Model 1 Basic Time & Distance using formula Practice Questions Answers Test with Solutions & More Shortcuts

Question : 16 [SSC CGL Prelim 1999]

A train is travelling at the rate of 45km/hr. How many seconds it will take to cover a distance of $4/5$ km ?

a) 120 sec.

b) 90 sec.

c) 64 sec.

d) 36 sec.

Answer: (b)

Using Rule 1,
Distance = Speed × Time
Speed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$
1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

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

= ${4/5}/45$ hour

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

Question : 17 [SSC SAS 2010]

A and B travel the same distance at speed of 9 km/hr and 10 km/ hr respectively. If A takes 36 minutes more than B, the distance travelled by each is

a) 66 km

b) 60 km

c) 54 km

d) 48 km

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

Question : 18 [SSC CGL Tier-I 2013]

A man rides at the rate of 18 km/ hr, but stops for 6 mins. to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is

a) 6 hrs. 24 min.

b) 6 hrs. 18 min.

c) 6 hrs. 12 min.

d) 6 hrs.

Answer: (b)

90 km = 12 × 7km + 6 km.

To cover 7 km total time taken = $7/18$ hours + 6 min. = $88/3$ min.

So, (12 × 7 km) would be covered in $(12 × 88/3)$ min.

and remaining 6km is $6/18$ hrs or 20 min.

Total time = $1056/3$ + 20

= $1116/{3 × 60}$ hours = 6$1/5$ hours

= 6 hours 12 minutes.

Question : 19 [SSC CGL Prelim 2004]

A man travelled a certain distance by train at the rate of 25 kmph. and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, the distance was

a) 15 km

b) 20 km

c) 30 km

d) 25 km

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.

Using Rule 5,
If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be $({2xy}/{x + y})$

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

Question : 20 [SSC CGL Tier-II 2014]

A man travelled a distance of 80 km in 7 hrs partly on foot at the rate of 8 km per hour and partly on bicycle at 16km per hour. The distance travelled on the foot is

a) 44 km

b) 36 km

c) 48 km

d) 32 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.

Using Rule 13,
Let a man take 't' hours to travel 'x' km. If he travels some distance on foot with the speed u km/h and remaining distance by cycle with the speed v km/h,then time taken to travel on foot.
Time = ${(vt - x)}/{(v - u)}$
Distance travelled on foot = Time × u

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

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