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

Answer: (a)

Let the required distance be x km.

Difference of time

= 6 + 6 = 12 minutes = $1/5$ hr.

According to the question,

$x/{5/2} - x/{7/2} = 1/5$

${2x}/5 - {2x}/7 = 1/5$

${14x - 10x}/35 = 1/5$

${4x}/35 = 1/5 ⇒ x = 35/20 = 7/4$ km.

Using Rule 10,

Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3{1}/2 , t_2$ = 6

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

= ${5/2 × 7/2 × (6 + 6)}/{7/2 - 5/2}$

= $35/4 × 12/60 = 7/4$ km

Answer: (c)

Let the required distance be x km.

According to the question,

$x/4 - x/5 = 18/60$

${5x - 4x}/20 = 3/10$

$x = 3/10 × 20$ = 6 km

Using Rule 10,

Here, $S_1 = 4, t_1 = 9, S_2 = 5, t_2$ = 9

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

= ${(4 × 5)(9 + 9)}/{5 - 4}$

= $20 × 18/60 = $6 km

Answer: (b)

Using Rule 1

Distance between stations X and Y

= Speed × Time = 55 × 4 = 220 km.

New speed = 55 + 5 = 60 kmph

Required time = $220/60$

= $11/3$ hours

= 3 hours 40 minutes.

Required answer

= 4 hours - 3 hours 40 minutes

= 20 minutes

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)

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: (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