Kavitha travelled a distance of 40 km in 6 hours . She travelled partly on foot at 4 kmph and partly on bicycle at 8 kmph . What is the distance travelled by foot ?4 km6 km8 km10 km2 km

Let's denote the time Kavitha spent walking as T1 and the time she spent cycling as T2. We know that T1 + T2 = 6 hours.

We also know that the distance she covered on foot (D1) plus the distance she covered on bicycle (D2) equals 40 km.

We can express D1 as the product of the speed on foot (4 km/h) and the time spent walking (T1), and D2 as the product of the speed on bicycle (8 km/h) and the time spent cycling (T2). So, we have the equation 4T1 + 8T2 = 40.

We can solve these two equations simultaneously to find the values of T1 and T2, and then use those to find D1 and D2.

From T1 + T2 = 6, we can express T1 as 6 - T2 and substitute this into the second equation:

4(6 - T2) + 8T2 = 40 24 - 4T2 + 8T2 = 40 4T2 = 16 T2 = 4

Substituting T2 = 4 into T1 + T2 = 6, we find T1 = 2.

Finally, we substitute T1 = 2 into D1 = 4T1 to find D1 = 8 km.

So, Kavitha travelled 8 km on foot.

Let's denote the time Kavitha spent walking as T1 and the time she spent cycling as T2. We know that T1 + T2 = 6 hours.

We also know that the total distance she covered is 40 km. The distance covered walking is 4T1 and the distance covered cycling is 8T2. Therefore, we have the equation 4T1 + 8T2 = 40.

We can solve these two equations to find the values of T1 and T2.

First, let's multiply the first equation by 4: 4T1 + 4T2 = 24.

Subtract this from the second equation: 4*T2 = 16, so T2 = 4 hours.

Substitute T2 = 4 into the first equation: T1 + 4 = 6, so T1 = 2 hours.

Therefore, the distance Kavitha travelled on foot is 4 kmph * 2 hours = 8 km.

Let's denote the time Kavitha spent walking as T1 and the time she spent biking as T2. We know that T1 + T2 = 6 hours.

We also know that the total distance she traveled is 40 km, which is the sum of the distance she walked and the distance she biked. We can express the distance as the product of speed and time, so we have 4T1 (the distance she walked) + 8T2 (the distance she biked) = 40 km.

We now have a system of two equations, and we can solve it step by step:

1. Express T2 from the first equation: T2 = 6 - T1.
2. Substitute T2 in the second equation: 4T1 + 8(6 - T1) = 40.
3. Simplify the equation: 4T1 + 48 - 8T1 = 40.
4. Combine like terms: -4*T1 = -8.
5. Solve for T1: T1 = 2 hours.

So, Kavitha spent 2 hours walking. To find the distance she walked, multiply the time by her speed: 4 kmph * 2 hours = 8 km.

Therefore, Kavitha travelled 8 km on foot.

Let's denote the time Kavitha spent walking as 't' and the time she spent cycling as '6-t' (since the total time spent is 6 hours).

We know that speed = distance/time. Therefore, distance = speed * time.

The distance Kavitha walked is therefore 4t and the distance she cycled is 8(6-t).

The total distance travelled is the sum of these two distances, which we know is 40 km. Therefore, we can set up the following equation:

4t + 8(6-t) = 40

Solving this equation will give us the value of 't', which we can then substitute back into the equation 4t to find the distance Kavitha walked.

Let's solve the equation:

4t + 48 - 8t = 40 -4t = -8 t = 2

Substituting t = 2 into the equation 4t gives:

4*2 = 8

Therefore, Kavitha walked 8 km.

Let's denote the time Kavitha spent walking as 't' and the time she spent cycling as '6-t' (since the total time spent is 6 hours).

We know that speed = distance/time. Therefore, distance = speed * time.

The distance Kavitha walked is therefore 4t and the distance she cycled is 8(6-t).

