@abhishek.2011 said:yar m getting 998??doing without pen may be thats why getting error??whats oa?? is it 997 0r 998
997 is the OA
@abhishek.2011 said:yar m getting 998??doing without pen may be thats why getting error??whats oa?? is it 997 0r 998
@albiesriram says it is 997
Solve this :
Nimai and Nitai start running from opposite ends of a 100m long racing track with speeds 4m/s and 7m/s towards each other simultaneously. They keep on moving to and fro between the two ends without stoppoing anywhere till they reach their initial positions simultaneously.
At how many different points, Nitai and Nimai meet after starting?
At how many different points, Nitai and Nimai meet after starting?
Team BV--Pratik Gauri
@bodhi_vriksha said:Solve this :Nimai and Nitai start running from opposite ends of a 100m long racing track with speeds 4m/s and 7m/s towards each other simultaneously. They keep on moving to and fro between the two ends without stoppoing anywhere till they reach their initial positions simultaneously. At how many different points, Nitai and Nimai meet after starting?Team BV--Pratik Gauri
nimal will run 800 m=8 rounds
and nitai 1400 m =14 rounds
14?
@albiesriram said:and
for first is it 400
root at every gap of 5 integers
from -998 to 997
in total 400 values
@Subhashdec2 said:@albiesriram says it is 997
f(999) = f(f(1004)) = f(1001) = 998
f(998) = f(f( 1003)) =f( 1000) = 997
f(997) = f(f( 1002)) = f( 999) = 998
f(996) = f(f(1001)) = f(998) = 997 . enough , pattern found.
f(k) = 998 if k is odd,
= 997 if k is even.
84 is even.
@Subhashdec2 said:nimal will run 800 m=8 roundsand nitai 1400 m =14 rounds14?
Different meeting points . All these points would be diff? Think again :)
Team BV--Pratik Gauri
@bodhi_vriksha said:Different meeting points . All these points would be diff? Think again Team BV--Pratik Gauri
6?
@bodhi_vriksha said:Solve this :Nimai and Nitai start running from opposite ends of a 100m long racing track with speeds 4m/s and 7m/s towards each other simultaneously. They keep on moving to and fro between the two ends without stoppoing anywhere till they reach their initial positions simultaneously. At how many different points, Nitai and Nimai meet after starting?Team BV--Pratik Gauri
total time=2d=200
time when they meet moving in same direction=200*(2k+1)/3=66.66*(2k+1)=11 points
time when they meet moving in opposite direction=200*(2k+1)/7=18.18*(2k+1)=3 points
11+3=14
@bodhi_vriksha said:Solve this :Nimai and Nitai start running from opposite ends of a 100m long racing track with speeds 4m/s and 7m/s towards each other simultaneously. They keep on moving to and fro between the two ends without stoppoing anywhere till they reach their initial positions simultaneously. At how many different points, Nitai and Nimai meet after starting?Team BV--Pratik Gauri
answer 6 hai kya sir??
approach :
distance travelled by them when they meet each time...
100, 300, 500 ....
also ratio of distance travelled is d*4/11 and d*7/11
so d*4/11 = 100A
and d*7/11 = 100 B
on solving this it comes out as distance travlled as 400 and 700m..
so they would have met 6 times..
@saurav205 said:answer 6 hai kya sir??approach :distance travelled by them when they meet each time...100, 300, 500 ....also ratio of distance travelled is d*4/11 and d*7/11so d*4/11 = 100Aand d*7/11 = 100 Bon solving this it comes out as distance travlled as 400 and 700m..so they would have met 6 times..
Yes you are correct but after making these equations , how can you deduce that they would have met at 6 distinct points????? Elaborate :)
Team BV--Pratik Gauri
@bodhi_vriksha sir total distance travelled by them is 400+700 = 1100 m
so when they meet for the 1st time together they travel 100m
the next time they meet together they travel = 100(last time ka) + 200 = 300m
3rd time = 500m
4th time = 700m
5th time = 900m
6th time = 1100m(together travelled: 400+700m)
@saurav205 said:@bodhi_vriksha sir total distance travelled by them is 400+700 = 1100 mso when they meet for the 1st time together they travel 100mthe next time they meet together they travel = 100(last time ka) + 200 = 300m3rd time = 500m4th time = 700m5th time = 900m6th time = 1100m(together travelled: 400+700m)
nice man
PS:- Happy Birthday Liverpool...YNWA....
@saurav205 said:@bodhi_vriksha sir total distance travelled by them is 400+700 = 1100 mso when they meet for the 1st time together they travel 100mthe next time they meet together they travel = 100(last time ka) + 200 = 300m3rd time = 500m4th time = 700m5th time = 900m6th time = 1100m(together travelled: 400+700m)
I was looking for a graphical approach . I have understood what you r trying to do . But saurav, How can you make sure that all points would repeat after they have collectiveely travelled 1100 m ?
Team BV--Pratik Gauri
@bodhi_vriksha said:I was looking for a graphical approach . I have understood what you r trying to do . But saurav, How can you make sure that all points would repeat after they have collectiveely travelled 1100 m ?Team BV--Pratik Gauri
Sir, I did it using the graphical method...as in a line worth 100 mts...
the reason they will be distinct points is because the distance traveled by each of them is something like 100k*4/11 and 100m*7/11...4 and 7 being co-prime will not give you repeated points..
I think after 2800 mts you will get repeated points...
Not too sure if this is the best way to explain it...plus will not be able to solve it and check...on the way back home from office...so not possible to solve it...not too sure of the logic either..but this is what i thought of..
If its wrong , please correct me..
@saurav205 said:@bodhi_vriksha sir total distance travelled by them is 400+700 = 1100 mso when they meet for the 1st time together they travel 100mthe next time they meet together they travel = 100(last time ka) + 200 = 300m3rd time = 500m4th time = 700m5th time = 900m6th time = 1100m(together travelled: 400+700m)
i am not getting this??
how r u deducing that after travelling 1100m the points will repeat??
why not add cases for 1300 ,1500,1700,1900,2100 ,as total distance is 2200
and add 5 to the solution?????
@saurav205 said:Sir, I did it using the graphical method...as in a line worth 100 mts...the reason they will be distinct points is because the distance traveled by each of them is something like 100k*4/11 and 100m*7/11...4 and 7 being co-prime will not give you repeated points..I think after 2800 mts you will get repeated points...Not too sure if this is the best way to explain it...plus will not be able to solve it and check...on the way back home from office...so not possible to solve it...not too sure of the logic either..but this is what i thought of..If its wrong , please correct me..
Yes you r right that points would repeat after 2800 m 😃 But you didnt check till 2800 m ..did u? 2800 m means 14 rounds of one person .. But you didnt check 😃 Think again :)
Team BV--Pratik Gauri
@bodhi_vriksha said:Yes you r right that points would repeat after 2800 m But you didnt check till 2800 m ..did u? 2800 m means 14 rounds of one person .. But you didnt check Think again Team BV--Pratik Gauri
no sir..didnt check till 2800m...
I thought that at 1100mts..they reach the initial position..but i guess yahan par galti ki hai..
at 2200mts..they both will travel 800 and 1400 mts respectively...getting them back to initial position..
at 1100 mts , they have traveled 400 and 700...so only one person is on his initial position..and the other 700 mts is at the opposite end..i.e. both are at the same point...
ab bhi kuch galat hai kya??
Usha bought a certain no of chocoltes at the rate of 16 chocolates for Rs 12 and the same no of chocolates at the rate of 24 chocolates for Rs 20. She sold all of them at the rate of 30 chocolates for RS 30. Find her gain/loss %.