Let's denote the speed of the man going from his house to the office as v1 and the speed coming back from the office to the house as v2.

Given that the ratio of v1 to v2 is 6:5, we can write this as v1/v2 = 6/5.

We also know that speed is equal to distance divided by time. So, we can write the time it takes to travel from the house to the office as t1 = 180/v1 and the time it takes to travel from the office to the house as t2 = 180/v2.

We are given that t2 is 1 hour more than t1, so we can write this as t2 = t1 + 1.

Substituting the expressions for t1 and t2 from above, we get 180/v2 = 180/v1 + 1.

We can simplify this equation by multiplying all terms by v1*v2 to get rid of the denominators: 180*v1 = 180*v2 + v1*v2.

Substituting v1/v2 = 6/5 into this equation gives us 180*6 = 180*5 + 6*5.

Solving this equation gives us v1 = 30 km/h.

So, the initial speed of the man when he was going from his house to the office was 30 km/h.

Question

Let's denote the speed of the man going from his house to the office as v1 and the speed coming back from the office to the house as v2.

Given that the ratio of v1 to v2 is 6:5, we can write this as v1/v2 = 6/5.

We also know that speed is equal to distance divided by time. So, we can write the time it takes to travel from the house to the office as t1 = 180/v1 and the time it takes to travel from the office to the house as t2 = 180/v2.

We are given that t2 is 1 hour more than t1, so we can write this as t2 = t1 + 1.

Substituting the expressions for t1 and t2 from above, we get 180/v2 = 180/v1 + 1.

We can simplify this equation by multiplying all terms by v1*v2 to get rid of the denominators: 180*v1 = 180*v2 + v1*v2.

Substituting v1/v2 = 6/5 into this equation gives us 180*6 = 180*5 + 6*5.

Solving this equation gives us v1 = 30 km/h.

So, the initial speed of the man when he was going from his house to the office was 30 km/h.

Knowee AI · Accepted Answer

Let's denote the speed of the man going from his house to the office as v1 and the speed coming back from the office to the house as v2.

Given that the ratio of v1 to v2 is 6:5, we can write this as v1/v2 = 6/5.

We also know that speed is equal to distance divided by time. So, we can write the time it takes to travel from the house to the office as t1 = 180/v1 and the time it takes to travel from the office to the house as t2 = 180/v2.

We are given that t2 is 1 hour more than t1, so we can write this as t2 = t1 + 1.

Substituting the expressions for t1 and t2 from above, we get 180/v2 = 180/v1 + 1.

We can simplify this equation by multiplying all terms by v1*v2 to get rid of the denominators: 180*v1 = 180*v2 + v1*v2.

Substituting v1/v2 = 6/5 into this equation gives us 180*6 = 180*5 + 6*5.

Solving this equation gives us v1 = 30 km/h.

So, the initial speed of the man when he was going from his house to the office was 30 km/h.

The ratio of the speed of a man, going from his house to the office andcoming back from the office to the house is 6 : 5. If he takes 1 hour more tocome back from the office, find the initial speed of that man if the distancebetween his house and office is 180 km

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