IPMAT Indore 2023 · Question Paper & Solutions

IIM Indore

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Worked solutions — IPMAT Indore 2023

Every question, solved step by step. Each answer is checked against the official key printed with the paper — 88 of 90 solutions reach it. Try each question before opening it.

1 answer corrects the official key, where the working settles it. The key's own answer is shown alongside each one.

The other 1 question is flagged. The official key is always shown; the flag says what the problem is.

1 no option is correct — the correct answer is not among the printed options

Quantitative Ability (Short Answer) 15 Quantitative Ability (MCQ) 30 Verbal Ability 45

Quantitative Ability (Short Answer)

Q1 If f(1) = 1 and f(n) = 3n − f(n−1) for n > 1, find f(2023).

Answer: 3034

From the rule, $f (n) + f (n - 1) = 3 n$ . Compute a few values: $f (1) = 1, f (2) = 5, f (3) = 4, f (4) = 8, f (5) = 7$ .

Subtracting consecutive relations: $f (n + 1) - f (n - 1) = 3 (n + 1) - 3 n = 3$ , so odd-indexed terms rise by 3 each step:
$f (2 k + 1) = 1 + 3 k$

For $2023 = 2 k + 1$ , $k = 1011$ , so $f (2023) = 1 + 3033 = 3034$ .

Q2 Amisha (20 days alone) works 4 days, is sick 4 days, then at half rate finishes in 12 days with a co-worker. Co-worker's solo time?

Answer: 24

Amisha's rate is $\frac{1}{20}$ per day. In 4 days she does $\frac{4}{20} = \frac{1}{5}$ ; remaining work $= \frac{4}{5}$ .

From day 9, for 12 days she works at $\frac{1}{40}$ per day: $\frac{12}{40} = \frac{3}{10}$ .

The co-worker does $\frac{4}{5} - \frac{3}{10} = \frac{1}{2}$ of the task in 12 days, so alone he needs $2 \times 12 = 24$ days.

Q3 f(n) = 1+2+…+(n+1) and g(n) = Σ 1/f(k) for k = 1…n. Least n with g(n) > 99/100?

Answer: 199

$f (k) = \frac{( k + 1 ) ( k + 2 )}{2}$ , so
$\frac{1}{f ( k )} = 2 (\frac{1}{k + 1} - \frac{1}{k + 2})$

The sum telescopes: $g (n) = 2 (\frac{1}{2} - \frac{1}{n + 2}) = 1 - \frac{2}{n + 2}$ .

$g (n) > \frac{99}{100} ⟺ \frac{2}{n + 2} < \frac{1}{100} ⟺ n + 2 > 200 ⟺ n > 198$ .

Least value: $n = 199$ .

Q4 Two-candidate election: 15% didn't vote, 50 invalid votes, winner got 44% of the list and won by 200. Voters on the list?

Answer: 5000

Let the list have $N$ voters. Votes cast $= 0.85 N$ ; valid votes $= 0.85 N - 50$ .

Winner $= 0.44 N$ , loser $= 0.44 N - 200$ . Valid votes are shared by the two:
$0.88 N - 200 = 0.85 N - 50 ⟹ 0.03 N = 150 ⟹ N = 5000$

Q5 Number of positive integer solutions of 21 ≤ a + b + c ≤ 25.

Answer: 1160

For a fixed sum $s$ , the number of positive integer solutions of $a + b + c = s$ is $(2 s - 1)$ .

$s = 21 \sum 25 (2 s - 1) = (2 20) + (2 21) + (2 22) + (2 23) + (2 24)$
$= 190 + 210 + 231 + 253 + 276 = 1160$

Q6 Three consecutive coefficients in the expansion of (x + y)^n are in the ratio 1 : 9 : 63. Find n.

Answer: 39

Let the coefficients be $(r - 1 n), (r n), (r + 1 n)$ .

