Model 1 Basic Time & Distance using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Distance covered by car in2hours

=${300 × 40}/100$ = 120km

Remaining distance

= 300 - 120 = 180 km

Remaining time= 4 - 2= 2 hours

Required speed

=$180/2$= 90kmph

Original speed of car

=$120/2$= 60kmph

Required increase in speed

= 90 - 60 = 30 kmph

Answer: (d)

Let the required distance be x km. Then,

$x/3 + x/2$ = 5

${2x + 3x}/6$ = 5

5x = 6 × 5

x = ${6 × 5}/5$ = 6 km

Using Rule 5,

Here, x = 3, y = 2

Average Speed = ${2 × x × y}/{x + y}$

= ${2 × 3 × 2}/{3 + 2} = 12/5$ km/hr

Total distance = $12/5 × 5$ = 12km

Required distance = $12/2$ = 6 km

Answer: (b)

If the required distance be = x km, then

$x/4 - x/5 = {10 + 5}/60$

${5x - 4x}/20 = 1/4$

$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 5 km.

Here, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10

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

=${(4 × 5)(5 + 10)}/{5 - 4}$

=20 × $15/60$ = 5 kms

Answer: (b)

Speed of train = 60 kmph

Time = 210 minutes

= $210/60$ hours or $7/2$ hours

Distance covered

= $60 × 7/2$ = 210 km

Time taken at 80 kmph

= $210/80 = 21/8$ hours

= 2$5/8$ hours

Using Rule 9,

Here, $S_1 = 60, t_1 = 210/60 hrs, S_2 = 80, t_2$ = ?

$S_1t_1 = S_2t_2$

60 × $210/60 = 80 × t_2$

$t_2 = 21/8$ hrs = 2$5/8$ hrs

Answer: (c)

If the required distance be x km, then

$x/3 - x/4 = 1/2$

${4x - 3x}/12 = 1/2$

$x/12 = 1/2$ ⇒ x = 6 km

Using Rule 9,

Here $S_1 = 4, t_1 = x, S_2 = 3, t_2 = x + 1/2$

$S_1t_1 = S_2t_2$

$4 × x = 3(x + 1/2)$

$4x - 3x = 3/2 x = 3/2$

Distance= $4 × 3/2$ = 6 kms

Answer: (b)

Using Rule 1,

Speed = $150/25$ = 6 m/sec

= $6 × 18/5 = 108/5$ = 21.6 kmph