A bucket, when full of water, weighs 18 kg.

Half of the water is poured away.

The bucket and water now weigh 11 kg.

What is the weight of the empty bucket?

\boxed{x=18-y} \\ x+\frac{y}{2}=11 \\\\ 18^{(2}-y^{(2}+\frac{y}{2}=11^{(2} \\\\ 36-2y+y=22 \\\\ 36-y=22 \\\\ -y=22-36 \\\\ -y=-14 \\\\ \boxed{y=14 kg} \\\\ x=18-14 \\\\ \boxed{x=4kg}"> x − w e i g h t s t h e b u c k e t y − w e i g h t s t h e w a t er x + y = 18 => x = 18 − y ​ x + 2 y ​ = 11 1 8 ( 2 − y ( 2 + 2 y ​ = 1 1 ( 2 36 − 2 y + y = 22 36 − y = 22 − y = 22 − 36 − y = − 14 y = 14 k g ​ x = 18 − 14 x = 4 k g ​

Answered by mariamikayla | 2024-06-10

18-11=7 to find out how much halve the water weight then subtract 7 from 11 11-7=4 so the emtpy bucket weighs 4 kg

Answered by conungi | 2024-06-10

The weight of the empty bucket is 4 kg. This can be determined by setting up equations based on the total weights given in the problem. By solving these equations, we find that the empty bucket weighs 4 kg.
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Answered by mariamikayla | 2024-10-13