C

Given distance = 100m
Speed of kamal= 18 kmph
Time taken by kamal= [latex]\frac {100}{(\frac {18 × 5}{18})}[/latex] = 20 sec
Time taken by Bimal = 20 + 5 = 25 sec
So, Bimal’s speed= [latex]\frac{100}{25}[/latex] = 4 m/s
[latex]\frac{4 × 18}{5}[/latex] = 14.4 kmph
Hence option 3 is correct.

C

Given
Distance= length of train = 240m
If the speed of train be x kmph then.
Its relative speed = (x + 3) kmph
Time = length of train/relative speed
= [latex]\frac {10}{3600}[/latex] = [latex]\frac{(\frac{240}{1000})}{(x + 3)}[/latex]
X + 3 = 86.4
X = 83.4 kmph
Hence option 3 is correct.

B

According to the question :
Let one side distance is D
T1 = [latex]\frac {D}{8}[/latex] ------------ (I)
T2 = [latex]\frac {D}{6}[/latex] ------ (II)
Adding (I) and (II)
T1 + T2 = D( [latex]\frac {1}{8} + \frac {1}{6}[/latex])
3 [latex]\frac {1}{2}[/latex] = D [latex]\frac {( 3 + 4)}{(24)}[/latex]
[latex]\frac {7}{2}[/latex] = D ([latex]\frac {7}{24}[/latex]) = D = 12 kms.
For both the sides = 12 + 12 = 24 km.

A

Distance = [latex]\frac {2 × Time × {S}_{1} × {S}_{2}}{{S}_{1} + {S}_{2}}[/latex]
Where [latex]{S}_{1}[/latex] = speed during first half
[latex]{S}_{2}[/latex] = speed during second half of journey
i.e, Distance = [latex]\frac {2 × 10 × 21 × 24}{21 + 24}[/latex] = 224 km

C

Average speed = [latex]\frac {2(xy)}{(x + y)}[/latex]
Where x = 120, y = 80
[latex]\frac {2(120 \times 80)}{(120 + 80)}[/latex] = [latex]\frac {2(120 \times 80)}{200}[/latex] = 96 km/h