Mesa MSAT · Paper 1

20 MCQs · Numerical Reasoning 10, Logical Reasoning 10

The numerical and logical reasoning sections.

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Section 1 · Numerical Reasoning

Q1Arithmetic › Ratio, Proportion & VariationMedium
Asked in Mesa 2024
Two individuals, P and Q, had money in the ratio $5 : 23$ . After Q gave some of his money to P, the ratio of the amount of money with P to that with Q became $1 : 3$ . What fraction of his original amount of money was given by Q to P?
Correct option B $\frac{2}{23}$
Take the amounts as $5k$ and $23k$ . The total $28k$ never changes when money just moves between them.
After the transfer the ratio is $1:3$ , so P has $\frac{1}{4} \times 28k = 7k$ and Q has $21k$ .
P went from $5k$ to $7k$ , so Q handed over $2k$ .
Fraction of Q's original money: $\frac{2k}{23k} = \frac{2}{23}$ .
The Fast Way
Money only moves between them, so the total stays fixed: put both ratios over the same total.
$5 + 23 = 28$ , and $1 : 3$ over a total of 28 is $7 : 21$ .
Q fell from 23 parts to 21, so he gave 2 of his 23 parts: $\frac{2}{23}$ .
Q2Logical Reasoning › PuzzlesMedium
Six students, Anya, Kabir, Meera, Dev, Riya, and Omar, live on different floors of a 6-floor hostel, with floor 1 at the bottom and floor 6 at the top. Meera lives on the 2nd floor. Kabir lives somewhere above Meera but not on the top floor. Exactly two floors are between Omar and Riya. Anya lives immediately above Omar. Dev lives below Anya. Who lives on the top floor?
Correct option DRiya
Meera is fixed on floor 2. Two floors between Omar and Riya means their floors differ by 3: $\{1,4\}$ , $\{2,5\}$ or $\{3,6\}$ .
$\{2,5\}$ is out (Meera is on 2). Anya is directly above Omar, so Omar can't be on 6, and can't be on 1 (Anya would land on floor 2).
Try Omar 4, Riya 1: Anya is 5, Kabir must take 3, and Dev is left with 6, which is not below Anya. Contradiction.
So Omar 3, Riya 6, Anya 4, Kabir 5, Dev 1, Meera 2. Every clue checks out.
Riya lives on the top floor.
The Fast Way
Kill the options instead of building the whole hostel.
Kabir is barred from the top, and Omar can't be there because Anya sits directly above him.
If Anya were on 6, Omar would be on 5 and Riya three floors away on 2, which is Meera's floor.
So Riya lives on the top floor.
Q3Modern Math › Set TheoryMedium
In a Mesa prep cohort of 120 students, 58 chose Marketing, 52 chose Finance, and 46 chose Analytics as interest areas. Also, 22 chose both Marketing and Finance, 18 chose both Finance and Analytics, 16 chose both Marketing and Analytics, and 8 chose all three. How many students chose exactly one of the three areas?
Correct option D68
Use the 'exactly one' formula: $S_1 - 2S_2 + 3S_3$ , where $S_1$ adds the single counts, $S_2$ adds the pair counts and $S_3$ is the all-three count.
$S_1 = 58 + 52 + 46 = 156$ , $S_2 = 22 + 18 + 16 = 56$ , $S_3 = 8$ .
Exactly one $= 156 - 2(56) + 3(8) = 156 - 112 + 24 = 68$ .
Check by regions: only Marketing $= 58 - 22 - 16 + 8 = 28$ , only Finance $= 20$ , only Analytics $= 20$ , total $68$ .
The Fast Way
Skip the Venn diagram: exactly one $= S_1 - 2S_2 + 3S_3$ , and the 120 is never used.
$S_1 = 58 + 52 + 46 = 156$ , $S_2 = 22 + 18 + 16 = 56$ , $S_3 = 8$ .
$156 - 112 + 24 = 68$ students chose exactly one area.
