Question

[Nguyễn Tấn Dũng]

Mr.Joseph drives car from A to B at a constant speed. If the speed of the car is increased by 20%, it takes him one hour less than the usual time. If he drives at the constant speed for the first 100km before increasing the speed by 30%, it also takes him one hour less than the usual. The distance of AB is ..........km.

Answer 1

[duy khang nguyễn]

Gọi S là quãng đường
v là vận tốc

t là thời gian Mr.J đi thường ngày

=>\(\dfrac{S}{120\%v}=t-1\)

\(\dfrac{100}{v}+\dfrac{S-100}{130\%v}=t-1\)

=>\(\dfrac{S}{120\%v}=\dfrac{100}{v}+\dfrac{S-100}{130\%v}\)

<=>\(\dfrac{130S}{156v}=\dfrac{15600+120S-12000}{156v}\)

<=>\(10S=3600\)

<=>\(S=360\)

Xong rầu đó khỏi cảm ơn

Answer 2

[Mỹ Duyên]

Đáp án: AB = 360 km