Question

A bus started 30 minutes later than the scheduled time from a place 195 km away from its destination. To
co
reach the destination at the scheduled time the driver had to increase the speed by 13 km/h. Find the speed (in
km/h) of the bus by which it has completed the journey.
AI​

Answer 1

Answer:

76,050

Step-by-step explanation:

i dunno just multiply

Answer 2

Correct - Question

A bus started 30 minutes later than the scheduled time from a place 195 km away from its destination to reach the destination at the scheduled time the driver had to increase the speed by 13 km/hr. Find the speed (in km/hr) of the bus by which it has completed the journey.

Answer:

The speed  of the bus by which it has completed the journey is 78km/hr.

Step-by-step explanation:

Given:

A bus started 30 minutes later than the scheduled time from a place 195 km away from its destination to reach the destination.

Increase in the speed by 13 km/hr by driver to reach in the scheduled time.

Formula Used:

Speed= Distance  covered /Time taken to cover distance

Solution:

Let  the usual time taken by  the bus to reach destination = p hr.

Destination distance = 195 km.

$Speed=\frac{195}{p}$   -------------- equation no.01

As given- A bus started 30 minutes later than the scheduled time from a place 195 km away from its destination to reach the destination.

Time taken by  the bus to reach destination  $=p-\frac{1}{2}$  hr.

$Speed=\frac{195}{(p-\frac{1}{2}) }$     --------------  equation no.02

As given- Increase in the speed by 13 km/hr by driver to reach in the scheduled time.

Therefore,  $\frac{195}{(p-\frac{1}{2}) }-\frac{195}{p }=13$

                 $\frac{195}{\frac{(2p-1)}{2} } -\frac{195}{p }=13$

                 $\frac{195\times2}{(2p-1)} -\frac{195}{p }=13$

               $\frac{195(2p-2p+1)}{(2p-1)p} =13$

               $\frac{195}{(2p-1)p} =13$

Dividing both  sides by 13 we get.

              $\frac{15}{(2p-1)p} =1$

            $2p^{2} -p=15$

             $2p^{2} -p-15=0\\2p^{2} -6p+5p-15=0\\2p(p-3)+5(p-3)=0\\(2p+5)(p-3)=0$

              $(p-3)=0\\p=3$

             $2p+5=0\\p=-\frac{5}{2}$(time can not be negative).

Putting value of p=3 in equation no.02.

Therefore ,the speed  of the bus   $=\frac{195}{(p-\frac{1}{2}) }$

                                                        $=\frac{195}{(3-\frac{1}{2}) }$

                                                        $=\frac{195 \times 2}{(6-1)}$

                                                          $=\frac{390}{5}=78 \ km/hr$

                             

Thus, the speed  of the bus by which it has completed the journey is 78km/h.