Q4Modern Math › ProbabilityEasy
A startup welcome hamper has 6 black notebooks, 5 blue notebooks, and 4 green notebooks. Two notebooks are picked one after another without replacement. What is the probability that both notebooks picked are blue?
Correct option D $\frac{2}{21}$
There are $6+5+4=15$ notebooks, and 5 of them are blue.
First pick blue: $\frac{5}{15}$ . Now only 4 blue are left out of 14.
Second pick blue: $\frac{4}{14}$ .
Multiply: $\frac{5}{15}\times\frac{4}{14}=\frac{20}{210}=\frac{2}{21}$ .
Quick check: $\frac{\binom{5}{2}}{\binom{15}{2}}=\frac{10}{105}=\frac{2}{21}$ .
The Fast Way
Without replacement: take 1 off the top and the bottom for the second pick, and cancel before you multiply.
$\frac{5}{15} \times \frac{4}{14} = \frac{1}{3} \times \frac{2}{7}$ .
So the probability is $\frac{2}{21}$ ; black and green only fix the total of 15.
Q5Algebra › Maxima & MinimaEasy
The daily profit from selling limited-edition college hoodies is modelled by $P(x)=-2x^2+80x-300$ , where $P(x)$ is in hundreds of rupees and $x$ is the number of hoodies sold. What is the maximum possible daily profit?
Correct option C₹50,000
A downward parabola $ax^2+bx+c$ peaks at $x=-\frac{b}{2a}=-\frac{80}{2(-2)}=20$ .
Plug in: $P(20)=-2(400)+80(20)-300=-800+1600-300=500$ .
But $P$ is in hundreds of rupees, so the profit is $500\times100$ = ₹50,000.
The Fast Way
Use the peak-value formula for a downward parabola: maximum $= c - \frac{b^2}{4a}$ .
$-300 - \frac{6400}{-8} = -300 + 800 = 500$ , and that is in hundreds of rupees.
Multiply by 100 for the units: ₹50,000.
Q6Arithmetic › PartnershipsEasy
Three friends run a campus merchandise stall. Aarav invests ₹50,000 for 12 months, Bina invests ₹70,000 for 8 months, and Chirag invests ₹40,000 for 10 months. If the total profit at the end of the year is ₹1,17,000, what is Bina’s share of the profit?
Correct option D₹42,000
Profit is shared in the ratio of capital $\times$ time.
Aarav: $50{,}000 \times 12$ , Bina: $70{,}000 \times 8$ , Chirag: $40{,}000 \times 10$ , i.e. $600 : 560 : 400$ (in thousands).
Simplify: $600 : 560 : 400 = 15 : 14 : 10$ , a total of $39$ parts.
One part $= 1{,}17{,}000 \div 39 = 3{,}000$ , so Bina gets $14 \times 3{,}000 =$ ₹42,000.
The Fast Way
Partnership: share in the ratio of capital × months, with the zeros dropped.
$5 \times 12 : 7 \times 8 : 4 \times 10 = 60 : 56 : 40 = 15 : 14 : 10$ , which is 39 parts.
$117 = 3 \times 39$ , so one part is ₹3,000 and Bina's 14 parts are ₹42,000.
Q7Algebra › Quadratic EquationsEasy
A campus café wants to set up a rectangular outdoor counter with an area of $150$ sq. ft. The length of the counter must be $5$ ft more than its width. What should be the width of the counter?
Correct option A10 ft
Let the width be $w$ , so the length is $w + 5$ and $w(w + 5) = 150$ .
$w^2 + 5w - 150 = 0$ factors as $(w + 15)(w - 10) = 0$ .
A width can't be negative, so $w = 10$ ft (length 15 ft, and $10 \times 15 = 150$ ).
The Fast Way
Back-solve: test each width with a length 5 more.
10 ft gives $10 \times 15 = 150$ on the first try.
So the width is 10 ft, no quadratic needed.
Q8Arithmetic › Simple & Compound InterestHard
A student can buy a laptop by paying two equal annual installments of ₹12,100 each, the first due after 1 year and the second due after 2 years. If the seller charges 10% compound interest per annum, what is the cash price of the laptop today?
