Bunuel wrote:
If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?
(1) c = 5a
(2) On the number line, a is closer to 10 than it is to 0.
Given \(0<a<10\)
To Find is \(c>\frac{(a+10)}{2}\) OR \(2c>(a+10)\) ---- (1)
Stat 1 \(c=5a\)
=> substituting in (1) we have
=> is \(10a>a+10\)
=> is \(9a>10\)
=> OR is \(a>\frac{10}{9}\) - No information
The above deduction suggest that if
Case 1 \(a>\frac{10}{9}\) let say \(a=2\)
=> then \(c=10\) , \(2c=20\) and \(a+10=12\)
=> So \(2c>(a+10)\)
Case 2 \(a<\frac{10}{9}\) let say \(a=1\)
=> then \(c=5\) , \(2c=10\) and \(a+10=11\)
=> So \(2c<(a+10)\)
Thus depending on the value of 'a' the answer to the question can be 'Y' or 'N'.
So NOT SUFFICIENT
Stat 2 On the number line, a is closer to 10 than it is to 0
=> Gives a>5 and No information of c
=> NOT SUFFICIENT
BOTH => \(a>5\) from Stat 2
=> then from Stat 1 for all values of 'a' \(c>25\)
=> Since \(\frac{(a+10)}{2}< 10\) AND \(c > 10\) (both ALWAYS as per the given conditions)
=> Therefore \(c>\frac{(a+10)}{2}\) OR \(2c>(a+10)\)
SUFFICIENT
Option 'C'
Regards
Dinesh