Masters' Union MUSAT · Paper 1

20 MCQs · Numerical Ability 7, Data Analysis 7, Logical Reasoning 6

No negative marking, as in the real test.

Tap an option to answer

Section 1 · Numerical Ability

Q1Number System › FactorisationEasy
Find the total number of positive factors of $1080$ .
Correct option A $32$
Prime factorise first: $1080 = 2^3 \times 3^3 \times 5^1$ .
For $N = p^a q^b r^c$ , the number of factors is $(a+1)(b+1)(c+1)$ .
So we get $(3+1)(3+1)(1+1) = 4 \times 4 \times 2 = 32$ , option A.
The Fast Way
Factor count: add 1 to each prime power and multiply.
Split $1080 = 8 \times 135 = 2^3 \times 3^3 \times 5^1$ in your head.
$(3+1)(3+1)(1+1) = 4 \times 4 \times 2 = 32$ , so $1080$ has 32 factors.
Q2Algebra › Quadratic EquationsEasy
A pop-up café at a college fest estimates that if it sells $x$ premium snack boxes, its profit in rupees will be $P = x(80 - x) - 300$ . What is the maximum possible profit?
Correct option A₹1,300
$P = 80x - x^2 - 300$ is a downward parabola, so it has a single peak.
$x(80 - x)$ is largest midway between its roots 0 and 80, at $x = 40$ .
$P = 40 \times 40 - 300 = 1600 - 300 = 1300$ .
So the maximum profit is ₹1,300.
The Fast Way
A product $x(k - x)$ peaks at $x = \frac{k}{2}$ , giving $\left(\frac{k}{2}\right)^2$ .
Here $k = 80$ : $40^2 = 1600$ , then take off the fixed 300.
So the maximum profit is ₹1,300 (₹1,600 forgets the 300, and ₹40 is the quantity).
Q3Modern Math › Set TheoryMedium
In a batch of 90 MUSAT aspirants, 48 have enrolled for Numerical Ability, 42 for Data Interpretation, and 35 for Logical Reasoning. Also, 18 enrolled for Numerical Ability and Data Interpretation, 15 for Data Interpretation and Logical Reasoning, 12 for Numerical Ability and Logical Reasoning, and 7 enrolled for all three. How many students have not enrolled for any of the three modules?
Correct option A $3$
Use inclusion-exclusion for three sets:
$n(N \cup D \cup L) = 48 + 42 + 35 - 18 - 15 - 12 + 7 = 125 - 45 + 7 = 87$ .
So $87$ students took at least one module, and $90 - 87 = 3$ took none, option A.
Sanity check: every Venn region is non-negative (only NA $= 25$ , only DI $= 16$ , only LR $= 15$ ), so the data is consistent.
The Fast Way
Inclusion-exclusion in one line: singles minus pairs plus the triple.
$125 - 45 + 7 = 87$ students took at least one module.
So $90 - 87 = 3$ took none.
Q4Number System › Number PropertiesMedium
A startup prints numbered access cards from 1 to 375 for an event. How many times does the digit 7 appear in these card numbers?
Correct option B $73$
Count the 7s one digit position at a time.
Units place: $7, 17, 27, \dots, 367$ gives $37$ sevens.
Tens place: $70$ to $79$ , $170$ to $179$ , $270$ to $279$ give $30$ , plus $370$ to $375$ give $6$ more, so $36$ .
Hundreds place: none, since the cards stop at $375$ .
Total: $37 + 36 = 73$ , option B.
The Fast Way
Every full hundred holds 20 sevens: 10 in the units place and 10 in the tens place.
1 to 299 is three hundreds, so $3 \times 20 = 60$ .
300 to 375 adds 307, 317, ..., 367 (7 sevens) and 370 to 375 (6 sevens), so $60 + 13 = 73$ .
Q5Algebra › Quadratic EquationsMedium
A pop-up food stall wants to create a rectangular seating area using 96 m of temporary fencing. One side of the rectangle will be along an existing campus wall, so fencing is needed only for the other three sides. What is the maximum possible area of the seating space?
Correct option D1,152 sq m
Let the two sides perpendicular to the wall be $x$ each and the side parallel to the wall be $y$ .
Fencing: $2x + y = 96$ , so the area is $A = x(96 - 2x)$ .
This parabola peaks at $x = \frac{96}{4} = 24$ , which gives $y = 96 - 48 = 48$ .
Maximum area $= 24 \times 48 = 1152$ sq m, option D.
The Fast Way
Wall on one side: the side along the wall gets half the fencing.
So that side is $\frac{96}{2} = 48$ m and the other two are $\frac{48}{2} = 24$ m each.
Maximum area $= 24 \times 48 = 1152$ sq m.
Q6Arithmetic › Simple & Compound InterestHard
A student buys a laptop on a no-down-payment plan for ₹21,000. The amount is to be repaid in 2 equal annual installments, with the first installment paid at the end of 1 year. If interest is charged at $10\%$ per annum, what should each installment be?
Correct option D₹12,100
The present values of the two instalments must add up to the price, at $10\%$ compound interest.
$\frac{X}{1.1} + \frac{X}{1.1^2} = 21000$ .
Multiply through by $1.21$ : $1.1X + X = 2.1X = 21000 \times 1.21 = 25410$ .
So $X = 12100$ , i.e. ₹12,100 per instalment, option D.
Check: after year 1 you owe $23100 - 12100 = 11000$ , which grows to $12100$ in year 2. Perfect.
The Fast Way
Move both instalments to the end of year 2: $1.1X + X = 2.1X$ must equal $21000 \times 1.21$ .
$\frac{21000}{2.1} = 10000$ , so $X = 10000 \times 1.21$ .
Each instalment is ₹12,100.
Q7Number System › FactorisationMedium
During a coding bootcamp, a student writes the product $1 \times 2 \times 3 \times \dots \times 150$ . How many trailing zeros will this product have?
Correct option A $37$
Each trailing zero needs a pair of $2$ and $5$ , and $5$ s are the scarcer factor.
Count the $5$ s in $150!$ : $\lfloor 150/5 \rfloor + \lfloor 150/25 \rfloor + \lfloor 150/125 \rfloor = 30 + 6 + 1 = 37$ .
So the product ends in $37$ zeros, option A.
The Fast Way
Trailing zeros: keep dividing by 5 and add the quotients.
$150 \to 30 \to 6 \to 1$ , so $30 + 6 + 1 = 37$ .
The product ends in 37 zeros; 36 forgets the extra 5 inside $125 = 5^3$ .

