ESAT Practice Questions

Why the other options tempt you:

A: Multiplied by 10 instead of 1000

C: Multiplied by 100 instead of 1000

D: Multiplied by 10000 instead of 1000

E: Divided by 1000 (converted in the wrong direction)

F: Confused milligrams with micrograms, multiplying by 1000000

Why the other options tempt you:

A: Combined under one root, $\sqrt{48+27}=\sqrt{75}=5\sqrt{3}$ (added radicands instead of adding like surds)

B: Added the radicands and dropped the root, $48+27=75$

C: Added the like surds to $7\sqrt{3}$ then stripped the surd, keeping only the coefficient

E: Summed the coefficients correctly ($4+3=7$) but also summed the radicands ($3+3=6$) instead of keeping $\sqrt{3}$

F: Simplified to $4\sqrt{3}$ and $3\sqrt{3}$ then multiplied the coefficients ($4\times3$) instead of adding them

Why the other options tempt you:

A: Answered the middle (3-part) share, $3\times50$, the wrong sub-question

B: Reported the value of one part, $450\div9$

D: Miscounted the parts as 10, $450\times\tfrac{4}{10}=180$

E: Answered the smallest (2-part) share, $2\times50$, instead of the largest

F: Divided the total by the ratio number directly, $450\div4$

Why the other options tempt you:

A: Near-miss conversion, divided by 3 instead of 3.6 (truncated the factor, 72/3)

B: Divided by 360 instead of 3600 (72*1000/360), a factor-of-10 conversion slip

C: Stopped-at-intermediate: reported the average speed in km/h (108/1.5) without converting to m/s

E: Converted km/h to m/s by dividing by 6 instead of 3.6 (72/6)

F: Ignored the 30 minutes, used t=1 h (108000 m / 3600 s = 30)

Why the other options tempt you:

A: Applied $+1$ to only two of the three primes, leaving the largest exponent raw: $(3)(2+1)(1+1)$

B: Multiplied the exponents themselves ($3\times2\times1$) instead of $(e+1)$ each

C: Added the incremented exponents $(3+1)+(2+1)+(1+1)$ instead of multiplying them

E: Omitted the $+1$ rule for the exponent-1 prime, counting $5^{1}$ as a single way: $(3+1)(2+1)(1)$

F: Multiplied the prime bases ($2\times3\times5$) instead of the incremented exponents

Why the other options tempt you:

A: Reported the constant $k=9\times2^2=36$ as the answer

B: Direct/inverse swap + dropped power: treated $y\propto x$, $k=\tfrac92$, $y=4.5\times3=13.5$

C: Skipped finding $k$, used $y=\tfrac{9}{x^2}=\tfrac{9}{9}=1$

D: Treated $y\propto\tfrac1x$, $k=9\times2=18$, $y=\tfrac{18}{3}=6$

F: Treated $y\propto x^2$, $y=9\times(\tfrac32)^2=20.25$

Why the other options tempt you:

A: Reported $2xy=42$ without halving

B: Sign_error + stopped-early: $2xy=58-100=-42$, reported without halving

C: Expanded as $x^2-2xy+y^2$, $100=58-2xy\Rightarrow xy=-21$ (opposite-sign partner of the key)

D: Computed $(x-y)^2=x^2+y^2-2xy=58-42=16$ and reported it as $xy$

F: Halved $x^2+y^2$, taking $xy=\tfrac{58}{2}=29$

Why the other options tempt you:

A: Sign_error in factorising: took $(3u-1)(u+4)$ giving $u=\tfrac13,-4$, hence $x=\tfrac23,-\tfrac32$ and sum $\tfrac23-\tfrac32$

C: Reported the sum of the $u$-roots ($-\tfrac13+4$) by Vieta on $3u^2-11u-4$, before un-substituting back to $x$

D: Computed the product of the $x$-roots ($\tfrac13\times\tfrac52$) instead of their sum

E: Summed the $x$-values but forgot to divide by 2 when reversing $u=2x-1$

F: Halved the $u$-root sum $\tfrac{11}{3}$ (an incomplete reversal of $u=2x-1$)

Why the other options tempt you:

A: Approximated tan 30 degrees as 1/3 when applying it to the half-base BM = 6.

C: Inverted the sine rule, $BC=\tfrac{12\sin120^\circ}{\sin30^\circ}=12\sqrt3$, then took $h=BC\sin30^\circ$

D: Used cosine instead of sine for the height, $BC\cos30^\circ=4\sqrt3\times\tfrac{\sqrt3}{2}$

E: Approximated tan 30 degrees as 1/3 and applied it to the full base AB = 12 instead of the half-base.

