IIM Bangalore UG and BBA DBE · Paper 1

20 MCQs · Quantitative Ability & Data Interpretation 10, Logical & Analytical Reasoning 5, Verbal Ability & Reading Comprehension 5

In the pattern of IIM Bangalore's on-campus UG test, which also prepares you for the online BBA DBE.

Tap an option to answer

Section 1 · Quantitative Ability & Data Interpretation

Q1Arithmetic › AveragesEasy
Asked in BBA DBE 2024
The average of three numbers is 19 while the average of two of these numbers is 21.25. What is the value of the third number?
Correct option B14.5
Sum of all three numbers $= 3 \times 19 = 57$ .
Sum of the two numbers $= 2 \times 21.25 = 42.5$ .
Third number $= 57 - 42.5 = 14.5$ .
The Fast Way
Use deviations from the average of 19.
The two numbers average $21.25$ , which is $2.25$ above 19, so together they carry a surplus of $2 \times 2.25 = 4.5$ .
The third must sit $4.5$ below 19 to balance it: $19 - 4.5 = 14.5$ .
Q2Number System › Unit DigitEasy
Asked in BBA DBE 2024
The digit in the units' place of $13^{2003} - 7^{291}$ is ___.
Correct option D4
Only units digits matter. Powers of 3 cycle $3, 9, 7, 1$ and powers of 7 cycle $7, 9, 3, 1$ (both repeat every 4).
$2003 = 4 \times 500 + 3$ , so $13^{2003}$ ends in the 3rd term of the 3-cycle: $7$ .
$291 = 4 \times 72 + 3$ , so $7^{291}$ ends in the 3rd term of the 7-cycle: $3$ .
$13^{2003}$ is far bigger, and $7 > 3$ so no borrowing: the difference ends in $7 - 3 = 4$ .
The Fast Way
Units digit decides it: take each exponent's remainder on division by 4.
2003 and 291 both leave remainder 3 (check the last two digits, 03 and 91), so use cubes: $3^3 = 27$ ends in 7 and $7^3 = 343$ ends in 3.
$13^{2003}$ is the bigger number and $7 > 3$ , so the difference ends in $7 - 3 = 4$ .
Q3Modern Math › Matrices & DeterminantsEasy
$A$ is a square matrix of order 3 with $|A| = -2$ . What is the value of $|3A^{T}|$ ?
Correct option B $-54$
For a square matrix of order $n$ , $|kA| = k^n|A|$ , and a matrix and its transpose have the same determinant.
Here $n = 3$ , so $|3A^{T}| = 3^3|A^{T}| = 27|A|$ .
$27 \times (-2) = -54$ .
Trap: $3 \times (-2) = -6$ forgets that the 3 multiplies every one of the 3 rows.
The Fast Way
Use $|kA| = k^n|A|$ ; transposing changes nothing.
$k^n = 3^3 = 27$ , and $27 \times (-2) = -54$ .
So the value is $-54$ .
Q4Geometry › CirclesMedium
Asked in BBA DBE 2024
PQRS is a cyclic quadrilateral whose diagonals PR and QS intersect at the point T. $\angle SQR$ measures $72^\circ$ , and $\angle QPR$ measures $28^\circ$ . What is the measure of $\angle QRS$ ?
Correct option A $80^\circ$
Angles standing on the same arc of a circle are equal. $\angle SQR$ and $\angle SPR$ both stand on arc SR, so $\angle SPR = 72^\circ$ .
So the whole angle at P is $\angle QPS = \angle QPR + \angle RPS = 28^\circ + 72^\circ = 100^\circ$ .
Opposite angles of a cyclic quadrilateral add up to $180^\circ$ , so $\angle QRS = 180^\circ - 100^\circ = 80^\circ$ .
The Fast Way
Same-segment angle, then the triangle angle sum.
$\angle QSR$ and $\angle QPR$ stand on the same arc QR, so $\angle QSR = 28^\circ$ .
In triangle QRS, $\angle QRS = 180^\circ - 72^\circ - 28^\circ = 80^\circ$ .
