Susan marks-off a 2-meter cloth into multiples of three and five. She

Question

Susan marks-off a 2-meter cloth into multiples of three and five. She then breaks the cloth along all the markings and removes one piece of every distinct length. What fraction of the original cloth remains?

A. 2/5
B. 7/15
C. 1/2
D. 8/15
E. 3/5

A. 2/5
B. 7/15
C. 1/2
D. 8/15
E. 3/5

chetan2u · Answer

Of course, the question is wrongly worded.  How can we cut a 2 units thing into MULTIPLES of 3 and 5. 
So, it is meant to be cut into thirds and fifths..

So basically we are cutting at 1/3 and 1/5 of 2..
Different locations of cut...
0, 2/5, 2/3, 4/5, 6/5, 4/3, 8/5, 1...
Distinct lengths...(2/5-0 or 2/5), (2/3-2/5 or 4/15), (4/5-2/3 or 2/15).. Rest all will again be 2/5, 2/15, 4/15

Sum of these lengths = 2/5+4/15+2/15=4/5...
Left is 2-4/5=6/5...
6/5 as a fraction of 2 is (6/5)/2=6/(5*2)=3/5..

E

abhishek31 · Answer

hi, can anyone offer an explanation for the answer?

btrg · Answer

Hey  abhishek31 , sorry to say this but there is no official explanation for this question. Please refer to prev post by  chetan2u

briantoth6 · Answer

The actual length is irrelevant, so pretend length = 1 for simplicity, and draw a line segment from 0-1.

Mark a tick at all the appropriate spots (1/3, 2/3) and (1/5, 2/5, 3/5, 4/5). You should see there are 3 distinct lengths that are formed by these cuts:

A = 1/5
B = 1/3 - 1/5 = 2/15
C = 2/5 - 1/3 = 1/15

*be careful and note that the middle section is simply a type A length (1/5).

Take the sum of A,B,C and subtract it from 1 (i.e. 100%) to get the answer.

Conversely, it is probably easier to just see it on the number line with A,B,C segments being next to each other and terminating at 2/5. (1-2/5 = 3/5)

KK12333 · Answer

someone pls explain this simply

PietroMeltzer · Answer

I only got after reading briantoth6 and chetan2u explanations

If you use 1 meter to simplify, as stated before

0 -> 1/5 -> 1/3 -> 2/5 -> 3/5 -> 2/3 -> 4/5 of the length
0 -> 0.2 -> 0.33 -> 0.4 -> 0.6 -> 0.66 -> 0.8 of the length

Now lets make some segments
A: 0 -> 1/5 = 1/5
B: 1/5 -> 1/3 = 1/3 - 1/5 = 2/15
C: 1/3 -> 2/5 = 2/5 - 1/3 = 1/15
D: 2/5 -> 3/5 = 3/5 - 2/5 = 1/5
E: 3/5 -> 2/3 = 2/3 - 3/5 = 1/15
F: 2/3 -> 4/5 = 4/5 - 2/3 = 2/15
G: 4/5 -> 1 = 1/5

So A (1/5), B (2/15) and C (1/15) form distinctive lengths
And D (1/5), E (1/15), F (2/15) and G (1/5) repeat those lengths

The questions asks "What fraction of the original cloth remains?"
We know that from 0 to C we have 2/5 of the original cloth cut into distinctive lengths, and then, the patterns repeat
Thus 1 - 2/5 = 3/5 or just D+E+F+G = 9/15 = 3/5

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Susan marks-off a 2-meter cloth into multiples of three and five. She

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