Question

A boy travels at the rate of 4 kmph and he reaches 15 min late to his school. If he travels at the rate of 5 kmph, he reaches 15 min earlier. The distance boy has to travels

Answer 1

Answer:

We know that, speed= distance/time and distance is same in both cases so let distance be x and the normal time it takes him to reach the school be y.

In first case,

4=x/y+15/60 [since speed is in kmph, we'll change 15 minutes into hours.

4(y+1/4)=x

4y+1=x

x-4y=1 (eq 1)

In second case,

5=x/y-15/60 [since speed is in kmph, we'll change 15 minutes into hours]

5(y-1/4)=x

5y-5/4=x

x-5y=-5/4 (eq 2)

Solve the simultaneous equations 1 & 2 by subtraction.

x-4y=1

(-)x-5y=-5/4

y=5/4+1

y=9/4

Substitute value of y in eq 1

x-4(9/4)=1

x=10

Distance= x= 10km

Hope it helps!

Answer 2

Answer:

Distance = 10km

Step-by-step explanation:

Let the distance = xy

Let the time = y hr

According to the question:-

The first equation:-

(4)(y+15/60) = x

=> 4y +1 = x

The other equation:-

(5)(y-15/60) = x

=>5y -5/4 = x

So,

4y+1 = 5y - 5/4

=> 9/4hr = y

So,

Time = 9/4 hr

Distance = 4*9/4 +1 = 10km

Plz mark it Brainliest!!!