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Question

Eileen had 3 times as much money as Jason at first. After Jason spent $290 and Eileen’s father gave her $446 more, Eileen had 7 times as much as what Jason had in the end. How much money did Jason have at first?

Answer

Sorry…no time to draw a nice model now.

At first

E -> [ u + 290 ][ u + 290 ][ u + 290 ]

J -> [ u + 290 ]

Change

E -> + $446

J -> – $290

After

E -> [ u ][ u ][ u ][ u ][ u ][ u ][ u ]

J -> [ u ]

But E in the end is also [ u + 290 ][ u + 290 ][ u + 290 ][ $446 ]

7u = (u + 290) + (u + 290) + (u + 290) + 446

7u = 3u + 1316

4u = 1316

u = 329

At first, Jason had (u + 290) = 329 + 290 = $619

1 Reply 1 Like

Hi BigDevil, found a little time and hope my model is acceptable. 🙂

 

For general note here: By P5, kids should know how to draw the first 2 sets “Before” and “After” well. It is the analysis part (circled above) that some kids need more practice with. So depending on the competency level the child is at, he may need to work on model drawing or analysis part or both.

But then again, some kids or parents give up model drawing totally and go for the algebra method. To me, it is a pity to give heuristics a complete miss and not enjoy the process. Most of us (parents) here had never touched model drawing in our old school and we learn anyway.

1 Reply 0 Likes

Of course it is acceptable.  🙂 

There is no right or wrong way of drawing a model. A model is simply a visual aid to help solve the problem. As long as the kids can understand and use it to solve the problem, it is a good model.  😀 

0 Replies 0 Likes

At first
Eileen : 1u + 1u + 1u = 3u
Jason : 1u

After
Eileen : 3u + 446
Jason : 1u – 290

(1u – 290) x 7 = 7u – 2030 ——-3u + 446
7u – 3u = 4u ——- 446 + 2030 = 2476
1u ——- 2476/4 = 619

Ans : $619.

0 Replies 1 Like