Model 6 Change in speed Vs Change with time travel Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Let the distance of the office be x km, then

$x/5 - x/6 = 8/60$

${6x - 5x}/30 = 2/15$

x = 2 × 2 = 4 km

Using Rule 10,

Here, $S_1 = 5, t_1 = 6, S_2 = 6, t_2$ = 2

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${(6 × 5)(6 + 2)}/{6 - 5}$

= $30 × 8/60$ = 4 km

Answer: (b)

Let the distance of office be x km.

$x/24 - x/30 = 11/60$

${5x - 4x}/120 = 11/60$

$x/120 = 11/60$

$x = 11/60$ × 120 = 22 km.

Using Rule 10,

Here, $S_1 = 24, t_1 = 5, S_2 = 30, t_2$ = 6

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${24 × 30(5 + 6)}/{30 - 24}$

= ${720 × 11}/{6 × 60}$ = 22 km

Answer: (b)

Time taken to cover 20 km at the speed of 5km/hr

= 4 hours.

Fixed time = 4 hours - 40 minutes

= 3 hour 20 minutes

Time taken to cover 20 km at the speed of 8 km/hr

= $20/8$ = 2 hours 30 minutes

Required time

= 3 hours 20 minutes - 2 hours 30 minutes

= 50 minutes

Answer: (b)

Distance of journey = x km

Difference of time

= 12 - 3 = 9 minutes

= $9/60$ hour = $3/20$ hour

$x/70 - x/80 = 3/20$

$x/7 - x/8 = 3/2$

${8x - 7x}/56 = 3/2$

$x/56 = 3/2$

$x = 3/2 × 56$ = 84 km

∴ Required correct time

= $84/70$ hours - 12 minutes

= $(84/70 × 60 - 12)$ minutes

= 72 - 12 = 60 minutes

= 1 hour

Answer: (d)

Let the distance between stations be x km, then speed of train

= $x/{45/60} = {4x}/3$ kmph

$x/{{4x}/3 - 5} = 48/60$

${3x}/{4x - 15} = 4/5$

16x - 60 = 15x

x = 60 km

Answer: (b)

Let the required distance be x km.

Then, $x/3 - x/5 = 24/60$

${5x - 3x}/15 = 2/5$

${2x}/3 = 2$ ⇒ 2x = 2 × 3

x = 3 km

Using Rule 10,

Here, $S_1 = 3, t_1 = 9, S_2 = 5, t_2$ = 15

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${(3 × 5)(9 + 15)}/{5 - 3}$

= $15/2 × 24/60$ = 3 km