# How many oranges were there in the morning?

#### anemone

##### MHB POTW Director
Staff member
There was a total of 748 oranges and apples at a fruit stall in the morning. By afternoon, the number of oranges sold was thrice the number of apples sold. The number of apples left was twice the number of oranges left. There were 22 more apples left than the apples sold.

How many oranges were there in the morning?

This primary math problem (as usual) can be solved without setting up equations and solve the simultaneous system of equations.

I welcome members to give it a try but nevertheless, I will post the solution tomorrow to showcase how Singapore model method works well in this kind of problem.

#### anemone

##### MHB POTW Director
Staff member
We can draw the following model based on the given information, and the top bar and lower bar are equal after we added another 5 units and 4 of 22 into them and this makes the sum of both bars becomes 748+5 units + 4(22).

\begin{tikzpicture}

\filldraw [fill=cyan, thick, draw=black] foreach \i in {-4,...,-3} { ({\i*1.8},0) rectangle ({(\i+2)*1.8},0.8) };
\filldraw [fill=cyan, thick, dotted, draw=black] foreach \i in {-1,...,2} { ({\i*1.8},0) rectangle ({(\i+2)*1.8},0.8) };
\filldraw [fill=magenta, thick, draw=black] (-7.2,-1) rectangle (-4.8,-0.2);
\draw[thick, dotted] (-4.8,0) -- (-4.8,-0.2);
\filldraw [fill=magenta, thick, draw=black] (-3.6,-1) rectangle (-1.2,-0.2);
\draw[thick, dotted] (-1.2,0) -- (-1.2,-0.2);
\filldraw [fill=magenta, thick, draw=black] (0,-1) rectangle (2.4,-0.2);
\draw[thick, dotted] (2.4,0) -- (2.4,-0.2);
\filldraw [fill=magenta, thick, draw=black] (3.6,-1) rectangle (6,-0.2);
\draw[thick, dotted] (6,0) -- (6,-0.2);
\draw[thick, dotted] (7.2,-1) -- (7.2,-0.2);
\draw[<->] (7.5,0.9) -- (7.5,-1);
\node at (9.4,-0) {748+5 units+4(22)};
\draw [<->] (6, -0.5) -- (7.2, -0.5);
\node at (6.6,-0.69) {22};
\draw [<->] (2.4, -0.5) -- (3.6, -0.5);
\node at (3,-0.69) {22};
\draw [<->] (-1.2, -0.5) -- (0, -0.5);
\node at (-0.6,-0.69) {22};
\draw [<->] (-4.8, -0.5) -- (-3.6, -0.5);
\node at (-4.2,-0.69) {22};
\node at (-6.3,0.5) {\small apple left};
\node at (-4.5,0.5) {\small apple left};
\node at (-2.7,0.5) {\small orange left};
\node at (-6,-0.6) {\small apple sold};
\node at (-2.4,-0.6) {\small orange sold};
\node at (1.2,-0.6) {\small orange sold};
\node at (4.8,-0.6) {\small orange sold};
\node at (-6.3,1.6) {\small 1 unit};
\draw [<->] (-7.2, 1.1) -- (-5.4, 1.1);
\node at (-6.1,-1.6) {\small 1 part};
\draw [<->] (-7.2,-1.4) -- (-4.8, -1.4);

\end{tikzpicture}

The problem is asking to find the sum of 1 unit and 3 parts, hence we have

\begin{align*} 8\text{units}&=\dfrac{748+5\text{units}+4(22)}{2}\\&=\dfrac{836+5\text{units}}{2}\\16\text{unit}&=836+5\text{units}\\ 11\text{units}&=836\\1\text{unit}&=76\end{align*}

\begin{align*} 1\text{part}+22&=2(76)\\1\text{part}&=130\end{align*}

Therefore there were $76+3(130)=466$ oranges in the morning.

#### Country Boy

##### Well-known member
MHB Math Helper
I don't see any reason NOT to set up equations- they help to organize one' thoughts.
Let x be the number of oranges and let y be the number of apples in the fruit store in the morning. Let u be the number of oranges sold by afternoon and let v be the number of

There was a total of 748 oranges and apples at a fruit stall in the morning.
So x+ y= 748

By afternoon, the number of oranges sold was thrice the number of apples sold.
So u= 4v

The number of apples left was twice the number of oranges left.
The number of apples left was the number of apples in the morning minus the number of apples sold, y- v, and the number of oranges left is the number of oranges in the morning minus the number of oranges sold, x- u.

So y- v= 2(x- u) or 2x- y- 2u+ v= 0.

There were 22 more apples left than the apples sold.
So y- v= v+ 22 or y= 2v+ 22.

How many oranges were there in the morning?

This primary math problem (as usual) can be solved without setting up equations and solve the simultaneous system of equations.

I welcome members to give it a try but nevertheless, I will post the solution tomorrow to showcase how Singapore model method works well in this kind of problem.
We have four equations
x+ y= 748,
u= 4v,
2x- y- 2u+ v= 0, and
y= 2v+ 22

They can be solved for all of x, y, u, and v.
The problem is asking for the value of x.

#### Fantini

MHB Math Helper
What is the Singapore model method?

#### Country Boy

##### Well-known member
MHB Math Helper
What is the Singapore model method?
It's now six days after he said he would post it "tomorrow"!

#### MarkFL

Staff member
It's now six days after he said he would post it "tomorrow"!
She posted the solution using the cited method in the second post.

#### anemone

##### MHB POTW Director
Staff member
What is the Singapore model method?
Hello Fantini!

Singapore model method is a child-friendly pictorial language designed to develop students' understanding of fundamental mathematics concepts and proficiency in solving basic mathematics word problems. The model method affords higher ability children without access to letter-symbolic algebra as a means to represent and solve algebraic word problems. In Singapore, primary school children are taught a visual and concrete approach to solve arithmetic and/or algebraic word problems involving whole numbers, fractions, ratios and percents.

The model method focuses on the importance of representation. It builds upon children's knowledge of part-whole relationships in numbers. Children are taught to use appropriately sized rectangles to represent the information presented in word problems. In arithmetic problems, the rectangles represent specific numbers. By changing the role of the rectangles and using them to represent unknown quantities, the model method also can be used to depict algebraic word problems. Such representations effectively serve as pictorial equations.

The model method has proved to be an effective teaching tool for helping elementary aged children solve more word problems successfully.

MHB Math Helper