$\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r} = 9 ⟹ n + 1 = 10 r$

$\frac{( r + 1 n )}{( r n )} = \frac{n - r}{r + 1} = 7 ⟹ n = 8 r + 7$

So $10 r - 1 = 8 r + 7 ⟹ r = 4$ , and $n = 39$ .

Q7 Polynomial 4x¹⁰ − x⁹ + … + cx⁶ + … + 6x − 2 leaves remainder 2 when divided by x−1. Find c+6.

Answer: 5

By the remainder theorem the remainder is $P (1)$ :
$P (1) = 4 - 1 + 3 - 5 + c + 2 - 1 + 1 - 4 + 6 - 2 = c + 3$

$c + 3 = 2 ⟹ c = - 1$ , so $c + 6 = 5$ .

Q8 Product of the roots of log₂(2^{(log₂x)²}) − 5 log₂x + 6 = 0.

Answer: 32

$lo g_{2} 2^{(l o g_{2} x)^{2}} = (lo g_{2} x)^{2}$ . Let $t = lo g_{2} x$ :
$t^{2} - 5 t + 6 = 0 ⟹ t = 2 or 3$

So $x = 4$ or $x = 8$ , and the product of the roots is $32$ .

Q9 Remainder when 1! + 2! + 3! + … + 95! is divided by 15.

Answer: 3

For $n \geq 5$ , $n!$ contains both 3 and 5, so it is divisible by 15.

Only $1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$ matters, and $33 = 2 \times 15 + 3$ .

Remainder $= 3$ .

Q10 Four-gear car (v₄ = 5v₁, v₃ = 2v₂) takes 585 min with 25 km per gear. Time for 4, 4, 32, 60 km?

Answer: 312

Let the first and second gear speeds be $v_{1}, v_{2}$ . Then $v_{4} = 5 v_{1}$ , and since second gear takes twice as long as third over the same distance, $v_{3} = 2 v_{2}$ .

Equal distances (25 km each):
$\frac{25}{v _{1}} + \frac{25}{v _{2}} + \frac{25}{2 v _{2}} + \frac{25}{5 v _{1}} = \frac{30}{v _{1}} + \frac{37.5}{v _{2}} = 585$

New distances:
$\frac{4}{v _{1}} + \frac{4}{v _{2}} + \frac{32}{2 v _{2}} + \frac{60}{5 v _{1}} = \frac{16}{v _{1}} + \frac{20}{v _{2}}$

Notice $\frac{30}{v _{1}} + \frac{37.5}{v _{2}} = \frac{15}{8} (\frac{16}{v _{1}} + \frac{20}{v _{2}})$ , so the required time is $585 \times \frac{8}{15} = 312$ minutes. (Two unknowns, but only this combination is needed.)

Q11 Positive integers with a+b+c+d = 2023, a:b = 2:5 and c:d = 5:2. Find the maximum of a+c.

Answer: 1442

Write $a = 2 p, b = 5 p, c = 5 q, d = 2 q$ with $p, q$ positive integers.

$7 p + 7 q = 2023 ⟹ p + q = 289$ .

$a + c = 2 p + 5 q = 2 (289) + 3 q$ , which is largest when $q$ is largest: $q = 288, p = 1$ .

$a + c = 2 + 1440 = 1442$ .

Q12 A = (−2, 0), B = (2, 0). Count points C on x² + y² = 4 for which triangle ABC has integer area.

Answer: 14

$A B = 4$ and the height from $C = (x, y)$ is $∣ y ∣$ , so area $= \frac{1}{2} \cdot 4 \cdot ∣ y ∣ = 2∣ y ∣$ . On the circle, $0 \leq ∣ y ∣ \leq 2$ , so area $\in [0, 4]$ .

Area 0 means $C = A$ or $B$ : no triangle, excluded.
Area 1, 2, 3: $∣ y ∣ = \frac{1}{2}, 1, \frac{3}{2}$ . Each gives 4 points (two values of $y$ , two of $x$ ): $3 \times 4 = 12$ .
Area 4: $∣ y ∣ = 2$ , i.e. $(0, \pm 2)$ : 2 points.

