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.
If the average of three different positive integers is 5, what is the
[#permalink]
17 Dec 2021, 07:00
17 Dec 2021, 07:00
Expert Reply
00:00
Question Stats:
79% (01:48) correct
20% (03:03) wrong based on 24 sessions
Hide Show timer Statistics
If the average of three different positive integers is 5, what is the greatest possible product of these three integers?
(A) 18
(B) 32
(C) 80
(D) 120
(E) 210
(A) 18
(B) 32
(C) 80
(D) 120
(E) 210
Re: If the average of three different positive integers is 5, what is the
[#permalink]
17 Dec 2021, 08:01
17 Dec 2021, 08:01
1
Three integers with avg \(5\)
Their sum \(= 15\)
Sum of the rest of the no \(= 10\)
These no can be: 1 & 9 , 2 & 8 , 3 & 7, 4 & 6
The highest product will be with the triplets of \(4, 5, 6 = 120\)
Answer D
Their sum \(= 15\)
Sum of the rest of the no \(= 10\)
These no can be: 1 & 9 , 2 & 8 , 3 & 7, 4 & 6
The highest product will be with the triplets of \(4, 5, 6 = 120\)
Answer D
If the average of three different positive integers is 5, what is the
[#permalink]
17 Dec 2021, 08:19
17 Dec 2021, 08:19
1
Let the 3 nos be x,y and z resp.
Given avg. of the three nos is 5 OR (x+y+z)/3 = 5 => x+y+z = 15
To obtain the maximum product of these nos, we first give equal numbers to x, y and z.
so x=5, y=5, z=5. For all other cases, product is smaller.
E.g. x=1, y=2, z=12, x*y*z = 24, and so on for other cases
Now we are told that x,y and z are distinct positive integers. So the numbers assigned to these variables will be different but all three would be fairly close to 5.
With that knowledge, we can safely assume that x*y*z must be near to 125.
Checking the options, only D stands out. (120 closer to 125)
Given avg. of the three nos is 5 OR (x+y+z)/3 = 5 => x+y+z = 15
To obtain the maximum product of these nos, we first give equal numbers to x, y and z.
so x=5, y=5, z=5. For all other cases, product is smaller.
E.g. x=1, y=2, z=12, x*y*z = 24, and so on for other cases
Now we are told that x,y and z are distinct positive integers. So the numbers assigned to these variables will be different but all three would be fairly close to 5.
With that knowledge, we can safely assume that x*y*z must be near to 125.
Checking the options, only D stands out. (120 closer to 125)
