D

speed = [latex](\frac{5 \times 5}{18})[/latex]m/sec
= [latex]\frac{25}{18}[/latex] m/sec.
Distance covered in 15 minutes = ([latex]\frac{25}{18} \times 15 \times 60[/latex])m
= 1250 m.

C

The hour hand moves from pointing to 12 to pointing to half way between 2 and 3. The angle covered between each hour marking on the clock is [latex]\frac{360}{12}[/latex] = 30. Since the hand has covered 2.5 of these divisions the angle moved through is 75.

C

Time taken by Akash = 4 h
Time taken by Prakash = 3.5 h
For your convenience take the product of times taken by both as a distance.
Then the distance = 14 km
Since, Akash covers half of the distance in 2 hours(i.e at 8 am)
Now, the rest half (i.e 7 km) will be covered by both Prakash and Akash
Time taken by them = [latex]\frac{7}{7.5}[/latex] = 56 min
Thus , they will cross each other at 8 : 56 am.

D

Let the speeds of two trains be 7X and 8X km/hr.
8x = [latex]\frac{400}{4}[/latex]
[latex]\Rightarrow[/latex]x= 12.5 km/hr
So speed of first train is 12.5 [latex]\times[/latex] 7 = 87.5 km/hr

A

Let the original speed be x km/h, then
[latex]\frac{600}{x}[/latex] - [latex]\frac{600}{(x + 20)}[/latex] = 1
[latex]\Rightarrow 600 \frac{x + 20 -x}{x(x + 20)} = 1[/latex]
[latex]{x}^{2}[/latex] + 120x - 12000 = 0
[latex]\Rightarrow (x - 120)(x - 100) = 0[/latex]
Therefore, original speed = 100 km/h