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Town X and Town Y were 980 km apart. At 8 a.m, Kay traveled from town X to Town Y while Jolin traveled from town Y to Town X. B...Asked by ken Raisyah wahono
Town X and town Y were 980 km apart. At 8 a.m, Kay traveled from town X to town Y while Jolin traveled Town Y to Town X. Both of them did not change their speeds throughout their journeys. Jolin's speed was 75 km/h. find the distance Kay traveled when they met at 3 p.m.
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Answered by
mathhelper
Assuming that they both left at 8:00 am, then by 3:00 pm, both went for
7 hours.
so Joline went 7(75) km, or 525 km
so that left 980-525 or 455 km for Kay.
or
let Kay's speed be x km/h, we know Jolin's speed was 75 km/h,
then
7x + 7(75) = 980
7x = 455
x = 65 km/h
Kay's speed was 65 km/h and he went 7(65) km, or 455 km
7 hours.
so Joline went 7(75) km, or 525 km
so that left 980-525 or 455 km for Kay.
or
let Kay's speed be x km/h, we know Jolin's speed was 75 km/h,
then
7x + 7(75) = 980
7x = 455
x = 65 km/h
Kay's speed was 65 km/h and he went 7(65) km, or 455 km
Answered by
ashton
how to do it in a simpler method
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