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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d

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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 09 Mar 2017, 22:42
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If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)

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Re: If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 10 Mar 2017, 00:19
ziyuen wrote:
If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)


I chose D.

1- This suggests both m and n are > 0.6 and since we know they are both positive, they can only add up to > 1.

2- I could not find any value where two decimals m and n ending up in a sum less than 1 if their multiple is greater than 0.5. I would greatly appreciate it if someone can provide some mathematical base to this rough calculation.
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Re: If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 31 Jul 2017, 05:40
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0.7X0.7=0.49 hence 2nd statement is sufficient.
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Re: If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 23 Sep 2017, 11:36
gps5441 wrote:
0.7X0.7=0.49 hence 2nd statement is sufficient.



Oh so it could be : 0.7 * 0.9 or 0.7*0.8 --> these give values more than 0.5
so st 2 suff
Ans will be D
Thank you for the hint :)
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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 23 Sep 2017, 20:44
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hazelnut wrote:
If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)


Statement 1: implies \(m>0.5\) & \(n>0.5\) Add the two you will get \(m+n>1\). Hence Sufficient

Statement 2: \(mn>0.5\) here you can logically deduce the values of \(m\) & \(n\) to know that both will be greater than \(0.5\), hence \(m+n>1\)

Alternatively, \(m>\frac{0.5}{n}\), it is given that \(m\) & \(n\) are decimals, so \(n>0.5\) because if \(n<0.5\), then \(\frac{0.5}{n}>1\), which is not possible

similarly \(m>\frac{0.5}{n}\), hence \(m>0.5\). this is equivalent to our statement 1. Hence Sufficient

Option D
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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 23 Sep 2017, 21:57
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1. Each decimal digit is greater than 5
m,n >0.6
This is sufficient.
Note: Give attention to "greater than". If it was greater than equal to 5, we could have m=n=0.5. In that scenario, m+n=1, not >1

2. For second statement, you should know this
AM >= GM

Arithmetic Mean of 2 numbers is greater than equal to Geometric Mean of 2 numbers.
So,
\(\frac{(m+n)}{2}\) >= \(\sqrt{mn}\)

We know that mn > \(\frac{1}{2}\)
So \(\sqrt{mn}\) > \(\frac{1}{\sqrt{2}}\)

Substitute this in our original equation
\(\frac{(m+n)}{2}\) >= \(\frac{1}{\sqrt{2}}\)

m+n >= \(\sqrt{2}\)
Hence, m+n >1
This is sufficient.

D.


Additional Lesson: AM >= GM >= HM (Harmonic Mean)

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Re: If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 24 Sep 2017, 10:08
jedit wrote:
ziyuen wrote:
If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)


I chose D.

1- This suggests both m and n are > 0.6 and since we know they are both positive, they can only add up to > 1.

2- I could not find any value where two decimals m and n ending up in a sum less than 1 if their multiple is greater than 0.5. I would greatly appreciate it if someone can provide some mathematical base to this rough calculation.


For the second part > 0.5 so let's say 0.63 which is 0.9 and 0.7
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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 25 Jan 2019, 19:38
hazelnut wrote:
If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)


(1) the least decimal with digits greater than 5 is 0,6. If m and n are equal than 0.6, m+n=1,2>1. SUFFICIENT

(2) If m is the greatest decimal less than 1, we can write it like (99...9)/(10^x) where 99...9 has "x" digits.
If mn were at least 1/2, n should be (10^x)/(2*99...9) or (5000...0000)/(99...9) where each number has "x digits. Hence n>1/2.
According to these premises m and n are greater than 1/2. Hence m+n>1. SUFFICIENT

ANSWER (D)

Thanks for reading!
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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d  [#permalink]

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New post 25 Jan 2019, 20:45
hazelnut wrote:
If m and n are positive decimals smaller than 1, is \(m+n>1\)?

1) All decimal digits of m and n are greater than 5

2) \(mn > \frac{1}{2}\)


Multiply carefully

Statement 1 is sufficient alone
All decimal digits of m and n are greater than 5

0.66 + 0.6 => 1.26

Statement 2 i sufficient alone
\(mn > \frac{1}{2}\)

mn > 0.5

0.8 * 0.7
0.8 * 0.8
0.9 * 0.7

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If m and n are positive decimals smaller than 1, is m+n>1? 1) All d   [#permalink] 25 Jan 2019, 20:45
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