Question

A taxi driver takes charge and some amount for every KM travelled. Take x as fixed charge and y as fare per KM. Write the expression in linear form, if a person who has travelled 25 KM has to pay Rs. 100. Draw its graph also. i will mark you as brainalist

Answer 1

Answer:

Let fixed charge be Rs x and charge per km be Rs y.

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1)

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5⇒15y=150

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5⇒15y=150⇒y=10

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5⇒15y=150⇒y=10Hence x=5    and  y=10

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5⇒15y=150⇒y=10Hence x=5    and  y=10Fixed charge is Rs.55 and the charge per kilometer is Rs.1010 

Let fixed charge be Rs x and charge per km be Rs y. ⇒x+10y=105...(1) ⇒x+15y=155..(2).Now From eq 2⇒15y=155−x⇒y=15155−x...(3)Substituting y from eq3 in eq1⇒x+1510(155−x)=105⇒15x+1550−10x=1575⇒5x=1575−1550⇒5x=25⇒x=5Substituting x in eq2⇒5+15y=155⇒15y=155−5⇒15y=150⇒y=10Hence x=5    and  y=10Fixed charge is Rs.55 and the charge per kilometer is Rs.1010 For 2525 km person have to pay = 5+10×25=255Rs.

Step-by-step explanation:

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Answer 2

Answer:

x + 25y = 100

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