Question

A person walks at 4km/hr for a particular duration t1 and 3km/hr for another duration t2 and covers a total distance of 36km. if he walks at 4km/hr for the duration t2 and at 3km/hr for the duration t1,then he covers only 34km. what will be the time taken by him to cover the one of the legs?

Answer 1

We know that Distance = Speed * Time.

(i)

Given that A person walks 4km/hr for a duration t1 and 3km/hr for duration t2 and covers a total distance of 36km.

36 = 4 * T1 + 3 * T2

36 = 4T1 + 3T2 -------- (1)

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(ii)

Given that he walks at 4km/hr for duration t2 and 3km/hr for t1 and covers only 34 km.

34 = 3 * T1 + 4 * T2

34 = 3T1 + 4T2 ----- (2)

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On adding (1) & (2), we get

= > 4T1 + 3T2 + 3T1 + 4T2 = 36 + 34

= > 7T1 + 7T2 = 70 ------- (3)

On subtracting (1) & (2), we get

= > 4T1 + 3T2 - 3T1 - 4T2 = 36 - 34

= > T1 - T2 = -2 ------ (4)

On solving (3) & (4) * 7, we get

= > 7T1 + 7T2 = 70

= > 7T1 - 7T2 = -14

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14T1 = 56

T1 = 4

Substitute T1 = 4 in (3), we get

= > 7T1 + 7T2 = 70

= > 7(4) + 7T2 = 70

= > 28 + 7T2 = 70

= > 7T2 = 70 - 28

= > 7T2 = 42

= > T2 = 42/7

= > T2 = 6.

So, Total time taken = 4 + 6 = 10.

Therefore, the total time taken by him to cover one of the legs = 10 hours.

Hope this helps!

Answer 2

Answer:

The total time taken by him to cover one of the legs is 10 hours.

Explanation:

As we all know, distance equals speed times time.

(i)Suppose that a person moves at a speed of 4 km/h for $T_{1}$ and 3km/hr for a duration $T_{2}$ and travels a total of 36 kilometers.

36 = 4 × $T_{1}$ + 3 ×$T_{2}$

36 = 4$T_{1}$+ 3$T_{2}$ -------- (1)

(ii) Considering that he moves along at a constant 4 km/hr for $T_{2}$ and 3km/hr for $T_{1}$ and covers only 34 km.

34 = 3 × $T_{1}$ + 4 × $T_{2}$

34 = 3$T_{1}$ + 4$T_{2}$ ----- (2)

On adding (1) & (2), we get

= > 4$T_{1}$ + 3$T_{2}$ + 3$T_{1}$ + 4$T_{2}$ = 36 + 34

= > 7$T_{1}$ + 7$T_{2}$ = 70 ------- (3)

On subtracting (1) & (2), we get

= > 4$T_{1}$ + 3$T_{2}$ - 3$T_{1}$- 4$T_{2}$ = 36 - 34

= > $T_{1}$ - $T_{2}$ = -2 ------ (4)

On solving (3) & (4) × 7, we get

= > 7$T_{1}$ + 7$T_{2}$ = 70

= > 7$T_{1}$ - 7$T_{2}$ = -14

14$T_{1}$ = 56

$T_{1}$= 4

Substitute $T_{1}$= 4 in (3), we get

= > 7$T_{1}$ + 7$T_{2}$ = 70

= > 7(4) + 7$T_{2}$ = 70

= > 28 + 7$T_{2}$= 70

= > 7$T_{2}$ = 70 - 28

= > 7$T_{2}$ = 42

= > $T_{2}$ = 42/7

= > $T_{2}$ = 6.

So, Total time taken = 4 + 6 = 10.

Therefore, the total time taken by him to cover one of the legs is 10 hours.

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