Nimm Hundert für Hundert

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Andreas Vohns

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4 years, 4 months ago

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Andreas Vohns 4 years, 4 months ago

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Andreas Vohns 4 years, 4 months ago

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Created this as a copy of Mehrdeutige Lösungen.

Name	Status	Author	Last Modified
Mehrdeutige Lösungen	Ready to use	Andreas Vohns	11/12/2020 07:44
Nimm Hundert für Hundert	Ready to use	Andreas Vohns	17/12/2020 07:32

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Question statement

Eine Abwandlung der "100-Vögel"-Aufgabe lautet:

Für 100 Sapek sollen 100 {strings[dice[0]][0]} gekauft werden. {strings[dice[0]][1]} kosten {costs[dice[1]][0]} Sapek, {strings[dice[0]][2]} kosten {costs[dice[1]][1]} Sapek und für einen Sapek erhält man {costs[dice[1]][2]} {strings[dice[0]][3]}.

Give any introductory information the student needs.

			Name	Type	Generated Value

n	number	7865
a	number	605
b	integer	13
costs	list	Nested 5×3 list
strings	list	Nested 4×4 list
dice	list	[ 0, 0 ]
solutions	list	Nested 5×3 list
Unnamed

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
a*b

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

7865

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap a Number entry
  Add
2. Gap b Number entry
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3. Gap c Number entry
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Ask the student a question, and give any hints about how they should answer this part.

Wie viele {strings[dice[0]][0]} von jeder Sorte soll man kaufen?

Geben Sie bitte eine mögliche ganzzahlige Lösung an.

{strings[dice[0]][1]}: a

{strings[dice[0]][2]}: b

{strings[dice[0]][3]}: c

Advice

Es seien $x$ die Anzahl der {strings[dice[0]][1]}, $y$ die Anzahl der {strings[dice[0]][1]} und $z$ die Anzahl der {strings[dice[0]][2]}.

(1) Aus den Anzahlen ermitteln wir dann die Gleichung $x+y+z=100$

(2) Aus den Kosten zudem die Gleichung $\var{costs[dice[1]][0]}\cdot x + \var{costs[dice[1]][1]}\cdot y+\frac{1}{\var{costs[dice[1]][2]}}\cdot z=100$

Durch Einsetzen von (1) in (2) erhalten wir die Gleichung $\simplify{{costs[dice[1]][0]}*x+{costs[dice[1]][1]}*y+1/{costs[dice[1]][2]}*(100-y-x)=100}$

Umformen ergibt: $\simplify{{costs[dice[1]][0]-1/{costs[dice[1]][2]}}*x+{costs[dice[1]][1]-1/{costs[dice[1]][2]}}*y=({100-100/costs[dice[1]][2]})}$ bzw. $\simplify{{costs[dice[1]][2]}*{costs[dice[1]][0]-1/{costs[dice[1]][2]}}*x+{costs[dice[1]][2]}*{costs[dice[1]][1]-1/{costs[dice[1]][2]}}*y=({costs[dice[1]][2]}*{100-100/costs[dice[1]][2]})}$ .

Man wählt nun $y$ so, dass sich für $x$ ein ganzzahliger Wert ergibt, der dann auch zu einem ganzzahligen Wert von $z$ führen muss, wobei man noch aufpassen muss, dass $z$ nicht negativ werden darf.

Eine mögliche Lösung ist $x=\var{solutions[dice[1]][0]}, y=\var{solutions[dice[1]][1]}, z=x=\var{solutions[dice[1]][2]}$ .

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Number entry		Hasn't run yet
Number entry
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Number entry
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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Error in variable testing condition

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Generated value: `number`

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Nimm Hundert für Hundert

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Andreas Vohns 4 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`