# 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 method^{TM}.

Suppose 1 village needs “L” litres of water per day.

Therefore 120 villagers need 120xL litres of water per day. <–This is our “speed”

120 villagers | |

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) =120L (200) |

## Run STD on last column

Speed x time =distance

80L x time =120L x 200

Time=(120L x 200) / 80L

Time=300 days.

**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

4000 villagers | |

Speed (Daily need) | 4000x 150L |

Time (water can last) | ?? |

Distance (amt of water) | 20x15x6 m^{3}=20x15x6x1000 lit. |

Run STD

Speed x time = distance

(4000 x 150) x time = 20 x 15 x 6 x1000 m^{3}

Time =(20 x 15 x 6 x 1000) / (4000 x 150) =3 days.

**Final answer**: water will last for 3 days

# Mock questions

- 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)*

sanjeevMISHRA14/05/2012 at 1:17 pmIIM 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, :-).

dr.rani14/05/2012 at 3:11 pmGot both answers easily and quickly!

Thank you very much

Gajendrasinh30/03/2013 at 2:54 pmHow do you find answers ? please explain.

HIRA15/05/2013 at 11:01 am1. 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

sanjeevMISHRA14/05/2012 at 5:27 pmTell 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.

mahendra15/05/2012 at 5:48 amin Mock question 2… is the question asking abt the bale of grass eaten by 10 buffaloes???

amit12/09/2012 at 12:10 amPlease elaborate the mock question answers.

Narayan31/03/2013 at 8:30 pm“Want to read my other articles on Aptitude? Just Click ME” -This Link isn’t working

Mrunal02/04/2013 at 10:53 pmgoto

Ashwin04/06/2013 at 11:59 amMrunal please tell how to decide what comes in speed what comes in time what comes in distance please reply

Dheerendra patnaik22/08/2013 at 7:58 amQ) 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 !!!

shrinivas13/10/2013 at 9:13 amhow to decide what is time distance speed ?plz reply

Mudasir26/11/2013 at 1:16 pmCost of 5 bales=5*150rs=750rs

total bufs=25

Amount spent on 1 buf=750/25=30rs

total days=12

Amnt on 1 buf/day=30rs/12=5rs/2

2nd case

bufs=10,days=18

amnt needed to feed 18 bufs in 1

days=5rs/2*10*18

=450rs

amrit03/07/2014 at 2:18 ami dnt get it how u solved. the part from amnt needed to feed 18 bufs in 1 days=

T A M07/08/2014 at 11:26 pmAmrit 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.

chandra prakash11/05/2014 at 7:21 pmanswer for the first question is 5.2 not 52

amrit03/07/2014 at 2:29 ami dnt think answer to secong question is correct. if it is, den plz explain.

why?

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.

Pawan Kumar21/08/2014 at 6:33 pmMrunal Sir u please tell me the Procedure of STD … dat How to decide which one we choose Speed and which one 2 Time , Distance

NehaS03/09/2014 at 12:42 amMrunal 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.

aspirant05/09/2014 at 5:39 pmPLS EXPLAIN THE MOCK QUESTIONS :(

aspirant05/09/2014 at 5:42 pmANS(2ND):-

25 buffalos in 12-days eat = 5 bales,

1-buffalo in 1-day eats = 5 / (25 * 12) bales,

SO,

10-buffalos in 18 days eat = 5*10*18 / 25 * 12 = 3 bales,

Cost of 3-bales = 3*150 = Rs.450 >=========================< ANSWER

shivangi17/09/2014 at 7:49 pmmrunal sir plz explain both solution by TSD METHOD