Case: Dying villagers and water supply
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]
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.
Case: Constant villagers and Water supply
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
- Rates of Wedding reception party: 8 food-items for 50 guests = total bill is Rs.1800. If Jethalal has total budget of Rs.21060 and planning to invite 90 guests for his son’s wedding, how many “food-items” can he have in the reception? (Ans.52 items)
- One bale of grass costs Rs.150. Lalu has 25 buffalos. They eat up 5 bales of grass in 12 days. If Lalu had only 10 buffalos, how much money will he need, to feed them for 18 days? (Ans.Rs.450)
IIM Ranchi results are out I suppose? I have read that you are converted or something. Does that mean you are selected?
If yes, congrats to IIM-Ranchi for getting a talent like you, :-).
Got both answers easily and quickly!
Thank you very much
How do you find answers ? please explain.
1. For Guest question,here is explanation:-
1st case: speed= 50*8 ; distance=1800 ;
2nd case: speed=90*x[x=no. of plates] ;distance=21060 ;
time in both cases is same…so just equate them by d1/s1=d2/s2
you will get x as 52
Tell me some good websites for free mock test.
I have given test in TesFunda, edooni and Tcyonline. TestFunda was great for paper-1, while paper-2 was similar to that asked in CSAT-2011. Edooni and tcyonline are not usefull.
in Mock question 2… is the question asking abt the bale of grass eaten by 10 buffaloes???
Please elaborate the mock question answers.
“Want to read my other articles on Aptitude? Just Click ME” -This Link isn’t working
Mrunal please tell how to decide what comes in speed what comes in time what comes in distance please reply
Q) Rates of Wedding reception party: 8 food-items for 50 guests = total bill is Rs.1800. If Jethalal has total budget of Rs.21060 and planning to invite 90 guests for his son’s wedding, how many “food-items” can he have in the reception? (Ans.52 items)
Q) One bale of grass costs Rs.150. Lalu has 25 buffalos. They eat up 5 bales of grass in 12 days. If Lalu had only 10 buffalos, how much money will he need, to feed them for 18 days? (Ans.Rs.450)
In the above questions, plz elaborate as to how you chose which parameter goes under Speed, Time & Distance – watz the logic behind. Your STD method is exceptionally great, but i am facing problem with initiation – as to what to consider as time, speed & distance respectively !!!
how to decide what is time distance speed ?plz reply
Cost of 5 bales=5*150rs=750rs
Amount spent on 1 buf=750/25=30rs
Amnt on 1 buf/day=30rs/12=5rs/2
amnt needed to feed 18 bufs in 1
i dnt get it how u solved. the part from amnt needed to feed 18 bufs in 1 days=
Amrit in the former case. it is given that 25 buffaloes consume 5 bales in 12 days. that implies in 25 buffaloes consume amount of 750 in 12 days. so we calculated it for 1 day which comes out to be 5/2. Now we simply need to calculate how much will 10 buffaloes consume in 18 days. so we multiplied our per buffalo per day calculation with the given conditions which is 10 x 18 x 5/2 = 450.
answer for the first question is 5.2 not 52
i dnt think answer to secong question is correct. if it is, den plz explain.
bcoz total expenditure on 25 buffaloes for feeding them for 12 days= 25*5*150= 25*750= 18750 RS.
nodoubt in second case we need to feed only 10 buffaloes but d days hv also increased frm 12 to 18, se expenditure cnt be 450 in any case.
Mrunal Sir u please tell me the Procedure of STD … dat How to decide which one we choose Speed and which one 2 Time , Distance
Mrunal Sir, Plz tell how to decide which should be in speed, time or distance column?
Describe the last mock ques! Its a request because this STD method is really time saving.
PLS EXPLAIN THE MOCK QUESTIONS :(
25 buffalos in 12-days eat = 5 bales,
1-buffalo in 1-day eats = 5 / (25 * 12) bales,
10-buffalos in 18 days eat = 5*10*18 / 25 * 12 = 3 bales,
Cost of 3-bales = 3*150 = Rs.450 >=========================< ANSWER
mrunal sir plz explain both solution by TSD METHOD