@gs4890 said:10 x 10 = 10014 x x = 105(10 - 105/14 ) / 10 = 25%
100 -------> 140
In order to maintane the expenditure he reduces
(140 - 100 ) / 140 * ( 100 ) = 28.56 % !!
What if reduce 5 % in this ??
@ani6 said:A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.a. 9km/h, 3km/hb. 9km/h, 6km/hc. 8km/h, 2km/hd. 12km/h, 3km/h.??plz provide elaborate answer
Let, u =speed of boat in still water & v = speed of river
---------------------------
28/u+v = x & 28/u-v = 2x
-----------------------------
also, 28/u+2v + 28/u-v = 11.2
-----------------------------
Solve three eqtns , u'll get , v=3 , u=9
@ani6 said:A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.a. 9km/h, 3km/hb. 9km/h, 6km/hc. 8km/h, 2km/hd. 12km/h, 3km/h.??plz provide elaborate answer
Let, u =speed of boat in still water & v = speed of river
---------------------------
28/u+v = x & 28/u-v = 2x
-----------------------------
also, 28/u+2v + 28/u-v = 11.2
-----------------------------
Solve three eqtns , u'll get , v=3 , u=9
@ani6 said:A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.a. 9km/h, 3km/hb. 9km/h, 6km/hc. 8km/h, 2km/hd. 12km/h, 3km/h.??plz provide elaborate answer
Let, u =speed of boat in still water & v = speed of river
---------------------------
28/u+v = x & 28/u-v = 2x
-----------------------------
also, 28/u+2v + 28/u-v = 11.2
-----------------------------
Solve three eqtns , u'll get , v=3 , u=9
------------------------------
Hint by Amresh is a shortcut though :)
@mani0303 Can u plz plz explain ur approach for finding the shortest distance? 😲 [Sums 2 & 3]. I can see that its 2*pi*r*(1/5), but why are we considering the circumference & what is this (1/5)? :O
@next_big_thing said:Can someone plz solve this sum with (approach) coz I am stuck -->Chiku & Charu are running at a circular track path having a diameter of 1050m at the speeds of 15m/s & 25m/s while Chaman is running along the path AB (diameter of the track) at a speed of 35m/s. Chaman turns back immediately after reaching B & again immediately after reaching A. All of them start from the same point, i.e. A, simultaneously.If Chiku & Charu are running in the same direction,1) What is the number of distinct points at which Chiku & Charu meet on the circular track? [2, 3, 7, 8, None of these]If Chiku & Charu are running in opposite direction,2) What is the shortest distance along the circular path between point A & the point at which Chiku & Charu meet for the 3rd time? [247.5, 412.5, 660, 1320, 2887.5]3) What is the shortest distance between Chiku & Charu along the circular path when Chaman reaches point B for the 5th time? [150, 270, 412.5, 900, 750]4)What is the number of distinct points on the circular path at which Chiku & Charu NOW meet? [2, 3, 7, 8, 10]Plz someone solve this sum....
same direction
1- chiku : charu = a : b = 15 : 25 = 3 : 5
number of distinct meeting pts = 5 - 3 = 2
opposite dir
2- time taken by chiku and charu to move around d circle = 220s and 132s
time of 1st meeting = T = 220x132 /( 220 - 132) = 330s
d meeting pts will be seperated by an angle of 45 degree on the circumference of circle. the 3rd meeting pt is adjacent to the point A,
distance = 45/360 (1050x22/7) = 412.5
3- chaman reaches B for the 5th time when t = 270s.
so distance by chiku and chaman can be calculated. i guess it should be 750..? not sure though
5- meeting pts = 5 + 3 = 8
@mani0303 said:1)if they are running in the same direction,then the number of distinct points that they would meet = Difference of ratio of their speeds..so in this case it should be 5 - 3 = 2 points2)This one,it should be 660:it's 2*22/7*(1050/2)*(1/5) 3)it's 150 4) 1)if they are running in the opposite direction,then the number of distinct points that they would meet = sum of ratio of their speeds..so in this case it should be 5 + 3 = 8 points
i think 2nd one should be 412.5m not 660
the meeting will be 45 degree adjacent to point A.
