Ramgarh had population of 120 people and water supply for 210 days.
But after 10 days, 40 villages die.
How long will the remaining water last?
We can solve it, using our Universal STD table methodTM.
Suppose 1 village needs “L” litres of water per day.
Therefore 120 villagers need 120xL litres of water per day. <–This is our “speed”
|Speed (Daily need)||120L|
|Time (water can last)||210 days|
|Distance (amt of water)|
Run STD on first column.
Speed x time = distance=(120 x“L”) x 210
Given: 40 villagers died after 10 days.
It means for the first 10 days, there were 120 villagers, right ?
Update the table
|120 villagers||10 days|
|Speed (Daily need)||120L||120L|
|Time (water can last)||210 days||10 days|
|Distance (amt of water)||120L x210||??|
Run STD on last column to find out the water consumed by 120 villagers in 10 days
Speed x time = distance
120L x 10 = distance.
Remaining water after 10 days
=total water minus amount of water used in 10 days
=[120L x 210] MINUS [120L x 10] =120L [210-10] =120L (200)
So, we’ve this much water left and there are 120 minus 40 = 80 villagers surviving.
They consume 80xL = 80L litres of water per day.
Make a new column and fill up the table.
|120 villagers||10 days||Remaining|
|Speed (Daily need)||120L||120L||80L|
|Time (water can last)||210 days||10 days||??|
|Distance (amt of water)||120L x210||120L x 10||=120L (210-10)|
Run STD on last column
Speed x time =distance
80L x time =120L x 200
Time=(120L x 200) / 80L
Final answer: the water will last for 300 days.
This one was actually asked in UPSC CSAT 2011.
A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. The water of this tank will last for how many days?
Tank capcity given in meters but water consumption given in litres, Otherwise, There is no “dum” in this sum.
But Units must be same.
1 cubic meter=1000 litres.
Fill up the STD table
|Speed (Daily need)||4000x 150L|
|Time (water can last)||??|
|Distance (amt of water)||20x15x6 m3|
Speed x time = distance
(4000 x 150) x time = 20 x 15 x 6 x1000 m3
Time =(20 x 15 x 6 x 1000) / (4000 x 150) =3 days.
Final answer: water will last for 3 days
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