Lets first do some calculations:
17280 = N1*N2*N3 = (2^7)*(3^3)*5
As we know that the N1, N2, N3 have a Highest common number of 12 we know that the can be showed as:
N1 = (2^2)*3*X
N2 = (2^2)*3*Y
N3 = (2^2)*3*Z
X,Y,Z are created from the left out number (2^1, 5) and 1 and therefore can hold the following values: 10, 1, 2, 5
(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.
(2) if we assume N3 to be the lower number - N1 and N2 must be multiplied by some factor, so one of them would be multiplied by 2 and the other by 5:
N1 = (2^2)*3*5
N2 = (2^2)*3*2
N3 = (2^2)*3
Therefore, sufficient.
IMO B
HCF /LCM
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sanju09
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shulapa wrote:Lets first do some calculations:
17280 = N1*N2*N3 = (2^7)*(3^3)*5
As we know that the N1, N2, N3 have a Highest common number of 12 we know that the can be showed as:
N1 = (2^2)*3*X
N2 = (2^2)*3*Y
N3 = (2^2)*3*Z
X,Y,Z are created from the left out number (2^1, 5) and 1 and therefore can hold the following values: 10, 1, 2, 5
(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.
(2) if we assume N3 to be the lower number - N1 and N2 must be multiplied by some factor, so one of them would be multiplied by 2 and the other by 5:
N1 = (2^2)*3*5
N2 = (2^2)*3*2
N3 = (2^2)*3
Therefore, sufficient.
IMO B
Good Reasoning. IMO DLet me contradict your this reasoning:
(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.
No, we can't take X = 10 here. Otherwise we will have to take Y = Z = 1, or N2 = N3, which may or may not be true.
Each statement alone is sufficient.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Hi nehakhas,
At a first glance of the question I had the same line of reasoning as yours, but then I figured there is nothing that prevents N2 from being the same number as N3. Am I missing something?
Eyal
At a first glance of the question I had the same line of reasoning as yours, but then I figured there is nothing that prevents N2 from being the same number as N3. Am I missing something?
Eyal
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coffee5251
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Out of curiosity does anyone know what level a problem like this is??nehakhas1 wrote:1. The HCF of three numbers is 12. Their product is 17280. What are the three numbers?
I. One of the numbers is greater than the other two.
II. One of the numbers is lesser than the other two.
Please solve
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krisraam
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aroon7 wrote:But the numbers need not be different so 10,1,1 or 5,2,1 both satisfies even if both the statements are taken together.
then why not E...
someone pl help me
Stmt 1 says one of the numbers is greater than other two
10,1,1 and 5,2,1 both satisfies the condition so not sufficient.
Stmt 2 says one of the number is lesser than other two.
Only 5,2,1 satisfies the condition. So sufficient
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sanju09
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IMO B now, I always want a discussion on points that could have room for errors if pondered carelessly. I did "IMO D" for the same reasons. Thanksshulapa wrote:Hi nehakhas,
At a first glance of the question I had the same line of reasoning as yours, but then I figured there is nothing that prevents N2 from being the same number as N3. Am I missing something?
Eyal
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
