DAT Practice Questions (2026) — Free Sample Questions with Answers

15 free DAT questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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Biology

Question 1

A phenotypically normal woman has a brother with hemophilia A, an X-linked recessive disorder. Her parents are unaffected. She has a child with an unaffected man. What is the probability that the child is a son with hemophilia A?

Show answer & explanation

Answer: A

  1. Unaffected parents with an affected son: the mother is a carrier (XᴴXʰ).
  2. The woman, a normal daughter, is a carrier with probability 1/2.
  3. If she is a carrier: P(son) × P(son gets Xʰ) = 1/2 × 1/2 = 1/4.
  4. Overall probability = 1/2 × 1/4 = 1/8.

1/8: Correct. Her mother must be a carrier, so the woman is a carrier with probability 1/2. A carrier’s child is an affected son with probability 1/2 (son) × 1/2 (gets the mutant X) = 1/4. Overall 1/2 × 1/4 = 1/8 (Genetics: human genetics).

1/6: Uses a carrier probability of 2/3, the value for an unaffected sibling in an autosomal recessive family: 2/3 × 1/4.

1/4: Treats the woman as a certain carrier and skips the 1/2 chance that she is a carrier.

1/2: Gives the chance that a son of a known carrier is affected, ignoring both the carrier probability and the chance of a daughter.

1/16: Multiplies in an extra factor of 1/2, for example by also applying a carrier probability to the unaffected father.

For a “son who is affected” question, multiply the carrier probability by 1/2 for sex and 1/2 for which X he inherits.

Question 2

A purified enzyme is assayed with and without compound X. The results are shown in the table. Which conclusion about compound X is best supported?

Kinetic constants of the enzyme
ConditionApparent Km (mM)Vmax (μmol/min)
No compound X2.050
With compound X (1 μM)6.050
Very high substrate concentrations restore the uninhibited reaction rate in the presence of X.
Show answer & explanation

Answer: C

It binds only to the enzyme–substrate complex.: An uncompetitive inhibitor lowers both apparent Km and Vmax; here Km rises and Vmax is unchanged.

It binds an allosteric site on both E and ES equally.: A pure noncompetitive inhibitor lowers Vmax and leaves Km unchanged, the opposite of these data.

It competes with the substrate for the active site.: Correct. A competitive inhibitor raises apparent Km while Vmax is unchanged, because enough substrate outcompetes it (Cell and Molecular Biology: enzymology).

It irreversibly inactivates part of the enzyme.: Removing active enzyme lowers Vmax, and excess substrate could not restore the rate.

It raises the enzyme’s affinity for its substrate.: Inverts the meaning of Km: a higher Km means lower apparent affinity, not higher.

Competitive inhibition: Km up, Vmax unchanged (overcome by substrate). Noncompetitive: Vmax down, Km unchanged. Uncompetitive: both down.

General Chemistry

Question 3

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH° = −92 kJ. Which single change increases both the equilibrium yield of NH₃ and the value of the equilibrium constant K?

Show answer & explanation

Answer: D

Adding an iron catalyst: A catalyst speeds the approach to equilibrium but changes neither the yield nor K.

Compressing the mixture to a smaller volume: Compression shifts the reaction toward fewer moles of gas and raises the yield, but K depends only on temperature.

Adding more N₂ at constant volume: Adding a reactant shifts the equilibrium right but leaves K unchanged.

Lowering the temperature: Correct. For an exothermic reaction, lowering T shifts the equilibrium toward products and increases K itself (Chemical Equilibria).

Raising the temperature: For an exothermic reaction, heating lowers K and the equilibrium yield.

Concentration, pressure and catalysts can shift the position of equilibrium; only temperature changes K.

Organic Chemistry

Question 4

A compound with molecular formula C₄H₈O₂ shows a strong IR band at 1740 cm⁻¹ and the ¹H NMR spectrum summarized in the table. What is the compound?

¹H NMR data (CDCl₃)
Chemical shift (δ, ppm)SplittingIntegration
1.26triplet3H
2.04singlet3H
4.12quartet2H
Show answer & explanation

Answer: E

Butanoic acid: A carboxylic acid would show a broad O–H signal near 11–12 ppm and a broad IR O–H band.

Methyl propanoate: Same splitting pattern, but its singlet (OCH₃) would be near 3.7 ppm and its quartet (CH₂C=O) near 2.3 ppm.

Propyl formate: A formate would show a 1H singlet near 8 ppm for the H–C=O.

