Final Exam Model Paper 3

i.

The answer, the one number left, is Rs 360.

• ii.3 × 120 + 2 × 25 − 50
• iii.500 120 − 2 ×

• i.The longest chain: Ask the members (2) → Design the screens (4) → Write the program (5) → Go live (1): 12 weeks. (Through the database it is 2 + 3 + 5 + 1 = 11; through training, 2 + 4 + 2 + 1 = 9.)
• ii.Design the database: 1 week. It can finish at week 5 or 6; writing the program can’t start before week 6 anyway. Train the staff: 3 weeks. It can finish at week 8, but going live starts only at week 11.
• iii.It now takes 5 weeks and finishes at week 7, a week later than the screens, so writing the program starts at week 7 and the project ends at week 13: one week late. Its 1 week of float absorbed the first week of the delay.

• i.On the line inside the loop that adds to total (or the line just after it). Each time it stops, look at total and the item in the variables, and check them against a total worked out by hand: the first stop where they differ shows the sale that goes wrong.
• ii.Step into when you suspect price_with_vat itself (it may work out VAT wrongly): it goes inside the function, line by line. Step over when you trust it: it runs the whole call at once and stops at the next line, so you stay in monthly_total.
• iii.The functions waiting, one inside another: price_with_vat on top, called from monthly_total, called from the main program, each with its own variables.

• i.Clerk 1’s transaction locks Hari’s account and waits for Gita’s; clerk 2’s has locked Gita’s and waits for Hari’s. Each holds the lock the other needs, so neither can ever go on.
• ii.It rolls one transaction back, releasing its locks, so the other can finish.
• iii.Try the rolled-back transaction again. A rollback undoes every part of it (atomicity), so no money left one account without reaching the other; the second try simply does the whole transfer.

• i.On average, students who study one hour more a week score 2.5 marks more. (It describes the students in the data; it doesn’t prove that the extra hour causes the marks.)
• ii.The line accounts for 36% of the spread in marks; the other 64% comes from everything else (ability, the teacher, sleep, luck). Many points are far from the line.
• iii.No. 40 hours is far outside the data (0 to 20 hours): an extrapolation. The line would say 30 + 2.5 × 40 = 130 marks, more than the full 100, which shows the line can’t be trusted there.

The claim is weak. It starts from 2079, the worst year in the table: comparing 2082 with 2077 shows almost no change (60% to 61%), so the “rise” is mostly a return from one bad year, picked as the starting point (cherry-picking). The rates go up and down by several points from year to year anyway. And with no comparison, anything else that changed (new teachers, an easier paper, pass rates rising across the country) could explain it.
Better evidence: compare the change at this school with the change at similar schools in wards without the classes over the same years; compare students who went to the classes with similar students who didn’t; or, best, offer the places by lot and compare the two groups (a randomized trial). Report every year, not only the convenient ones.

• i.φ(n) = (7 − 1)(11 − 1) = 60, and 7 × 43 = 301 = 5 × 60 + 1: one more than a multiple of φ(n), so 43 undoes 7.
• ii.Work out 26⁷ mod 77: 26² = 676 ≡ 60 (676 − 616); 26⁴ ≡ 60² = 3,600 ≡ 58 (3,600 − 3,542); 26⁷ = 26⁴ × 26² × 26 ≡ 58 × 60 × 26. 58 × 60 = 3,480 ≡ 15 (3,480 − 3,465); 15 × 26 = 390 ≡ 5 (390 − 385). It equals the hash, 5: the signature is good.
• iii.No. The check still gives 26⁷ mod 77 = 5, which doesn’t equal the new hash, 12. To make a signature that fits the new hash, the forger would need the private power, 43.

• A moral agent is someone who can be held responsible for what they do: an adult person.
• A moral patient is someone who can be wronged, whose suffering counts: a baby, a dog, and any adult too.
• Whether a machine could be a moral patient depends on whether it can feel anything (consciousness), not on how clever it is.

i. The total for a junction is its distance so far + its estimate; the junction with the smallest total is taken next.

ii. 16 m, by S → B → D → G (G came from D, D from B, B from S). The way through A and C, which heads straight at the flag, is 18 m.

iii. Dijkstra’s algorithm takes junctions in order of distance from the start alone, so it takes H (4 m) and K (5 m) as well, before reaching G at 16. A* never took them: they lie behind the start, so their estimates are large, and their totals (17) were never smaller than the flag’s 16. The estimate kept the search pointed at the flag.

iv. With the estimates tripled: from S, A’s total is 5 + 21 = 26, B’s 5 + 24 = 29, K’s 5 + 36 = 41, H’s 4 + 39 = 43, so A is taken. From A, C’s total is 9 + 15 = 24, so C is taken; then G, at 18 + 0 = 18, is smaller than B’s 29, so G is taken: the route S → A → C → G, 18 m. It is not the shortest. The estimates are now too high (C’s is 15, but C is only 9 m from the flag), so the long line from C looked better than it was, and the way through B and D was never tried. A* is sure to be right only when the estimate is never more than the real distance left.

i. TP = 48, FN = 2, FP = 57, TN = 893.

