If the sum of the three different numbers is 54, what is the largest

If the sum of the three different numbers is 54, what is the largest number?

Given: \(x<y<z\) and \(x+y+z=54\).
Question: \(z=?\)

(1) The largest number is twice the smallest number --> \(x+y+2x=54\) --> \(3x+y=54\). Now we should find \(x\) so that \(x<y\) and \(3x+y=54\):
\(x=11\), \(y=21\), \(z=22\);
\(x=12\), \(y=18\), \(z=24\);
\(x=13\), \(y=15\), \(z=26\);

Also notice that we are not told that unknowns represent integers only so non-integer values are also valid. For example: \(x=12.5\), \(y=16.5\), and \(z=25\).
Not sufficient.

(2) The sum of the two smaller numbers is 30 --> x+y=30 --> \(30+z=54\) --> z=24. Sufficient.

Answer: B.
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1)
2x + y + x = 54
3x + y = 54

2x = 24
y = 18
x = 12

2x = 22
y = 21
x = 11

NS

How do we know z is largest

Hi,

the statement tells us that 30 is the sum of two smallest numbers..
so the remaining difference will have to be the third number, which will be the largest
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From Statement 2 how do we know that x is not 29 and y=1, or x=28 and y=2?
In those cases x would be the largest number.

Hi,
the answer lies within the statement 2 itself..
(2) The sum of the two smaller numbers is 30

now this tells us that 30 is the sum of two smaller numbers..
so x cannot be 28 or 29 it has to be less than the largest number 54-30=24..
hope it helps
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When you modify the original condition and the question, based on a<b<c, you only need to know a+b
from 1) a+b+c=54, c=54-(a+b).
Since 2) a+b=30, c=54-30=24,
which is unique and sufficient.
Therefore the answer is B.

\(\left\{ \matrix{\\
a > b > c \hfill \cr \\
a + b + c = 54\,\,\,\,\,\, \hfill \cr} \right.\,\,\,\,;\,\,\,\,\,\,\,\,\,\,?\,\, = \,\,a\)

Let´s start with statement (2) not only because it´s easier, but also because it will help us finding (uniqueness or) a BIFURCATION of the other statement!

\(\left( 2 \right)\,\,\,b + c = 30\,\,\,\,\, \Rightarrow \,\,\,\,\,? = a = 54 - 30 = 24\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)

\(\left( 1 \right)\,\,a = 2c\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {24,18,12} \right)\,\,\,\,\, \Rightarrow \,\,\,\,a = 24\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {26,15,13} \right)\,\,\,\,\, \Rightarrow \,\,\,\,a = 26\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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