Total $= 12 + 2 = 14$ .

Q13 Length of the segment joining the two intersection points of y = 4970 − |x| and y = x².

Answer: 140

For $x \geq 0$ : $x^{2} = 4970 - x ⟹ x^{2} + x - 4970 = 0$ .

$x = \frac{- 1 + 1 + 19880}{2} = \frac{- 1 + 141}{2} = 70$

By symmetry the other point is at $x = - 70$ ; both have $y = 4900$ . The segment is horizontal, with length $70 - (- 70) = 140$ .

Q14 Ankita deposits yearly income (10,000, rising by 10,000) at year-end at simple interest starting 0.5%, rising 0.5% yearly. Total interest in five years?

Answer: 2500

Deposits are made at the end of each year; interest counts until the end of year 5.

End of year 1: 10,000 at 0.5% for 4 years = 200
End of year 2: 20,000 at 1% for 3 years = 600
End of year 3: 30,000 at 1.5% for 2 years = 900
End of year 4: 40,000 at 2% for 1 year = 800
End of year 5: 50,000 at 2.5% for 0 years = 0

Total interest $= 200 + 600 + 900 + 800 = 2500$ .

Q15 Four equal groups; players meet own group once and the other three groups 2, 3, 4 times. Over 1000 matches: minimum players per group?

Answer: 8

Let each group have $n$ players.

Within groups: $4 (2 n) = 2 n (n - 1)$ matches.

Between groups: there are 6 pairs of groups, and every group must see the multiplicities 2, 3 and 4 exactly once. This means splitting the 6 pairs into three "matchings" of 2 pairs each (e.g. G1–G2 and G3–G4 at 2 times, G1–G3 and G2–G4 at 3, G1–G4 and G2–G3 at 4). Each pair of groups has $n^{2}$ player pairs:
$n^{2} (2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4) = 18 n^{2}$

Total $= 2 n^{2} - 2 n + 18 n^{2} = 20 n^{2} - 2 n$ .

$n = 7$ : $980 - 14 = 966 \leq 1000$
$n = 8$ : $1280 - 16 = 1264 > 1000$

Minimum $n = 8$ .

Quantitative Ability (MCQ)

Q16 If log_{cos x}(sin x) + log_{sin x}(cos x) = 2, find x.

Answer: A —

$2 nπ + \frac{π}{4}$ , $n$ an integer

Let $t = lo g_{c o s x} (sin x)$ . Then $lo g_{s i n x} (cos x) = \frac{1}{t}$ , and
$t + \frac{1}{t} = 2 ⟹ (t - 1)^{2} = 0 ⟹ t = 1$

So $sin x = cos x$ . Both must be valid log bases/arguments: positive and not equal to 1. That forces $x$ into the first quadrant, i.e. $x = \frac{π}{4}$ plus full turns.

$x = 2 nπ + \frac{π}{4}$ .

Option B ( $nπ + \frac{π}{4}$ ) also includes $\frac{5 π}{4}$ , where $sin x = cos x < 0$ , so the logarithms are undefined.

Q17 Walk 32 m east, turn right 16 m, turn right 8 m, and so on forever. Final distance from the start?

Answer: B —

$64/ 5$ meters

Right turns cycle E → S → W → N, and each leg is half the previous one.

East–west: $32 - 8 + 2 - \dots = \frac{32}{1 + \frac{1}{4}} = \frac{128}{5}$

North–south: $- 16 + 4 - 1 + \dots = \frac{- 16}{1 + \frac{1}{4}} = - \frac{64}{5}$

$d = (\frac{128}{5})^{2} + (\frac{64}{5})^{2} = \frac{64}{5} 4 + 1 = \frac{64 5}{5} = \frac{64}{5} m$

Q18 What does x² + y² − 2x − 4y + 5 = 0 represent?

