Mr. Raju travelled a certain distance at a certain speed. Had he moved 4 kmph faster, he would have taken
30 minutesless to cover the distance. If he had moved 2kmph slower, he would have taken 30 minutes
more to cover the distance. What was the distance travelled by him?
12 km
9 km
8 km
4 km
Answers
Answer:
ANSWER
Let the distance = x km and usual rate = y km/hr
Then
x−
y+3
x
=
60
40
⇒
y(y+3)
x(y+3)−xy
=
3
2
⇒
y(y+3)
3x
=
3
2
⇒2y(y+3)=9x...............(i)
and
y−2
x
−
y
x
=
60
40
⇒
y(y−2)
xy−x(y−2)
=
3
2
⇒
y(y−2)
2x
=
3
2
⇒y(y−2)=3x...............(ii)
On dividing eqn (i) by eqn (ii) we get
y−2
2(y+3)
=3⇒2y+6=3y−6⇒y=12
∴ Putting the value of y in (i) we get 2 x 12 x 15 = 9x ⇒ x = 40 km/hr
Answer:
Distance travelled by him was 24 km.
Step-by-step explanation:
Suppose Raju has covered 's' km distance with 'v' km/hr speed in 't' hr time.
So, we can write, s = vt ...................(1)
- According to 1st condition: In speed is increased by 4 km/hr, then new speed = (v + 4) km/hr.
If time is decreased by 30 min = 0.5 hr, then time becomes = (t - 0.5) hr.
Distance covered, s = speed × time = (v + 4)(t - 0.5)
vt = (v + 4)(t - 0.5) [from (1)]
vt = vt + 4t - 0.5v - 2
4t - 0.5v = 2 .................(2)
- According to 2nd condition: In speed is decreased by 2 km/hr, then new speed = (v - 2) km/hr.
If time is increased by 30 min = 0.5 hr, then time becomes = (t + 0.5) hr.
Distance covered, s = speed × time = (v - 2)(t + 0.5)
vt = (v - 2)(t + 0.5) [from (1)]
vt = vt - 2t + 0.5v - 2
2t - 0.5v = -2 .................(3)
- Distance calculation: Subtracting equation number (3) from (2) we get, (4t - 0.5v) - (2t - 0.5v) = 2 - (-2)
4t - 0.5v - 2t + 0.5v = 2 + 2
2t = 4, or, t = 2 hr.
Putting value of 't' in equation (2), 4(2) - 0.5v = 2
0.5v = 8 - 2 = 6
v = 12 km/hr
Distance = vt = 12 × 2 = 24 km
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