Car parking is a major problem in urban areas in both developed and
developing countries .In Jaipur city GT mall is shopping place charged the parking
lot is ₹30 for the first two hours and ₹10 per hour for subsequent hours, then answer
the following questions.
(i) Taking total parking time to be x hour, and total charges as ₹ 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 the charges of a car in a parking lot is ₹ 50 for first two hours, then find the
linear equation.
(a) 10x-y+30=0 (b) 10x+y+30=0 (c) 10x+y-30=0 (d)
30x+y-10=0
(v) If charges for subsequent hours is ₹ 8 per hour, then write linear equation.
(a) 10x-Y+8=0 (b) 10x-Y+10=0 (c) 8x-y+14=0 (d) 8x-y+8=0
Answers
Given : ₹30 for the first two hours and ₹10 per hour for subsequent hours,
Taking total parking time to be x hour, and total charges as ₹ y.
To Find : Write a linear equation.
(a) 10x+y-10=0 (b) 10x-y+10=0 (c) 8x+y-10=0 (d) 8x-y+10=6
Solution:
₹30 for the first two hours
₹10 per hour for subsequent hours,
Charges = 30 + 10 (x - 2) where x ≥ 2
=> y = 30 + 10(x - 2)
=> y = 10 x + 10
=> 10x - y + 10 = 0
10x - y + 10 = 0 is the linear equation.
y = 20 =>
10x - 20 + 10 = 0
=> x = 1 but x ≥ 2 hence not a feasible solution
If x=2, then value of y is
=> 10(2) - y + 10 = 0
=> y = 30
charges for subsequent hours is ₹ 8 per hour, then write linear equation.
y = 30 +8(x - 2)
=> y = 8x + 14
=> 8x - y + 14 = 0
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(i) Taking total parking time to be x hour, and total charges as ₹ 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
Answer :-
→ fixed price for first 2 hours = Rs.30
and,
→ Price after that per hour = Rs.10
and,
→ Total price = Rs. y
→ Total hours = x .
then,
→ 30 + 10(x - 2) = y
→ 30 + 10x - 20 = y
→ 10x - y + 10 = 0 (b) (Ans.)
(ii) If y=20, then value of x
(a) 1 (b) 2 (c) 3 (d) 4
Answer :-
when y is 20,
→ 10x - y + 10 = 0
→ 10x - 20 + 10 = 0
→ 10x - 10 = 0
→ 10x = 10
→ x = 1 (a) (Ans.)
(iii) If x=2, then value of y is
(a) 10 (b) 20 (c) 30 (d) 40
Answer :-
when x = 2,
→ 10x - y + 10 = 0
→ 10*2 - y + 10 = 0
→ 20 + 10 - y = 0
→ y = 30 (c) (Ans.)
(iv) If the charges of a car in a parking lot is ₹ 50 for first two hours, then find the linear equation.
(a) 10x-y+30=0 (b) 10x+y+30=0 (c) 10x+y-30=0 (d) 30x+y-10=0
Answer :-
→ 50 + 10(x - 2) = y
→ 50 + 10x - 20 = y
→ 10x - y + 30 = 0 (a) (Ans.)
(v) If charges for subsequent hours is ₹ 8 per hour, then write linear equation.
(a) 10x-Y+8=0 (b) 10x-Y+10=0 (c) 8x-y+14=0 (d) 8x-y+8=0
Answer :-
→ 30 + 8(x - 2) = y
→ 30 + 8x - 16 = y
→ 8x - y + 14 = 0 (c) (Ans.)
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