3-Hydroxybutan-2-one: Would show an exchangeable O–H signal and a 3H doublet, and its ketone IR band would be near 1715 cm⁻¹.

Ethyl acetate: Correct. The 4.12 ppm quartet is an OCH₂ next to a CH₃ (1.26 triplet), and the 2.04 singlet is a CH₃ on the C=O; 1740 cm⁻¹ is an ester carbonyl (Chemical & Physical Properties: spectroscopy).

In esters, the group on the oxygen side appears downfield (δ 3.7–4.1) and the group on the carbonyl side near δ 2.0–2.3.

Question 5

1-Methylcyclohexene is treated with (1) BH₃·THF and then (2) H₂O₂, NaOH. What is the major product?

Show answer & explanation

Answer: A

trans-2-Methylcyclohexan-1-ol: Correct. Hydroboration is anti-Markovnikov and syn: B and H add to the same face, with B on the less substituted carbon; oxidation keeps that configuration, so OH ends up trans to CH₃ (Chemical Synthesis: alkenes).

cis-2-Methylcyclohexan-1-ol: Right regiochemistry, wrong stereochemistry: syn addition puts H, not CH₃, cis to the new OH.

1-Methylcyclohexan-1-ol: Markovnikov hydration product, as from oxymercuration–demercuration.

2-Methylcyclohexan-1-one: Basic H₂O₂ replaces boron with OH; it does not oxidize the alcohol to a ketone.

Cyclohexylmethanol: OH on the methyl carbon would require a different alkene (methylenecyclohexane).

Hydroboration–oxidation: anti-Markovnikov, syn addition of H and OH, with no rearrangement.

Perceptual Ability

Question 6

Which figure can be formed from the pattern?

Pattern and figures A–D
ABCD
The pattern shows the outside surface of the figure.
Show answer & explanation

Answer: B

Figure A: The faces are placed correctly, but the stripe lies along the wrong edge; in the pattern it runs along the edge shared with the blank face.

Figure B: Correct. Folding the pattern so the markings stay on the outside gives a cube on which these marked faces meet exactly as shown: each mark keeps the same neighbours, and the corner or stripe marks point to the same neighbouring faces as in the pattern.

Figure C: The dark face and dot can meet at one corner, but here they run the opposite way around it: a mirror image of the folded pattern.

Figure D: The faces are placed correctly, but the stripe lies along the wrong edge; in the pattern it runs along the edge shared with the blank face.

Pick one marked face, note which faces touch each of its edges in the pattern (faces that touch when folded include faces two squares apart around a corner), and check that every figure keeps those neighbours and their clockwise order; a mirror image keeps the neighbours but reverses the order.

Question 7

In Figure 2, how many cubes have only four of their exposed sides painted?

Figure 2
Identical cubes cemented together and painted on every exposed side except the bottom. Hidden cubes are only those needed to support the cubes shown.
Show answer & explanation

Answer: D

  1. Count the stack: 13 cubes, including the hidden cubes that support the ones on top.
  2. For each cube, count the top (if nothing is on it) and each of the four sides not touching another cube; never the bottom.
  3. Tally: 2 painted: 4; 3 painted: 5; 4 painted: 4.
  4. Cubes with exactly 4 painted sides: 4.

1 cube: Too few: 1 misses 3 of the cubes with 4 painted sides (look again at hidden back faces and the supporting cubes).

2 cubes: Too few: 2 misses 2 of the cubes with 4 painted sides (look again at hidden back faces and the supporting cubes).

3 cubes: Too few: 3 misses 1 of the cubes with 4 painted sides (look again at hidden back faces and the supporting cubes).

4 cubes: Correct. 4 cubes have exactly 4 painted sides (full tally: 2 painted: 4; 3 painted: 5; 4 painted: 4; total 13 cubes).

5 cubes: Too many: 5 includes 1 cube that actually has a different number of painted sides.

Work cube by cube: a painted side is any exposed face except the bottom, including back faces you cannot see. Tally every cube once by its number of painted sides and check that the tally adds up to the total.

Question 8

Which pattern shows the holes when the paper is unfolded?

Folds, punch and unfolded patterns A–E
Fold 1Fold 2Fold 3ABCDE
Broken lines show the original position of the square sheet; solid lines show the folded paper. The black circle in the last fold panel is the punched hole.
Show answer & explanation

Answer: E

  1. The punch goes through 4 layers of paper, so the unfolded sheet has 4 holes.
  2. Open fold 3 (horizontal fold at the centre line): reflect across that fold line → 2 holes at row 4, column 2; row 1, column 2.
  3. Open fold 2 (vertical fold at x = 3): reflect across that fold line → 2 holes at row 4, column 2; row 1, column 2.
  4. Open fold 1 (vertical fold at x = 1): reflect across that fold line → 4 holes at row 4, column 2; row 4, column 1; row 1, column 2; row 1, column 1.

