Question

there are 60 passengers in a bus, some with rs. 5 tickets, some with rs. 10 and remaining with rs. 15 tickets. the passenger with rs. 5 tickets is twice the passenger with rs 15 tickets. if the total collection from these passengers is rs. 525, find how many passengers have tickets worth a rs. 10 ​

Answer 1

Answer:

Let the number of passengers bought Rs 3 tickets = x

and the number of passengers bought Rs 10 tickets = y

Given, there are 40 passengers in a bus

=> x + y = 40 ...............1

Again, total collection from the passenger is Rs 295

=> 3x + 10y = 295 .........2

Multiply by 10 in equation 1, we get

10x + 10y = 400 ............3

Now, subtract equation 3 and 2, we get

10x + 10y - 3x - 10y = 400 - 295

=> 7x = 105

=> x = 105/7

=> x = 15

So, the number of passengers bought Rs 3 tickets is 15

Answer 2

Answer:

Number of passengers with ₹10 ticket is 15

Step-by-step explanation:

Let the number of passengers with ₹5 tickets be     a

Let the number of passengers with ₹10 tickets be    b

Let the number of passengers with ₹15 tickets be    c

Since there are 60 passengers on the bus

$a+b+c=60 \ ...(1)$

The passenger with ₹5 tickets is twice the passenger with ₹15 tickets (Given)

Therefore,

$a=2c \ ... (2)$

Total collection = ₹525 (Given)

$5a +10b+15c= 525 \ ... (3)$

Solving Equation 3

$5a +10b+15c= 525\\5(a+2b+3c) = 525\\a+2b+3c=\frac{525}{5} = 105\\a+2b+3c = 105\\ Putting \ value \ of \ a \ from \ Eq \ 2\\2c+2b+3c = 105\\2b+5c=105 \ ...(4)$

Putting value of a from 2 in equation 1

$a+b+c=60\\2c+b+c=60\\b+3c=60\\b=60-3c \ ...(5)$

Putting value of b from Eq 5 in Eq 4

$2b+5c=105\\2(60-3c)+5c=105\\120-6c+5c=105\\-c=105-120\\c=120-105=15$

Putting value of c in Eq 2

$a = 2c\\a = 2*15 = 30$

Putting value of a and c in Eq 1

$a+b+c = 60\\30+b+15=60\\b+45=60\\b=15$

Therefore number of passengers with

₹5 ticket = a = 30

₹10 ticket = b = 15

₹15 ticket = c = 15