F: Stopped at the sine rule and reported the side $BC$ itself instead of the perpendicular distance from C

Why the other options tempt you:

A: Missing-case (kept-only-one-case): counted only red-red as a same-colour outcome and dropped the blue-blue case, so P(same)=3/6 x 2/5 = 1/5 and P(different) = 1 - 1/5 = 4/5

B: Mixed/with-replacement model: drew the second counter with replacement, so P(same) = (3/6)²+(2/6)²+(1/6)² = 7/18 and P(different) = 1 - 7/18 = 11/18

C: Doubly-compounded slip (complement-of-the-wrong-event + missing-case): reported the same-colour probability AND omitted the blue-blue case, giving P(RR) = 3/6 x 2/5 = 1/5

E: Assume-symmetry / equally-likely-categories: treated the three colour categories as equally likely, so P(same) = 1/3 and P(different) = 1 - 1/3 = 2/3, ignoring the actual counts

F: Complement-of-the-wrong-event: computed P(same colour) = P(RR)+P(BB) = 4/15 and reported it instead of taking 1 - 4/15 for 'different colours'

Why the other options tempt you:

A: Sign_error compounded: $(-2)^2\to-4$ and $-3(-2)\to-6$, giving $-8-6+1=-13$

B: Squared $-2$ as $-4$, giving $2(-4)+6+1=-1$

D: Order-of-operations: squared $2x$ as $(2\cdot-2)^2=16$, giving $16+6+1=23$

E: Read $-3(-2)$ as $-6$, giving $8-6+1=3$

F: Dropped the $+1$ constant, $2(4)+6=14$

Why the other options tempt you:

A: Multiplied $a$ by $n$ as well, $5\times...$; used $5n=50$ confusing term index with $a\times n$

B: Gave $(n-1)d=9\times3=27$, the intermediate value, without adding $a$

C: Used $a+(n-2)d=5+8\times3=29$ (off-by-one the other way)

D: Used $a+nd=5+10\times3$ instead of $a+(n-1)d$

E: Computed $10\times3$ and forgot to add the first term $5$ then... took $(n-1)d=27$ but wrote $3\times10=30$

Why the other options tempt you:

B: Added coordinates instead of subtracting, $\dfrac{11+2}{4+1}=\dfrac{13}{5}$

C: Gave the perpendicular gradient $-1/m$ instead of the line's own gradient

D: Used the rise only, $y_2-y_1=11-2=9$, without dividing by the run

E: Computed $\dfrac{y_1-y_2}{x_2-x_1}=\dfrac{2-11}{3}=-3$, flipping the numerator order on one point only

F: Inverted the gradient as $\dfrac{x_2-x_1}{y_2-y_1}=\dfrac{3}{9}$

Why the other options tempt you:

B: Dropped the constant $-4$, solving $x^2=3x$ to get $x=0,3$, then substituted into $y=3x$

C: Found correct $x$-values but forgot to compute $y$, leaving $y=0$

D: Kept only one root and discarded $x=4$

E: Mis-factorised $x^2-3x-4$ as $(x+4)(x-1)$, roots $x=-4,1$, then substituted into $y=3x$

F: Substituted $x=-1$ into $y=3x$ but dropped the minus, $y=+3$

Why the other options tempt you:

A: Used the gradient of $AB$ itself ($\tfrac12$) through the midpoint instead of the perpendicular gradient

C: Point-slope sign slip on the midpoint $x$, $y-2=-2(x+4)$

D: Missing_constraint/wrong-point: used the perpendicular gradient through endpoint $A(2,1)$ instead of the midpoint, $y-1=-2(x-2)$

E: Dropped the minus on the perpendicular gradient, used $+2$ through the midpoint

F: Used the midpoint $y$-value $2$ directly as the intercept without point-slope

Why the other options tempt you:

A: Gave cos x = adjacent/hypotenuse (4/5) instead of sin x

B: Inverted the given tangent (adjacent over opposite, 4/3), the reciprocal pair to the 3/4 copy slip

C: Inverted the sine ratio, used hypotenuse over opposite (5/3)

D: Used hypotenuse over adjacent (5/4), inverted cosine

F: Copied tan x = 3/4 straight through as the answer

Why the other options tempt you:

B: Reports the sum of the $y$-roots, $\tfrac12+4=\tfrac92$ (equivalently $-b/a=9/2$), instead of the sum of the recovered $x$-values

C: Confuses the substituted root $y=2^{x}=\tfrac12$ with $x$ itself, never recovering $x$

D: Keeps only the root $y=\tfrac12$ (giving the negative solution $x=-1$) and discards $y=4$

E: Missing_constraint / domain_confusion: keeps only $y=4$ (giving $x=2$) and discards $y=\tfrac12$ as if invalid, so reports a single solution instead of the sum

F: Sums the absolute values $|x|$ of the two solutions, $2+1=3$, instead of the signed sum $2+(-1)$

Why the other options tempt you:

A: Computed a/(1-r) mixing branches (a=4 with r=1/2)

B: Used the rejected divergent root r=2 with the correct a=16 in a/(1-r)=16/(1-2)=-16

D: Quoted the first term a=16 as if it were the sum to infinity (stop-at-intermediate)

E: Summed just the first three terms (16+8+4) instead of using the sum-to-infinity formula

F: Used the rejected divergent root r=2 (with a=4) in a/(1-r)=4/(1-2)=-4 instead of taking |r|<1

Why the other options tempt you:

A: Keeps only the obtuse triangle $A=120^\circ$, $c=2$, discarding the acute solution.