Pattern A: Unfolds the earlier folds but leaves out the reflection across fold 3, so the holes from the folded-over layer are missing.

Pattern B: Reflects the holes across the wrong line at fold 3 (vertical instead of horizontal).

Pattern C: Left–right mirror image of the correct pattern: the holes are reflected to the wrong side.

Pattern D: Unfolds correctly but from a punch one square away from where it is shown, so every hole is shifted.

Pattern E: Correct. Unfolding the last fold first and reflecting the hole across each fold line in turn gives 4 holes: row 1, column 1; row 1, column 2; row 4, column 1; row 4, column 2.

Unfold in reverse order: open the last fold first and reflect every hole across that fold line, then repeat for each earlier fold. The number of holes equals the number of paper layers at the punch.

Question 9

Rank the four angles from SMALLEST to LARGEST.

Angles 1–4
1234
Four interior angles, labelled 1 to 4.
Show answer & explanation

Answer: B

1-3-2-4: Swaps angles 4 (37.5°) and 2 (40°); angle 4 is 2.5° smaller, so it comes first.

1-3-4-2: Correct. The angles measure 1 = 33°, 2 = 40°, 3 = 36°, 4 = 37.5°, so from smallest to largest the order is 1-3-4-2.

1-4-3-2: Swaps the closest pair: angle 3 (36°) is smaller than angle 4 (37.5°) by only 1.5°.

3-1-4-2: Swaps angles 1 (33°) and 3 (36°); angle 1 is 3° smaller, so it comes first.

Judge each opening, not the arm length: compare the angles in pairs, settle the clearly smallest and largest first, then look hard at the closest pair.

Question 10

Choose the correct FRONT VIEW.

Two views given; choose the FRONT VIEW
TOPFRONTENDABCD
Views are drawn without perspective at the same scale: TOP upper left, FRONT lower left, END lower right. Dotted lines are edges hidden from that view.
Show answer & explanation

Answer: A

Figure A: Correct. The TOP view shows the wall along the back with a centred notch; the END view shows the wall's height and, as a dotted line, how deep the notch goes. Seen from the front, the notch is open to view, so its edges are solid.

Figure B: Shifts the notch off-centre; the TOP view shows it centred on the wall.

Figure C: Makes the notch too shallow; the dotted line in the END view puts the notch bottom lower.

Figure D: Draws the notch dotted, but nothing stands in front of the wall, so the notch edges are visible from the front.

Line the views up: TOP and FRONT share width, FRONT and END share height, TOP and END share depth (front edge at the bottom of the TOP view and at the left of the END view). Every feature must appear at matching positions, and an edge is dotted only when material hides it from that viewing direction.

Question 11

Which aperture would the object pass through?

Object and apertures A–E
ABCDE
The object (top) and five apertures drawn to the same scale.
Show answer & explanation

Answer: A

Aperture A: Correct. Looking straight down the length of the three-step staircase block, its outline is the shaded end face; Aperture A is that exact outline at full size, only turned or flipped (allowed: the object may be rotated before insertion or enter from its far end), so the object passes through end-first. Its long sides are too long to fit any opening.

Aperture B: Adds a notch in the bottom edge that the object does not have, so the solid base would hit it.

Aperture C: Proportions are off: the outline is wider but too short, so the full height of the tallest step cannot pass.

Aperture D: Two steps instead of three: the middle step is missing, so the corner of the second tread has no room.

Aperture E: The lowest riser is only half as tall as the object's, so the corner of the bottom step would hit.

Find the object's narrowest projection (here, its cross-section seen end-on), then reject apertures that have a step, notch or groove moved, resized, added or missing, or that have the wrong proportions. A mirrored outline is still acceptable, because the object can enter from its other end.

Reading Comprehension

Question 12

The primary purpose of the passage is to

Counting Neighbors: Quorum Sensing in Bacteria

1A single bacterium is tiny, and acting alone it can accomplish little. Some activities make sense only when many cells act together: producing light bright enough to be seen, releasing toxins in amounts large enough to overwhelm a host’s defenses, or building the slimy, layered communities called biofilms. For a long time it was assumed that bacteria simply responded one by one to conditions around them. It is now clear that many species can sense how many of their own kind are nearby and change their behavior accordingly. The process is called quorum sensing, after the minimum number of members a committee needs before it can make decisions.

