It is common that Governments revise travel fares from time to time based on various factors such as inflation (a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:
Name of the City
Distance travelled (km)
Amount paid (Rs.)
City A
10 75
15 110
City B
18 91
14 145
Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110.
Situation 2: In a city B, for a journey of 8km, the charge paid is Rs91 and for a journey of 14km, the charge paid is Rs 145.
Refer situation 1:
1. If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing the situation is
a) x + 10y =110, x + 15y = 75
b) x + 10y = 75, x + 15y = 110
c) 10x + y = 110, 15x + y = 75
d) 10x + y = 75, 15x + y = 110
2. A person travels a distance of 50 km. The amount he has to pay is:
a) Rs.155
b) Rs.255
c) Rs.355
d) Rs.455
Refer situation 2:
3. What will a person have to pay for travelling a distance of 30km?
a) Rs.185
b) Rs.289
c) Rs.275
d) Rs.305
4. The graph of lines representing the conditions is: (situation 2)
a) Intersecting Lines
b) Parallel Lines
c) Coincident Lines
d) None of these
Answers
Answer:
I don't know
Solution :-
Let the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr .
so,
→ charges for 10 km = Rs.75
→ x + 10y = 75 ------ Eqn.(1)
and,
→ charges for 15 km = Rs.110
→ x + 15y = 110 ----- Eqn.(2)
therefore, the pair of linear equations representing the situation is Option (B) x + 10y = 75, x + 15y = 110 .
now, subtracting Eqn.(1) from Eqn.(2) ,
→ (x + 15y) - (x + 10y) = 110 - 75
→ 5y = 35
→ y = 7
putting value of y in Eqn.(1),
→ x + 10*7 = 75
→ x + 70 = 75
→ x = 75 - 70
→ x = 5
then,
→ charges for 50 km = x + 50y = 5 + 7 * 50 = 5 + 350 = Rs.355 (C) .
now, situation (2) :-
→ x + 8y = 91 ------- Eqn.(3)
→ x + 14y = 145 ------ Eqn.(4)
subtracting Eqn.(3) from Eqn.(4) ,
→ (x + 14y) - (x + 8y) = 145 - 91
→ 6y = 54
→ y = 9
putting value of y in Eqn.(3),
→ x + 8*9 = 91
→ x + 72 = 91
→ x = 91 - 72
→ x = 19
then,
→ charges for 30 km = x + 30y = 19 + 9 * 30 = 19 + 270 = Rs.289 (b) .
and,
→ x + 8y - 91 = 0
→ x + 14y - 145 = 0
so,
→ a1/a2 = 1/1
→ b1/b2 = 8/14 = 4/7
since,
→ a1/a2 ≠ b1/b2
therefore, lines intersect at a point and have unique solution .
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