Question

A dealer finds that 48 refined oil cans (of 5litres each) can be packed in eight cartons of the same size. How many such cartons will he require to pack 216 cans?

Answer 1

Given :

• 48 refined oil cans (of 5litres each) can be packed in eight cartons of the same size.

To Find :

• How many such cartons will he require to pack 216 cans

Solution :

Let the number of cartons be x

This is the case of Direct Variation. If refined oil cans increases,the no. of cartons will definitely increase in the same ratio.

[If two quantities are related in such a way that an increase in one quantity results a corresponding increase in the other and vice-versa, then such a variation is called Direct Variation ]

$\longrightarrow \sf \: \dfrac{48}{8} = \dfrac{216}{x}$

$\longrightarrow \sf \: 48 \times x = 216 \times 8$

$\longrightarrow \sf \: 48 x = 1728$

$\longrightarrow \sf \: x = \cancel \dfrac{1728}{48}$

$\longrightarrow \red{\sf \: x =36}$

$\therefore$ 36 cartons will be required to pack 216 cans.

Answer 2

Answer:

$\begin{tabular}{|c|c|c|}\cline{1-3}\sf Oil Cans &\sf 48&\sf 216\\\cline{1-3}\sf Cartons &\sf 8&\sf n\\\cline{1-3}\end{tabular}$

$\underline{\bigstar\:\textbf{Ratio of Oil Cans \& Cartons :}}$

$:\implies$ 48 Cans : 8 Cartons

• Dividing both term by 8

$:\implies$ 6 : 1

$\rule{150}{1}$

So we can see that, 1 carton contains 6 cans. Therefore According to this :

$\dashrightarrow\:\:\textsf{6 Cans are in = 1 carton}\\\\\\\dashrightarrow\:\:\sf1\:Can\:will\:in=\dfrac{1}{6} \:carton\\\\\\\dashrightarrow\:\:\sf 216\:Cans\:will\:be\:in =\dfrac{1}{6} \times 216 \:carton\\\\\\\dashrightarrow\sf\:\:216\:Cans\:will\:be\:in =36\:carton$

⠀

$\therefore\:\underline{\textsf{Hence, Number of cartons required are \textbf{36}}}.$