2The phenomenon was first worked out in a marine bacterium, Vibrio fischeri, which lives freely in seawater and also in the light organs of certain squid and fish. Floating in the open ocean, where cells are widely scattered, the bacteria are dark. Packed at high density inside a light organ, they glow. The light is produced by enzymes encoded by a cluster of genes, and the puzzle was how scattered cells “knew” not to switch on those genes while crowded cells did.

3The answer lies in a small signaling molecule that each cell makes continuously and releases into its surroundings. The molecule, called an autoinducer, passes freely across cell membranes. When cells are few, it drifts away and its concentration stays low. When cells are many and confined in a small space, the autoinducer accumulates. Once it passes a threshold concentration, it binds to a receptor protein inside each cell, and the receptor–signal complex activates the genes for light production. Among the activated genes is the one encoding the enzyme that makes the autoinducer itself, so the signal amplifies its own production. This positive feedback ensures that once the population crosses the threshold, the whole group switches on together rather than gradually.

4The partnership between the bacterium and one of its hosts shows why such a switch is useful. The Hawaiian bobtail squid hunts at night in shallow water, where moonlight from above would cast a shadow visible to predators below. Light from the bacteria in its light organ, emitted downward, matches the light from above and erases the squid’s silhouette. The squid supplies nutrients to the bacteria; the bacteria supply camouflage. Each morning the squid expels most of the bacteria into the sea, and the remaining cells multiply during the day, reaching the density needed to glow again by nightfall.

5Since that early work, quorum sensing has been found in a wide range of bacteria. Many species use signals chemically related to the Vibrio autoinducer, while others use short protein fragments or entirely different molecules. Some bacteria produce more than one signal: one may be recognized only by members of the same species, while another, more widely shared signal may let different species gauge the total number of bacteria around them. Quorum sensing, in other words, can serve not only as a census of relatives but also as a rough reading of the whole community.

6The discovery has drawn strong interest from medicine, because several disease-causing bacteria use quorum sensing to coordinate their attacks. In some species, the genes for toxins and other harmful products stay silent while the population is small, when an early immune response could still eliminate it, and are switched on only when the population is large enough to overwhelm the host. Quorum sensing also helps regulate the formation of biofilms, which shield bacteria from antibiotics and immune cells and are a major cause of persistent infections on medical devices.

7This has suggested a different strategy for treating infections. Conventional antibiotics kill bacteria or stop them from growing, which creates intense pressure for resistant strains to emerge. Drugs that instead block quorum-sensing signals, a strategy sometimes called quorum quenching, would leave bacteria alive but keep them from acting as a group, in principle disarming them rather than destroying them. Because such drugs do not threaten bacterial survival directly, some researchers predicted that resistance to them would develop slowly. Laboratory studies have identified compounds that interfere with signaling, and some organisms, including certain plants and other bacteria, naturally produce substances that do the same.

8Whether this promise will be realized is uncertain. Bacteria that ignore or overproduce signals have already appeared in laboratory populations exposed to quorum-quenching compounds, suggesting that resistance is possible after all. And because different pathogens use different signals, no single drug is likely to work against all of them. Still, the study of quorum sensing has changed how microbiologists think about their subjects. Bacteria are no longer viewed only as isolated cells but also as members of communities that exchange chemical information and act on it together.

Show answer & explanation

Answer: E

argue that quorum-quenching drugs should replace conventional antibiotics: Overstatement: paragraph 8 calls their promise uncertain.

describe the bobtail squid’s use of bacterial light as camouflage: Too narrow: the squid appears in one paragraph as an example.

compare the chemical signals that different bacterial species use: Paragraph 5 mentions different signals but does not compare them in detail.

explain why some marine bacteria evolved the ability to make light: Light production is the opening example, not the main subject.

explain how bacteria sense their own numbers and why that matters: Correct. The passage explains the autoinducer mechanism (paragraphs 2–3), its wider occurrence (5), and its medical implications (6–8).

Choose the purpose that covers the mechanism, the examples and the implications together.

Question 13

Which statement best expresses the main idea of the passage?

