Question

4 22. A man cycles at the speed of 8km/hr and reaches office at 11 am and when he cycles at the speed of 12 km/hr he reaches office at 9 am. At what speed should he cycle so that he reach- es his office at 10 am? (1) 9.6 kmph. (2) 10 kmph. (3) 11.2 kmph. (4) Cannot be determined (SSC CPO SI, ASI Online ​

Answer 1

(1) 9.6 kmph

Step-by-step explanation:

Given -

• Man cycles at the speed of 8km/hr and reaches office at 11 am.
• When he he cycles at the speed of 12 km/hr he reaches office at 9 am.

To find -

• Speed at which he should cycle so that he reaches his office at 10 am.

Solution -

Let 'x' be the distance to be travelled.

Let 't' be the time required to reach the office.

According to the first statement,

$\mathrm{\dfrac{x}{t} = 8} \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{{ \large(}time = {\dfrac{distance}{speed} {\large)} }}$

$\longmapsto \mathrm{ {x}= 8t \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)}$

Now , according to the second statement,

$\mathrm{ \dfrac{x}{t - 2} = 12}$

Here, we took time as (t - 2) hours because the man took 2 hours less to reach the office then in the first case.

$\longmapsto \mathrm{ {x}= 12(t - 2)}$

$\longmapsto \mathrm{ {x}= 12t - 24} \: \: \: \: \: \: \: \: \: \: \: (2)$

On equating (1) and (2) , we get ,

$\longmapsto \mathrm{ 8t = 12t - 24}$

$\longmapsto \mathrm{ 24 = 12t - 8t}$

$\longmapsto \mathrm{ 24 = 4t}$

$\longmapsto \mathrm{ \dfrac{ \cancel{24}}{ \cancel{4}} = \large t}$

$\longmapsto \mathrm{ \bf \underline \green{ 6 \: hours \: = t}}$

Putting 't' in (1),

$\longmapsto \mathrm{ {x}= 8(6)}$

$\longmapsto \mathrm{ \bf \underline{ \pink{ {x}= 48 \: kilometers}}}$

When the man has to reach at 10 a.m ,

$\mathrm{speed = \dfrac{distance}{time} }$

$\longmapsto \mathrm{speed = \dfrac{x}{t - 1} }$

Here, we took time as (t - 1) hours because the man has 1 hour less to reach the office then in the first case.

$\longmapsto \mathrm{speed = \dfrac{48}{6 - 1} }$

$\longmapsto \mathrm{speed = \dfrac{ \cancel{48}}{ \cancel5} }$

$\longmapsto \mathrm{ \bf \underline { \orange{speed = 9.6 \: km/h}} }$

Therefore, the man has to cycle at the speed of 9.6 km/hr to reach office at 10 am.

{excellent question}