Answer: C —

a point

Complete the squares:
$(x - 1)^{2} + (y - 2)^{2} = 1 + 4 - 5 = 0$

A sum of squares is zero only when both are zero, so the only solution is $(1, 2)$ . It is a single point (a circle of radius 0).

Q19 A ₹10 lakh gold ball is melted into beads of one-fifth the radius, sold at 20% discount for 20% total profit. List price per bead?

Answer: A —

12000

Volume scales as $r^{3}$ : each bead has $(\frac{1}{5})^{3} = \frac{1}{125}$ of the volume, so there are 125 beads.

Total selling price $= 1.2 \times 10, 00, 000 = 12, 00, 000$ . Per bead: $\frac{12 , 00 , 000}{125} = 9600$ .

Selling price is 80% of list price: $LP = \frac{9600}{0.8} = 12, 000$ .

Q20 A helicopter flies the four 100 km sides of a square at 100, 200, 300, 400 km/h. Average speed?

Answer: C —

192 kmph

Total distance $= 400$ km.

Total time $= \frac{100}{100} + \frac{100}{200} + \frac{100}{300} + \frac{100}{400} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12}$ h.

Average speed $= \frac{400}{25/12} = 192$ km/h.

Q21 Probability that a random positive divisor of 10^2023 is a multiple of 10^2001.

Answer: D —

$\frac{2 3 ^{2}}{202 4 ^{2}}$

$1 0^{2023} = 2^{2023} 5^{2023}$ . Its divisors are $2^{a} 5^{b}$ with $0 \leq a, b \leq 2023$ : $202 4^{2}$ divisors.

A multiple of $1 0^{2001}$ needs $2001 \leq a, b \leq 2023$ : $23$ choices each, so $2 3^{2}$ divisors.

$P = \frac{2 3 ^{2}}{202 4 ^{2}}$

Q22 Triangle ABC with AB:BC:CA = 2:4:3; D is midpoint of BC, AM the altitude. Find BM : AD.

Answer: A —

$11 : 410$

Take $A B = 2, B C = 4, C A = 3$ .

BM (projection of $A B$ on $B C$ ): $B M = \frac{A B ^{2} + B C ^{2} - C A ^{2}}{2 B C} = \frac{4 + 16 - 9}{8} = \frac{11}{8}$ .

AD (median, Apollonius): $A D^{2} = \frac{2 A B ^{2} + 2 C A ^{2} - B C ^{2}}{4} = \frac{8 + 18 - 16}{4} = \frac{10}{4}$ , so $A D = \frac{10}{2}$ .

$B M : A D = \frac{11}{8} : \frac{10}{2} = 11 : 410$

Q23 A = [[1, 2], [3, a]] and det(A³ − 3A² − 5A) = 0. One possible value of a?

Answer: C —

$A^{3} - 3 A^{2} - 5 A = A (A^{2} - 3 A - 5 I)$ , so the determinant is $det A \cdot det (A^{2} - 3 A - 5 I)$ .

$det A = a - 6$ , which is 0 when $a = 6$ . So $a = 6$ works.

Check the other options with Cayley–Hamilton, $A^{2} = (1 + a) A - (a - 6) I$ :
- $a = 5$ : $A^{2} - 3 A - 5 I = 3 A - 4 I = [- 1 9 611]$ , determinant $- 65$ ; and $det A = - 1$ . Not zero.
- $a = 4$ : $2 A - 3 I = [- 1 6 45]$ , determinant $- 29$ ; $det A = - 2$ . Not zero.
- $a = 1$ : $- A$ , determinant $det A = - 5$ . Not zero.

So $a = 6$ .

Q24 Which pair of lines are both tangents to x² + y² − 6x + 4y − 12 = 0?

Answer: B —

$4 x + 3 y + 19 = 0, 4 x + 3 y - 31 = 0$

Centre $(3, - 2)$ , radius $9 + 4 + 12 = 5$ .

