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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