Section :D 12. Rohan, a student of Class IX booked a la cab for travelling the tourist places of Jaipur city. The taxi fare in a city is as follows. Fare for the first km is Rs.8 and for the subsequent distance is Rs.5 per km. Then, choose the correct answers for the following questions. (i) Consider the distance covered as x km and the total fare as Rs.y, write a linear equation for this information. (a) 5x + y + 3 = 0 (b) 5x - y -3 = 0 (c) 5x – y + 3 = 0 (d) 9x + y + 3 = 0 (ii) Compare the above equation with ax + by + c = 0, find the value of (a + b + c). (a) 6 (b) 7 (C) 8 (d) 9 (iii) If total fare is Rs.93, then find the total distance. (a) 16 km (b) 17 km (c) 18 km (d) 19 km (iv) If the distance covered is 11 km, then find total fare. (a) Rs.58 (b) Rs.68 (c) Rs.48 (d) Rs.88 (v) If fare for first kilometer is Rs. 12 and fare for the subsequent distance is Rs.8 per km, then the linear equation for this information is (a) 4x + y + 4 = 0 (b) 6x – y + 6 = 0 (c) 8x - y + 4 = 0 (d) 7x - y + 7 = 0
Answers
1. According to the question,
Total distance covered x km = 1 + (x-1) km.
Total fare Rs. y
The fare of first kilometre Rs. 8.
The subsequent fare Rs. 5 per km.
So, we can write these statements in the form of equation as:
8 (1) + 5 (x - 1) = y
=> 8 + 5x - 5 = y
=> 5x - y + 3 = 0 (ans)
2. 5x - y + 3 = 0 compare this equation with ax + by + c = 0 we get, a= 5, b= (-1), C= 3
so, (a + b + C) = (5 - 1 + 3) = 7 (ans)
3. total fare is Rs.93
put the value in equation 5x - y + 3 = 0 we get,
5x - 93 + 3 = 0
=> 5x = 90
=> x = 18
so, the total distance = 18 km (ans)
4. If the distance covered is 11 km,
put the value in equation 5x - y + 3 = 0 we get,
5(11) - y + 3 = 0
=> 55 + 3 = y
=> y = 58
so, the total fare is Rs.58 (ans)
5. If fare for first kilometer is Rs. 12 and fare for the subsequent distance is Rs.8 per km, then the linear equation for this information is
12 (1) + 8 (x - 1) = y
=> 12 + 8x - 8 = y
=> 8x - y + 4 =0
The equation will 8x - y + 4 =0 (ans)
Answer:
1.5x-y+3=0
2.7
3.18
4.58
5.8x-y+4=0