10 boys plan a camp and have sufficient food for 14 days, if 3 boys are ill and cannot go, howlong will the food now last?

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Im working on ratios for GCSE, I've become comfortable increasing/decreasing ratios and sharing values based on ratios but Im not sure how to proceed with this type of question.

Intially I thought along the lines of 10/14 to calculate the amount of 'foods' for each child. Then multiple that by 3 to get the value of food for the boys that dont come and add that to the 14 however, my materials tell me the answer should be 20days. So this method is clearly not correct.

Any help on the correct method would be appreciated

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First, notice that $\frac{10}{14}$ is not food per child, but rather child per day.

The important ratio here is not food per child, or day per child or anything like that. Its the amount of children before and after the illness. It is also important to note that number of children and days are inversely proportional. Meaning that if the number of boys goes up, number of days go down. So: $$\frac{days\ after\ illness}{days\ before\ illness}=\frac{boys\ before\ illness}{boys\ after \ illness}$$ $$\frac{days\ after\ illness}{14}=\frac{10}{7}$$ by $14$.

Simply put, you want to scale the ratio $\frac{10}{7}$ $$days\ after\ illness=14\cdot\frac{10}{7}=20$$

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At the rate 10 people eat it would've lasted 14 days. But now there are only 7 to feed:

$\dfrac{10}{7}\times14=20$

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10 boys * 3 meal/boy-day * 14 day = 420 meals

7 boys * 3 meal/boy-day * x day = 420 meals

21x =420

x = 20 days

Cheers

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"UNIT_OF_FOOD"...that supports one boy for one day

"ALL"... all the food that we have available

"X"... number of days we'll last now

so...

14 * 10 * UNIT_OF_FOOD = ALL

X * 7 * UNIT_OF_FOOD = ALL

14 * 10 * UNIT_OF_FOOD = X * 7 * UNIT_OF_FOOD

14 * 10 = X * 7

X = 20