Question

The maximum baggage weight allowed on the ferry is 9 kg. The laptop weighs $\frac{5}{3}$ kg and the bag itself weighs is $\frac{3}{2}$ kg.

What is the maximum amount of tea ($x$) that can be carried on the ferry?

• A. $x+\frac{5}{3}<\frac{19}{6}$
• B. $x<\frac{35}{3}$
• C. $x>\frac{35}{2}$
• D. $x<\frac{35}{6}$

Answer 1

The correct option is D $x<\frac{35}{6}$

The maximum weight of tea that can be carried is $x$ kg.

Weight of laptop = $\frac{5}{3}$ kg
Weight of bag itself = $\frac{3}{2}$ kg
Maximum baggage weight allowed = 9 kg

So, the ineqality that we get is,
$x+\frac{5}{3}+\frac{3}{2}<9$
$\Rightarrow x+\frac{10+9}{6}<9$
$\Rightarrow x+\frac{19}{6}<9$
$\Rightarrow x+\frac{19}{6}-\frac{19}{6}<9-\frac{19}{6}$
$\Rightarrow x<\frac{35}{6}\text{kg}$