Q5Algebra › Progression & SeriesMedium
Asked in BBA DBE 2024
Let $\dfrac{m}{n} = \dfrac{90^2+5 \times 90+4}{90^2+5\times 90+6} \times \dfrac{91^2+5\times 91+4}{91^2+5\times 91+6} \times \cdots \times \dfrac{98^2+5\times 98+4}{98^2+5\times 98+6}$ where $m$ and $n$ are co-primes.
What is the value of $|m - n|$ ?
Correct option A3
Factor: $x^2 + 5x + 4 = (x+1)(x+4)$ and $x^2 + 5x + 6 = (x+2)(x+3)$ .
So the product is $\prod_{x=90}^{98} \dfrac{x+1}{x+2} \times \prod_{x=90}^{98} \dfrac{x+4}{x+3}$ .
Both telescope: $\dfrac{91}{100} \times \dfrac{102}{93}$ .
$= \dfrac{91 \times 34}{100 \times 31} = \dfrac{3094}{3100} = \dfrac{1547}{1550}$ , and $1547 = 7 \times 13 \times 17$ shares no factor with $1550$ .
$|m - n| = 1550 - 1547 = 3$ .
The Fast Way
Two quadratics differing only in the constant: factor into neighbouring brackets and telescope.
$\frac{(x+1)(x+4)}{(x+2)(x+3)}$ : the $\frac{x+1}{x+2}$ chain from 90 to 98 collapses to $\frac{91}{100}$ and the $\frac{x+4}{x+3}$ chain to $\frac{102}{93}$ , giving $\frac{9282}{9300}$ .
The gap is 18 and $\text{HCF}(9282, 9300) = \text{HCF}(9282, 18) = 6$ , so in lowest terms the gap is $18 \div 6 = 3$ .
Q6Algebra › PolynomialsMedium
Asked in BBA DBE 2026
A polynomial $P(x)=x^4+3x^3-5x^2+7x+9$ is divided separately by $(x-2)$ and $(x+1)$ , giving remainders $R_1$ and $R_2$ , respectively. Find the value of $\dfrac{R_1+R_2}{2}$ .
Correct option A19
Remainder theorem: dividing $P(x)$ by $(x - a)$ leaves remainder $P(a)$ .
$R_1 = P(2) = 16 + 24 - 20 + 14 + 9 = 43$ .
$R_2 = P(-1) = 1 - 3 - 5 - 7 + 9 = -5$ .
$\dfrac{R_1 + R_2}{2} = \dfrac{43 - 5}{2} = 19$ .
Watch the signs at $x = -1$ : odd powers turn negative, even powers stay positive.
The Fast Way
Remainder theorem: just evaluate $P(2)$ and $P(-1)$ .
$P(2) = 16 + 24 - 20 + 14 + 9 = 43$ ; $P(-1)$ is even-power coefficients minus odd-power ones: $(1 - 5 + 9) - (3 + 7) = -5$ .
So $\frac{R_1 + R_2}{2} = \frac{43 - 5}{2} = 19$ .
Q7Calculus › Derivatives & Their ApplicationsMedium
Air is pumped into a spherical balloon at a constant rate of 900 cm³ per second. At what rate is the surface area of the balloon increasing at the instant its diameter is 30 cm?
Correct option D120 cm²/s
$V = \frac{4}{3}\pi r^3$ , so $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ . The diameter is 30 cm, so $r = 15$ .
$900 = 4\pi(15)^2\frac{dr}{dt} = 900\pi\frac{dr}{dt}$ , so $\frac{dr}{dt} = \frac{1}{\pi}$ cm/s.
$S = 4\pi r^2$ , so $\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi(15)\cdot\frac{1}{\pi} = 120$ cm²/s.
Trap: using 30 (the diameter) as the radius gives 60.
The Fast Way
Divide the two rates so that $\frac{dr}{dt}$ and $\pi$ cancel: $\frac{dS/dt}{dV/dt} = \frac{8\pi r}{4\pi r^2} = \frac{2}{r}$ .
With $r = 15$ : $\frac{dS}{dt} = \frac{2}{15} \times 900 = 120$ cm²/s.
Q8Modern Math › ProbabilityMedium
Bag I has 4 white and 2 black balls. Bag II has 3 white and 5 black balls. Two balls are drawn at random from Bag I and put into Bag II without being seen. A ball is then drawn at random from Bag II and is found to be white. What is the probability that the two balls moved were one white and one black?
