تصحيح نسخة رقم 27: منهج حرّ، خطأ حسابي، وقراءة محاور
A corrector is grading copy #27 of a 20-mark Brevet blanc problem set in a stationery shop. Part A (8 marks): solve the system 3x + 2y = 47 and 5x + 3y = 76, where x is the price in thousands of Lebanese pounds of a large notebook and y that of a small one. The instruction reads «méthode au choix». The pupil substitutes, reaches x = 11 and y = 7, writes «donc 11 000 LL et 7 000 LL» and stops: no vérification, no concluding sentence beyond the units. Part B (6 marks): «A customer buys 4 large and 3 small notebooks and pays with a 70 000 LL note. Does she receive change?» The pupil writes only «Non, il manque 500 LL». Part C (6 marks): the two lines are drawn on the same grid and cross at (11 , 7); the pupil writes «le point d'intersection est (7 ; 11) car on lit les x sur l'axe vertical». The school's marking charter states that partial credit must reward valid reasoning even when the final answer is wrong, and that an error carried forward from an earlier part is not penalised twice.
1. How many of the 8 marks should Part A receive? Address the «méthode au choix» instruction and the missing vérification.
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Award 7/8. The instruction «méthode au choix» means the objective is solving, not naming a method; substitution is a legitimate algebraic method and the pupil used it correctly:
- correct transcription of the system into two equations - 1 mark;
- isolating one unknown correctly, y = (47 − 3x)/2 - 2 marks;
- substituting into the second equation and reducing to one equation in x - 2 marks;
- correct values x = 11, y = 7 - 2 marks;
- vérification and a full concluding sentence with units - 1 mark, which is lost.
The missing vérification is not a mathematical error, so it must not touch the calculation marks; but the charter's final point covers exactly that habit, and 11 000 LL substituted into 3x + 2y = 47 costs ten seconds. The deduction teaches the habit without punishing the algebra.
Note the unit discipline: «11 000 LL et 7 000 LL» is acceptable, but the marker should write on the copy that x and y are prices, so the answer sentence should name which notebook costs what - a pupil who inverts them in the sentence loses nothing here, because the system's meaning was fixed by the question.
2. Part B is answered «Non, il manque 500 LL». Score it and explain how the carried-forward-error rule applies.
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The correct computation: 4 × 11 + 3 × 7 = 44 + 21 = 65 (thousands), so the bill is 65 000 LL and the change from a 70 000 LL note is 5 000 LL. The pupil's answer is wrong in sign (there is change) and in magnitude (5 000, not 500).
Score: 1/6. The pupil shows no calculation line at all, so the marker cannot award marks for a method that was never written; one point is given for having recognised that Part B depends on the prices found in Part A - the only defensible element visible on the copy.
The carried-forward rule. It applies when a pupil's earlier result, although wrong, is used consistently and correctly afterwards. Here the earlier result is right, so the rule is irrelevant: the error is entirely in Part B. The marker should still write the rule on the copy, because the pupil needs to know that a wrong x in Part A would not have destroyed Part B.
Feedback: require one written line per step («4 × 11 000 + 3 × 7 000 = …») and a comparison with the note. This is a reading-and-checking failure, not an algebra failure.
3. Part C gives the intersection as (7 ; 11). How many marks, and what exactly should the feedback say?
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Award 3/6. The pupil has understood the essential mathematical idea - the intersection point of the two lines is the solution of the system - and names it, which is worth the reasoning marks. The coordinates are swapped, which is worth zero, because reading a graph is a skill being tested in its own right.
What the feedback must say: the horizontal axis is read first, so the coordinates of a point are (x ; y) in that order; the pupil has read y first, then x. There is a built-in check the pupil missed: the point read on the grid must be the same pair as the solution found in Part A, (11 ; 7). Substituting confirms it: 3(11) + 2(7) = 47 and 5(11) + 3(7) = 76. A pupil who cross-checks the graph against the algebra cannot leave (7 ; 11) on the copy.
Total for the copy: about 11/20 (7 + 1 + 3). The overall comment should name two habits rather than list errors: verify each result in the other equation, and label the axes before plotting. Both are worth more, over a whole exam, than the two lost points here.