Grade 6 word question

Question

if a car speeds up 2/10 of its original speed it will arrive in its destination an hour before, if a car also drives on normal speed for []/[] (fraction) amount of the track then speeds up 3/10 of its speed it will also arrive in its destination an hour before, ( i need to answer to the []/[] )

Answer 1

s = normal speed

d = distance

d= st

s = d/t

The following is a correct calculation of the fraction, if the problem were asking for the fraction of time on the track the car went at normal speed, before speeding up to 3/10 more.

But the problem asks for the fraction of distance on the track at normal speed before speeding up. The answer will = 5/18, which takes up 1/3 of the time on the track, which almost 1/3 of the distance but slightly different. In the following when it reaches 1 and 2.6, take the ratio, 1/(1+2.6) = 1/3.6 = 5/18

speeding up 2/10 of original speed means 2/10 = .2

and s + .2 = 1.2s

speeding up 3/10 means 3/10 = .3 and s+.3 = 1.3s

1.2s = d/(t-1)

s = d/t

xs + (1-x)1.3s = d/(t-1) = 1.2s

xs + 1.3s - 1.3xs = 1.2s

x + 1.3 -1.3x = 1.2

.3x = .1

x = .1/.3 = 1/3

go for 1/3 of the time on the track at normal speed, then go 3/10 faster for the remaining 2/3 of the time on the track, and arrive one hour earlier

go for 2/10 faster and arrive one hour earlier

for example,

if s = 5 mph and d= 30 miles

then d =st and t = 6 hours

d=st, 30 = 5(6)

30= (5)(6)

and

30 = 1.2(5)(6-1) if speed upped by 2/10 = 20% = 0.2

30 = [(1/3)(5) + (2/3)(1.3)(5)](6-1) partly upped by 3/10= 0.3

set last two equations equal and divide both sides by (6-1)(5)

1.2 = 1/3 + (2/3)(1.3) factor out 1/3

1.2 = 1/3(1 + 2(1.3))

multiply both sides by 3

3.6 = 1+2.6

Now take the 1 and 2.6, and take the ratio. 1/(1+2.6) = 5/(3.6)(5) = 5/18 = the fraction of distance on the track before the car speeds up. 1/3 is the fraction of time on the track before speeding up.