KCC's Quizzes AQQ301 - Counting Bicycles and Tricycles

10 Bicycles and 20 Tricycles

Students N = 30
Wheels W = 80
Number of Bicycles B
Number of Tricycles T = N - B

W= B*2+T*3 = 2B + 3N - 3B = 3N-B

Therefore Number of Bicycles is
B= 3N - W = 90 -80 = 10

Number of Tricyles is
T= N -B = 30-10 = 20

or more easy

T = W - 2N  = 80 -60  =  20

;-)

Number of Wheels W has to be in range

W >= 2N  

W <= 3N

1.    20.

2.   2N <= W <= 3N

The solution is a simple set of two linear simultaneous  equations with two unknowns.

Oops, where did that -ve sign come from?  Silly me!  I've edited.

1. B= nr of Bicycles, T= nr of Tricycles. Then you have 2 equations and solving them you get B=10 and T=20.
2B+3T=W
B+T=N

2. If you have no bicycles, then W=3N. If you have no tricycles, then W=2N. So the general condition between N and W must be:   2N <= W <= 3N.

x equals the number of bike riders => 2x+3(N-x) = W or x = 3N-W

y equals the number of trike riders => 2(N-y)+3y = W or y = W-2N

1. y = 80-60 or 20 tricycle riders

2. To avoid negative x, 3N-W ≥ 0

    To avoid negative y, W-2N ≥ 0

The general condition between W and N is: 2N ≤ W ≤ 3N

T=Trikes, B=Bikes

3T+2B=80

T+B=30  -> 3T+3B=90.  Can solve by inspection since the extra factor of B=10.

So Ten Bikes and 20 Trikes.

In case 1,  if all student ride trikes W>=3N.   If all ride bikes, then W>=2N.  The stronger condition is W>=3N.

Q1: We will apply the so-called “false position” method:

1°) First application: If all students used a bicycle, there would be 60 wheels. Therefore, in reality, 20 students use a tricycle and 10 use a bicycle.

2°) Second application: If all students used a tricycle, the number of wheels would be 90. Therefore, in reality, 10 of them use a bicycle.

The two ways of applying the false position method naturally lead to the same solution: 20 students use a tricycle and 10 students use a bicycle.

Q2:

Let N be the number of students. Since each of them uses at least one bicycle, the number of wheels W must be greater than or equal to 2N.

If all of them used a tricycle, the number of wheels W would be 3N.

In reality, all intermediate situations are possible.

Therefore, if we first choose the number N of students, we must then choose, to avoid negative values for the number of wheels W, a number W of wheels such that:

2N ≤ W ≤ 3N

20 student use tricycle

1. 20 students use a tricycle. 

2. 3 N >= W >= 2 N >= 0.

X = Peoplr who uses Bicycle
Y = People who uses Tricycle
X+Y = 30
X *2 + Y * 3 = 80 wheels
X = 10  Y = 20 --> 20 students use tricycles