• Accuracy = (48 + 893) ÷ 1,000 = 94.1%
• Precision = 48 ÷ (48 + 57) = 48 ÷ 105 = 46%: more than half of the “absent” marks are wrong.
• Recall = 48 ÷ (48 + 2) = 96%: it finds nearly every real absence.

ii. Grades 1–3: 42 of 285 present student-days, 14.7%. Grades 4–10: 15 of 665, 2.3%. A young child in school is marked absent about one day in seven, more than six times as often as an older student. A likely cause: young children’s faces change fast, and the system compares them with photos from enrollment, months or years old: the world has drifted away from the data it was given. (Smaller faces, or children who don’t stand still at the gate, could add to it.)

iii. One well-argued case (other conclusions earn marks with good reasons): Not for scholarships, as it is.

• The cost of each mistake. A wrong “absent” costs a family a worrying SMS, and, through the 80% rule, it can stop a scholarship that keeps a child in school. Those mistakes fall mostly on the youngest children. 94.1% accuracy hides this, because most days are present days anyway.
• What must change first. Retake the photos regularly (say, every term) for the youngest grades and measure again by grade; keep checking each group after release, since faces keep changing.
• How it could be used. A teacher checks the “absent” list each morning before any SMS is sent or any record kept (human review); families can see the record and ask for a correction (appeal); no scholarship is stopped on the camera’s marks alone; the old register is kept ready if the numbers fall. A paper register, which the teacher already calls, may be cheaper and fairer for Grades 1–3.

The first run:

PASS test_no_changesPASS test_sports_dayFAIL test_next_day_as_usual period 2 is Sports2 passed, 1 failed

After the sports day, the next day’s plan still has Sports in period 2. Line 6, plan = TIMETABLE, doesn’t copy the list: it gives the same list a second name, so line 8 changes the usual timetable itself, and every later day gets the change. Make a copy, so each day changes only its own list:

TIMETABLE = ["Nepali", "English", "Maths", "Science", "Social", "Computer"]  def day_plan(changes):    """Today's periods: the usual timetable with today's changes (period number: subject)."""    plan = list(TIMETABLE)          # a copy: the usual timetable stays as it is    for period, subject in changes.items():        plan[period] = subject    return plan  def test_no_changes():    assert day_plan({}) == ["Nepali", "English", "Maths", "Science", "Social", "Computer"]  def test_sports_day():    assert day_plan({2: "Sports"})[2] == "Sports"  def test_next_day_as_usual():    assert day_plan({})[2] == "Maths", f"period 2 is {day_plan({})[2]}"  passed = failed = 0for name, test in list(globals().items()):    if name.startswith("test_"):        try:            test()            print("PASS", name)            passed += 1        except AssertionError as problem:            print("FAIL", name, problem)            failed += 1print(passed, "passed,", failed, "failed")

PASS test_no_changesPASS test_sports_dayPASS test_next_day_as_usual3 passed, 0 failed

(Keeping the usual timetable as a tuple, which can’t be changed, would also stop the mistake: line 8 would then fail at once instead of quietly changing it.)

truth = "--------------F-------F-------FFFF---F-F---F-F-----F-----F--"said  = "----F---------F------------F--FFFF---F-FF-----F----------F--"note  = "555151115155151515515515551511555551555555151115155151511115"  def measure(name, which):    """Counts the outcomes for the notes whose digit is in `which`."""    tp = fp = fn = tn = 0    for real, model, n in zip(truth, said, note):        if n not in which:            continue        if real == "F" and model == "F":            tp += 1        elif real == "F":            fn += 1        elif model == "F":            fp += 1        else:            tn += 1    precision = tp / (tp + fp)    recall = tp / (tp + fn)    print(f"{name:9} TP {tp}  FP {fp}  FN {fn}  TN {tn:2}  precision {precision:.0%}  recall {recall:.0%}")  measure("All", "51")measure("Rs 500", "5")measure("Rs 1,000", "1")

All       TP 8  FP 4  FN 4  TN 44  precision 67%  recall 67%Rs 500    TP 6  FP 3  FN 1  TN 25  precision 67%  recall 86%Rs 1,000  TP 2  FP 1  FN 3  TN 19  precision 67%  recall 40%

One limit: “The machine caught 6 of 7 fake Rs 500 notes, but only 2 of 5 fake Rs 1,000 notes (40%), the notes that cost you most; and it was tested on only 12 fakes, too few to be sure of any of these numbers.” Until it is tested on many more fake Rs 1,000 notes, the shopkeeper should still check those by hand.

• i.The cell in row i, column j holds the edit distance between the first i letters of one word and the first j letters of the other. Each cell is worked out once, from three cells already filled (above, left and diagonal), and kept in the table, so every smaller problem is solved once and reused: dynamic programming.
• ii.They ran before the bug did any harm: the first test made no change, and the second only looked at the day it changed. The damage, a changed usual timetable, shows only on a later call, which is why a test of the next day was needed.
• iii.Most of the notes are real, so a machine that called every note real would still be right about 48 of the 60 (80%). Accuracy hides that it missed 3 of the 5 fake Rs 1,000 notes. Test it on many more fake Rs 1,000 notes, measure each kind of note apart, and keep checking Rs 1,000 notes by hand until it does better.