All 10 SA previous year questions (PYQs) from the IPMAT Indore 2020 past year paper, with answers and full solutions.
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Q1:ipmat indore 2020QA › HCF & LCMEasySA · TITA
In a division problem, the product of the quotient and the remainder is 24 while their sum is 10. If the divisor is 5 then the dividend is __________.
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The Setup: We are dealing with the foundational division algorithm: Dividend=(Divisor×Quotient)+Remainder. The absolute golden rule of division is that the remainder must be strictly non-negative and less than the divisor (0≤r<d). We need to min-max the given sum and product to find the exact values before plugging them in.
Step 1: Formulate the system of equations. Let the quotient be q and the remainder be r. The problem gives us their product q×r=24 and their sum q+r=10.
Step 2: Solve the quadratic substitution. You can substitute r=10−q into the product equation:
q(10−q)=2410q−q2=24q2−10q+24=0
Factoring this gives (q−6)(q−4)=0. The roots are 6 and 4. So, our variables {q,r} are exactly {6,4} - no other pair of numbers has this sum and product.
Step 3: Enforce the remainder rule. We are told the divisor is 5. By the laws of division, r<5. Between 6 and 4, only 4 satisfies this condition. Therefore, it is a hard lock that r=4 and q=6.
Step 4: Execute the final calculation. Substitute everything back into the core algorithm:
Dividend=(5×6)+4Dividend=30+4=34Final Answer: 34
Q2:ipmat indore 2020QA › CirclesMediumSA · TITA
The shortest distance from the point (−4,3) to the circle (x2+y2=1) is __________.
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The Setup: This is a coordinate geometry problem finding the shortest path from an external point to a circle's boundary. Think of the circle as a forcefield; the shortest route is always a straight line aimed dead center, and we just subtract the forcefield's radius from the total distance.
Step 1: Deconstruct the circle's equation. The given equation is x2+y2=1. This perfectly matches the standard form (x−h)2+(y−k)2=r2. Thus, the center is the origin (0,0) and the radius is r=1=1.
Step 2: Calculate absolute distance to center. We use the Euclidean distance formula between the point (−4,3) and the center (0,0):
d=(x2−x1)2+(y2−y1)2d=(−4−0)2+(3−0)2d=(−4)2+(3)2d=16+9=25=5Step 3: Confirm the point is outside. This step is not optional - the *subtract the radius* move only works for an external point. Since d=5>r=1, the point sits outside the circle. (Had it been inside, the shortest distance would have been r−d instead.)
Step 4: Find the boundary distance. The total distance to the core is 5 units. Because the circle's radius occupies 1 unit of that path, the remaining gap from the point to the edge is simply the difference: 5−1=4.
Final Answer: 4
The value of (0.04log5(41+81+161+...)) is __________.
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The Setup: This expression looks incredibly toxic, but we can beat it by breaking it down into two phases: resolving the infinite geometric progression (GP) first, and then applying logarithm exponent rules to clean up the base.
Step 1: Evaluate the infinite GP. Look at the sequence inside the log: 41+81+161+…
This is an infinite GP where the first term a=41 and the common ratio r=21. Since ∣r∣<1 the series converges, so the sum to infinity S=1−ra applies:
S=1−2141=2141=41×12=21Step 2: Reconstruct the base and the logarithm. The original expression's base is 0.04, which translates to the fraction 1004=251. We can write this as an exponent: 5−2.
The logarithm's base is 5, which is 521.
So the exponent is log51/2(21).
Step 3: Simplify the exponent using log properties. We use the rule logbn(x)=n1logb(x):
log51/2(21)=211log5(21)=2log5(21)
Bring the multiplier inside as a power: log5((21)2)=log5(41).
