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If m and n are the positive integers whose sum is 64,

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If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 18:20
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GMATBusters’ Quant Quiz Question -5


If m and n are the positive integers whose sum is 64, then ratio of m to n cannot be:
A. 1/7
B. 1/3
C. 5/11
D. 1/2
E. 3/5

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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 18:20
Let the ratio of m/n = a/b
hence the number can be represented as ax, bx where x is a positive integer.
Sum = 64
So, ax+bx= 64
a+b = 64/x
hence the sum of numerator and denominator in ratio must be a factor of 64.
out of given options, 1:2, sum = 1+2= 3 which is not a factor of 64. hence it is the required answer.
Answer = D
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 18:29
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If m and n are the positive integers whose sum is 64, then ratio of m to n will be such that if we add Numerator and Denominator it will divide 64...
A. 1/7 , here x+ 7x= 8x can divide 64
B. 1/3, here x+ 3x = 4x can divide 64
C. 5/11, here 5x+11x= 16x can divide 64
D. 1/2, here x+2x= 3x cannot divide 64
E. 3/5, here 3x+5x=8x can divide 64

hence D is the answer
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 18:30
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If m and n are the positive integers whose sum is 64, then ratio of m to n cannot be:
A. 1/7
B. 1/3
C. 5/11
D. 1/2
E. 3/5

\(m+n = 64\)

It can be \(16 + 48 = 64\) => \(\frac{16}{48}\) => \(\frac{1}{3}\)
Therefore if the addition of Numerator and Denominator is divisibile by 64 then that ratio is possible.

A. \(\frac{1}{7}\) - 1+7 = 8 divisible by 64 => This ratio is possible
B. \(\frac{1}{3}\) - 1+3= 4 divisible by 64 => This ratio is possible
C. \(\frac{5}{11}\) - 5+11 = 16 divisible by 64 => This ratio is possible
D. \(\frac{1}{2}\) - 1+2 = 3 not divisible by 64 => This ratio is not possible - Answer
E. \(\frac{3}{5}\) - 3+5 = 8 divisible by 64 => This ratio is possible

Answer - D
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 18:38
1
Going back from solution,
A) 1/7, means m and n are 1 & 7, whose sum is 8 and 8 is factor of 64. So out
B) 1/3, means m and n are 1 & 3, whose sum is 4 and 4 is factor of 64. So out
C) 5/11, means m and n are 11 & 5, whose sum is 16 and 16 is factor of 64. So out,
D) 1/2, means m and n are 1 & 2, whose sum is 3 and 3 is not factor of 64. So correct.
E) 3/5, means m and n are 3 & 5, whose sum is 8 and 8 is factor of 64. So out,

So, D is correct.
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 20:12
m+n=64 and n:m=xa:ya=x:y => (x+y)*a=m+n=64 => 64 must be divided by X+y
x:y is ratio of n to m
Now let's check
A. 1/7: 1+7=8 64:8=8 ok
B. 1/3: 1+3=4 64:4=16 ok
C. 5/11: 5+11=16 64:16=4 ok
D. 1/2: 1+2=3 64:3=21.333 RIGHT ANSWER
E. 3/5: 3+5=8 64:8=8 ok

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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 20:23
It is told that m,n > 0 and m,n are integers.
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 20:40
if m and n are positive integers that mean
m+n = 64
now 64 is factor of 2 with no odd factor
thus the sum of m and n should be of the form 2^nk
looking a options
A. 1/7 = 7k + k = 64 = 8k = 64 (valid)
B. 1/3 = 3k+k = 4k = 64 (valid_
C. 5/11 = 16k = 64 (valid)
D. 1/2 = 3k = 64 (invalid)
E. 3/5 = 8k = 64 (valid)
thus D answer
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 15 Feb 2020, 22:38
If m and n are the positive integers whose sum is 64, then ratio of m to n cannot be:

A. \(\frac{1}{7}\)
m = t and n = 7t
m+n = 8t = 64 (possible)
m= 8 n = 56

B. \(\frac{1}{3}\)
m = t and n = 3t
m+n = 4t = 64 (possible)
m = 16 n = 48

C. \(\frac{5}{11}\)
m = 5t and n = 11t
m+n = 16t = 64 (possible)
m = 20 n = 44


D. \(\frac{1}{2}\)
m = t and n = 2t
m+n = 3t (this must be integer since m and n are integers)
64 is not divisible by 3 so this is not possible (correct)

E. \(\frac{3}{5}\)
m = 3t and n = 5t
m+n = 8t = 64
m = 24 n = 40 (possible)
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 16 Feb 2020, 00:01
Solution:

Question 5: If m and n are the positive integers whose sum is 64, then ratio of m to n cannot be:

A. 1/7
B. 1/3
C. 5/11
D. 1/2
E. 3/5

In this question, the most important point to note is that we need to select the option that CANNOT be value of \(\frac{m}{n}\).

According to the information in the question, both m and n are positive integers and \(m + n = 64\).

Considering each of the options:

Option A: \(\frac{1}{7}\) In this fraction, since denominator is much greater than numerator, we would think of values close to 64 that have 7 as one of the factors for the denominator. \(8*7 = 56\) is closest to 64. Then, numerator is \(64 - 56 = 8\). Thus, option A can be eliminated as \(\frac{1}{7}\) can be written as \(\frac{8}{56}\) and \(8 + 56 = 64\). Thus, it can be the value of \(\frac{m}{n}\) and hence it is not the right answer.

Option B: Similarly, \(\frac{1}{3}\) can also be written as \(\frac{16}{48}\). Thus, option B can also be eliminated.

Option C: In, \(\frac{5}{11 }\) the numerator and denominator can be multiplied by same positive integer to find the equivalent fraction.\(\frac{(5*4)}{(11*4)}\) = \(\frac{20}{44}\) and \(44 + 20 = 64\). Thus, option C is also eliminated.

Option D: \(\frac{1}{2}\). Fraction closest to \(\frac{1}{2}\) is \(\frac{21}{42 }\) in which \(21 + 42 = 63\). Thus, option D is the correct answer.

Option E: 3/5 can be written as \(\frac{(3*8) }{ (5*40)} = \frac{24}{40}\) and \(24 + 40 = 64\). Thus, option E is not the answer.

Option D \(\frac{1}{2}\) is the correct answer.
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Re: If m and n are the positive integers whose sum is 64,  [#permalink]

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New post 16 Feb 2020, 05:46
Solution:

Given: m+n=64

Now lets do it option wise:

A. 1/7 => m=x, n=7x => 8x, which is divisible by 64, so it can be the ratio.
B. 1/3 => m=x, n=3x => 4x, which is divisible by 64, so it can be the ratio.
C. 5/11 => m=5x, n=11x => 16x, which is divisible by 64, so it can be the ratio.
D. 1/2 => m=x, n=2x => 3x, which is not divisible by 64, so it cannot be the ratio.
E. 3/5 => m=3x, n=5x => 8x, which is divisible by 64, so it can be the ratio.

So answer will be D.
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Re: If m and n are the positive integers whose sum is 64,   [#permalink] 16 Feb 2020, 05:46

If m and n are the positive integers whose sum is 64,

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