Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 27 Oct 2017
Posts: 1783
WE: General Management (Education)

If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
15 Feb 2020, 18:20
Question Stats:
80% (01:13) correct 20% (02:27) wrong based on 35 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Retired Moderator
Joined: 27 Oct 2017
Posts: 1783
WE: General Management (Education)

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



Manager
Joined: 16 Sep 2011
Posts: 128

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
15 Feb 2020, 18:29
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



Director
Joined: 14 Dec 2019
Posts: 657
Location: Poland
WE: Engineering (Consumer Electronics)

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
15 Feb 2020, 18:30
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



Director
Joined: 22 Feb 2018
Posts: 737

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
15 Feb 2020, 18:38
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.



Manager
Joined: 04 Jun 2019
Posts: 79
Location: United States

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
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
Posted from my mobile device



Intern
Joined: 13 Nov 2019
Posts: 33
Location: India
Concentration: International Business, Marketing
GPA: 4

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
15 Feb 2020, 20:23
It is told that m,n > 0 and m,n are integers.
Attachments
New Doc 20200216 08.09.13_5.jpg [ 771.24 KiB  Viewed 388 times ]



VP
Joined: 28 Jul 2016
Posts: 1019
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)

Re: If m and n are the positive integers whose sum is 64,
[#permalink]
Show Tags
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
_________________
Keep it simple. Keep it blank



Senior Manager
Joined: 31 May 2018
Posts: 432
Location: United States
Concentration: Finance, Marketing

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



Intern
Joined: 25 Mar 2013
Posts: 11

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



Manager
Joined: 05 May 2016
Posts: 142
Location: India

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




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