Section 2 · Data Analysis

Q8Data Interpretation › Pie ChartsEasy
A pie chart shows the preferred UG specialisation of 720 MUSAT applicants. The sectors are: Product Management - $90^\circ$ , Finance - $72^\circ$ , Data Analytics - $108^\circ$ , Entrepreneurship - $54^\circ$ , and Marketing - $36^\circ$ . How many more applicants preferred Data Analytics than Marketing?
Correct option C $144$
Quick check: $90 + 72 + 108 + 54 + 36 = 360^\circ$ , so the chart is complete.
Each degree stands for $\frac{720}{360} = 2$ applicants.
The gap between Data Analytics and Marketing is $108^\circ - 36^\circ = 72^\circ$ .
So $72 \times 2 = 144$ more applicants, option C.
The Fast Way
Subtract the angles first, convert to people once.
$108^\circ - 36^\circ = 72^\circ$ , and each degree is $\frac{720}{360} = 2$ applicants.
So $72 \times 2 = 144$ more chose Data Analytics (72 is the unconverted trap).
Q9Arithmetic › AveragesEasy
A B-school club conducted a mock pitch round for three batches. Batch A had 20 students with an average score of 70, Batch B had 30 students with an average score of 80, and Batch C had 50 students with an average score of 76. What was the overall average score of all students together?
Correct option B $76$
The overall average is the weighted average by batch size.
Total score $= 20 \times 70 + 30 \times 80 + 50 \times 76 = 1400 + 2400 + 3800 = 7600$ .
There are $100$ students, so the average is $\frac{7600}{100} = 76$ , option B.
The Fast Way
Deviation method: measure every batch from 76.
$20 \times (-6) + 30 \times (+4) + 50 \times 0 = -120 + 120 = 0$ .
Nothing is left over, so the overall average is 76 (75.33 ignores the batch sizes).
Q10Arithmetic › AveragesEasy
A data table shows scores in a startup simulation round. The morning batch had 24 students with an average score of 68. The evening batch had 36 students with an average score of 73. After adding a night batch of 20 students, the combined average score of all 80 students became 72. What was the average score of the night batch?
Correct option C $75$
Total for all 80 students: $80 \times 72 = 5760$ .
Morning plus evening: $24 \times 68 + 36 \times 73 = 1632 + 2628 = 4260$ .
Night batch total: $5760 - 4260 = 1500$ , so its average is $\frac{1500}{20} = 75$ , option C.
The Fast Way
Deviation method: measure from the final average 72.
Morning $24 \times (-4) = -96$ , evening $36 \times (+1) = +36$ , net $-60$ .
The 20 night students must add $+60$ , i.e. $+3$ each, so their average is 75.
Q11Arithmetic › AveragesEasy
Three batches appeared for a startup aptitude test. Batch A had 24 students with an average score of 68, Batch B had 36 students with an average score of 78, and Batch C had 30 students. If the overall average score of all 90 students was 75, what was the average score of Batch C?
Correct option C $77$
Total for all 90 students: $90 \times 75 = 6750$ .
Batches A and B: $24 \times 68 + 36 \times 78 = 1632 + 2808 = 4440$ .
Batch C total: $6750 - 4440 = 2310$ , so its average is $\frac{2310}{30} = 77$ , option C.
