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smodak
Detailed video solution explaining the method to approach question rightly is mentioned below
The primary method to be used to answer the question is explained after 2:18 minutes though it's recommended that viewers watch it completely.
Answer: Option E
18 Jan 2021, 23:33
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Bunuel
Bunuel what if the tree was 2 meters? would we still take a line and divide it with the LCM 15? And once we get 3/5 we would do 3/5 * 15 = 9meters?
kaladin123
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20 Aug 2021, 19:00
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I got this question wrong in an MGMAT CAT.
However, I was able to solve it later during review.
My approach was to draw a number line and mark off the thirds and the fifths, as shown below:
nl.jpg [ 8.46 KiB | Viewed 1445 times ]
I used the decimal equivalents to help me arrange the fractions on the number-line.
Now, it is tempting to calculate the differences of the decimal values. The challenge with this approach is that fractions like 1/3, 2/3, etc. might lead to rounding errors later.
In summary, I think that Bunuel's approach using LCM is the the perfect way to solve such a problem.
The same number line, now on a scale of [0-15]:
nl2.jpg [ 4.94 KiB | Viewed 1451 times ]
Difference={3, 2, 1, 3, 1, 2, 3}
We need to take away the 1st piece of each distinct length.
Fraction of original branch left=\(\frac{3+1+2+3}{15}=\frac{9}{15}=\frac{3}{5}\)
However, I was able to solve it later during review.
My approach was to draw a number line and mark off the thirds and the fifths, as shown below:
Attachment:
nl.jpg [ 8.46 KiB | Viewed 1445 times ]
Now, it is tempting to calculate the differences of the decimal values. The challenge with this approach is that fractions like 1/3, 2/3, etc. might lead to rounding errors later.
In summary, I think that Bunuel's approach using LCM is the the perfect way to solve such a problem.
The same number line, now on a scale of [0-15]:
Attachment:
nl2.jpg [ 4.94 KiB | Viewed 1451 times ]
Difference={
We need to take away the 1st piece of each distinct length.
Fraction of original branch left=\(\frac{3+1+2+3}{15}=\frac{9}{15}=\frac{3}{5}\)