Correct option A $\frac{32}{65}$
Ways to move 2 of the 6 balls: $\binom{6}{2} = 15$ . Both white: $\binom{4}{2} = 6$ ; one of each: $4 \times 2 = 8$ ; both black: $1$ .
Bag II then has 10 balls, of which 5, 4 or 3 are white in these three cases.
$P(\text{white}) = \frac{6}{15}\cdot\frac{5}{10} + \frac{8}{15}\cdot\frac{4}{10} + \frac{1}{15}\cdot\frac{3}{10} = \frac{30 + 32 + 3}{150} = \frac{65}{150}$ .
By Bayes' theorem, $P(\text{one of each} \mid \text{white}) = \frac{32/150}{65/150} = \frac{32}{65}$ .
Trap: $\frac{8}{15}$ is the chance before the white ball is seen.
The Fast Way
Bayes by weights: (ways to move) $\times$ (whites then in Bag II).
Both white $6 \times 5 = 30$ , one of each $8 \times 4 = 32$ , both black $1 \times 3 = 3$ .
So the probability is $\frac{32}{30 + 32 + 3} = \frac{32}{65}$ .
Q9Arithmetic › Time & WorkHard
Asked in BBA DBE 2024
A, B and C work in a workshop for creating identical idols. A can craft 10 idols in 150 days. Similarly, B and C can individually create 16 idols in 192 days and 320 days, respectively. B works every day, A works only on alternate days starting from the first day, i.e. 1st March, and C also works only on alternate days starting from the second day, i.e. 2nd March. On which date will the construction of 2 idols be completed?
Correct option B15 March
Rates in idols per day: A $= \frac{10}{150} = \frac{1}{15}$ , B $= \frac{16}{192} = \frac{1}{12}$ , C $= \frac{16}{320} = \frac{1}{20}$ .
In every 2-day block B works twice, A once and C once: $\frac{2}{12} + \frac{1}{15} + \frac{1}{20} = \frac{10 + 4 + 3}{60} = \frac{17}{60}$ .
Target $= 2 = \frac{120}{60}$ , and $120 = 7 \times 17 + 1$ .
After 7 blocks (14 days) the work done is $\frac{119}{60}$ , just $\frac{1}{60}$ short.
Day 15 is an A + B day worth $\frac{1}{15} + \frac{1}{12} = \frac{9}{60}$ , more than enough, so the 2nd idol is finished on 15 March.
The Fast Way
LCM units: make one idol 60 units, so A does 4 a day, B 5 and C 3.
A 2-day block gives $5 + 5 + 4 + 3 = 17$ units, and two idols need $120 = 7 \times 17 + 1$ .
Seven blocks end on 14 March with 1 unit left, which day 15 covers, so the idols are done on 15 March.
Q10Algebra › Progression & SeriesHard
Asked in BBA DBE 2025
What is the sum of the infinite series: $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots$ ?
Correct option C $\frac{2}{7}$
The signs repeat in the pattern $+, -, -$ , so group the terms in threes.
First group: $1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ . Each later group is the previous one multiplied by $\frac{1}{8}$ .
So the sum is a GP: $\dfrac{\frac{1}{4}}{1 - \frac{1}{8}} = \frac{1}{4} \times \frac{8}{7} = \frac{2}{7}$ .
Sanity check: $\frac{2}{7} \approx 0.286$ , and the partial sums $1, 0.5, 0.25, 0.375, \dots$ settle there.
The Fast Way
The signs repeat every 3 terms, so group in threes, then let the options finish it.
First group: $1 - \frac12 - \frac14 = \frac14$ , and each later group is $\frac18$ of the one before, so the sum is $\frac{1/4}{1 - 1/8}$ , only a little above 0.25.
Only $\frac27 \approx 0.286$ sits just above 0.25, so the sum is $\frac{2}{7}$ .

