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