If one bus has 100 passengers carring capacity .in one travel childs pay 10 cents,adults pay 2 birr and youngsters pay 1 birr . the bus got in one travel 100 birr so how many childrens , adults,youngsters does the bus carry
May 26, 2024- Let C = children, A = adults, Y = youngster we can write the following equation based on the information provided: 0.10C + 2A + Y = 100 Since the bus has a capacity of 100 passengers, we know that the total number of passengers on the bus is given by: C + A + Y = 100. By equating the above two equations you can get the solution.
May 26, 2024
We have 3 unkowns, But you give us only one equation... 0.1x+2y+1z=100 So it is Unsolved Unless u guess it..
May 26, 2024you cant solve for all three since you have 2 equations with three unkowns.
May 26, 2024I see some m.r smarty pants in the comments trying to answer lmao. 💀 sijemer the question lacks a lot of information. zmblachu atkebatru
May 27, 2024To solve this problem, we need to set up equations based on the given information. Let's define: - \( c \) as the number of children - \( a \) as the number of adults - \( y \) as the number of youngsters We have the following constraints: 1. The total number of passengers is 100. \[ c + a + y = 100 \] 2. The total fare collected is 100 birr. \[ 0.1c + 2a + 1y = 100 \] Now, we can solve these equations step-by-step. First, let's multiply the second equation by 10 to eliminate the decimal: \[ c + 20a + 10y = 1000 \] Now we have: \[ 1. \quad c + a + y = 100 \] \[ 2. \quad c + 20a + 10y = 1000 \] Subtract equation 1 from equation 2: \[ (c + 20a + 10y) - (c + a + y) = 1000 - 100 \] \[ 19a + 9y = 900 \] Let's simplify this equation: \[ 19a + 9y = 900 \] Now, we solve this simplified equation. Since \(a\) and \(y\) must be whole numbers, we can look for integer solutions to this equation. Let's solve for \(a\) in terms of \(y\): \[ a = \frac{900 - 9y}{19} \] \(900 - 9y\) must be divisible by 19. Let's find the value of \(y\) that satisfies this: \[ 900 - 9y \equiv 0 \mod 19 \] \[ 900 \equiv 9y \mod 19 \] \[ 900 \mod 19 = 7 \] \[ 9y \equiv 7 \mod 19 \] We need to find the modular inverse of 9 modulo 19. The inverse of 9 mod 19 is 17 (since \(9 \times 17 \equiv 1 \mod 19\)). Now multiply both sides by 17: \[ y \equiv 7 \times 17 \mod 19 \] \[ y \equiv 119 \mod 19 \] \[ y \equiv 5 \mod 19 \] So \(y = 5\). Substitute \(y = 5\) back into the equation for \(a\): \[ a = \frac{900 - 9 \times 5}{19} \] \[ a = \frac{900 - 45}{19} \] \[ a = \frac{855}{19} \] \[ a = 45 \] Now, substitute \(a = 45\) and \(y = 5\) back into the equation for the total number of passengers: \[ c + 45 + 5 = 100 \] \[ c = 100 - 50 \] \[ c = 50 \] So, the bus carries: - 50 children - 45 adults - 5 youngsters
May 26, 2024By iteration 40 children = 4 birr 36 Adults = 72 birr 24 youngsters = 24
May 26, 2024100 youngsters 🤷♂
May 26, 2024stupid question, it can carry 100 youngsters , no childrens and no adult
May 26, 2024