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

Answer: (c)

Let the distance be x km.

Total time = 5 hours 48 minutes

= $5 + 48/60 = (5 + 4/5)$ hours

= $29/5$ hours

$x/25 + x/4 = 29/5$

${4x + 25x}/100 = 29/5$

5 × 29x = 29 × 100

$x = {29 × 100}/{5 × 29}$ = 20 km.

Here, x = 25, y = 4

Average speed = ${2xy}/{x + y}$

= ${2 × 25 × 4}/{25 + 4} = 200/29$

Total Distance = $200/29 × 5{4}/5$

= $200/29 × 29/5$ = 40 km

Required distance = 20 km

Answer: (d)

Journey on foot=x km

Journey on cycle = (80 –x)km

$x/8 + {80 - x}/16 = 7$

${2x + 80 - x}/16 = 7$

x + 80 = 16 × 7 = 112

x= 112 - 80 = 32 km.

Here, x = 80, t = 7, u = 8, v = 16

Time = $({vt - x}/{v - u})$

=$({16 × 7 - 80}/{16 - 8})$

=$({112 - 80}/8) = 32/8$ = 4 hrs

Distance travelled

= 4 × 8 = 32 kms

Answer: (c)

Using Rule 1,

Remaining time

= $2/5 × 15 = 6$ hours

Required speed

= $60/6$ = 10 kmph

Answer: (b)

Let the distance between A and B be x km, then

$x/9 - x/10 = 36/60 = 3/5$

$x/90 = 3/5$

$x = 3/5 × 90$ = 54 km.

Using Rule 9,

Here, $S_1 = 9, t_1 = x, S_2 = 10, t_2 = x - 36/60$

$S_1t_ 1 = S_2t_ 2$

$9 × x = 10(x - 36/60)$

9x = 10x - 6 = 6

Distance travelled = 9 × 6 = 54 km

Answer: (b)

Time taken = $\text"Distance"/ \text"time"$

= ${4/5}/45$ hour

= ${4 × 60 × 60}/{5 × 45}$ sec. = 64 seconds

Answer: (a)

Difference of time

= 4.30 p.m - 11.a.m.

= $5{1}/2$ hours $11/2$ hours

Distance covered in $11/2$ hrs

= $5/6 - 3/8 = {20 - 9}/24 = 11/24$ part

Since, $11/24$ part of the journey is covered in $11/2$ hours

$3/8$ part of the journey is covered in

= $11/2 × 24/11 × 3/8$

= $9/2$ hours = 4$1/2$ hours.

Clearly the person started at 6.30 a.m.