The Fast Way
Deviation method: measure from the overall average 75.
A gives $24 \times (-7) = -168$ , B gives $36 \times (+3) = +108$ , net $-60$ .
C's 30 students must add $+60$ , i.e. $+2$ each, so Batch C averaged 77.
Q12Data Interpretation › TablesEasy
A college fest team ran four marketing campaigns. The table shows total spend and leads generated: Nexus: ₹90,000 and 300 leads; Orbit: ₹1,20,000 and 500 leads; Pulse: ₹75,000 and 250 leads; Zest: ₹1,00,000 and 400 leads. Which campaign had the lowest cost per lead?
Correct option BOrbit
Cost per lead $=$ spend $\div$ leads.
Nexus: $\frac{90000}{300} = 300$ . Orbit: $\frac{120000}{500} = 240$ .
Pulse: $\frac{75000}{250} = 300$ . Zest: $\frac{100000}{400} = 250$ .
Orbit is cheapest at ₹240 per lead, option B.
The Fast Way
Spot the equal pair first, then compare only what is left.
Nexus ( $90000 = 300 \times 300$ ) and Pulse ( $75000 = 250 \times 300$ ) both cost ₹300 a lead.
Orbit $\frac{1200}{5} = 240$ against Zest $\frac{1000}{4} = 250$ , so Orbit has the lowest cost per lead.
Q13Algebra › Linear EquationsMedium
At a Masters’ Union pitch day, 102 people registered as founders, investors, or student volunteers. The number of founders was twice the number of investors. The number of student volunteers was 18 fewer than the number of founders. Also, 12 of the registered people were speakers, and every speaker was a founder already counted in the founders category. How many founders were not speakers?
Correct option D $36$
Let investors $= I$ . Then founders $= 2I$ and volunteers $= 2I - 18$ .
Total: $I + 2I + (2I - 18) = 102$ , so $5I = 120$ and $I = 24$ .
Founders $= 48$ and volunteers $= 30$ (check: $24 + 48 + 30 = 102$ ).
All 12 speakers are founders, so founders who are not speakers $= 48 - 12 = 36$ , option D.
The Fast Way
Add the missing 18 back so the volunteers match the founders.
$102 + 18 = 120$ is $1 + 2 + 2 = 5$ investor-sized parts, so one part is 24 and founders are 48.
Speakers sit inside the founders, so $48 - 12 = 36$ founders are not speakers (48 is the trap).
Q14Data Interpretation › TablesMedium
A B-school club tracked webinar registrations by specialisation. The table shows total registrations and attendance rate: Finance: 180 registrations, $60\%$ attended; Marketing: 150 registrations, $80\%$ attended; Analytics: 120 registrations, $75\%$ attended. Among those who attended, the paid-to-free attendee ratios were Finance $= 2:1$ , Marketing $= 3:2$ , and Analytics $= 4:5$ . How many paid attendees were there in Finance and Marketing together?
Correct option B $144$
Attendees first: Finance $= 60\%$ of $180 = 108$ , Marketing $= 80\%$ of $150 = 120$ .
Finance paid $= \frac{2}{3} \times 108 = 72$ .
Marketing paid $= \frac{3}{5} \times 120 = 72$ .
Together: $72 + 72 = 144$ , option B.
The Fast Way
Ignore Analytics and fold the two fractions into one per row.
Finance: $60\% \times \frac{2}{3} = 40\%$ of 180 is 72. Marketing: $80\% \times \frac{3}{5} = 48\%$ of 150 is 72.
Together that is $72 + 72 = 144$ paid attendees.