Section 2 · Logical & Analytical Reasoning

Q11Logical Reasoning › Mathematical OperationsEasy
Asked in BBA DBE 2026
If $+$ means $-$ , $-$ means $\times$ , $\times$ means $\div$ , $\div$ means $+$ , then what will come in place of the question mark (?) in the following equation? $17 \div 63 - 3 + 84 \times 12 = ?$
Correct option A199
Swap every symbol: $17 \div 63 - 3 + 84 \times 12$ becomes $17 + 63 \times 3 - 84 \div 12$ .
Apply BODMAS: $63 \times 3 = 189$ and $84 \div 12 = 7$ .
$17 + 189 - 7 = 199$ .
Tip: rewrite the whole expression with the new symbols first, then calculate; mixing both steps is where most slips happen.
The Fast Way
Swap the symbols in one pass: $17 + 63 \times 3 - 84 \div 12$ .
Units digits are enough: $63 \times 3$ ends in 9 and $84 \div 12 = 7$ , so $7 + 9 - 7$ ends in 9.
Only 199 ends in 9, so the value is 199.
Q12Logical Reasoning › SeriesEasy
Asked in BBA DBE 2026
What should come in place of the question mark ‘?’ in the given series? 81, 84, 90, 99, 111, ?
Correct option C126
Differences: $84 - 81 = 3$ , $90 - 84 = 6$ , $99 - 90 = 9$ , $111 - 99 = 12$ .
The gaps are consecutive multiples of 3, so the next gap is $15$ .
$111 + 15 = 126$ .
The Fast Way
Gentle growth, so check the first differences.
The gaps are 3, 6, 9, 12, the 3-times table, so the next gap is 15.
So the missing term is $111 + 15 = 126$ .
Q13Logical Reasoning › Coding-DecodingEasy
In a certain code, FRIEND is written as HUMJTK. How is CASTLE written in that code?
Correct option CEDWYRL
Compare each letter with its code: F to H is +2, R to U is +3, I to M is +4, E to J is +5, N to T is +6, D to K is +7.
Apply the same shifts to CASTLE: C+2 = E, A+3 = D, S+4 = W, T+5 = Y, L+6 = R, E+7 = L.
So CASTLE is written as EDWYRL.
The Fast Way
The shift grows by one at each letter: +2, +3, +4 and so on.
The options differ only in letters 2 to 4: A+3 = D, S+4 = W, T+5 = Y.
Only EDWYRL has D, W and Y there.
Q14Logical Reasoning › Seating ArrangementMedium
Asked in BBA DBE 2026
A, B, C, D, E, F and G are sitting around a circular table facing the centre. B sits third to the right of E. C sits second to the left of B. A is the immediate neighbour of D and E. G sits to the immediate left of D. How many people sit between F and D when counted from the left of D?
Correct option BTwo
Facing the centre, \"right\" is anticlockwise and \"left\" is clockwise. Number the seats $0$ to $6$ going right (anticlockwise).
Put E at $0$ . B is third to the right of E: seat $3$ . C is second to the left of B: seat $1$ .
A neighbours both D and E. Seat $1$ is taken, so A is at $6$ and D at $5$ .
G is immediately left of D: seat $4$ . F takes the only free seat, $2$ .
Going left from D you meet G, then B, then F: two people sit between them.
Going right instead you would pass A, E and C (three people), which is exactly the trap in option A.
The Fast Way
Facing the centre, right is anticlockwise: pin E and number the seats going right.
E 0, B 3, C 1; A must take 6 (seat 1 is gone) with D at 5, then G at 4 and F at the last seat, 2.
Going left from D you pass G and B before F, so two people sit between them.
Q15Logical Reasoning › Blood RelationsMedium
A, B, C, D, E and F are six members of a family spanning three generations, with exactly two married couples. B is the father of D and the grandfather of F. C is the mother-in-law of E, and E is the mother of F. A is D's unmarried sister. How is A related to E?
Correct option ASister-in-law
E is F's mother and B is F's grandfather, so B is not E's husband; C is E's mother-in-law, so C is not E's spouse either.
A is unmarried and F is the youngest generation, so the two couples must come from B, C, D and E: they can only be D with E, and B with C.
So D is E's husband and C's son, and A, D's sister, is E's husband's sister.
A is E's sister-in-law.
Trap: A is F's aunt, but the question asks about E.
The Fast Way
Fix the couples first: E, a mother, must be married to C's son, and that is D.
A is D's sister, so A is E's husband's sister.
So A is E's sister-in-law.

