arjunshn wrote:If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

hi please check the question again..!!!

a^2b^2c^3=4500, let's perform prime factorization of 4500
4500 (2) 2250
2250 (2) 1125
1125 (3) 375
375 (3) 125
125 (5) 25
25 (5) 5
5 (5) 1, 2^2 * 3^2 * 5^3 -> a=|2| or |3|, b=|3| or |2|, c=5 and b+c=7

st(1)a,b,c >0; 2+5=7 BUT 3+5=8 Not Sufficient;
st(2) a>b therefore a=|3| or |2|, b=|2| or |3|; b+c=|2|+5=7 or 3 Not Sufficient, as 'a' or 'b' can be -ve/+ve

Combined st(1&2) Sufficient

IOM c

arjunshn wrote:If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

hi..even i chose c.but the correct answer is E.

Even I got C

hard question to me, but i inclined to vote for E, my rationale:
if a^2*b^2*c^3=45000=2^2*3^2*5^3
is b+c=7
(1)a=2,b=3,c=5, b+c=8-no
a=3,b=2, c=5 b+c=7-yes
insuff
(2)a>b a=3,b=2.c=5, b+c=7-yes
a=3^2*2^2=(3*2)^2, b=1, c=5, here 45000=1^2*((3*2)^2)*(5^3)
b+c=1+5=6-no
so also insuff
even both are insuff

Nice thinking clock...

arjunshn wrote:If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

Insight: from the fact that 4500 = 5*9*4*25 = 5^3*4*9 and 4 and 9 are perfect squares, you should have in mind there could be many (positive integer) possibilities, especially for a and b.

More explicitly: Take c=5 ("naturally") and a=3 (a^2=9) and b=2 (b^2=4) or a=6 (a^2 = 4*9) and b=1.

Regards,
Fabio.