Life Without Water

1Tardigrades, often called water bears, are microscopic animals rarely more than half a millimeter long. They have eight stubby legs tipped with claws, and they are found almost everywhere: in ocean sediments, in hot springs and, most famously, in the thin films of water that cling to mosses and lichens. That last habitat is an unstable one. A clump of moss on a sunny wall can be soaked in the morning and dry by afternoon, and most animals in such a place would simply die. Many tardigrade species instead enter a state called anhydrobiosis, literally “life without water,” in which they lose nearly all of their body water, show no measurable metabolism, and can remain for months or years before reviving within hours of being wetted again.

2The transition begins with a change in shape. As its surroundings dry, a tardigrade pulls in its head and legs and contracts into a compact, barrel-shaped form known as a tun. The tun has a much smaller surface area than the active animal, so it loses water more slowly. That slowness turns out to matter a great deal. In laboratory experiments, tardigrades dried gradually over many hours usually recover, whereas animals dried within minutes usually do not, even when both groups end up with the same final water content. The difference suggests that survival is not a passive consequence of becoming dry. Rather, the animal appears to need time to prepare, and researchers have spent several decades trying to work out what that preparation involves.

3To appreciate the problem, it helps to consider what water does inside a cell. Water is not simply a filler in which molecules float. Its hydrogen bonds help hold proteins in their folded, working shapes, and its interactions with the charged heads of membrane lipids keep membranes flexible and fluid. When most of the water is removed, proteins tend to unfold and stick to one another in useless clumps, membranes can become leaky or fuse together, and DNA strands can break. Any organism that survives drying must therefore protect its molecules against all of these forms of damage at once, and it must be able to undo whatever damage does occur once water returns.

4For many years the leading explanation centered on a sugar. Trehalose, a disaccharide made of two glucose units, accumulates to high levels in several organisms that tolerate drying, including the dormant cysts of brine shrimp, some nematode worms and baker’s yeast. Two complementary ideas were proposed to explain how trehalose protects cells. According to the water-replacement hypothesis, the sugar forms hydrogen bonds with proteins and membranes in place of the departing water, helping them keep their shapes. According to the vitrification hypothesis, the sugar turns the drying cytoplasm into a glass. A glass, in the physical sense, is a solid that keeps the disordered molecular arrangement of a liquid rather than the orderly lattice of a crystal. Because a glass is so viscous, molecular motion within it almost stops, and the chemical reactions that would ordinarily destroy unfolded proteins are slowed to a crawl.

5Tardigrades complicated this tidy picture. When researchers measured trehalose in a range of tardigrade species, they found that many produce only small amounts, and in some species it could not be detected at all. Whatever allows these animals to survive drying, then, trehalose could not be the whole story, and in many species it seemed to play little part. The search therefore shifted from measuring known protective molecules to asking a broader question: which genes do tardigrades switch on when they begin to dry out?

6The answer included a set of proteins found in no other group of animals. Most proteins fold into a characteristic three-dimensional shape, but these tardigrade-specific proteins are intrinsically disordered: in solution they have no fixed structure and behave more like flexible, shifting chains. They were grouped into families according to where they occur, for example cytosolic abundant heat soluble (CAHS) proteins in the cytoplasm and secreted abundant heat soluble (SAHS) proteins outside cells. Two kinds of experiment pointed to their importance. When the genes for CAHS proteins were silenced in tardigrades using RNA interference, fewer animals survived drying. And when the same genes were inserted into yeast and bacteria, which normally tolerate drying poorly, the modified microbes survived desiccation markedly better than unmodified ones. A single protein from a tardigrade, in other words, conferred part of the tardigrade’s resilience on an unrelated organism.

7How a protein without a fixed shape could protect other proteins is still being worked out. One proposal is that the disorder is itself the point. As water is lost and the proteins become more concentrated, CAHS proteins appear to gain structure and link together, forming a gel and eventually a glass-like solid that immobilizes the cell’s other contents. In this view, tardigrades achieve with proteins what brine shrimp achieve largely with sugar. Not all of the evidence fits neatly, however. Some studies have found that the degree of vitrification correlates well with how much protection a protein provides, while others have found protective proteins that form glasses no better than unprotective ones. Many researchers now suspect that the proteins work in several ways at once, perhaps also acting as molecular shields that keep unfolding proteins from clumping together.

