Speed of train relative to boy= `(60 + 6)` km/hr = 66 km/hr = [66 xx 5/18]` m/sec = `[55/3]` m/sec.

Time taken to pass the boy=` [120 xx 3/55]` sec = 6.54 seconds

Time taken to pass the boy=` [120 xx 3/55]` sec = 6.54 seconds

Let be the 4% of a is 4a/100. Since this equals 8, we have 4a/100=8. Solving for a yields a=8×`(100/4)`=200. Also, 8% of b equals 8b/100, and this equals 4.

Hence, we have `(8/100)`×b=4. Solving for b yields b = 50. Now, c=`b/a`=`50/200`=`1/4.`

Hence, we have `(8/100)`×b=4. Solving for b yields b = 50. Now, c=`b/a`=`50/200`=`1/4.`

`speed =( 6**5/18)m/sec) = (30/18) m/sec`

Time taken by P to cover 100 m = `(100 ** 18/30)`m/sec =60 sec

Time taken by Q to cover 92 m = (60 + 8) = 68 sec.

Q's speed =`("Distance"/"Time"**18/5) kmph =(92/68 ** 18/5)` kmph = 4.86kmph.

Time taken by P to cover 100 m = `(100 ** 18/30)`m/sec =60 sec

Time taken by Q to cover 92 m = (60 + 8) = 68 sec.

Q's speed =`("Distance"/"Time"**18/5) kmph =(92/68 ** 18/5)` kmph = 4.86kmph.

Speed = `[142/6]` m/sec = `[23.6 ** 18/5]` km/hr = 84.9 km/hr

Sum of present ages of A, B and C is = 92 years

Therefore , Sum of their ages 4 years ago = 92 – (4 * 3)= 80 years.

4 years ago ratio of the ages of A , B and C was = 1:2:3

Therefore, A’s age four years ago = `1/6` * 80 = 13.3 years.

So , A’s present age =13.3 + 4 = 17.3 years

Therefore , Sum of their ages 4 years ago = 92 – (4 * 3)= 80 years.

4 years ago ratio of the ages of A , B and C was = 1:2:3

Therefore, A’s age four years ago = `1/6` * 80 = 13.3 years.

So , A’s present age =13.3 + 4 = 17.3 years