Every option has lines $4 x + 3 y + c = 0$ . Tangency needs distance from the centre $= 5$ :
$\frac{∣12 - 6 + c ∣}{5} = 5 ⟹ ∣6 + c ∣ = 25 ⟹ c = 19 or c = - 31$

So the tangents are $4 x + 3 y + 19 = 0$ and $4 x + 3 y - 31 = 0$ .

Q25 a₁, a₂, a₃ distinct, in GP; a₁x²+2a₂x+a₃ = 0 and b₁x²+2b₂x+b₃ = 0 share a root. What must hold?

Answer: B —

$\frac{b _{1}}{a _{1}}, \frac{b _{2}}{a _{2}}, \frac{b _{3}}{a _{3}}$ are in arithmetic progression

(The stem's " $a_{1} a_{2}, a_{2}$ " is a typo for $a_{1}, a_{2}, a_{3}$ .)

Since $a_{2}^{2} = a_{1} a_{3}$ , the first equation has discriminant $4 a_{2}^{2} - 4 a_{1} a_{3} = 0$ , so its only root is $x = - \frac{a _{2}}{a _{1}}$ . This must satisfy the second equation:
$b_{1} \frac{a _{2}^{2}}{a _{1}^{2}} - 2 b_{2} \frac{a _{2}}{a _{1}} + b_{3} = 0$

Use $a_{2}^{2} = a_{1} a_{3}$ in the first term, then divide by $a_{3}$ :
$\frac{b _{1}}{a _{1}} - \frac{2 b _{2} a _{2}}{a _{1} a _{3}} + \frac{b _{3}}{a _{3}} = 0 ⟹ \frac{b _{1}}{a _{1}} + \frac{b _{3}}{a _{3}} = \frac{2 b _{2}}{a _{2}}$

So $\frac{b _{1}}{a _{1}}, \frac{b _{2}}{a _{2}}, \frac{b _{3}}{a _{3}}$ are in AP.

Q26 A rabbit climbs a 10-step staircase one or two steps at a time. Number of ways to reach the top?

Answer: D —

Let $w_{n}$ be the number of ways to reach step $n$ . The last move is 1 or 2 steps, so $w_{n} = w_{n - 1} + w_{n - 2}$ with $w_{1} = 1, w_{2} = 2$ .

$1, 2, 3, 5, 8, 13, 21, 34, 55, 89$

$w_{10} = 89$ .

Q27 Minimum number of fair-coin tosses so that P(at least one head) exceeds 0.8.

Answer: D —

$P (at least one head) = 1 - (\frac{1}{2})^{n} > 0.8 ⟺ (\frac{1}{2})^{n} < 0.2$ .

$n = 2$ : $0.25$ (not enough). $n = 3$ : $0.125 < 0.2$ .

Minimum $n = 3$ .

Q28 Ways to place 8 rooks on an 8 × 8 board with no two in the same row or column.

Answer: C —

Place one rook per row. Row 1 has 8 column choices, row 2 has 7 remaining, and so on:
$8 \times 7 \times \dots \times 1 = 8!$

Q29 a, b, c > 1, n > 0, n ≠ 1, with log_n(log₂a) = 1, log_n(log₂b) = 2, log_n(log₂c) = 3. Which relation holds?

Answer: B —

$(a^{n} + b)^{n} = a c$

$lo g_{2} a = n, lo g_{2} b = n^{2}, lo g_{2} c = n^{3}$ , so $a = 2^{n}, b = 2^{n^{2}}, c = 2^{n^{3}}$ .

Then $a^{n} = 2^{n^{2}} = b$ , so $a^{n} + b = 2 b = 2^{n^{2} + 1}$ and
$(a^{n} + b)^{n} = 2^{n^{3} + n} = 2^{n} \cdot 2^{n^{3}} = a c$

Checks on the others with $n = 2$ ( $a = 4, b = 16, c = 256$ ): A gives $20 \neq = 256$ ; C gives $144 \neq = 240$ ; D gives $16 + 256 \neq = 65536$ .

