Question 131
Let $$a = 2^{129} \times 3^{81} \times 5^{128}, b = 2^{127} \times 3^{81} \times 5^{128}, c = 2^{126} \times 3^{82} \times 5^{128}$$, and $$d = 2^{125} \times 3^{82} \times 5^{129}$$. Then
Solution
$$a = 2^{129} \times 3^{81} \times 5^{128}$$
$$b = 2^{127} \times 3^{81} \times 5^{128}$$
$$c = 2^{126} \times 3^{82} \times 5^{128}$$
$$d = 2^{125} \times 3^{82} \times 5^{129}$$
Eliminating common factors $$(2)^{125}\times(3)^{81}\times(5)^{128}$$
=> $$a=16,b=4,c=6,d=15$$
$$\therefore$$ $$b<c<d<a$$
=> Ans - (B)
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