Step 4: Combine the final expression. Substitute the cleaned-up base and exponent back together:
(5−2)log5(41)=5−2log5(41)
Move the −2 inside the log as a power:
5log5((41)−2)
Using the identity blogb(x)=x, the 5 and log5 cancel out. We are left with:
(41)−2=42=16Final Answer: 16
Suppose abca2b2c2a3−1b3−1c3−1=0, where a,b and c are distinct real numbers. If a=3, then the value of abc is:
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The Setup: This is a top-tier matrix determinant manipulation. When a determinant's column contains a sum or difference, it can be split into two separate determinants. The given a=3 is literal clickbait meant to throw you off; it has absolutely zero impact on the algebraic proof.
Step 1: Split the original determinant. We separate the third column (a3−1,b3−1,c3−1) into two determinants:
Δ=abca2b2c2a3b3c3−abca2b2c2111=0Step 2: Factor the first determinant. In the first matrix, notice that row 1 shares an a, row 2 shares a b, and row 3 shares a c. Let's factor them out completely:
abc111abca2b2c2Step 3: Align the second determinant. We want the second matrix to match the first, so we walk its last column to the front by two swaps.
Swap Column 3 with Column 2 (sign flips):
abca2b2c2111=−abc111a2b2c2
Now swap Column 2 with Column 1 (sign flips back):
=+111abca2b2c2
Two swaps, two sign flips, so the second determinant is exactly the same object as the first one's cofactor - no leftover minus sign.
Step 4: Factor the entire equation. Let the aligned matrix be V (a Vandermonde determinant). Our original equation Δ=0 now looks like this:
abc(V)−V=0⟹V(abc−1)=0
The Vandermonde determinant factors cleanly as V=(b−a)(c−a)(c−b). Because the problem states a, b and c are distinct real numbers, every one of those three brackets is non-zero, so V=0. Therefore the other multiplier must be zero:
abc−1=0⟹abc=1Final Answer: 1
Q5:ipmat indore 2020QA › ModulusMediumSA · TITA
The minimum value of f(x)=∣3−x∣+∣2+x∣+∣5−x∣ is equal to __________.
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The Setup: We are tasked with finding the global minimum of the function f(x)=∣3−x∣+∣2+x∣+∣5−x∣. In absolute value geometry, this represents the sum of distances from a coordinate x to three specific points on a number line, and the minimum lands at the median of those points. Rather than just quoting that rule, we can lock the answer down with a two-line proof.
Step 1: Standardize the terms. First, use the property ∣a−b∣=∣b−a∣ to rewrite the function so x is always first. This prevents sign errors.
f(x)=∣x−3∣+∣x−(−2)∣+∣x−5∣
The roots (the points on the number line) are exactly 3, −2 and 5. In ascending order: −2,3,5 - so the median is 3.
Step 2: Bound the outer pair. For the two extreme points, the triangle inequality gives a floor that no x can beat:
∣x+2∣+∣x−5∣≥∣(x+2)−(x−5)∣=7
with equality for every x in the interval [−2,5].
Step 3: Bound the middle term. Obviously ∣x−3∣≥0, with equality only at x=3. Adding the two bounds gives f(x)≥7+0=7 for all x.
Step 4: Show the floor is actually reached. The two equality conditions must hold at once, and they do: x=3 lies inside [−2,5]. Substituting:
f(3)=∣3−3∣+∣3−(−2)∣+∣3−5∣=0+5+2=7
The bound is achieved, so 7 is the true minimum and not merely a lower estimate.
Final Answer: 7
Ashok purchased pens and pencils in the ratio 2:3 during his first visit and paid Rs. 86 to the shopkeeper. During his second visit, he purchased pens and pencils in the ratio 4:1 and paid Rs. 112. The cost of a pen as well as a pencil in rupees is a positive integer. If Ashok purchased four pens during his second visit, then the amount he paid in rupees for the pens during the second visit is __________.
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The Setup: This is a system of linear equations hiding behind ratios, with a strict Diophantine constraint: the prices of the pens and pencils must be positive integers. We have to decode two separate shopping trips, set up the algebra, and pin down the only integer solution.