Q30 CI minus SI for 3 years at 10% p.a. is ₹527. Find the principal.

Answer: C —

17000

$CI = P (1. 1^{3} - 1) = 0.331 P$ and $SI = 0.3 P$ .

$0.031 P = 527 ⟹ P = 17, 000$ .

Q31 The harmonic mean of the roots of (5 + √2)x² − bx + 8 + 2√5 = 0 is 4. Find b.

Answer: D —

$4 + 5$

For roots $α, β$ : $HM = \frac{2 α β}{α + β}$ .

$α β = \frac{8 + 2 5}{5 + 2}$ and $α + β = \frac{b}{5 + 2}$ , so
$HM = \frac{2 ( 8 + 2 5 )}{b} = 4 ⟹ b = 4 + 5$

Q32 For natural n, sum of all x satisfying 2[x] = x + n{x}, where [x] is the floor and {x} the fractional part.

Answer: C —

$\frac{n ( n + 2 )}{2}$

Write $x = I + f$ with $I = [x]$ and $0 \leq f < 1$ . Then
$2 I = I + f + n f ⟹ I = (n + 1) f$

$I$ is an integer and $0 \leq (n + 1) f < n + 1$ , so $I = k$ and $f = \frac{k}{n + 1}$ for $k = 0, 1, \dots, n$ .

$x = k + \frac{k}{n + 1} = \frac{k ( n + 2 )}{n + 1}$ . Sum:
$\frac{n + 2}{n + 1} k = 0 \sum n k = \frac{n + 2}{n + 1} \cdot \frac{n ( n + 1 )}{2} = \frac{n ( n + 2 )}{2}$

Q33 If (a+b)/(b+c) = (c+d)/(d+a), which statement is always true? Key corrected

Answer: C —

$a = c$ , or $a + b + c + d = 0$

Official key: D — $a = c$, and $b = d$

Cross-multiply: $(a + b) (d + a) = (c + d) (b + c)$ .
$a^{2} + a d + ab + b d = c^{2} + b c + c d + b d$
$(a^{2} - c^{2}) + b (a - c) + d (a - c) = 0 ⟹ (a - c) (a + b + c + d) = 0$

So the condition means exactly: $a = c$ or $a + b + c + d = 0$ , which is option C.

The official key gives D ( $a = c$ and $b = d$ ). That is sufficient but not necessary. Counterexample: $a = c = 1, b = 2, d = 3$ gives $\frac{3}{3} = \frac{4}{4}$ , yet $b \neq = d$ . So D is not "always true" under the given condition.

Why we differ from the key:

The condition reduces to (a − c)(a + b + c + d) = 0, which is option C. The official key's "a = c and b = d" is sufficient but not necessary: a = c = 1, b = 2, d = 3 satisfies the condition with b ≠ d.

Q34 Solve x²(x+1) / ((x−1)(2x+1)³) > 0.

Answer: A —

$(- 1, - \frac{1}{2}) \cup (1, + \infty)$

$x^{2} > 0$ for $x \neq = 0$ (and $x = 0$ gives 0, not $> 0$ ). $(2 x + 1)^{3}$ has the same sign as $2 x + 1$ . So we need
$\frac{x + 1}{( x - 1 ) ( 2 x + 1 )} > 0, x \neq = 0$

Critical points $- 1, - \frac{1}{2}, 1$ :
- $x < - 1$ : $\frac{-}{( - ) ( - )} < 0$
- $- 1 < x < - \frac{1}{2}$ : $\frac{+}{( - ) ( - )} > 0$ ✓
- $- \frac{1}{2} < x < 1$ : $\frac{+}{( - ) ( + )} < 0$
- $x > 1$ : $\frac{+}{( + ) ( + )} > 0$ ✓

$x = 0$ is not in either interval anyway. Solution: $(- 1, - \frac{1}{2}) \cup (1, \infty)$ .

Q35 Five players, round robin, no draws, 1 point per win. Which score sequence is impossible?