Correct option B₹21,000
The cash price today is the present value of both instalments at 10% compound interest.
First instalment (1 year away): $\frac{12100}{1.1} = 11000$ .
Second instalment (2 years away): $\frac{12100}{1.1^2} = \frac{12100}{1.21} = 10000$ .
Cash price $= 11000 + 10000 =$ ₹21,000.
The Fast Way
Present value: divide each instalment by 1.1 for every year it is away.
$12100 = 1.21 \times 10000$ , so the 2-year one is worth 10000 today and the 1-year one $12100 \div 1.1 = 11000$ .
Cash price = ₹21,000; ₹24,200 just adds the instalments and ignores the interest.
Q9Algebra › Maxima & MinimaEasy
A campus merchandise stall estimates that if it prices a hoodie at ₹ $x$ , it can sell $(80 - x)$ hoodies in a day, where $x$ is between $0$ and $80$ . Ignoring cost, what is the maximum possible daily revenue?
Correct option A₹1,600
Revenue $= x(80 - x) = 80x - x^2$ , a downward-opening parabola.
Its roots are $x = 0$ and $x = 80$ , so the peak is midway, at $x = 40$ .
Revenue $= 40 \times 40 = 1600$ , so the maximum is ₹1,600.
The Fast Way
Fixed sum, biggest product: $x$ and $80 - x$ always add to 80, so make them equal.
$x = 40$ gives $40 \times 40 = 1600$ .
The maximum revenue is ₹1,600; ₹40 is the best price, not the revenue.
Q10Arithmetic › Mixture & AlligationMedium
A café has an 80-litre jar containing a drink mix that is 75% juice and 25% water. The manager removes 16 litres of the mixture and replaces it with 16 litres of water. He repeats the same process once more. What is the final percentage of juice in the jar?
Correct option B48%
Juice at the start: $0.75 \times 80 = 60$ L.
Each round removes $\frac{16}{80} = \frac{1}{5}$ of the mixture and adds back only water, so the juice is multiplied by $\frac{4}{5}$ .
After two rounds: $60 \times \left(\frac{4}{5}\right)^2 = 60 \times \frac{16}{25} = 38.4$ L.
Percentage: $\frac{38.4}{80} \times 100 = 48\%$ .
The Fast Way
Repeated replacement: each round keeps $1 - \frac{16}{80} = \frac{4}{5}$ of the juice.
Stay in percentages: $75\% \times 0.8 = 60\%$ , then $60\% \times 0.8 = 48\%$ .
So the jar ends at 48% juice.

Section 2 · Logical Reasoning

Q11Data Interpretation › Data SufficiencyEasy
In a campus coding contest, it is claimed that “Riya scored more marks than Kunal.” Consider the following statements: Statement I: Riya scored more than Mehul, and Mehul scored more than Kunal. Statement II: Kunal scored less than Tanya, and Tanya scored less than Riya. Which of the following is correct?
Correct option CEither Statement I alone or Statement II alone is sufficient
We want to know if Riya scored more than Kunal.
Statement I gives a chain: Riya > Mehul > Kunal, so Riya > Kunal. That settles it.
Statement II says Kunal < Tanya < Riya, which is the same chain read from the other end, so again Riya > Kunal.
Each statement answers the question on its own, so either one alone is sufficient.
The Fast Way
Test II even after I works, or you miss 'either alone'.
I: Riya > Mehul > Kunal. II: Riya > Tanya > Kunal.
Both chains put Riya above Kunal, so either statement alone is sufficient.
Q12Logical Reasoning › Direction SenseEasy
Riya starts from the admissions desk and walks 9 m east, then 12 m north, then 4 m west, and finally 24 m south to reach the interview waiting area. What is her shortest straight-line distance from the starting point, and in which direction?
Correct option A13 m, south-east
Net east-west: $9 - 4 = 5$ m east. Net north-south: $24 - 12 = 12$ m south.
Distance $= \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ m.
She ends up east and south of the start, so the direction is south-east.
