A family of six requires 30 kg of food per month. If six guests stayed over at their house for 10 days, how much extra food (in kg) would be required?

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Solution

The correct option is C 10
Six members of a family need 30 kg of food for a period of 30 days. If the number of members in the family increases, the amount of food required also increases.
Amount of food required by the family per day = $\frac{Q u a n t i t y o f f o o d r e q u i r e d}{N u m b e r o f d a y s i n a m o n t h} = \frac{30}{30}$ = 1 kg.
Amount of food consumed by one family member in a day = $\frac{A m o u n t r e q u i r e d b y t h e f a m i l y p e r d a y}{N u m b e r o f f a m i l y m e m b e r s} = \frac{1}{6}$ kg.

Amount of food required by 6 guests = $\frac{1}{6} \times$ 6 = 1 kg.
$∴$ Amount of food required by 6 guests for 10 days = $1 \times 10 = 10$ kg.

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