Question

The auto-rickshaw fare consists of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is ₹85 and for a journey of 15 km, the charge paid is ₹120. The fare for a journey of 25 km will be

Answer 1

Answer:

Step-by-step explanation:

Let use assume the fixed charge = ₹ a

and charge for 1 km is ₹ = b

According to question,

for 10 KM journey charge paid = 85

a + 10 x b = 85

a + 10b = 85 .........................(1)

for 15 KM journey charge paid = 120

a + 15b = 120.........................(2)

Subtract the equation (1) from equation (2). we will get,

a + 15b - a - 10b = 120 - 85

5b = 35

b = 7

Put the value of b in equation (1). we will get

a + 10 x 7 = 85

a = 85 - 70

a = 15

Charges for 25 km = a + 25 x b

Put the value of a and b in above equation.

Charges for 25 km =15 + 25 x 7 = 15 + 175 = 190

Charges for 25 km =₹190

Answer 2

AnswEr :

$\bullet\:\textsf{Let the Fixed Charge = Rs. x}\\\\\bullet\:\textsf{and, Charge for per kilometre = Rs. y}$

$\bigstar\:\underline\textsf{According to the Question Now :}$

$\longrightarrow\textsf{Charge for Journey of 10 kilometre = Rs. 85}\\\\\\\longrightarrow\textsf{Fixed Distance Charge + 10 km Charge = Rs. 85}\\\\\\\longrightarrow\sf Rs. \:x + (Rs. \:y\times 10) = Rs. \:85\\\\\\\longrightarrow\sf Rs. \:x + Rs.\:10y = Rs.\:85\\\\\\\longrightarrow\sf \cancel{Rs.} (x + 10y) = \cancel{Rs.}\:85 \\\\\\\longrightarrow\sf x + 10y = 85 \qquad\quad\dfrac{\quad}{}\:eq.(1)$

$\rule{160}{1}$

$\longrightarrow\textsf{Charge for Journey of 15 kilometre = Rs. 120}\\\\\\\longrightarrow\textsf{Fixed Distance Charge + 15 km Charge = Rs. 120}\\\\\\\longrightarrow\sf Rs. \:x + (Rs. \:y\times 15) = Rs. \:120\\\\\\\longrightarrow\sf Rs. \:x + Rs.\:15y = Rs.\:120\\\\\\\longrightarrow\sf \cancel{Rs.} (x + 15y) = \cancel{Rs.}\:120 \\\\\\\longrightarrow\sf x + 15y = 120 \qquad\quad\dfrac{\quad}{}\:eq.(2)$

$\rule{250}{1}$

$\bigstar\:\underline\textsf{Subtracting eq.(1) from eq.(2) :}$

$:\implies\tt x + 15y = 120\\\\:\implies\tt x + 10y = 85\\\dfrac{\qquad\qquad\qquad \qquad \qquad}{}\\:\implies\tt (15y - 10y) = (120 - 85)\\\\\\:\implies\tt5y = 35\\\\\\:\implies\tt y = \cancel\dfrac{35}{5}\\\\\\:\implies\blue{\tt y = Rs.\:7}$

$\bigstar\:\underline\textsf{Putting the value of y in eq.(2) :}$

$:\implies\tt x + 15y = 120\\\\\\:\implies\tt x + 15(7) = 120\\\\\\:\implies\tt x + 105 = 120\\\\\\:\implies\tt x = 120 - 105\\\\\\:\implies \blue{\tt x = Rs.\:15}$

$\rule{250}{2}$

$\bf{\dag}\:\underline\textsf{Fare for Journey of 25 kilometres :}$

$\leadsto\texttt{Fare = (Fixed Charge) + (Distance Charge)}\\\\\\\leadsto\tt Fare = x + (25 \times y)\\\\\\\leadsto\tt Fare = Rs.\:15 + (25 \times Rs.\:7)\\\\\\\leadsto\tt Fare = Rs.\:15 + Rs.\:175\\\\\\\leadsto \large\boxed{\red{\tt Fare = Rs.\:190}}$

⠀

∴ Fare for 25 kilometres will be Rs. 190.