Question

Ello ❤❤

plzz solve this problem !!
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Two brothers start at the same time from two places 20 km apart. if they walk in the same direction, the elder brother overtakes the younger in 10 hours but if they walk in opposite directions, they meet in 2 hours. find the rate of which they walk.
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its urgent !!!

Answer 1

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Rate of A =6


Rate of B=4

Answer in attachment!!


Thanks!!

Answer 2

Heya ❗

Here's ur answer ❗
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Let A and B be the two places 20km apart, and suppose the elder brother start from A and the younger brother from B.

Let x km/h be the speed of the elder brother and y km/h that of the younger. when they walk in the same direction, i.e., to the right, the elder brother will overtake the younger at some place M to the right of B.

Since elder brother walks x km and younger y km in 1 hr, therefore, the elder gains ( x - y ) km on the younger in 1 hr, i.e., he gains 10(x - y) km in 10 hr. but he has to gain 20 km altogether, which he does in 10 hours.

Therefore,
$\bold{10(x - y) = 20 \: \: \: \: \: \: \: \: \: \: ...(1)}$

when they walk in opposite directions, they will meet at some point N between A and B. now they approach each other both with a speed of (x + y) km/hr.

Therefore,
$\bold{2(x + y) = 20 \: \: \: \: \: \: \: \: \: \: \: \: ...(2)}$

solving (1) and (2), we obtain
$\boxed{ \bold{x = 6 \: \: \: \: \: \: \: \: y = 4.}}$

Therefore, the elder, and the younger brother walk at the rate of 6km/h and 4km/h respectively.
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$\bold{ \blue{method \: ll.}}$

Let the elder brother overtakes the younger brother at M in 10 hours.

Then,

$\bold{10x - 10y = 20 = > x - y = 2 \: \: \: \: ...(1)}$

In the second case, we get
$\bold{2x + 2y = 20 = > x + y = 10 \: \: \: \: \: ...(2)}$

Solving (1) and (2) simultaneously, we get
$\boxed{ \bold{x = 6 \: \: \: \: \: \: \: \: \: y = 4.}}$

Hence, the elder brother and the younger brother walk at the speed of 6km/h and 4km/h respectively.
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@vaibhavhoax