Model 1 Basic Time & Distance Using Formula Practice Questions Answers Test With Solutions & More Shortcuts

Question : 11 [SSC CPO S.I.2005]

A boy runs 20 km in 2.5 hours. How long will he take to run 32 km at double the previous speed ?

a) 5 hours

b) 4$1/2$ hours

c) 2$1/2$ hours

d) 2 hours

Answer: (d)

Using Rule 1,

The boy covers 20 km in 2.5 hours.

Speed = $20/{2.5}$ = 8 km/hr.

New speed = 16 km/hr

Time = $32/16$ = 2 hours.

Question : 12 [SSC CGL Prelim 2003]

A car travelling at a speed of 40 km/hour can complete a journey in 9 hours. How long will it take to travel the same distance at 60 km/hour ?

a) 4$1/2$ hours

b) 4 hours

c) 3 hours

d) 6 hours

Answer: (d)

Total distance covered

= Speed × Time

= 40 × 9 = 360 km.

The required time at 60 kmph

= $360/60 = 6$ hours.

Using Rule 9,
Speed(s) ∝ $1/{time (t)}$ ⇒ s ∝ $1/t$
$s_1t_1 = s_2t_2$(Provided distance is constant)

Here, $S_1 = 40, t_1 = 9, S_2 = 60, t_2$ = ?

$S_1t_1 = S_2t_2$

40 × 9 = 60 × $t_2$

$t_2 = {4 × 9}/6$ = 6 hours

Question : 13 [SSC MTS 2011]

Walking at the rate of 4 km an hour, a man covers a certain distance in 3 hours 45 minutes. If he covers the same distance on cycle, cycling at the rate of 16·5 km/hour, the time taken by him is

a) 45.55 minutes

b) 55.44 minutes

c) 54.55 minutes

d) 55.45 minutes

Answer: (b)

Using Rule 1,

Distance covered on foot

= 4 × 3$3/4$ km. = 15 km.

Time taken on cycle

= $\text"Distance"/\text"Speed" = 15/{16.5}$ hour

= ${15 × 60}/{16.5}$ minutes

= 54.55 minutes

Question : 14 [SSC CGL Tier-I 2013]

A man covers $2/15$ of the total journey by train, $9/20$ by bus and the remaining 10 km on foot. His total journey (in km) is

a) 12.8

b) 16.4

c) 24

d) 15.6

Answer: (b)

Let the total journey be of x km, then

${2x}/15 + {9x}/20 + 10 = x$

$x - {2x}/15 - {9x}/20$ = 10

${60x - 8x - 27x}/60$ = 10

${25x}/60$ = 10

$x = {60 × 10}/25$ = 24 km

Question : 15 [SSC CGL Prelim 2007]

A person started his journey in the morning. At 11 a.m. he covered $3/8$ of the journey and on the same day at 4.30 p.m. he covered $5/6$ of the journey. He started his journey at

a) 6.30 a.m.

b) 7.00 a.m.

c) 3.30 a.m.

d) 6.00 a.m.

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.

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