The Fast Way
Net displacement: cancel the opposite moves first.
East $9 - 4 = 5$ and south $24 - 12 = 12$ : the 5-12-13 triple.
So she is 13 m away, south-east.
Q13Critical Reasoning › Statement & AssumptionEasy
Statement: A campus café has announced, “From next Monday, we will accept only digital payments at the evening snack counter to reduce billing time.” Which of the following assumptions is implicit in the statement?
Correct option BDigital payments can help reduce billing time at the evening counter
An implicit assumption is something the café must believe for its announcement to make sense.
Its stated aim is to cut billing time by going digital-only, so it must assume digital payments help do that: option B.
A is too extreme ('every student'), C goes beyond the statement (it covers only the evening counter, not all counters for good), and D has nothing to do with billing time.
The Fast Way
The assumption is the bridge from the action to its stated purpose.
Going digital-only is meant to cut billing time, so the café must believe digital payments do that.
'Every student' and 'all counters permanently' over-claim, and the best-selling item is off-topic.
So the assumption is: digital payments can help reduce billing time at the evening counter.
Q14Logical Reasoning › AnalogyEasy
In a campus startup fair, the relation between the first pair is based on the exact purpose of the item. Pitch deck is to investor as product catalogue is to ?
Correct option BCustomer
A pitch deck is made to present a startup to an investor; the investor is its intended audience.
A product catalogue is made to present products to a customer; the customer is its intended audience.
A designer makes it, a warehouse stores the goods, and a competitor is not who it is for.
The Fast Way
Bridge sentence: a pitch deck is made to win over an investor.
Fill 'a product catalogue is made to win over a ___': only one option fits.
Designer is the maker trap, so the answer is Customer.
Q15Logical Reasoning › Coding-DecodingEasy
In a certain code, each letter is replaced by the letter that comes two places after it in the alphabet, and then the coded word is written in reverse order. For example, TEAM is coded as OCGV. How will PITCH be coded?
Correct option AJEVKR
Check the rule on TEAM: T, E, A, M shift two places to V, G, C, O, and reversed that reads OCGV, which matches.
For PITCH: P to R, I to K, T to V, C to E, H to J gives RKVEJ.
Reverse it: JEVKR.
The Fast Way
Reverse first, then shift: the two steps give the same result in either order.
PITCH reversed is HCTIP, and moving each letter 2 places gives J, E, V, K, R.
So the code is JEVKR; the third letter V already rules out JEWKR.
Q16Algebra › InequalitiesMedium
In a coded inequality system, the symbols mean: $A @ B$ means $A \ge B$ , $A \# B$ means $A > B$ , $A * B$ means $A < B$ , and $A \$ B$ means $A = B$ . Given: $M @ N$ , $N \# P$ , $P \$ Q$ , and $Q * R$ . Which of the following conclusions definitely follows? Conclusions: I. $M \ge R$ II. $R > M$
Correct option CEither conclusion I or II follows
Decode: $M \ge N$ , $N > P$ , $P = Q$ , $Q < R$ .
Chain them: $M \ge N > P = Q < R$ . The sign changes direction at $Q < R$ , so nothing fixes how $M$ compares with $R$ . Neither I nor II follows on its own.
But I ( $M \ge R$ ) and II ( $R > M$ ) cover every possibility for $M$ and $R$ , and they can't both be true.
So exactly one of them must hold: either I or II follows.
The Fast Way
Chain the decoded signs and look for a change of direction.
$M \ge N > P = Q < R$ : the signs turn at $Q < R$ , so nothing fixes M against R.
I ( $M \ge R$ ) and II ( $R > M$ ) cover every case for the same pair, so either I or II follows.
Q17Data Interpretation › Data SufficiencyMedium
Five analysts, Asha, Dev, Mehul, Tara, and Rohan, are sitting in a straight line facing north. The question is: Is Asha sitting immediately to the right of Dev? Statement I: Asha is second from the right end. Statement II: Dev is not at an end, and exactly one person sits between Dev and the person at the extreme right end. Choose the correct option.
