Question

Let the A meets ( B ) and ( C ) in ( X & Y ) min.
The distance covered by A in ( 5 mathrm{min}=1 mathrm{km} ) Then, the distance covered by A in ( 1 mathrm{min}=1000 / 5=200 mathrm{m} ).
The distance covered by B in 8 ( mathrm{min}=1 mathrm{km} ) Then, the distance covered by B in ( 1 mathrm{min}=1000 / 8=125 mathrm{m} ).
The distance covered by ( mathrm{C} ) in ( 10 mathrm{min}=1 mathrm{km} ) Then, the distance covered by ( mathrm{C} ) in ( 1 mathrm{min}=1000 / 10=100 mathrm{m} )
According to the question, Distance covered by ( mathrm{C} ) in ( (mathrm{X}+2) mathrm{min}= ) the distance covered by in ( mathrm{X} ) min Then, ( 100(x+2)=200 x )
( 100 x+200=200 x )
( 200=200 x-100 x )
( 100 x=200 )
( x=200 / 100 )
( mathrm{X}=2 mathrm{min} )
Now, for ( mathrm{A} & mathrm{B} )
Distance covered by B in ( (mathrm{Y}+1) mathrm{min}= ) Distance covered by A in Y min Then, ( 125(Y+1)=200 times Y )
( 125 mathrm{Y}+125+200 mathrm{Y} )
( 125=200 mathrm{Y}-125 mathrm{Y} )
( 125=75 mathrm{Y} )
( Y=125 / 75 )
( mathrm{Y}=5 / mathrm{min} )
Hence, A meets B and ( mathrm{C} ) at ( 2 mathrm{min} ), ( 5 / 3 mathrm{min} ).

# 5. A, B and C walk 1 km in 5 min, 8 min and 10 min, respectively. C starts walking from a point, at a certain time, B starts from the same point 1 min later and A starts from the same point 2 min later than C. Then, A meets B and C at times (a) 2 min, 3 min (b) min, 3 min (c) 2 min, - min (d) 1 min, 2 min

Solution