If a man walks to his office at 5/4 of his usual rate, he reaches office 30 min. early than usual. What is his usual time to reach office?
@Raghu5jan said:If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office?
Pls let me know how much time does it takes to solve lod 1 of arun sharma of arithematic Geometry,P n C and probability
Lets say distance = 10kms
and his speed s x kmph
so time = 10/x
now if he goes with 3/4th of speed
he will take = 40/3x
40/3x-10/x = 1/3
10/3x = 1/3
so x = 10
so he takes 1 hr???
@chandrakant.k said:Lets say distance = 10kmsand his speed s x kmphso time = 10/x now if he goes with 3/4th of speed he will take = 40/3x40/3x-10/x = 1/310/3x = 1/3 so x = 10so he takes 1 hr???
Ya dude its right..cud u pls tell me how much time it takes to cover arun sharma lod 1 ?? of all imp chap?
@Raghu5jan said:Ya dude its right..cud u pls tell me how much time it takes to cover arun sharma lod 1 ?? of all imp chap?
i haven't done from this book at all 😞 so not a right person to answer this question 😃
@Raghu5jan said:Ya dude its right..cud u pls tell me how much time it takes to cover arun sharma lod 1 ?? of all imp chap?
It depends person to person and how much time u r giving for Quant. practice
If ur basics are clear it wont take more than a day according to me 😃
@chandrakant.k said:i haven't done from this book at all so not a right person to answer this question
by the way from which material u r practicing ? 😃
@shattereddream said:It depends person to person and how much time u r giving for Quant. practiceIf ur basics are clear it wont take more than a day according to meby the way from which material u r practicing ?
no proper material 😁 😁 have solved from TIME . for more problems i am here 😁 :D
Two trains for mumbai leave delhi at 6 am and 6:45 am and travel at 100 kmph and 136 kmp respectively.. How many kilometres from delhi will the two trains be together ??
Pls help with solution
@Raghu5jan said:Two trains for mumbai leave delhi at 6 am and 6:45 am and travel at 100 kmph and 136 kmp respectively.. How many kilometres from delhi will the two trains be together ??
Pls help with solution
nothing else??? when the distance is not given or the time reached is not given we cannot calculate AFAIK
@chandrakant.k said:nothing else??? when the distance is not given or the time reached is not given we cannot calculate AFAIK
frame an equation
d1 = (100)(t)
d2 = (136)(t) - (136)(3/4) [since this train started later]
now equate these two
d2 = (136)(t) - (136)(3/4) [since this train started later]
now equate these two
koi plz bata do Number System ki theory k liye kya prefer kiya jaye . I want to strong my Number System . thanks in advance .......
Naryan Murthy walking at a speed of 20kmph reaches his college 10 mint late..Next time he increases his speed by 5kmph but finds tat he is still late by 4 mints..What is the distance of his college from his house.. ??
Looks like an easy sum but nothing getting the approch
@Raghu5jan said:Naryan Murthy walking at a speed of 20kmph reaches his college 10 mint late..Next time he increases his speed by 5kmph but finds tat he is still late by 4 mints..What is the distance of his college from his house.. ??Looks like an easy sum but nothing getting the approch
D/20=t+1/6------------------1
D/25=t+1/15------------------2
now 2 equations 2 variables solve them you will get the answer
@gudda1122 said:D/20=t+1/6------------------1D/25=t+1/15------------------2now 2 equations 2 variables solve them you will get the answer
dude one doubt wen they say he was late by 10mint shudn't we suppose to minus instead of plus ?? I mean this equation t+1/6??
@Raghu5jan said:dude one doubt wen they say he was late by 10mint shudn't we suppose to minus instead of plus ?? I mean this equation t+1/6??
buddy "He was late by 10 mins means he tooh 10 mins more than the Usual Time to travel that distance"
@gudda1122 said:buddy "He was late by 10 mins means he tooh 10 mins more than the Usual Time to travel that distance"cheers tat is there..thanks for the explanation mate 😃
@Raghu5jan said:Naryan Murthy walking at a speed of 20kmph reaches his college 10 mint late..Next time he increases his speed by 5kmph but finds tat he is still late by 4 mints..What is the distance of his college from his house.. ??Looks like an easy sum but nothing getting the approch
One more approach
At 20 kmph------------ 10 min (Late)
At 25 kmph------------ 4 min (Late)
Distance constant
take the lcm of of his speeds 20 & 25 which is 100 .this 100 Km. is our assumed distance
first day to travel 100 km he takes 5hrs & next day he takes 4 hrs.
Now the difference between the time is one hr. ie 60 min.
suppose he reaches office at 9:00 Daily but when he is late by 10 min he reaches at 9:10 & when he is late by 4 min he reaches at 9:04 So the difference comes to 6 mins.
Now 60/6= 10 min & similarly 100/10 = 10 km.
Similar approach u can apply for if he reaches office early. 😃
At 20 kmph------------ 10 min (Late)
At 25 kmph------------ 4 min (Late)
Distance constant
take the lcm of of his speeds 20 & 25 which is 100 .this 100 Km. is our assumed distance
first day to travel 100 km he takes 5hrs & next day he takes 4 hrs.
Now the difference between the time is one hr. ie 60 min.
suppose he reaches office at 9:00 Daily but when he is late by 10 min he reaches at 9:10 & when he is late by 4 min he reaches at 9:04 So the difference comes to 6 mins.
Now 60/6= 10 min & similarly 100/10 = 10 km.
Similar approach u can apply for if he reaches office early. 😃
@shattereddream said:One more approach At 20 kmph------------ 10 min (Late) At 25 kmph------------ 4 min (Late)Distance constanttake the lcm of of his speeds 20 & 25 which is 100 .this 100 Km. is our assumed distancefirst day to travel 100 km he takes 5hrs & next day he takes 4 hrs.Now the difference between the time is one hr. ie 60 min.suppose he reaches office at 9:00 Daily but when he is late by 10 min he reaches at 9:10 & when he is late by 4 min he reaches at 9:04 So the difference comes to 6 mins. Now 60/6= 10 min & similarly 100/10 = 10 km.Similar approach u can apply for if he reaches office early.
Nice one buddy thanks
@Raghu5jan A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter if the parameters are inversely proportional.- This is also one f the short cut method for these questions
@shattereddream said:@Raghu5janA 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter if the parameters are inversely proportional.- This is also one f the short cut method for these questions
@Raghu5jan application for the above mentioned method. Consider this example
A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office?
Let original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ‚Ë?. increase in speed will result in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hours. Ans.= 4 x 1.5=6 km
A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office?
Let original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ‚Ë?. increase in speed will result in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hours. Ans.= 4 x 1.5=6 km