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

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: (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: (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

Answer: (a)

Let the required distance be x km.

$x/{5/2} - x/3 = 16/60$

${2x}/5 - x/3 = 4/15$

${6x - 5x}/15 = 4/15 ⇒ x = 4$ km.

Using Rule 10,

Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3, t_2$ = 10

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

= ${5/2 × 3(6 + 10)}/{3 - 5/2}$

= $15 × 16/60$ km = 4 km