Answer: C —

4, 4, 1, 1, 0

There are $(2 5) = 10$ games, so scores must add to 10. All four options do.

In C, two players have 4 points, meaning each won all 4 games. But they played each other, and only one can win that game. So C is impossible.

The others are achievable, e.g. B: arrange players in a circle and let each beat the next two.

Q36 If cos α + cos β = 1, find the maximum of sin α − sin β.

Answer: D —

$3$

Let $S = sin α - sin β$ . Square and add to $(cos α + cos β)^{2} = 1$ :
$S^{2} + 1 = 2 + 2 (cos α cos β - sin α sin β) = 2 + 2 cos (α + β)$

So $S^{2} = 1 + 2 cos (α + β) \leq 3$ , i.e. $S \leq 3$ .

It is attained at $α = \frac{π}{3}, β = - \frac{π}{3}$ : $cos α + cos β = 1$ and $sin α - sin β = 3$ .

Q37 P(x) leaves remainder 2 on division by x−1 and 1 by x−2. Remainder on division by (x−1)(x−2)?

Answer: A —

$3 - x$

The remainder is linear, $R (x) = p x + q$ , with $R (1) = P (1) = 2$ and $R (2) = P (2) = 1$ .

$p + q = 2, 2 p + q = 1 ⟹ p = - 1, q = 3$ .

$R (x) = 3 - x$ .

Q38 120 students: 80 Science, 40 Commerce; 70 support Mumbai Indians. Number of Science MI supporters?

Answer: A —

30 or more

At most all 40 Commerce students support MI, so at least $70 - 40 = 30$ MI supporters are Science students. At most $70$ can be (all MI fans being Science students is allowed, since $70 \leq 80$ ).

Range: $30$ to $70$ , i.e. 30 or more. B, C and D all allow values below 30.

Q39 The unit digit of p³ is 4 (p a positive integer). Possible unit digits of (p + 3)³? No option is correct

Official key: A —

1, 3

Cube unit digits by last digit: $0 \to 0, 1 \to 1, 2 \to 8, 3 \to 7, 4 \to 4, 5 \to 5, 6 \to 6, 7 \to 3, 8 \to 2, 9 \to 9$ .

The only last digit whose cube ends in 4 is 4, so $p$ ends in 4.

Then $p + 3$ ends in 7, and $7^{3} = 343$ ends in 3. The only possible unit digit is 3.

Option A (1, 3) is the only option that contains 3, which is why the key picks it. But a unit digit of 1 cannot occur.

Why no option fits:

p must end in 4, so p + 3 ends in 7 and (p+3)³ ends in 3. Only the unit digit 3 is possible, and every option pads it with a second digit that cannot occur.

Q40 Probability that a random three-digit number is divisible by neither 3 nor 4.

Answer: D —

$\frac{1}{2}$

There are 900 three-digit numbers (100–999).

Divisible by 3: $333 - 33 = 300$
Divisible by 4: $249 - 24 = 225$
Divisible by 12: $83 - 8 = 75$

Divisible by 3 or 4: $300 + 225 - 75 = 450$ . Neither: $900 - 450 = 450$ .

$P = \frac{450}{900} = \frac{1}{2}$ .

Q41 Drug-organism caselet: drug D works on which organism(s)?

Answer: D —

P and Q

Deduction for the caselet. Given: A works on R; B and C work on Q.

1. Wherever A works, B works, so B works on R. With Q, B already has 2 organisms: B = {Q, R}. So A can only work on Q or R.
2. C works on Q, so D works on Q.
3. P needs at least 2 drugs, and A and B can't work on P, so they must come from C, D, E. D and E can't share an organism, and C on P forces D on P. The only valid pair is P = {C, D}. So C = {P, Q} and D = {P, Q}.
4. E can't be on P (not chosen) or Q (D is there), so E = {R}.
5. Q already has B, C, D (maximum 3), so A is not on Q: A = {R}.