From a bus stand in Bangalore , if we buy 2 tickets to Malleswaram and 3 tickets to
Yeshwanthpur, the total cost is Rs 46; but if we buy 3 tickets to Malleswaram and 5
tickets to Yeshwanthpur the total cost is Rs 74. Find the fares from the bus stand to
Malleswaram, and to Yeshwanthpur.
Answers
Given:
Cost of 2 tickets to malleshwaram and 3 tickets to yeshwantpur =RS. 46
Cost of 3 tickets to malleshwaram and 5 tickets to yeshwantpur =Rs.74
To find:
Fares to malleshwaram and yeshwantpur?
Solution:
So as per the question we can talk that,
⪼ 2x + 3y = 46 -eq(1)
⪼ 2x + 3y = 46 -eq(1)⪼ 3x + 5y = 74 -eq(2)
Multiply equation 1 by 3 and equation 2 by 2
➝ 3(2x + 3y = 46)
➝ 6x + 9y = 138
Similarly,
»» 2(3x + 5y = 74)
»» 6x + 10y = 148
Solving by elimination method
6x + 9y = 138
- 6x + 10y = 148
------------------------
-y = -10
------------------------
➝ y = 10
Hence fare for yeshwantpur is 10 rupees
Substitute the value of y in eq(1)
➳ 2x + 30 = 46
➳ 2x = 16
➳ x = 8
Hence fare for malleshwaram is 8 rupees
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Hello!
Step-by-step explanation:
Given:
Cost of 2 tickets to malleshwaram and 3 tickets to yeshwantpur =RS. 46
Cost of 3 tickets to malleshwaram and 5 tickets to yeshwantpur =Rs.74
To find:
Fares to malleshwaram and yeshwantpur?
Solution:
So as per the question we can talk that,
⪼ 2x + 3y = 46 -eq(1)
⪼ 2x + 3y = 46 -eq(1)⪼ 3x + 5y = 74 -eq(2)
Multiply equation 1 by 3 and equation 2 by 2
➝ 3(2x + 3y = 46)
➝ 6x + 9y = 138
Similarly,
»» 2(3x + 5y = 74)
»» 6x + 10y = 148
Solving by elimination method
6x + 9y = 138
- 6x + 10y = 148
------------------------
-y = -10
------------------------
➝ y = 10
Hence fare for yeshwantpur is 10 rupees
Substitute the value of y in eq(1)
➳ 2x + 30 = 46
➳ 2x = 16
➳ x = 8
Hence fare for malleshwaram is 8 rupees
Hope It Helps u :)