1.
Four antibodies — P, Q, R, and S — each caught some germs. Q caught more than P. R caught fewer than P. S caught more than Q. List all four from most to least.
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2.
Three scientists — Nadia, Yusra, and Tariq — each study one antibody shape: Y-shape, T-shape, or disc. Nadia does not study the disc. Tariq studies the Y-shape. Who studies each shape?
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3.
A dish starts with 4 antibodies at 2:00. The number doubles every 20 minutes. How many antibodies are in the dish at 3:00?
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4.
Vials are numbered 1 to 24. Every 3rd vial gets a green dot. Every 4th vial gets a yellow dot. Which vials get both dots? And how many vials get no dot at all?
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Answer:
1.
Four researchers — Amina, Bilal, Hafsa, and Idris — each matched an antibody to one germ: flu, measles, tetanus, or cholera. Amina matched neither flu nor cholera. Bilal matched measles. Hafsa did not match cholera. Who matched which germ?
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2.
Nine sealed vials look identical, but one is heavier than the other eight. You have a balance scale. What is the fewest number of weighings that is guaranteed to find the heavy vial? Explain why it always works.
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3.
A counting log reads 5, then 11, then 23, then 47. What comes next, and what is the rule?
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4.
A sample must be checked every 12 minutes, starting at 9:00 and finishing at 11:00, with a check at both the start and the finish. How many checks are made altogether?
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Answer:
3rd Grade
1. S, Q, P, R.
Q > P and R < P gives Q, P, R. S > Q puts S on top. Chain them one clue at a time rather than guessing a whole order.
2. Tariq — Y-shape; Nadia — T-shape; Yusra — disc.
Tariq is given. That leaves Nadia and Yusra for T-shape and disc; Nadia is not the disc, so Nadia takes the T-shape and Yusra the disc. A "not" clue is as useful as a "yes" clue.
3. 32 antibodies.
2:00 → 3:00 is 60 minutes, so three doublings at 2:20, 2:40 and 3:00. 4 → 8 → 16 → 32. The trap is doubling only twice, or multiplying 4 by 3.
4. Both dots: 12 and 24. No dot: 12 vials.
Both dots means a multiple of 12. Green (multiples of 3) = 8 vials; yellow (multiples of 4) = 6 vials; 2 of them are counted twice, so 8 + 6 − 2 = 12 vials get at least one dot, leaving 24 − 12 = 12 with none.
4th Grade
1. Bilal — measles; Amina — tetanus; Hafsa — flu; Idris — cholera.
Bilal is given (measles). Amina is not flu, not cholera, and measles is taken, so Amina is tetanus. Hafsa is not cholera and the other two are gone, so Hafsa is flu. Idris takes what is left, cholera. A grid with rows for people and columns for germs makes this fall out.
2. Two weighings.
Split into three groups of three. Weigh group A against group B: if one side sinks, that group holds it; if they balance, it is group C. That is one weighing to find the group of three. Then weigh one vial against another from that group: if one sinks it is the heavy one, if they balance it is the third. Guaranteed, not lucky — every outcome is covered. One weighing cannot do it: a single weighing has only three outcomes and there are nine possible vials.
3. 95. Rule: double, then add 1.
5 × 2 + 1 = 11; 11 × 2 + 1 = 23; 23 × 2 + 1 = 47; 47 × 2 + 1 = 95. The gaps (6, 12, 24) double too, which is the other way to see it.
4. 11 checks.
9:00 to 11:00 is 120 minutes; 120 ÷ 12 = 10 gaps. Checks happen at both ends, so it is gaps + 1 = 11. Fenceposts: 10 gaps need 11 posts.