this one is a personal favourite::TSD::
In TSD solving the questions with equations is a great optionbut what is more great is learning the application of the single formula
DISTANCE=SPEED * TIME
1st part--> distance constant.
Understanding the part that time --> 1 / speed when distance is constant has got many advantages....
t1/t2 = v2/v1. thus wen the speed of travelling is doubled the time is halved and when the speed is 2/3rd time is 3/2th...let us see its practical application::
Q.I travel to my swimming club and take 1 hour for the journey,one day i travelled first half of the distance at 2/3rd of the normal speed.What should be my speed for the second half so that i reach the club in time.
Ans: SInce here distance is constant..so speed *time = constant...
Now travelling 1st half with 2/3rd speed takes time 3/2th of the original time...i.e 45mins...as earlier for half of the journey time was 30mins...
Now in order to reach in time..i have 15 mins left for next half..so need to travel with double speed....coz with normal speed i took 30 mins..so with double speed ill take 15 mins......
Q.Now if suppose i travel half with 2/3rd speed and next half with 3/2th speed..how much time will i take to reach the club..???
for first half:: time taken=3/2th=45 mins
for second half::time taken=2/3rd=20 mins
hence total time=65 mins...
Q. walking at 3/4th of his usual speed,a man is 1.5 hours late.find his usual travel time..??
Ans. distance is constant...since 3/4th speed hence time taken 4/3rd...1+1/3rd....n this 1/3rd accounts for late time....thus
t/3=3/2..hence original time=9/2 hours...
train problem..met with an accident smthng....can also be solved using the same concept..::::
Q.A train travelling from A to B,met with an accident at distance of 150 km and after the accident it was travelling at 3/4th speed and was 3 hours late.If the accident would have occured at 210 km then it would have reached 2.5 hours late.find the original speed of the train.
Ans:: the whole prob can be understood as....
A--------------------C--------D--------------B
A--------------------C--------D--------------B In first case accident took place at C and in second case at D.Thus CD is 60km....In first case CD is travelled with 3/4th speed and in second case CD is travelld with normal speed...
Delay is 3 hours in first case and 2.5 hours in second case,hence diff=30 mins..
this 30 mins delay is because of 60km....Suppose time taken to travel CD is t....
first case time taken=4/3t..coz 3/4th speed..
second case time taken=t..coz normal speed...
Hence 4/3t-t=30/60
t=1.5 hours....
hence 60km in 1.5 hours....thus 40km/hr....
A little compicated,but once inherited will be very easy....
For practise::
Q1. If a man cycles at 10k/hr,den he reaches some place at 1 pm..if he cycles at 15k/hr den he reaches the same place at 11 am...at what speed must he cycle to reach at noon..??/
Q2. A train travels from A to B.One day it travels with 4/5th of its speed because of some trouble at A and reached B at 6.15 pm..If it wud have travelled 100km further and the travelled with 4/5th speed,it wud have reached at 6:00 pm.what is its original speed..????
IN overtake problems,same concept is applied,the moment A overakes B it means that both A and B travel same distance....
CAt06 prob::
Q. Arun,Barun and kiranmala start from the same place and travel in the same direction with speeds of 30k/hr,40k/hr and 60k/hr.Barun starts 2 hours after Arun,if Barun and kiranmala overtake Arun at the same intant,how many hours after arun did kiranmala start..???
Ans:: at the momnt B and K overtake A,the distance travlled will be same...
so time*speed=contant...
Now suppose K starts x hours after A..thus 3
time for which A travels::t
B travels:: t-2
K travels::t-x
t/t-2=40/30 also t/t-x=60/30
solve and x=4hrs....
Another concept:: time constant::
since time is constant so distance-->speed....
Suppose two cars travel towards each other with speed 40km/hr ans 50 K/hr.The distance between them is 1800 km.at what dstance do they meet..???
since time is constant...hence distance from A-40/90*1800 and distance from B=50/90*1800...the distance will be tarvelled in ratio of speeds only....
hence 800 km from A and 1000 km from B///
NOw the cars dont stop and keep moving..reach their respective destis and return immediately...calculate the distance from A where they meet for the second time..???
Suppose they meet at distance D from A...thus distance from B 1800-D...
hence car from A travels:: 1800+1800-D=3600-D
CAr from B travels:: 1800 + D.....(why..???)
total distance:: 5400.....which will be divided in the same ratio of speeds....
calculate please..?????? and get D=1200....
REmember:: the only reason this concept is applied bcz the car started at same time..for diff times see the prob below::
Here is a prob taking into account both the above cases from CAt05.....
Q3.Ram and Shyam run a race between A and B,5km apart.Ram starts at 9:00 am from A and shyam starts from 9.45 am (delay of 45mins).ram speed is 5km/hr and shyam speed is 10k/hr.Both wen reach B return immediately....Find..
A>At what time do they cross fro the first time..????
B> AT what time do shyam overtakes RAm..????
solution later...m sure many of you would be able to sove it..bt try applying the above approaches....
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