Question

A city bus carrying 70 passengers,some with 3 rupee ticket and remaining with 5 rupee tickets.If the amount recieved by conductor was rs.300,find the number of passengers of each type

Answer 1

Hey

your answer is ----

Let, those passenger whose ticket price is ₹3 be x and those passenger whose ticket price is ₹5 be y .

since, total number of passenger in bus is 70.

so, x+y = 70 .......(1)

now, according to given condition

3x + 5y = 300 .....(2)

now , multiply equation (1) by 3 and (2) by 1 , we get

3x + 3y = 210 ...(3)

3x + 5y = 300 ....(4)

now, subtract equ(3) from (4), we get

3x+5y -(3x+3y) = 300-210

=> 3x+5y-3x-3y = 90

=> 2y = 90

=> y = 90/2

=> y = 45

now, put this value in equation (3), we get

3x + 3×45 = 210

=> 3x + 135 = 210

=> 3x = 210-135 = 75

=> x = 75/3

=> x = 25

hence, total number of passenger whose ticket price is ₹3 is 25 and total number of passenger whose ticker price is ₹5 is 45

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HOPE IT HELP YOU
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Answer 2

In one variable-


Let the number of passengers with a ₹3 ticket be x. Number of passengers with a ₹5 ticket=70-x

According to question,

3x+5(70-x)=300

3x+350-5x=300

-2x=-50

x=25

Therefore, 25 passengers are travelling with a ₹3 ticket and 45 passengers are travelling with a ₹5 ticket.