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

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: (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)

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: (a)

Using Rule 1,

Speed of train = $\text"Distance"/\text"Time"$

= $10/{12/60}$ kmph

= ${10 × 60}/12 = 50$ kmph

New speed = 45 kmph

∴ Required time = $10/45$ hour

= $2/9 × 60$ minutes

= $40/3$ minutes or 13 minutes 20 seconds

Answer: (c)

Let the distance be x km and initial speed be y kmph.

According to question,

$x/y - x/{y + 3} = 40/60$ ...(i)

and, $x/{y - 2} - x/y = 40/60$ ...(ii)

From equations (i) and (ii),

$x/y - x/{y + 3} = x/{y - 2} - x/y$

$1/y - 1/{y + 3} = 1/{y - 2} - 1/y$

${y + 3 -y}/{y(y + 3)} = {y - y + 2}/{y(y - 2)}$

3 (y - 2) = 2 (y + 3)

3y - 6 = 2y + 6

y = 12

From equation (i),

$x/12 - x/15 =40/60$

= ${5x - 4x}/60 = 2/3$

$x = 2/3 × 60$ = 40

Distance = 40 km.

Answer: (a)

Let the distance of school be x km, then

$x/3 - x/4 = 20/60$

$x/12 = 1/3 ⇒ x =12/3 = 4$ km

Using Rule 10,

Here, $S_1 = 3, t_1 = 10, S_2 = 4, t_2$ = 10

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

= ${(3 × 4)(10 + 10)}/{4 - 3}$

= $12 × 20/60$ = 4 km