Section 3 · Verbal Ability & Reading Comprehension

Q16Grammar › TensesEasy
Asked in BBA DBE 2024
Select the CORRECT option to replace the bold words ("has improved") in the below sentence.
It's no secret that artificial intelligence (AI), specifically generative AI, has improved all the time.
Correct option Cis improving
"All the time" describes a process that is going on continuously right now, so the verb must be in the present continuous: "is improving all the time".
"Has improved" is a completed-result tense and clashes with an ongoing phrase like "all the time".
"Will improving" is ungrammatical, "will have improved" points to a finished moment in the future, and "had improved" needs an earlier past event that the sentence never gives.
The Fast Way
Let the time marker pick the tense.
'All the time' means the change is going on right now, so the verb must be present continuous (is + -ing).
'Will improving' is broken grammar and the other two are perfect tenses, so the answer is 'is improving'.
Q17Vocabulary › Confusing WordsEasy
Asked in BBA DBE 2026
Fill in the blanks with the most appropriate options.
The _______ scene at the gory accident site deeply _____ the young boy.
Correct option Dheart-wrenching; affected
A gory accident scene is distressing, so "heart-wrenching" (causing deep sorrow) fits; "heart-warming" means pleasantly touching, the opposite mood.
"Affected" means influenced or moved emotionally, which is what the scene did to the boy. "Effected" means brought about, as in "effected a change".
The Fast Way
Do the surer blank first: affect or effect.
Memory hook, A for Action: affect is the verb for moving someone, effect is the result. 'Deeply ___ the boy' means moved him, so both 'effected' pairs go.
A gory scene is not heart-warming, so the answer is 'heart-wrenching; affected'.
Q18Vocabulary › Phrasal VerbsEasy
Asked in BBA DBE 2026
Fill in the blank with the most appropriate phrasal verb.
During the transnational policy symposium on ethical governance, constitutional scholars attempted to _____ the socioeconomic ramifications of artificial intelligence.
Correct option Cdelve into
"Delve into" is the fixed phrasal verb meaning to examine or investigate something in depth, so the scholars tried to delve into the ramifications.
"Delve in" does not take an object this way, and "delve onto" and "delve inside" are not idiomatic English.
The Fast Way
Fixed phrasal verb: you always delve into a subject.
'Into' carries the going-deep sense of examining something closely; 'onto', 'inside' and a bare 'in' are never used with delve before an object.
So the blank is 'delve into'.
Q19Vocabulary › Odd Word OutMedium
Asked in BBA DBE 2026
Based on the meanings, choose the odd one out.
Correct option Btaciturn
Voluble, loquacious and garrulous all describe a person who talks a lot, easily and at length.
Taciturn means reserved and saying very little, the opposite of the other three, so it is the odd one out.
The Fast Way
Use roots and look for the opposite charge.
'Loqu' means speak (loquacious, eloquent), and voluble and garrulous are its talkative cousins; 'tacit' means unspoken, and taciturn grows from it.
Three mean talkative and one means silent, so the odd one out is taciturn.
Q20Vocabulary › AntonymsMedium
Asked in BBA DBE 2025
Match the words in Column A with their antonyms in Column B.
Column A: A. Diligent, B. Concise, C. Just, D. Abolish
Column B: 1. Negligent, 2. Vague, 3. Biased, 4. Establish
Correct option CA-1, B-2, C-3, D-4
Diligent (hard-working) vs negligent, just (fair) vs biased, abolish vs establish: these three pairs are clear.
That leaves concise with vague. Strictly the best antonym of concise is verbose, but once A-1, C-3 and D-4 are fixed only option C fits.
The Fast Way
Lock the pair you are surest of and scan the options for it.
Diligent vs negligent (A-1) is certain, and only one option has A-1.
It also gives just-biased (C-3) and abolish-establish (D-4), so the match is 'A-1, B-2, C-3, D-4'.