Correct option CBoth statements together are sufficient, but neither alone is sufficient
Number the seats 1 to 5 from left to right. Facing north, your right is the higher number.
Statement I puts Asha in seat 4, but Dev could be anywhere, so it is not enough.
Statement II: one person sits between Dev and seat 5, so Dev is in seat 3. We still don't know where Asha is.
Together: Dev is in seat 3 and Asha in seat 4, so Asha is immediately to Dev's right. Yes.
So both statements together are needed, and neither works alone.
The Fast Way
Each statement fixes one person, but the question needs two.
I puts Asha in seat 4. II puts Dev in seat 3 (one seat between Dev and seat 5).
Together seat 4 is just right of seat 3: both statements together are sufficient, neither alone.
Q18Data Interpretation › Data SufficiencyMedium
Four startup teams, Veda, Nexo, Orion and Pixel, present in four different slots numbered 1 to 4. Question: Is Nexo presenting in slot 2? Statement I: Veda presents immediately before Nexo. Statement II: Nexo presents immediately before Orion, and Orion is not in slot 4.
Correct option CBoth statements together are sufficient, but neither alone is sufficient
Statement I: Veda just before Nexo allows Nexo in slot 2, 3 or 4. Not sufficient.
Statement II: Orion follows Nexo and isn't in slot 4, so Orion is in 2 or 3 and Nexo is in 1 or 2. Not sufficient.
Together: Veda, Nexo, Orion take three consecutive slots with Orion not in 4, so the order is Veda 1, Nexo 2, Orion 3, Pixel 4.
Nexo is in slot 2, so both statements together are sufficient, but neither alone is.
The Fast Way
Reject each statement with one counter-case, then combine.
I alone: Veda, Nexo fits slots 1, 2 (yes) or 2, 3 (no). II alone: Nexo, Orion fits 2, 3 (yes) or 1, 2 (no).
Together Veda, Nexo, Orion run in a row with Orion off slot 4, so slots 1, 2, 3: both together are sufficient, neither alone.
Q19Logical Reasoning › Blood RelationsEasy
In a family, names are to be treated as gender-neutral. Rivan says, “Sam is the child of my mother’s only brother. Sam’s spouse is Noor. Noor’s only sibling is Ira. Ira’s child is Zed. Pihu is the mother of Noor and Ira, and Kabir is Pihu’s spouse.” How is Zed related to Sam?
Correct option ANephew/niece (by marriage)
Only the chain from Sam to Zed matters; Rivan, Pihu and Kabir are background detail.
Sam's spouse is Noor, Noor's sibling is Ira, and Zed is Ira's child.
So Zed is the child of Sam's spouse's sibling: Zed is Sam's nephew or niece by marriage.
Since names are gender-neutral, the answer is written as nephew/niece rather than one of them.
The Fast Way
Trace only the link between the two people asked about.
From Sam: spouse Noor, then Noor's sibling Ira, then Ira's child Zed.
A spouse's sibling's child is a nephew or niece by marriage, so Zed is Sam's nephew/niece (by marriage).
Q20Logical Reasoning › Cubes & DiceMedium
A large wooden cube used as a display prop at a campus fest is painted blue on all six faces and then cut into $4 \times 4 \times 4$ equal smaller cubes. The organisers remove all corner cubes for decoration. Among the remaining smaller cubes, how many have exactly two painted faces?
Correct option D24
In a painted $n \times n \times n$ cube, the small cubes with exactly two painted faces sit on the 12 edges, excluding corners: $12(n - 2)$ .
For $n = 4$ : $12 \times 2 = 24$ .
The corner cubes have three painted faces, so removing them doesn't change this count; the faces they uncover on neighbouring cubes were inside the block and are unpainted.
So 24 of the remaining cubes have exactly two painted faces.
The Fast Way
Painted-cube rule: exactly two painted faces means an edge cube that is not a corner, $12(n-2)$ .
$12 \times (4 - 2) = 24$ .
Corners have three painted faces, so removing them changes nothing: 24.