Simple Algebra Equations with 3 Unknowns – Cars Bicycles & Tricycles

There are a total of 17 bicycles, tricycles & cars in a parking lot. On counting, there are 45 wheels in total. There are thrice as many bicycles as cars & 4 more bicycles than tricycles. Find out the number of bicycles, tricycles & cars in the parking lot.

Solving:

Let’s note the clues in a logical order:

1. Each bicycle has 2 wheels
2. Each tricycle has 3 wheels
3. Each car has 4 wheels
4. Total of bicycles, tricycles & cars is 17
5. Total wheels in parking lot is 45
6. Number of Bicycles = 3 x (Number of Cars)
7. Number of Bicycles = 4 + (Number of Tricycles)

Let us assume that:

8. total number of bicycles is B  
9. total number of tricycles is T
10. total number of cars is C in the parking lot

Here B, T & C are variables (or unknowns) to help us solve the problem. (You can also assume them to be any other variables like x, y & z.)

Since there are 3 unknowns, we need to build at least 3 independent equations from the clues marked 1, 2, 3, 4, 5, 6 & 7.

Let’s look at clues (4):

Number of Bicycles + Number of Tricycles + Number of Cars = Total number of wheels 

So, from (4): 

B + T + C = 17 ————— lets label this equation as number (11)

Now let us look at clue (5):

(Wheels on 1 bicycle * number of bicycles) + (Wheels on 1 tricycle * number of tricycles) + (Wheels on 1 car * number of cars) = Total number of Wheels

So, from (5):

(2 * B) + (3 * T) + (4 * C) = 45

⇒ 2B + 3T + 4C = 45 ———– lets label this equation as number (12)

Let us look at clues (6) & (7):

B = 3C —- lets label this equation as number (13)

B = (4 + T) – —- lets label this equation as number (14)

So from equations 11, 12, 13 & 14 we have to solve for unknowns B, T & C.

Let us substitute / replace the value of B from (13) in the equation (11) & we get:

3C + T + C = 17

=> T + 4C = 17 —- lets label this equation as number (15)

Let us substitute / replace the value of B from (14) in the equation (12) & we get:

2B + 3T + 4C = 45

=> 2*(4 + T) + 3T + 4C = 45

=> 8 + 2T + 3T + 4C = 45

⇒ 5T + 4C = (45 – 8)

=> 5T + 4C = 37 —- lets label this equation as number (16)

Subtracting (15) from (16), we get:

(5T + 4C) – (T + 4C) = (37 – 17)

=> 5T + 4C – T – 4C = 20

=> 4T = 20

=> T = 20/4 

=> T = 5 —- lets label this equation as number (17)

So the number of Tricycles in the parking lot is 5.

Putting the value of T from (17) in the equation (15) to solve for the value of C:

T + 4C = 17

=> 5 + 4C = 17

=> 4C = (17 – 5)

=> 4C = 12

=> C = 12/4

=> C = 3 —- lets label this equation as number (18)

So the number of Cars in the parking lot is 3.

To find the value of B, let us substitute the value of T from (17) in equation (14):

B = (4 + T)

=> B = (4 + 5)

=> B = 9

So the number of Bicycles in the parking lot is 9.

Answer: Number of Tricycles is 5, Bicycles is 9 & Cars is 3.