Are you sure you transcribed the question correctly?enokhroot wrote:Can solve help with this one? Thanks in advance!
Given 0 < r < s < t where r, s, and t are integers, which of the
following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
Here's an ALGEBRAIC approach.enokhroot wrote: Given 0 < r < s < t where r, s, and t are integers, which of the following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
I had the same reaction that Brent had had and so I am going to use the following as the answer choices.enokhroot wrote:Can solve help with this one? Thanks in advance!
Given 0 < r < s < t where r, s, and t are integers, which of the
following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
Here's the PLUGGING in values approach:enokhroot wrote:Can solve help with this one? Thanks in advance!
Given 0 < r < s < t where r, s, and t are integers, which of the
following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
Sure, if you change the answer choices, the correct answer will change.Marty Murray wrote:I had the same reaction that Brent had had and so I am going to use the following as the answer choices.enokhroot wrote:Can solve help with this one? Thanks in advance!
Given 0 < r < s < t where r, s, and t are integers, which of the
following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) r(s - 1)t
e) rs(t + 1)
As far as how to do it goes, I found two ways.
One is to realize that changing a smaller number and a larger number by equal amounts changes the smaller number more proportionally.
In this case, changing r by 1 changes r more proportionally than does changing s or t. In the extreme case, if r were to equal 1, then subtracting 1 from r reduces it by 100 percent, and makes (r-1)st = 0, which, given the answer choices, we have to figure is the worst possible approximation of rst.
So given that all the answer choices involve changing one of the factors by the same number, 1, then the way to most closely approximate rst is to change the biggest factor, t.
So choose e.
You could also do this, or illustrate this by plugging in numbers.
Let's keep it simple and go with r = 1, s = 2 and t = 3. So rst = 6
For the problem to make sense, whatever holds true for those must hold true for all possibilities such that 0 < r < s < t where r, s, and t are integers.
Going through the answer choices:
a) (1 + 1)(2)(3) = 12 Difference From rst: +100%
b) (1 - 1)(2)(3) = 0 Difference From rst: -100%
c) (1)(2 + 1)(3) = 9 Difference From rst: +50%
d) (1)(2 - 1)(3) = 3 Difference From rst: -50%
e) (1)(2)(3 + 1) = 8 Difference From rst: +33.33%
The smallest difference is created via choice e.
Brent@GMATPrepNow wrote:Here's an ALGEBRAIC approach.enokhroot wrote: Given 0 < r < s < t where r, s, and t are integers, which of the following best approximates rst:
a) (r + 1)st
b) (r - 1)st
c) r(s + 1)t
d) rs(t - 1)
e) rs(t + 1)
Let's EXPAND each answer choice.
a) (r + 1)st = rst + st. The DIFFERENCE between this value and rst is st
b) (r - 1)st = rst - st. The DIFFERENCE between this value and rst is st
c) r(s + 1)t = rst + rt. The DIFFERENCE between this value and rst is rt
d) rs(t - 1) = rst - rs. The DIFFERENCE between this value and rst is rs
e) rs(t + 1) = rst + rs. The DIFFERENCE between this value and rst is rs
As we can see, the 5 answer choices differ from rst by st, st, rt, rs and rs
Since 0 < r < s < t, we can see that rs < rt < st
In other words, rs is the SMALLEST difference, which means the correct answer is EITHER D or E
Cheers,
Brent