8Drying is not the only insult tardigrades withstand. Some species survive doses of ionizing radiation hundreds of times greater than would kill a human. Part of the explanation may lie in another tardigrade-specific protein, named Dsup, for “damage suppressor.” Dsup binds to chromatin, the complex of DNA and proteins in the nucleus, and when it was produced in cultured human cells, those cells suffered roughly 40 percent fewer DNA breaks after exposure to X-rays than cells without it. Because drying itself breaks DNA, many biologists regard radiation tolerance not as an adaptation to radiation, which tardigrades rarely meet at such levels in nature, but as a by-product of the machinery that evolved to cope with desiccation. Notably, Dsup has been found in only some tardigrade lineages, a reminder that different species may have arrived at their toughness by different routes.

9These discoveries have attracted interest well beyond zoology. Many vaccines, enzymes and protein-based drugs must be kept refrigerated from factory to clinic, a requirement that is costly and difficult to meet in regions with unreliable electricity. If tardigrade proteins, alone or mixed with sugars, could stabilize such products in a dry state at room temperature, this “cold chain” might be shortened or avoided. Early laboratory studies have reported that CAHS proteins can help preserve the activity of a dried enzyme and of a blood-clotting protein. Even so, the step from a promising test-tube result to a reliable product is long. The tardigrade itself relies on a coordinated response involving many molecules, and it would be surprising if a single borrowed protein reproduced all of its protection.

10Perhaps the most lasting lesson of tardigrade research concerns proteins in general. For much of the twentieth century, biochemists assumed that a protein’s function depends on its folding into a precise shape, the way a key depends on its cut. Intrinsically disordered proteins, now known to be common in many organisms, challenge that assumption, and the tardigrade proteins offer an especially vivid example: molecules whose usefulness seems to emerge precisely as their surroundings change. A creature smaller than a grain of sand has thus helped to reshape a basic idea about how the molecules of life can work.

Show answer & explanation

Answer: E

Tardigrades survive drying chiefly by curling into a tun that slows the loss of water.: Too narrow: the tun is one detail of paragraph 2, which adds that slow drying matters because the animal needs time to prepare.

Trehalose is the main molecule that protects tardigrades from damage while they dry out.: Contradicted by paragraph 5: many tardigrades make little or no trehalose.

Disordered proteins, not sugars, are what protect most drying-tolerant organisms.: Overgeneralizes: paragraph 7 notes that brine shrimp rely largely on sugar.

Tardigrade proteins will probably soon make refrigerated vaccine storage unnecessary.: Overstatement: paragraph 9 calls the idea promising but says the path to a product is long.

Work on how tardigrades survive drying has revealed proteins that challenge old views.: Correct. The passage moves from the puzzle of anhydrobiosis (paragraphs 1–5) to the disordered proteins (6–9) and ends with their broader lesson about protein function (paragraph 10).

A main-idea answer must cover the whole passage: here, both the survival mechanism and the closing lesson about proteins.

Quantitative Reasoning

Question 14

The number of patients at Kendall Dental Clinic grew by 10% in the first year, 20% in the second year and 30% in the third year. By what percent did the number of patients increase over the three years?

Show answer & explanation

Answer: E

  1. Yearly growth factors: 1.10, 1.20 and 1.30.
  2. 1.10 × 1.20 = 1.32; 1.32 × 1.30 = 1.716.
  3. Increase = 1.716 − 1 = 0.716 = 71.6%.

20%: Averages the three yearly rates.

56%: Compounds only the second and third years: 1.2 × 1.3 = 1.56.

60%: Adds the three yearly percentages instead of compounding them.

62%: Compounds the first two years (1.32) but then adds the third-year 30% instead of multiplying by 1.3.

71.6%: Correct. The growth factors multiply: 1.1 × 1.2 × 1.3 = 1.716, an increase of 71.6%.

Successive percent changes multiply as factors (1 + r); they do not add.

Question 15

A 3-person committee is chosen from 5 students and 4 faculty members. The committee must include exactly one faculty member. How many different committees are possible?

Show answer & explanation

Answer: A

  1. Faculty: C(4, 1) = 4.
  2. Students: C(5, 2) = (5 × 4)/2 = 10.
  3. Committees = 4 × 10 = 40.

40: Correct. Choose 1 of 4 faculty (4 ways) and 2 of 5 students (C(5, 2) = 10 ways): 4 × 10 = 40.

74: Counts committees with at least one faculty member, C(9, 3) − C(5, 3) = 84 − 10, rather than exactly one.

80: Counts the two students in order (5 × 4) instead of as a combination.

84: Ignores the restriction: C(9, 3) = 84.

240: Treats the whole committee as ordered: 4 × 5 × 4 × 3.

Split a restricted selection into independent choices, multiply the combinations, and use combinations (not permutations) when order does not matter.

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