Question

Car parking is a major problem in urban areas in both developed and developing Countries. In the Jaipur City GT mall, a shopping place, charged the parking. The charges of a car in parking lot is Rs. 30 for the first two hours and Rs.10 per hour for subsequent hours, then answer the following questions.


Taking total parking time to be x hour, and total charges as Rs.y, write a linear equation.

(a) 10x+y-10=0 (b) 10x-y+10=0 (c) 8x+y-10=0 (d) 8x-y+10=6

(II) If y=20, then value of x

(a) 1 (b) 2 (c) 3 (d) 4

(III) If x=2, then value of y is

(a) 10 (b) 20 (c) 30 (d) 40

(IV) If charges of a car in a parking lot is Rs.50 for the first two hours, then find the linear equation.

(a) 10x-y+30=0 (b) 10x+y+30=0 (c) 30x+y-10=0 (d) 30x+y-10=0

(V) If charges for subsequent hours is Rs 8 per hour, than write linear equation.

(a) 10x-y+8=0 (b) 10x-y+10=0 (c) 8x-y+14=0 (d) 8x-y+8=0
​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that Car parking is a major problem in urban areas in both developed and developing countries. In the Jaipur City GT mall, a shopping place, charged the parking. The charges of a car in parking lot is Rs. $30$ for the first two hours and Rs. $10$ per hour for subsequent hours.

To Find:-

We have to find that answer the following questions. Taking total parking time to be $x$ hour, and total charges as Rs. $y$, write a linear equation.

Solution:-

According to the problem

Total parking time $=x$ hours

Total charges $=$  $y$

Charges for first $2$ hours $=$Rs $30$

charges for remaining hours

i.e. $(x-2)$ hours $=10$ hour

Total charges for remaining hours $=10(x-2)$

Total parking charges $=$ Rs $30+10(x-2)$

$\therefore y &=30+10(x-2) \\y &=30+10 x-20 \\y &=10+10 x\\10x-y+10=0$

$(ii)$ By putting $y=20$ in the above equation, we get

$10x-y+10=0\\\Rightarrow 10x-20+10=0\\\Rightarrow 10x-10=0\\\Rightarrow 10x=10\\\Rightarrow x=1$

$(iii)$ By putting $x=2$ in the above equation, we get

$10x-y+10=0\\\Rightarrow (10 \times2)-y+10=0\\\Rightarrow 20-y+10=0\\\Rightarrow 30-y=0\\\Rightarrow y=30$

$(iv)$$\therefore y &=50+10(x-2) \\y &=50+10 x-20 \\y &=30+10x\\10x-y+30=0$

$(v)$

$\therefore y &=30+8(x-2) \\y &=30+8x-16 \\y &=14+8x\\8x-y+14=0$

Final Answer:-

The correct option are $(b) 10x-y+10=0$, $(ii)$$(a) 1$, $(iii)$ $(c) 30$,$(iv)$$(a) 10x-y+30=0$ and $(v)$ $(c) 8x-y+14=0$.

#SPJ3