The distance of the college and the home of Rajeev is 80km. one day he

1.	The distance of the college and the home of Rajeev is 80km. one day he was late by 1 hour than the normal time to leave for the college,so he increased his speed by 4kmph and thus he reached to college at the normal time. The increased speed of Rajeev is.
Answer» Answer: The increased speed of Rajeev = 20 km/h Step-by-step EXPLANATION: Distance between home and college = 80km Let the speed is V and time t Distance = Speed × Time 80 = vt .......... (1) When he increases the speed by 4 KMPH he still reaches the college at time while starting 1 HOUR late Hence 80 = (v+4)(t-1) ⇒ 80 = vt + 4t - v - 4 ⇒ 80 = 80 + 4t - v - 4 ⇒ 4t = v + 4 ⇒ t = (v + 4)/4 Putting this value in eq (1) 80 = v (v + 4)/4 $v^2+4v=320$ $v^2+4v-320=0$ $\implies v^2+20v-16v-320=0$ $\implies v(v+20)-16(v+20)=0$ $\implies (v+20)(v-16)=0$ $\implies v=16, -20$ v cannot be negative Thus, v = 16 km/h The increased speed of Rajeev = 16 + 4 = 20 km/h Hope the answer is helpful.

The distance of the college and the home of Rajeev is 80km. one day he was late by 1 hour than the normal time to leave for the college,so he increased his speed by 4kmph and thus he reached to college at the normal time. The increased speed of Rajeev is.

Answer»

Answer:

The increased speed of Rajeev = 20 km/h

Step-by-step EXPLANATION:

Distance between home and college = 80km

Let the speed is V and time t

Distance = Speed × Time

80 = vt .......... (1)

When he increases the speed by 4 KMPH he still reaches the college at time while starting 1 HOUR late

Hence

80 = (v+4)(t-1)

⇒ 80 = vt + 4t - v - 4

⇒ 80 = 80 + 4t - v - 4

⇒ 4t = v + 4

⇒ t = (v + 4)/4

Putting this value in eq (1)

80 = v (v + 4)/4

$v^2+4v=320$

$v^2+4v-320=0$

$\implies v^2+20v-16v-320=0$

$\implies v(v+20)-16(v+20)=0$

$\implies (v+20)(v-16)=0$

$\implies v=16, -20$

v cannot be negative

Thus, v = 16 km/h

The increased speed of Rajeev = 16 + 4 = 20 km/h

Hope the answer is helpful.