Question

If it has rained 276mm in the last 3 days, how much cm of rain will fall in one full week (7days) ? Assume that the rain continues to fall at the same rate.

Answer 1

Hint: Use unitary method to first find out the rate at which the rain falls. Then multiply it with the number of days to get the total amount asked for in the question.

It is given to us that it has rained 276mm in 3 days. Let’s first find out the rate at which the rain falls, in terms of mm/day.
To do so, we can simply divide the amount of rain received by the number of days it was received over. We already know that the amount of rain received = 276mm and the period it was received over = 3days, since this is already given in the question.
Therefore, the rate at which the rain falls = $\dfrac{amount}{days}=\dfrac{276mm}{3days}=92mm/day$.
Therefore, we now have the rate of rainfall, or the amount of rain that falls in one day. The rate is thus = 92 mm/day, or we get 92 mm of rainfall in one day.
Now, we can simply use a unitary method, to find out the amount of rain received in 7 days, or a week.
We know that in one day, the amount of rainfall received = 92mm.
Therefore, the amount of rainfall received in $x$ days = $92x$ mm.
Here, $x$ is equal to 7. Therefore, the amount of rainfall received in a week, or in 7 days = $92x=92\times 7mm=644mm$
Therefore, the amount of rainfall received in a week = 644mm.
However, we have been asked the amount in cm. Thus, we have to divide the amount by 10, since the amount we have got is in mm.
Therefore, the amount of rainfall we will receive in one week, assuming that the rate remains constant = $\dfrac{644}{10}cm=64.4cm$.
Thus, we will receive 64.4 cm of rainfall in a week, assuming the rate remains unchanged.

Note: This is a pretty straightforward question involving basic unitary method, to find out the answer. However, in this question we had to assume that the rate of rainfall remains the same everyday. We could also have a question where the rate changes everyday, and in that case, we’d have to be careful while calculating the actual answer.