Q) 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
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 8 km, the charge paid is Rs 91 and for a journey of 14 km, 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, 15 x + y =110
2. A person travels a distance of 50km. 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 traveling a distance of 30km?
a) Rs.185 b) Rs.289 c) Rs.275 d) Rs.305
4. The graph of lines representing the conditions are: (situation 2)
Ans:
1. Pair of linear equations in Situation 1:
Given that the fixed charge is X and running charge is Y,
therefore the charges for A km, charges will be given by: X + A Y
Hence, for 10 km travel, charges = X + 10 Y
Since it is given the charges paid for this travel is 75
∴ X + 10 Y = 75 ………. (i)
Similarly, for 15 km, charges paid are Rs. 110
∴ X + 15 Y = 110 ………. (ii)
These equations (i) and (ii) make the required pair of linear equations representing the situation 1.
Therefore, option (b) is correct.
2. Charges for 50 km travel:
Let’s solve equation (i) and (ii) and find values of X and Y.
By deducting equation (ii) from (ii), we get:
(X + 15 Y) – (X + 10 Y) = 110 – 75
∴ 15 Y – 10 Y = 35
∴ 5 Y = 35
∴ Y = 
∴ Y = 7
By substituting the value of Y in equation (ii), we get:
X + 15 (7) = 110
∴ X + 105 = 110
∴ X = 110 – 105
∴ X = 5
For distance of 50 km, charges will be: X + 50 Y
= 5 + 50 (7)
= 355
Therefore, option (c) is correct.
3. Charges paid for 30 km travel in Situation 2:
Given that the fixed charge is X and running charge is Y,
therefore the charges for B km, charges will be given by: X + B Y
By given conditions: for 8 km travel, charges paid are Rs. 91
∴ X + 8 Y = 91 ………. (iii)
Similarly, for 14 km, charges paid are Rs. 145
∴ X + 14 Y = 145 ………. (iv)
Let’s solve equation (iii) and (iv) and find values of X and Y.
By deducting equation (iii) from (iv), we get:
(X + 14 Y) – (X + 8 Y) = 145 – 91
∴ 14 Y – 8 Y = 54
∴ 6 Y = 54
∴ Y = 
∴ Y = 9
By substituting the value of Y in equation (iv), we get:
X + 14 (9) = 145
∴ X + 126 = 145
∴ X = 145 – 126
∴ X = 19
Now, for distance of 30 km, charges will be: X + 30 Y
= 19 + 30 (9)
= 289
Therefore, option (b) is correct.
4. Graph for situation 2:
The pair of linear equations in situation 2 is:
X + 8 Y = 91
X + 14 Y = 145
Let’s check if the given intersection point lies on the lines. If it lies on a line, it should satisfy the line (LHS should be equal to RHS)
In graph (i), (20,25) is the intersection point of the two lines, hence:
X + 8 Y = 91
20 + 8 (25) = 91
220 ≠ 91
Therefore, graph (i) is not the correct option.
Let’s check graph (ii). (12.5,0) is the intersection point of the two lines, hence:
X + 8 Y = 91
12.5 + 8 (0) = 91
12.5 ≠ 91
Therefore, graph (ii) is not the correct option.
Let’s check graph (iv). (15,10) is the intersection point of the two lines, hence:
X + 8 Y = 91
15 + 8 (10) = 91
95 ≠ 91
Therefore, graph (iv) is not the correct option.
Graph (iii) doesn’t have any intersection point, hence, we will try by given points
Let’s check for (11,10) on both lines:
X + 8 Y = 91
11 + 8 (10) = 91
91 = 91
Similarly, we check for (19,9) on both lines:
X + 8 Y = 91
19 + 8 (9) = 91
91 = 91
Therefore, upper line in the graph is for X + 8 Y = 91
Next, let’s check (5,7) lies on 2nd line or not:
X + 14 Y = 145
5 + 14 (7) = 145
103 ≠ 145
Similarly, let’s check (15,6) lies on 2nd line or not:
X + 14 Y = 145
15 + 14 (6) = 145
99 ≠ 145
4th graph doesn’t represent both the lines.
Therefore, none of the graph represents situation 2.
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