GATE2018 CE-2: GA-8

Question

The annual average rainfall in a tropical city is $1000$ mm. On a particular rainy day ($24$-hour period), the cumulative rainfall experienced by the city is shown in the graph. Over the $24$-hour period, $50$% of the rainfall falling on a rooftop, which had an obstruction-free area of $50$ $m^2$, was harvested into a tank. What is the total volume of water collected in the tank liters?

• A. $25,000$
• B. $18, 750$
• C. $7,500$
• D. $3.125$

Comments

time	rainfall (mm)
0-9	(100-0) =100
9-12	(200-100)=100
12-15	200-200=0
15-18	300-200=100
18-24	300-300=0
total	300

since range over an interval =cummulative(final) value - initial value on range .

so rainfall accumulated on rooftop = 1/2 *300 = 150mm

now , volume collected in rooftop =  50m^2 *  0.15 m

                                                     =7.5 m3

                                                      = 7.5 *1000 L (since 1cu.m =1000L)

                                                      =7500L of water collected in the tank.

 

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best answer :)

Answer 1

The graph is plotting cumulative rainfall and so in $24$ hrs $300$ mm of rainfall was received.

$300$ mm means the rainfall received in $1$ square meter is $300$ liters.

So, in $50\;m^2$ the amount of rainfall received is $300\times 50 = 15,000 \; l$

$50\%$ of this is harvested in tank. So,

Tank capacity $ = 0.5 \times 15,000\;l = 7,500\;l$

Option C.