SSC CPO SI and ASI 2011 Paper

Question : 102 of 200

Marks: +1, -0

If (2000)

^{10 }

= 1.024 × 10

^{k}

, then the value of k is

Solution:

(1)

(2000)^{10}=1.024 \× 10^{k},

⇒2^{10} × 10^{30}=\ {2^{10}}/{1000} × 10^{k}=2^{10}

10^{k-3}

and hence,

k-3=30 \ {So}, \ {k}=33

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