Free DAT Exam Questions That Helps Candidates Pass

Annual depreciation = 20% of 500 = $100. After 3 years: 500 − (3 × 100) = $200. Correct answer is A, not B. Re-check carefully: straight-line reduces by fixed $100/year → after 3 years = $200. Thus Option A. DAT QR depreciation/percent problems test careful distinction between straight-line vs exponential decay.

x = 81 ÷ 9 = 9. Correct answer = C. DAT QR often includes simple linear equations.

SN1S_N1 rate depends primarily on carbocation stability because the rate-determining step is ionization to form a carbocation. Tertiary carbocations are strongly stabilized by hyperconjugation and inductive effects, making tert-butyl chloride exceptionally fast for SN1S_N1, especially in polar protic solvents like methanol that stabilize ions. Primary substrates (A, C, D) form unstable carbocations and thus are disfavored for SN1S_N1; they typically react via SN2S_N2 under appropriate conditions. While benzyl and allyl halides are also very reactive in SN1S_N1 due to resonance stabilization, none are listed here. Methanol enhances SN1S_N1 by solvating the leaving group and the carbocation intermediate. Additionally, a good leaving group (Cl⁻) supports ionization. Therefore, tertiary > secondary » primary ≈ methyl for SN1S_N1. Recognizing solvent effects and carbocation stability trends is a staple of DAT organic chemistry: tertiary substrates in protic media generally dominate SN1S_N1 reactivity.

Starch (amylose and amylopectin) stores glucose in plants. Cellulose (A) is structural, glycogen (B) is storage in animals, and chitin (D) is structural in fungi/exoskeletons. On the DAT, polysaccharide functions are common: starch = plant storage; glycogen = animal storage; cellulose/chitin = structure.

Peroxisomes contain catalase and peroxidase enzymes that neutralize hydrogen peroxide (H₂O₂) into water and oxygen. Lysosomes digest macromolecules, mitochondria produce ATP, and Golgi modifies proteins. DAT biology often tests peroxisomes’ detoxifying role.