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.
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.
@next_big_thing said:@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
first of all bro, don't call me sir 
see the 8 meeting points will be distributed along the circumference equally. so they will be seperated by an angle of 360/8 = 45 degrees.
chiku : charu = a : b = 3 : 5
=> when a covers 3x , b covers 5x
now from the fig attached, let the distance a->a1 be x. so when a moves a->a1->a2->a3 , it covers 3x. so b covers 5x counterclockwise. from the fig u will see 5x comes to the a3, labelled in green as 1, the 1st meeting point. similarly u can find the 2nd and 3rd meeting points. hope that's clear. :)
Can someone plz try this DS sum? :O
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.
@techgeek2050 said:think of it as a problem of relative speedlet there be n steps speed of escalator be e ; initial speed of woman be x , time t1 since she takes 10 steps , x = 10/t1upstairs moving speed = 25/t2 = 5x = 50/t1=> t1 = 2t2 ........(1)(e + x)t1 = n(5x - e)t2 = n=> solving these 2 using (1) , n = 20
Hi..i just don understand the steps (e + x)t1 = n and (5x - e)t2 = n . can u pls explain those?
a certain number "C" when divided by N1 it leaves remainder of 13 and when it is divided by n2 it leaves the remainder 1 where N1 and N2 are the positive integers.then the value N1+N2 is i f N1/N2=5/4:
@j.gayu said:a certain number "C" when divided by N1 it leaves remainder of 13 and when it is divided by n2 it leaves the remainder 1 where N1 and N2 are the positive integers.then the value N1+N2 is i f N1/N2=5/4:
25?
@j.gayu said:a certain number "C" when divided by N1 it leaves remainder of 13 and when it is divided by n2 it leaves the remainder 1 where N1 and N2 are the positive integers.then the value N1+N2 is i f N1/N2=5/4
25??
@j.gayu said:a certain number "C" when divided by N1 it leaves remainder of 13 and when it is
27 or 36
@j.gayu said:a certain number "C" when divided by N1 it leaves remainder of 13 and when it is divided by n2 it leaves the remainder 1 where N1 and N2 are the positive integers.then the value N1+N2 is i f N1/N2=5/4:
getting two values 27 and 36??
73 mod 15 = 13
73 mod 12 = 1
33 mod 20 = 13
33 mod 16 = 1
the first two terms of g.p add up to 12.the sum of 3rd & 4th term is 48.if the terms of the g.p r alternatively positive and negative,then what is the first term???? ans is -12. how to solve this???
Rajesh walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at
speed of 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he
returns to home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours?
speed of 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he
returns to home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours?
@ananyboss said:Question ---Price of petrol has increased to 40% . If MR banerjee wanted to limit his expenditure to 5% What should be the reduction in consumption ?????????????Please help with approach OA - 25
let original price be 100 expenditure be 100%
new price = 140
new expenditure = 105
therefore, reduction in consumption is 35/140*100=25%
@ananyboss said:Question ---Price of petrol has increased to 40% . If MR banerjee wanted to limit his expenditure to 5% What should be the reduction in consumption ?????????????Please help with approach
let original price be 100 expenditure be 100%
new price = 140
new expenditure = 105
therefore, reduction in consumption is 35/140*100=25%
@j.gayu said:a certain number "C" when divided by N1 it leaves remainder of 13 and when it is divided by n2 it leaves the remainder 1 where N1 and N2 are the positive integers.then the value N1+N2 is i f N1/N2=5/4:
i guess it cannot be determined. multiple values are possible
C = 33, 73, ....
(N1,N2) = (30,24) (20,16) (15,12)....
i stopped calculating after this but think there are more values
@priyank333333 said:Rajesh walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at speed of 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he returns to home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours?
20 kmph ?
@MANC-LONDON said:the first two terms of g.p add up to 12.the sum of 3rd & 4th term is 48.if the terms of the g.p r alternatively positive and negative,then what is the first term???? ans is -12. how to solve this???
a(1 + r) =12
ar^2(1 + r) = 48
=> r = -2
=> a = -12