 2/15

Piece 7 = 1 - 4/5 = 1/5

We see that 3 of the 7 pieces have a length of 1/5, 2 have a length of 1/15 and the remaining 2 have a length of 2/15. Therefore, the maximum number of pieces which are equal in length is 3.

Answer: B

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A magician marks off on a stick of length 1 yard in thirds and fifths

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