MHB Solve Math Problem: Number of Passengers on a Bus

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The math problem discusses the number of passengers on a bus before and after two stops. Initially, there were x passengers, with a fraction alighting and others boarding at each stop. After calculations, it was determined that there were 500 passengers on the bus before the first stop. The discussion also mentions the Singapore model method as a potential approach to solving such problems. Additional resources on the Singapore method were shared for those interested in exploring this problem-solving technique further.
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Hi all, this is another math problem that one can use the Singapore model method to solve. All are welcome to post a solution using the method you would like, if this problem interests you. :)

There were some passengers on a bus. At the first stop, $\dfrac{1}{5}$ of the passengers alighted and 80 boarded the bus. At the second stop, 240 passengers got off and 60 passengers boarded the bus. The bus now had $\dfrac{5}{8}$ of the number of passengers when it left the first stop. How many passengers were on the bus before the first stop?
 
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anemone said:
Hi all, this is another math problem that one can use the Singapore model method to solve. All are welcome to post a solution using the method you would like, if this problem interests you. :)
Standard algebra:
There were some passengers on a bus.
Let the number passengers initially be x.

At the first stop, $\dfrac{1}{5}$ of the passengers alighted and 80 boarded the bus.
Now there are $x- \dfrac{1}{5}x+ 80= \dfrac{4}{5}x+ 80$ passengers on the bus.

At the second stop, 240 passengers got off and 60 passengers boarded the bus.
Now there are $\dfrac{4}{5}x+ 80- 240+ 60= \dfrac{4}{5}x- 100$ passengers on the bus.
The bus now had $\dfrac{5}{8}$ of the number of passengers when it left the first stop. How many passengers were on the bus before the first stop?
So $\dfrac{4}{5}x- 100= \dfrac{5}{8}\left(\dfrac{4}{5}x+ 80\right)= \dfrac{1}{2}x+ 50$.

Solve the equation $\dfrac{4}{5}x- 100= \dfrac{1}{2}x+ 50$ for x.

Subtract $\dfrac{1}{2}x$ from both sides- $\dfrac{4}{5}- \dfrac{1}{2}= \dfrac{8}{10}- \dfrac{5}{10}= \dfrac{3}{10}$.
$\dfrac{3}{10}x- 100= 50$.

Add 100 to both sides. $\dfrac{3}{10}x= 150$.

Multiply both sides by $\dfrac{10}{3}$.
$x= 500$.

There were originally 500 passengers on the bus.
(That's a large bus!)
Check:
(When I first did this problem, I made a trivial arithmetic mistake. Always check!)

At the first stop, $\dfrac{1}{5}$ of the passengers alighted and 80 boarded the bus.
$\dfrac{1}{5}$ of 500 is 100. 100 of the 500 passengers got off, leaving 500- 100= 400 passengers and another 80 boarded so there are now 480 passengers on the bus.
At the second stop, 240 passengers got off and 60 passengers boarded the bus.
480- 240= 240. 240+ 60= 300.

The bus now had $\frac{5}{8}$ of the number of passengers when it left the first stop.
When it left the first stop it had 480 passengers. $\frac{5}{8}$ of that is 5(60)= 300. Yes, that is what we got.

Before the first stop the bus had 500 passengers,

Now, how would the "Singapore model method" solve this problem?
 
anemone said:
Hi all, this is another math problem that one can use the Singapore model method to solve. All are welcome to post a solution using the method you would like, if this problem interests you. :)

There were some passengers on a bus. At the first stop, $\dfrac{1}{5}$ of the passengers alighted and 80 boarded the bus. At the second stop, 240 passengers got off and 60 passengers boarded the bus. The bus now had $\dfrac{5}{8}$ of the number of passengers when it left the first stop. How many passengers were on the bus before the first stop?

