Question : In Delhi, taxi/cabs are running on compressed gas. To hire a taxi, one has to go to the location called taxi stand or book it on Mobile App. Shaila wanted to hire a taxi. So, on enquiring the taxi charges from the stand, she got the following information :
"For the first kilometer, the fare is Rs 20 and the subsequent distance it is Rs 12 per km."
Read the above information and answer the following by taking distance covered as x km and total fare Rs y.
(i) Write a linear equation for the above mentioned information. [2 Marks]
(ii) If shaila has hired the taxi for 26 km, find the amount paid by her. [2 Marks]
OR
If Ruby paid Rs 188 to the driver, find the distance travelled by her. [2 Marks]
Doubt by Lakshay
Solution :
i)
Total distance covered = x km
Total fare = Rs y
According to Questions (ATQ)
y=20+(x-1)12
y=20+12x-12
y=8+12x
y=12x+8
0=12x-y+8
12x-y+8=0
y=20+(x-1)12
y=20+12x-12
y=8+12x
y=12x+8
0=12x-y+8
12x-y+8=0
ii) Distance Covered = 26 km
i.e. x=26
y=12x+8
y=12(26)+8
y=312+8
y=320
y=12x+8
y=12(26)+8
y=312+8
y=320
Hence, amount paid by Shaila will be Rs 320.
OR
Amount Paid by Ruby = Rs 188
y=12x+8
188=12x+8
188-8=12x
180=12x
180/12=x
15=x
y=12x+8
188=12x+8
188-8=12x
180=12x
180/12=x
15=x
x=15
Hence, distance travelled by Ruby is 15 km.
Similar Case Study for Practice :
In Delhi, taxi/cabs are running on compressed gas. To hire a taxi, one has to go to the location called taxi stand or book it on Mobile App. Shaila wanted to hire a taxi. So, on enquiring the taxi charges from the stand, she got the following information :
"For the first kilometer, the fare is Rs 15 and the subsequent distance it is Rs 10 per km."
Read the above information and answer the following by taking distance covered as x km and total fare Rs y.
(i) Write a linear equation for the above mentioned information. [2 Marks]
(ii) If shaila has hired the taxi for 26 km, find the amount paid by her. [2 Marks]
OR
If Ruby paid Rs 185 to the driver, find the distance travelled by her. [2 Marks]
Ans :
i)
y=15+(x-1)10
y=15+10x-10
y=5+10x
0=10x-y+5
10x-y+5=0
y=15+(x-1)10
y=15+10x-10
y=5+10x
0=10x-y+5
10x-y+5=0
ii) x=26
y=5+10x
y=5+10x
y=5+10(26)
y=5+260
y=265
Rs 265
y=5+260
y=265
Rs 265
OR
y=185
y=5+10x
185=5+10x
185-5=10x
180=10x
180/10=x
18=x
x=18
18 km
y=185
y=5+10x
185=5+10x
185-5=10x
180=10x
180/10=x
18=x
x=18
18 km