@mani0303 said:1)if they are running in the same direction,then the number of distinct points that they would meet = Difference of ratio of their speeds..so in this case it should be 5 - 3 = 2 points2)This one,it should be 660:it's 2*22/7*(1050/2)*(1/5) 3)it's 150 4) 1)if they are running in the opposite direction,then the number of distinct points that they would meet = sum of ratio of their speeds..so in this case it should be 5 + 3 = 8 points
i think it should be 412.5m
@techgeek2050 said:i think it should be 412.5m
I'm not sure too..ll share the method if it's right@next_big_thing
@techgeek2050 said:i think it should be 412.5m
I'm not sure too...ll wait for or solve it if it's right
@next_big_thing - can you please share the answers,please?
@next_big_thing said:@mani0303@techgeek2050Can u plz how u found out the shortest distance in sums 2 & 3???
@mani0303 said:I'm not sure too...ll wait for or solve it if it's right@next_big_thing - can you please share the answers,please?
bhai i just typed the whole solution but as soon as i posted, i got this - "the post no longer exists". 
in the 2nd one, the number of meeting pts will be 8 equally distributed along the circumference. separated by an angle of 45 degrees.
seeing the figure, the 3rd meeting will be 45 degrees adjacent to A. so the distance will
45/360 (1050x22/7) = 412.5m
@techgeek2050 said:bhai i just typed the whole solution but as soon as i posted, i got this - "the post no longer exists". in the 2nd one, the number of meeting pts will be 8 equally distributed along the circumference. separated by an angle of 45 degrees.seeing the figure, the 3rd meeting will be 45 degrees adjacent to A. so the distance will45/360 (1050x22/7) = 412.5m
Yeah,You are right - i thought we were talking about the 3rd one
I could understand your frustration;),but o be on safer side,just copy ur post content in the notepad or something:)
@techgeek2050 Sir I am facing the same problem while posting too 😞 Plus notifications are not getting highlighted. Dont understand what is happening :(
Coming to the sum, did'nt quite understand what u said 😞 Yes, they meet at 8 distinct points as they are running in opp. direction. But did'nt understand the angle part :(
@techgeek2050 Sir I am facing the same problem while posting too 😞 Plus notifications are not getting highlighted. Dont understand what is happening 😞 Coming to the sum, did'nt quite understand what u said 😞 Yes, they meet at 8 distinct points as they are running in opp. direction. But did'nt understand the angle part :(
@techgeek2050 Sir I am facing the same problem while posting too 😞 Plus notifications are not getting highlighted. Dont understand what is happening 😞 Coming to the sum, did'nt quite understand what u said 😞 Yes, they meet at 8 distinct points as they are running in opp. direction. But did'nt understand the angle part :(
I posted this after 4 repeated attempts 😞 [Changed browser to mozilla from google chrome]
Can someone plz try this DS sum? 😲
A & B start from two ends of a swimming pool 50m long simultaneously. What is the total distance distance covered by them?
A. They are meeting 2nd time after start
B. A's speed is twice that of B's.
I know that this sum looks very simple. I just cant makeout if this book has a printing mistake in the answer keys or is there any hidden factor that I am missing. Plz help.
A & B start from two ends of a swimming pool 50m long simultaneously. What is the total distance distance covered by them?
A. They are meeting 2nd time after start
B. A's speed is twice that of B's.
I know that this sum looks very simple. I just cant makeout if this book has a printing mistake in the answer keys or is there any hidden factor that I am missing. Plz help.
Can someone plz try this DS sum? 😲
A & B start from two ends of a swimming pool 50m long simultaneously. What is the total distance distance covered by them?
A. They are meeting 2nd time after start
B. A's speed is twice that of B's.
I know that this sum looks very simple. I just cant makeout if this book has a printing mistake in the answer keys or is there any hidden factor that I am missing. Plz help.
A & B start from two ends of a swimming pool 50m long simultaneously. What is the total distance distance covered by them?
A. They are meeting 2nd time after start
B. A's speed is twice that of B's.
I know that this sum looks very simple. I just cant makeout if this book has a printing mistake in the answer keys or is there any hidden factor that I am missing. Plz help.