Question

( a+b+c=18, quad frac{a^{2} b^{3} c^{4}=2^{19} times 3^{3}}{1} )
From this eqn. we can cleculy assume that ( a, b, c ) are multiples of 2 and 3 only. so most switable values for these ( operatorname{eqn} operatorname{arcs} 8,6,4 )
And it also satisfy they ( operatorname{con} a+b+c=18 )
[
begin{array}{l}
text { So let } a=4, b=6, quad c=8
qquad begin{array}{rl}
frac{1}{2}^{2} & 2 times 3 & 1
left(2^{2}right)^{3} times(2 times 3)^{3} timesleft(2^{3}right)^{4} & 2^{3}
=2^{6} times 2^{3} times 3^{3} times 2^{12}
end{array}
end{array}
]
( =2^{19} times 3^{3}left(begin{array}{l}2 t text { als of satisfy } text { this eqn } )end{array}right. )
Mence the values we assumed
ar correct. so ( a^{2}+b^{2}=36+16= ) ine 52
well in these use options d 5 weur we can types of question and 5ee that option only option which are sum of two squares ret 16 , ( 100 rightarrow 64+36 ). ( 52 rightarrow 36+16=cos b e cos u y d theta )
so other options

# 180. If a, b and c are positive numbers such that a + b + c = 18. Also if a-b3c4 = 219 x 33 then the a+ b2 is equal to (A) 50 (B) 52 (C) 100 (D) 48

Solution