My Attempt

After if left the 1st stop $\dfrac{5}{8}$ remained so $\dfrac{3}{8}$ was the number of more persons alighted than boarded

The number is 180

so $\dfrac{3}{8}$ of number of persons when it left the $1^{st}$ stop is 180

so number of persons when it left $1^{st}$ stop =$\dfrac{180}{{\frac{3}{8}}} = 480$

That was the number after 80 boarded

So the number before 80 boarded was 480-80 = 400

after $\dfrac{1}{5}$ got of remaining was $\dfrac{4}{5}$ which is 400

so number started with $\dfrac{400}{{\frac{4}{5}}} = 500$
 
[TIKZ]
\filldraw [fill=yellow,thick] foreach \i in {-4,...,-1} { ({\i*2.4},0) rectangle ({(\i+2)*2.4},1) };
\filldraw [fill=green, thick, dotted] foreach \i in {3,...,9} { ({\i*0.8},0) rectangle ({(\i+2)*0.8},1) };
\node at (-8.5,1.4) {\small 1 unit = 2 parts};
\draw [<->] (-9.5, 1.2) -- (-7.2, 1.2);
\node at (-8.4,0.6) {\tiny alighted passengers};
\node at (-8.4,0.2) {\tiny at 1st stop};
\draw[thick, dotted] (-7.2, 1) -- (-9.5, 0);
\draw[thick, dotted] (-7.8, 1) -- (-9.6, 0.2);
\draw[thick, dotted] (-8.4, 1) -- (-9.65, 0.4);
\draw[thick, dotted] (-7.2, 0.7) -- (-9, 0);
\draw[thick, dotted] (-7.2, 0.4) -- (-8.2, 0);
\draw [<->] (-7.2, 1.2) -- (-6, 1.2);
\node at (-6.6,1.4) {\small 1 part};
\draw [<->] (-7.2, 2) -- (8.8, 2);
\node at (0,2.2) {\small Total passengers on the bus after 1st stop and before 2nd stop};
\draw [<->] (-1.2, 1.2) -- (4.8, 1.2);
\node at (2,1.5) {\small Number of passengers got off the bus at 2nd stop = 240 - 60 = 180};
\draw[thick, dotted] (4.8, 0.2) -- (3.6, 0);
\draw[thick, dotted] (4.8, 0.4) -- (2.4, 0);
\draw[thick, dotted] (4.8, 0.6) -- (1.2, 0);
\draw[thick, dotted] (4.8, 0.8) -- (-0, 0);
\draw[thick, dotted] (4.8, 1) -- (-1.2, 0);
\draw[thick, dotted] (3.6, 1) -- (-1.2, 0.2);
\draw[thick, dotted] (2.4, 1) -- (-1.2, 0.4);
\draw[thick, dotted] (1.2, 1) -- (-1.2, 0.6);
\draw[thick, dotted] (0, 1) -- (-1.2, 0.8);
\draw[thick, dotted] (-6,0) -- (-6,1);
\draw[thick, dotted] (-3.6,0) -- (-3.6,1);
\draw[thick, dotted] (-1.2,0) -- (-1.2,1);
\draw[thick, dotted] (1.2,0) -- (1.2,1);
\node at (2.8,0.5) {\small 10};
\node at (3.6,0.5) {\small 10};
\node at (4.4,0.5) {\small 10};
\node at (5.2,0.5) {\small 10};
\node at (6,0.5) {\small 10};
\node at (6.8,0.5) {\small 10};
\node at (7.6,0.5) {\small 10};
\node at (8.4,0.5) {\small 10};
[/TIKZ]

Represent the given information in the model diagram above, the question wanted us to find the value for 10 parts.

From the model diagram,

$\begin{align*}3 \text{ parts} + 30 &=180\\3 \text{ parts}&=150\\ 1\text{ part}&=50 \\ \therefore 10\text{ parts}&=500\end{align*}$
 
Hi anemone,

Is there information available somewhere about the Singapore method ?
 
Hi castor28!

Thanks for showing your interest in Singapore model method in solving primary math word problems.

I just found a few insightful articles that discuss about the effectiveness to use model method to solve word problems, I hope you or anyone who are interested (@Fantini) find those articles meaningful to you!

https://bsrlm.org.uk/wp-content/uploads/2016/02/BSRLM-IP-35-3-20.pdf
http://people.math.harvard.edu/~engelwar/MathS305/Singapore Model Method Text.pdf
https://files.eric.ed.gov/fulltext/EJ1115069.pdf
 
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