A bus carries $67$ travelers of three types

Question

A bus carries $67$ travelers of three types:

• travelers who pay the entire ticket, which costs $\$3200$.

• students who have a $43$% discount.

• local retirees who only pay $23$% of the ticket price.

The bus collection on that trip was $\$6,292,000$. Calculate the number of travelers in each class knowing that the number of retirees was the same as the number of other travelers.

Solution:

• $x$: travelers who pay the entire ticket, $\$3200$

• $y$: students who have a $43\%$ discount, that is, they pay the $57\%$ of the total ticket. ($3200*57\%=1824$)

• $z$: local retirees who only pay $23$% of the ticket price. ($3200*23\%=736$)

The equations are

$$\begin{aligned} x + y +z &= 67 \\ 100x + 57y +23z &= 196625 \\ x+y-z &= 0 \nonumber \end{aligned}$$

The solutions are $x = \frac{193945}{43}$, $y = -\frac{385009}{86} $ and $z = 33.5$. However, $y$ is negative. If they ask about the number of travelers, what does it tell me in the result?

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equations: 1) $x+y+z=0$

2) $3200x+1824y+736z=6292000$

3) $z=x+y$

Answer 1

You wrote "Yes, I am sure that the exercise data are correct" but it cannot be.

Let $n$ be the number of "normal" travelers, $s$ the number of students and $r$ the number of retirees and put numbers to have $$ 3200 n+1824 s+736 r=T\tag 1$$ $$n+s+r=P\tag 2$$ $$r=n+s\tag 3$$ where $T$ is the bus collection and $P$ the number of passengers.

Using $(2)$ and $(3)$ makes $r=\frac P 2$ so $P$ cannot be odd.

Second remark : if everyone pays the full price, it would be $3200P$ and $67\times 3200=214400$ which does not have anything to do with the $6292000$ given. To get such a collection, the bus would be bigger than a train !

Just solving the three equations would give $$n=\frac{T-1280 P}{1376}\qquad s=\frac{1968P-T}{1376}\qquad r=\frac P2$$ and each of $n,s,r$ must be an integer (again with $T \leq 3200 P$).

Answer 2

Following Claude Leibovici's answer, we have $$s-n=\frac{3248r - T}{688}.$$ Noting that $32\mid T,$ let $T=32k,$ then $$s-n=\frac{203r - 2k}{43}.$$

If we set $T=6292000$ as in the original problem, then $k = 196625.$ Integer solutions occur for $r = 43m + 1966$ and $s-n = 203m + 136$ where $m$ is an integer. But we also have $s+n=r,$ so $$s = \tfrac12(r + (s-n)) = 123 m + 1051$$ and $$n = \tfrac12(r - (s-n)) = 915 - 80 m = 5 (183 - 16 m).$$

Since $n$ and $s$ must be non-negative, this limits the possible solutions to $-8\leq m \leq 11.$ Hence the minimum number of retirees occurs when $m = -8$ and $r = 43(-8)+1966 = 1622.$ There are then $67$ students and $1555$ other passengers.

The problem is therefore solvable if $67$ is the number of students rather than the total number of passengers.