Section 3 · Logical Reasoning

Q15Number System › Number PropertiesEasy
Find the odd one out among the following number groups.
Correct option D $196, 225, 255$
Each group looks like three consecutive perfect squares.
A: $5^2, 6^2, 7^2$ . B: $8^2, 9^2, 10^2$ . C: $11^2, 12^2, 13^2$ .
D: $196 = 14^2$ and $225 = 15^2$ , but $16^2 = 256$ , not $255$ .
So D breaks the pattern and is the odd one out.
The Fast Way
Consecutive squares: check only the last number of each group.
$49 = 7^2$ , $100 = 10^2$ and $169 = 13^2$ fit, but after $196 = 14^2$ and $225 = 15^2$ comes $16^2 = 256$ .
So the group $196, 225, 255$ is the odd one out.
Q16Logical Reasoning › PuzzlesMedium
Seven students, Aisha, Bharat, Chetan, Diya, Emaan, Farah, and Gaurav, live on different floors of a 7-floor hostel, with floor 1 being the lowest and floor 7 being the highest. Chetan lives on the 5th floor. Aisha lives exactly two floors below Chetan. Diya lives immediately below Bharat. Farah lives above Emaan but not on the top floor. Gaurav lives below Emaan. Who lives on the top floor?
Correct option ABharat
Fix the sure facts: Chetan is on floor $5$ , so Aisha is on floor $5 - 2 = 3$ .
Free floors: $1, 2, 4, 6, 7$ . Bharat sits directly on top of Diya, so the pair is either (Diya $1$ , Bharat $2$ ) or (Diya $6$ , Bharat $7$ ).
Gaurav, Emaan and Farah fill the other three floors in the order $G < E < F$ .
If Bharat is on $2$ , those three floors are $4, 6, 7$ and Farah lands on $7$ , which is not allowed.
So Diya is on $6$ and Bharat on $7$ , with Gaurav $1$ , Emaan $2$ , Farah $4$ .
Bharat is on the top floor, option A.
The Fast Way
Ask only who can live on floor 7: anyone with somebody above them is out.
Diya has Bharat above her, Gaurav has Emaan above, Emaan has Farah above, Farah is barred from the top, and Chetan and Aisha are on 5 and 3.
Only Bharat is left, so Bharat lives on the top floor.
Q17Critical Reasoning › SyllogismsMedium
Statement: The campus incubator will approve only those student teams that submitted a working prototype by Friday evening. Team Orion submitted its working prototype on Thursday afternoon. Conclusions: I. Team Orion will definitely be approved by the incubator. II. Any team that did not submit a working prototype by Friday evening will not be approved by the incubator.
Correct option BOnly conclusion II follows
"Approve only those who submitted" means: approved $\Rightarrow$ submitted on time.
So submitting is a necessary condition, not a guarantee of approval.
I: Orion met the condition, but that does not force approval. The team could still fail on other grounds, so I does not follow.
II: this is just the contrapositive, not submitted $\Rightarrow$ not approved, which is exactly what the statement says. II follows.
Answer: only II follows.
The Fast Way
'Only those who X' makes X necessary, never sufficient.
I treats the prototype as a guarantee of approval, so it fails.
II is the contrapositive (no prototype by Friday, no approval), so only conclusion II follows.
Q18Logical Reasoning › Blood RelationsEasy
