150
100
90
120
Two cars, one faster than the other, travel either towards each other or in the same direction. The slower car has a speed of 60 km/h. They meet after:
What is the distance covered by the slower car in the first case?
Effective speed = \( v + 60 \) km/h
Time = 1.5 hours
\[ \text{Distance} = (v + 60) \times 1.5 \]
Effective speed = \( v - 60 \) km/h
Time = 10.5 hours
\[ \text{Distance} = (v - 60) \times 10.5 \]
\[ (v + 60) \times 1.5 = (v - 60) \times 10.5 \]
\[ 1.5v + 90 = 10.5v - 630 \Rightarrow 90 + 630 = 10.5v - 1.5v \Rightarrow 720 = 9v \Rightarrow v = 80 \]
So, the speed of the faster car is \( \boxed{80 \text{ km/h}} \)
Using the time in the first case (1.5 hours) and speed of the slower car (60 km/h): \[ \text{Distance} = 60 \times 1.5 = \boxed{90 \text{ km}} \]
\[ \boxed{90 \text{ km}} \quad \text{(Correct Option: C)} \]
Two cars travel towards each other and meet after 1.5 hours. The speed of the slower car is 60 km/h. Find the distance traveled by the slower car before they meet.
Using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \]
\[ \text{Distance} = 60 \times 1.5 = \boxed{90 \text{ km}} \]
\[ \boxed{90} \]
When $10^{100}$ is divided by 7, the remainder is ?