Question

If 7/8 part of a container is full of water, then the weight is 21 kg and if 3/4 part is full, the weight is 19 kg. What is the weight of the container?

Answer 1

Answer:

$x \times \frac{7}{8} = 63 \\ x = 21 \times \frac{8}{7 } \\ x = 24$

this is your answer

Answer 2

$\large{\red{ \bf{ \underline {\underline{Answer}}}}} \\ \\ \purple{ \sf{\mapsto Weight \: of \: container \: is \: 7 \: kg}} \\ \\ \mathrm{\green{ \underline{Given}}} \\ \\ \sf{ \rightsquigarrow A \: container \: \frac{7}{8}\: full \: of \: water \: weight \: 21 \: kg.} \\ \sf{When \: it \: i \: \frac{3}{4}th \: full \: of \: water \: it \: weight \: 19 \: kg.} \\ \\ \mathrm {\blue{ \underline{ To \: Find}}} \\ \\ \sf{ \rightsquigarrow The \: weight \: of \: empty \: container =\:?}$

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• According to given question

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$\sf{Let \: the \: weight \: of \: container \: be \: x \: kg} \\ \\ \sf{And \: weight \: of \:full \: container \: value \: be \: x} \\ \\ \sf{By \: the \: problem } \\ \\ \sf{ \implies x + \frac{7}{8}\:y = 21 \: - - - - (1)} \\ \\ \sf{Similarly,} \\ \\ \sf{ \implies x + \frac{3}{4}\:y = 19 \: - - - - (2) } \\ \\ \rm{Substractin \: equation \: (1) \: and \: equation \: (2) } \\ \\ \sf{ \implies \left(x + \frac{7}{8}\:y = 21 \right) - \left( x + \frac{3}{4}\:y = 19 \right) } \\ \\ \sf{ \implies \frac{1}{8} \:y = 2} \\ \\ \sf{ \implies y = 8 \times 2} \\ \\ \sf{ \implies y = 12} \\ \\ \rm{Putting \: value \: of \: y \: in \: equation \: (1)} \\ \\ \sf{ \implies x + \frac{7}{ \cancel{8}} \times \cancel{16} = 21 } \\ \\ \sf{ \implies x + 7 \times 2 = 21} \\ \\ \sf{ \implies x + 14 = 21} \\ \\ \sf{ \purple{ \implies}} \purple{ \underline{ \boxed{ \sf{x = 7 \: kg}}}}$

$\sf{\purple{\therefore\:\: Weight\: of\: Container\: is\: 7 \:Kg.}}$