Shobha has exactly two children: her son Rajat and her daughter Naina. Naina is unmarried and has no children. Shobha's only grandchild is a boy, Ayaan, whose parents are a married couple, and Ayaan's mother is Diya. How is Diya related to Shobha?
Correct option CDaughter-in-law
Ayaan is Shobha's grandchild, so one of his parents is a child of Shobha: Rajat or Naina.
Naina has no children, so Ayaan must be Rajat's son.
Ayaan's parents are married and his mother is Diya, so Diya is Rajat's wife.
The wife of Shobha's son is Shobha's daughter-in-law.
Trap: 'Daughter' would need Diya to be Naina, who has no children, and 'Sister-in-law' is Diya's relation to Naina.
The Fast Way
Strike the impossible parent first: childless Naina cannot be Ayaan's parent.
That makes Rajat the father, so Ayaan's mother Diya is Rajat's wife.
The wife of a son is a daughter-in-law, so Diya is Shobha's daughter-in-law.
Q19Logical Reasoning › Direction SenseMedium
A boat leaves a jetty and sails 8 km north-east. It then turns $90^\circ$ to its right and sails another 8 km. How far is the boat from the jetty now, and in which direction must it sail to go straight back?
Correct option C $8\sqrt{2}$ km, West
Turning $90^\circ$ right from north-east points the boat south-east, so the two legs are at right angles.
Each 8 km leg at $45^\circ$ splits into $\frac{8}{\sqrt{2}} = 4\sqrt{2}$ km east plus $4\sqrt{2}$ km north (first leg) or south (second leg).
The north and south parts cancel, and the east parts add up to $8\sqrt{2}$ km.
So the boat is $8\sqrt{2} \approx 11.3$ km due east of the jetty, and to go straight back it must sail West.
The Fast Way
Two equal legs at right angles make an isosceles right triangle, so the gap is one leg times $\sqrt{2}$ .
North-east and south-east mirror each other about the east line, so the boat ends due east of the jetty.
The way back is $8\sqrt{2}$ km, West (East is where the boat is, not where it must go).
Q20Logical Reasoning › Seating ArrangementMedium
Six startup founders, Aria, Bhav, Cia, Dev, Ema, and Farid, are sitting around a circular table facing outward. Aria sits opposite Dev. Bhav sits second to the left of Aria. Cia sits immediately to the right of Bhav. Ema is not adjacent to Aria. Who sits opposite Cia?
Correct option DEma
Number the seats $0$ to $5$ clockwise and put Aria at $0$ , so Dev (opposite) is at $3$ .
Facing outward, your left is anticlockwise and your right is clockwise.
Bhav is second to Aria's left: seat $0 - 2 \equiv 4$ . Cia is immediately to Bhav's right: seat $5$ .
Seats $1$ and $2$ remain. Seat $1$ touches Aria, so Ema takes $2$ and Farid takes $1$ .
Opposite Cia (seat $5$ ) is seat $5 - 3 = 2$ , which is Ema, option D.
The Fast Way
Pin Aria at seat 0 and Dev at 3, then follow the steps without worrying about left and right.
Bhav is 2 seats from Aria and Cia steps back toward her, so Cia is Aria's neighbour; the free seats are Aria's other neighbour and the seat opposite Cia.
Ema cannot sit next to Aria, so Ema sits opposite Cia.