GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Given Kudos: 81
Concentration: , International Business
If [m]a*c=0.0075[/m] and [m]b*c=0.09[/m],
[#permalink]
26 Jun 2021, 04:57
26 Jun 2021, 04:57
00:00
Question Stats:
20% (04:38) correct
80% (02:49) wrong based on 5 sessions
Hide Show timer Statistics
If \(a*c=0.0075\) and \(b*c=0.09\),where a,b and c are positive integers. What is the least possible value of c?
A. 100
B. 200
C. 400
D. 800
E. Can not be determined.
Kudos for the right solution.
A. 100
B. 200
C. 400
D. 800
E. Can not be determined.
Kudos for the right solution.
If [m]a*c=0.0075[/m] and [m]b*c=0.09[/m],
[#permalink]
26 Jun 2021, 07:37
26 Jun 2021, 07:37
1
This can be done with trying out options.
The LEAST value of \(c\) for which both \(a\) & \(b\) will be integers as given in the stem.
Answer C
The LEAST value of \(c\) for which both \(a\) & \(b\) will be integers as given in the stem.
Answer C
Re: If [m]a*c=0.0075[/m] and [m]b*c=0.09[/m],
[#permalink]
26 Jun 2021, 09:32
26 Jun 2021, 09:32
2
rx10 wrote:
This can be done with trying out options.
The LEAST value of \(c\) for which both \(a\) & \(b\) will be integers as given in the stem.
Answer C
The LEAST value of \(c\) for which both \(a\) & \(b\) will be integers as given in the stem.
Answer C
Hello!
can you please demonstrate your work? how you "tried out the options"? If a, b, and c are all positive integers, then (by the integer multiplication closure property (google "proofwiki integer multiplication is closed")) a*b, a*c, and b*c should all be positive integers as well.
I'm thinking that c cannot be an integer if a & b are positive integers and a = 75/(10000*c) and b = 9/(100*c)
kumarneupane4344 wrote:
If \(a*c=0.0075\) and \(b*c=0.09\),where a,b and c are positive integers. What is the least possible value of c?
A. 100
B. 200
C. 400
D. 800
E. Can not be determined.
Kudos for the right solution.
A. 100
B. 200
C. 400
D. 800
E. Can not be determined.
Kudos for the right solution.
If a and b are positive integers, and a = 3/(2^4 * 5^2 * c) and b = 3^2 / (2^2 * 5^2 * c), then the greatest possible value of c would be: c = (3 * 2^(-4) * 5^(-2)) = 3/400:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (3 * 2^(-4) * 5^(-2))) = 1
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (3 * 2^(-4) * 5^(-2))) = 3 * 2^2 = 12
The least possible value of c would be (k is any positive integer): c = (2^(-4) * 5^(-2) * k^(-1)) = 1/(400 * k):
-> let k = 1, c = 1/400:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 1^(-1))) = 3 * 1 = 3
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 1^(-1))) = 3^2 * 2^2 * 1 = 36
-> let k = 2, c = 1/800:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 2^(-1))) = 3 * 2 = 6
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 2^(-1))) = 3^2 * 2^2 * 2 = 72
-> let k = 3, c = 1/1200:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 3^(-1))) = 3 * 3 = 9
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 3^(-1))) = 3^2 * 2^2 * 3 = 108
------------
so because k can be any positive integer, and as k increases from 1, c decreases from 1/(400*1), the least value of c that makes a & b positive integers and a = 75/(10000*c) and b = 9/(100*c) cannot be determined.
