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

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: (d)

Since man walks at $2/3$ of usual speed, time taken wil be $3/2$ of usual time.

$3/2$ of usual time

= usual time + 1 hour.

$(3/2 –1)$ of usual time = 1

usual time = 2 hours.

Answer: (b)

Let x km. be the required distance.

Difference in time

= 2.5 + 5 = 7.5 minutes

= $7.5/60$ hrs. = $1/8$ hrs.

Now, $x/8 - x/10 = 1/8$

= ${5x - 4x}/40 = 1/8$

$x = 40/8$ = 5 km.

Here, $S_1 = 8, t_1 = 2.5, S_2 = 10, t_2 = 5$

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

= ${(8 × 10)(2.5 + 5)}/{10 - 8}$

= $40 × {7.5}/60$ = 5 km

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