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

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: (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)

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

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