Question

A man moves from a to b at the rate of 4 km an hour. had he moved at the rate of 11/3 km an hour , he would have taken 3 hours more to move that distance . find the distance between a and b
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Answer 1

$\purple\Large{\boxed{\tt{ \blue{Solution:- }}}}$

$\rm\red{Let\: distance \:between \:AB = x}$

$\bold\green\rightarrowtail$$\rm{t_{1}=\frac{x}{4}}$

$\bold\green\rightarrowtail$$\rm{t_{2}=\frac{3x}{11}}$

$\purple\looparrowright$$\rm{t_{1} = t_{2} - 3}$(A.T.Q)

$\purple\looparrowright$$\rm{\frac{x}{4} = \frac{3x}{11} - 3}$

$\purple\looparrowright$$\rm{\frac{x}{4}=\frac{3x-33}{11}}$

$\purple\looparrowright$$\rm{11x = 12x - 132}$

$\purple\looparrowright$$\rm\bold\blue{x= 132}$

$\tt\red{Hope\:it\:helps:)}$

Answer 2

Answer:

$\huge \tt \underline \orange{Solution:}$

$\fbox \red{Distance Between AB= \: x}$

$\red{↪} \tt \: t_1 = \frac{x}{4}$

$\red{↪} \tt \: t_{2} = \frac{3x}{11}$

$\red{↪}\tt \: t_1 = t_2 - 3 \: \red{( A.T.P)}$

$\red{↪} \tt \frac{x}{4} = \frac{3x}{11} - 3$

$\red{↪} \tt \frac{x}{4} = \frac{3x - 33}{11}$

$\red{↪} \tt \: 11x = 12x - 132$

$\red{↪} \tt \fbox \red { x = 132}$