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555-605 Level|   Fractions and Ratios|                        
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Walkabout
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

This problem is best solved by setting up a diagram to represent the markings. We are given that we are marking the pipe into 3rds and 4ths. So we will have:

1/3, 2/3, 3/3, and ¼, 2/4, ¾, 4/4

However, for these “markings” to be more meaningful, we should create a common denominator. Because the denominators are 3 and 4, we know that the common denominator is 12. Converting each fraction, we have:

4/12, 8/12, 12/12 = 1, and 3/12, 6/12, 9/12, 12/12 = 1

Putting these in order we have:

3/12, 4/12, 6/12, 8/12, 9/12, 1

We now label this on a diagram:



We are then asked, 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?

To determine this, we need to calculate the difference between each consecutive pair of markings. This is represented in another diagram below.



1) 3/12 – 0 = 3/12 = ¼

2) 4/12 – 3/12 = 1/12

3) 6/12 – 4/12 = 2/12 = 1/6

4) 8/12 – 6/12 = 2/12 = 1/6

5) 9/12 – 8/12 = 1/12

6) 1 – 9/12 = 3/12 = ¼

Thus, the different lengths are 1/12, 1/6, and ¼.

The answer is D.
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Walkabout
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


Generally fast way to solve such problem is writing the different marks in ascending/descending order with same denominator:

Here 4th : 0/4, 1/4, 2/4, 3/4, 4/4
and 3rd : 0/3, 1/3, 2/3, 3/3

Now with understood common denominator 12 write the numbers : for 4th : 0,3,6,9,12 and for 3rd : 0,4,8,12

Now comine : 0,3,4,6,8,9,12

Now find the cut with denominator 12 (Substracrt adjacent terms : 3/12, 1/12, 2/12, 1/12,3/12 i.e. 1/4, 1/12 and 1/6 after removing duplicates.
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Bunuel
Walkabout
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

Since we want to find the fractions, we can assume some other length of the pipe which will make calculation easier. Take the length of the pipe to be 12-meter long (the least common multiple of 3 and 4.

In this case the branch would be cut at 3, 4, 6, 8, and 9 meters (in black are given fourths of the length and in red thirds of the length).

Distinct lengths would be: 3=3/12=1/4, 4-3=1=1/12 and 6-4=2=2/12=1/6 meters long pieces.

Answer: D.


Hope it helps.

Hi,

Request you could explain the answer in more detail? I did not understand how the branch would be cut at 3,4,6,8,9.

since there are two markings, there would be three distinct pieces of the branch right? 1/4th part, 1/12th part [1/3 - 1/4] and i cannot understand about how the third part is 1/6th?
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Bunuel
Walkabout
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

Since we want to find the fractions, we can assume some other length of the pipe which will make calculation easier. Take the length of the pipe to be 12-meter long (the least common multiple of 3 and 4.

In this case the branch would be cut at 3, 4, 6, 8, and 9 meters (in black are given fourths of the length and in red thirds of the length).

Distinct lengths would be: 3=3/12=1/4, 4-3=1=1/12 and 6-4=2=2/12=1/6 meters long pieces.

Answer: D.


Hope it helps.

Hi,

Request you could explain the answer in more detail? I did not understand how the branch would be cut at 3,4,6,8,9.

since there are two markings, there would be three distinct pieces of the branch right? 1/4th part, 1/12th part [1/3 - 1/4] and i cannot understand about how the third part is 1/6th?

Imagine that we have 12-meter long pipe.

Cut in fourths means that it's cut at 1/4th, at 2/4th and at 3/4th. Thus at 3, 6, and 9 meters.
Cut in thirds means that it's cut at 1/3rd, and at 2/3rd Thus at 4 and 8 meters.

So, it would be cut at 3, 4, 6, 8, and 9 meters.

Does this make sense?

In my post above there are similar questions to practice. Please go through them.
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maibhihun
Walkabout
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


Generally fast way to solve such problem is writing the different marks in ascending/descending order with same denominator:

Here 4th : 0/4, 1/4, 2/4, 3/4, 4/4
and 3rd : 0/3, 1/3, 2/3, 3/3

Now with understood common denominator 12 write the numbers : for 4th : 0,3,6,9,12 and for 3rd : 0,4,8,12

Now comine : 0,3,4,6,8,9,12

Now find the cut with denominator 12 (Substracrt adjacent terms : 3/12, 1/12, 2/12, 1/12,3/12 i.e. 1/4, 1/12 and 1/6 after removing duplicates.


Now with understood common denominator 12 write the numbers : for 4th : 0,3,6,9,12 and for 3rd : 0,4,8,12 ( Where is 0,3,6,9,12) and 0,4,8,12 coming from for four and 3 it seems like it should be reverse to me, obviously i am mistaken, but why is this done like this?
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maibhihun
Walkabout
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


Generally fast way to solve such problem is writing the different marks in ascending/descending order with same denominator:

Here 4th : 0/4, 1/4, 2/4, 3/4, 4/4
and 3rd : 0/3, 1/3, 2/3, 3/3

Now with understood common denominator 12 write the numbers : for 4th : 0,3,6,9,12 and for 3rd : 0,4,8,12

Now comine : 0,3,4,6,8,9,12

Now find the cut with denominator 12 (Substracrt adjacent terms : 3/12, 1/12, 2/12, 1/12,3/12 i.e. 1/4, 1/12 and 1/6 after removing duplicates.


Now with understood common denominator 12 write the numbers : for 4th : 0,3,6,9,12 and for 3rd : 0,4,8,12 ( Where is 0,3,6,9,12) and 0,4,8,12 coming from for four and 3 it seems like it should be reverse to me, obviously i am mistaken, but why is this done like this?

Imagine that we have 12-meter long pipe.

Cut in fourths means that it's cut at 1/4th, at 2/4th and at 3/4th. Thus at 3, 6, and 9 meters.
Cut in thirds means that it's cut at 1/3rd, and at 2/3rd Thus at 4 and 8 meters.

So, it would be cut at 3, 4, 6, 8, and 9 meters.

Does this make sense?

In my post above there are similar questions to practice. Please go through them.
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I don't think you need to calculate anything here, unless I found the correct answer in random. This was the process I followed:

1) I drew a straight line for the pipe.
2) I marked the 1/4ths. This means I marked it in 3 places, as you will see in the drawing below (|).
3) I marked the 1/3rds. This means I marked it in 2 places, as you will see in the drawing below (!).

|_______|__!_____|__!_____|_______|

So, now we see the thirds and the fourths. What you see it that the "whole" pieces you see are of 3 different lenghts.
The 1/4 is seen in the begining and the end.
There is no whole 1/3 anywhere. But, there are 2 other lengths dividing the fourths: a smaller one and a larger one, marked by the !.

In other words, we need 3 different lengths:
One will be the 1/4
None will be 1/3
There will be 2 other lengths.

Only D satisfies this observation, so ANS D.
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