The total distance travelled is the sum of these two distances, which we know is 40 km. Therefore, we can set up the following equation:

4t + 8(6-t) = 40

Solving this equation will give us the value of 't', which we can then substitute back into the equation 4t to find the distance Kavitha walked.

Let's solve the equation:

4t + 48 - 8t = 40 -4t = -8 t = 2

Substituting t = 2 into the equation 4t gives:

4*2 = 8

Therefore, Kavitha walked 8 km.

Let's denote the time Kavitha spent walking as T1 and the time she spent cycling as T2. We know that T1 + T2 = 6 hours.

We also know that the distance she covered on foot (D1) plus the distance she covered on a bicycle (D2) equals 40 km.

We can express D1 as the product of the speed on foot (4 km/h) and the time spent walking (T1), and D2 as the product of the speed on a bicycle (8 km/h) and the time spent cycling (T2). So, we have the equation 4T1 + 8T2 = 40.

We can solve these two equations simultaneously to find the values of T1 and T2.

First, let's express T2 from the first equation: T2 = 6 - T1.

Now we can substitute T2 in the second equation: 4T1 + 8(6 - T1) = 40.

Solving this equation gives us T1 = 4 hours.

So, the distance travelled on foot is 4 km/h * 4 hours = 16 km.

Let's denote the time Kavitha spent walking as 't' and the time she spent cycling as '6-t' (since the total time spent is 6 hours).

We know that speed = distance/time. Therefore, distance = speed * time.

The distance Kavitha walked is therefore 4t and the distance she cycled is 8(6-t).

The total distance travelled is the sum of these two distances, which we know is 40 km. Therefore, we can set up the following equation:

4t + 8(6-t) = 40

Solving this equation will give us the value of 't', which we can then substitute back into the equation 4t to find the distance Kavitha walked.

Let's solve the equation:

4t + 48 - 8t = 40 -4t = -8 t = 2

Substituting t = 2 into the equation 4t gives:

4*2 = 8

Therefore, Kavitha walked 8 km.

Let's denote the time Kavitha spent walking as T1 and the time she spent cycling as T2. We know that T1 + T2 = 6 hours.

We also know that the total distance she covered is 40 km. This distance is made up of the distance she walked and the distance she cycled. We can express the distance walked as (speed * time) = 4 kmph * T1 and the distance cycled as 8 kmph * T2.

So, we have the equation 4T1 + 8T2 = 40.

We can solve these two equations simultaneously to find the values of T1 and T2.

Let's substitute T2 = 6 - T1 into the second equation:

4T1 + 8(6 - T1) = 40 4T1 + 48 - 8T1 = 40 4T1 - 8T1 = 40 - 48 -4T1 = -8 T1 = 2 hours

So, Kavitha spent 2 hours walking. The distance she covered on foot is therefore 4 kmph * 2 hours = 8 km.

Let's denote the time Kavitha spent walking as T1 and the time she spent cycling as T2. We know that T1 + T2 = 6 hours.

We also know that the distance she covered on foot (D1) plus the distance she covered on a bicycle (D2) equals 40 km.

We can express D1 as T1 * 4 km/h and D2 as T2 * 8 km/h.

So, we have the following system of equations:

T1 + T2 = 6 D1 + D2 = 40

Substituting D1 and D2, we get:

4T1 + 8T2 = 40

We can simplify this equation by dividing all terms by 4:

T1 + 2T2 = 10

Now we have a system of two equations:

T1 + T2 = 6 T1 + 2T2 = 10

Subtracting the first equation from the second, we get:

T2 = 4

Substituting T2 = 4 into the first equation, we get:

T1 + 4 = 6 T1 = 2

So, Kavitha spent 2 hours walking and 4 hours cycling.

The distance she covered on foot is D1 = T1 * 4 km/h = 2 hours * 4 km/h = 8 km.

So, Kavitha travelled 8 km on foot.