Ten trucks,numbered 1 to 10,are carrying pack | vedclass

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(D) Let $w_i$ be the weight of a packet in truck $i$,where $w_i \in \{999, 1000\}$.If all trucks carried $1000 \ g$ packets,the total weight would be:

Vedclass · Accepted Answer

$2, 8$

Vedclass · Answer

(D) Let $w_i$ be the weight of a packet in truck $i$,where $w_i \in \{999, 1000\}$.If all trucks carried $1000 \ g$ packets,the total weight would be:$W_{max} = 1000(1 + 2 + 2^2 + \dots + 2^9) = 1000(2^{10} - 1) = 1000(1024 - 1) = 1023000 \ g$.The actual total weight is $1022870 \ g$.The difference is $1023000 - 1022870 = 130 \ g$.Since each lighter packet weighs $1 \ g$ less than $1000 \ g$,the difference $130$ must be represented as a sum of powers of $2$ corresponding to the trucks with lighter bags.$130 = 128 + 2 = 2^7 + 2^1$.This corresponds to the $2^{nd}$ truck $(2^1)$ and the $8^{th}$ truck $(2^7)$.Thus,the trucks with the lighter bags are $2$ and $8$.

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