Time and Distance - 04

Explanation:-

3+7+11+15+ . . . . . (1)
1+9+17+25+ . . . . . (2)
The condition is satisfied for the 4 seconds in 2^nd journey, ant covered 17 m more than 1^st one.

Explanation:-

$$\eqalign{ & {\text{Speed of first Jogger}} \cr & = \left[ {\frac{{\left( {x - 1} \right)}}{{42}}} \right] \times 60\,{\text{kmph}} \cr & {\text{Speedof}}\,{2^{nd}}\,{\text{jogger}} \cr & = \left[ {\frac{{\left( {x - 2} \right)}}{{52}}} \right] \times 60\,{\text{kmph}} \cr & {\text{Then}}, \cr & \left( {\frac{{x - 2}}{{{s_b}}}} \right) - \left( {\frac{{x - 1}}{{{s_b}}}} \right) \cr & \cr}$$ Now, we use option checking method which gives us that option (b) is correct.

Explanation:-

The sum of speeds would be 0.8 m/s (relative speed in opposite direction).
Also if we go by option (b) the speeds will be 0.03 and 0.05 m/s respectively.
At this speed the overlapping would occur in every 60 second.

Explanation:-

A _______18miles_____B ___x(let)___C
Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h.
Let time t is taken to cover the distance x from point B to point C by ship.
Hence, in time t plane will cover (18+x) miles.

Now,
^x/₁ = ^(18+x)/₁₀
Or, 10x = 18+x
Or, x = 2 miles.
Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time.

Short method:
Relative speed = (10-1) = 9 miles/h.
hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.

Explanation:-

In the final minute of collision, the two boats travel ⁵/₆₀ km and ¹⁰/₆₀ km.
As they moves towards each other relative distance covered in 1 minute,
= (¹/₁₂)+(¹/₆) = ¹/₄ km.

Explanation:-

A__________300 km_______B
Let speed of the 1^st ship is s kmph and speed of 2^nd ship is (s+10) kmph.
Let 2^nd ship takes time t to cover the distance 300 km then 1^st ship takes (t+8) hours.

As distance is constant, we have,
^s/_(s+10) = ^t/_(t+8)
Or, st+8s = st+10t
Or, 8s = 10t

If the slower ship took 20 hours (option d) the faster ship would take 12 hours and their respective speeds would be 15 and 25 kmph.

Explanation:-

Since, the two horses meet after 200 minutes; they cover 0.5% of the distance per minute (combined) or 30% distance per hour.
This condition is satisfied only if the slower rider takes 10 hours (thereby covering 10% per hour) and the faster rider takes 5 hours.

Explanation:-

Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec.
Let the speed of train be x kmph.
Now,
Or, x * ¹¹/₆₀ = (⁴⁵/₃₆₀₀) * 330 * ¹⁸/₅
Or, x = 81 kmph.

Explanation:-

Distance covered in one complete revolution $$\eqalign{ & = 2\pi r \cr & = \frac{{2 \times 100 \times 22}}{7} \cr & = 628\,m \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{628}}{2} \cr & = 314\,m/\min \cr}$$

Explanation:-

In 6 minutes, the car goes ahead by 0.6 km.
Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph.
Hence, Net speed of the car is 8 kmph.