Step 1: Formalize the second visit. On the second visit, the ratio of pens to pencils is 4:1, and the total cost is Rs. 112. We are explicitly told Ashok bought 4 pens. By the ratio, he must have bought exactly 1 pencil.
Let x be the price of a pen, and y be the price of a pencil.
4x+1y=112
Isolate y so we can use it as a substitution later:
y=112−4xStep 2: Formalize the first visit. The ratio of pens to pencils bought is 2:3, costing Rs. 86. We don't know the exact quantities, just the ratio. So, let the quantities be 2k pens and 3k pencils, where k is a positive integer constant.
2kx+3ky=86⟹k(2x+3y)=86Step 3: Substitute, then kill every wrong branch. Plug the isolated y from Step 1 into the Step 2 equation:
k(2x+3(112−4x))=86k(2x+336−12x)=86⟹k(336−10x)=86
So 336−10x=k86, which forces k to be a positive factor of 86. The factors are 1,2,43,86, giving k86∈{86,43,2,1} respectively. Now the kill shot: the left side 336−10x is 6 short of a multiple of ten, so it always ends in the digit 6 - and of those four candidates, only 86 does. That single observation eliminates k=2, k=43 and k=86 outright, with no trial-and-error, leaving k=1 as the one surviving branch.
336−10x=86⟹10x=250⟹x=25
Then y=112−4(25)=112−100=12. Both prices are positive integers, and since only one branch survived, this pair is unique.
Step 4: Reconcile against the first visit. Uniqueness is worth nothing if the numbers don't actually satisfy the original trip. With k=1 he bought 2 pens and 3 pencils: 2(25)+3(12)=50+36=86. That is exactly what he paid, so the pair holds.
Step 5: Calculate the final request. The prompt asks for the total amount paid *specifically for the pens* during the second visit.
He bought 4 pens at Rs. 25 each: 4×25=100.
Final Answer: 100
In a four-digit number, the product of the thousands digit and units digit is zero while their difference is 7. Product of the middle digits is 18. The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit. The four-digit number is __________.
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The Setup: This is a cryptographic math puzzle. We need to build a four-digit integer, let's call it ABCD (where A is thousands, B is hundreds, C is tens, D is units). We will decode the rules one by one to lock in each place value.
**Step 1: Solve the outer digits (A and D).** We are given two conditions: A×D=0 and the absolute difference is 7.
For the product to be zero, either A=0 or D=0. Since it's a four-digit number, the leading digit A cannot be 0. Therefore, D=0.
Because A must be positive, the difference of 7 means A−0=7, locking in A=7.
**Step 2: Analyze the middle digits (B and C).** We are told their product is 18: B×C=18.
Since B and C are single digits, the only possible ordered pairs that multiply to 18 are (2,9), (9,2), (3,6) and (6,3).
Step 3: Apply the relational constraint. The prompt states: 'The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit.'
Translated to algebra, this means A−D=B−C.
We already know A−D=7−0=7, so we need B−C=7.
Run that filter across the four candidate pairs: 2−9=−7, 9−2=7, 3−6=−3, 6−3=3. Exactly one survives - 9−2=7 - which locks in B=9 and C=2. Note the filter is a signed difference, not an absolute one, so (2,9) is genuinely rejected rather than being a second answer.
Step 4: Assemble the final integer. We have A=7, B=9, C=2 and D=0. Concatenate them in order.
Final Answer: 7920
Q8:ipmat indore 2020QA › Set TheoryMediumSA · TITA
Out of 80 students who appeared for the school exams in Mathematics (M), Physics (P) and Chemistry (C), 50 passed M, 30 passed P and 40 passed C. At most 20 students passed M and P, at most 20 students passed P and C, and at most 20 students passed C and M. The maximum number of students who could have passed all three exams is __________.
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The Setup: We are min-maxing a 3-circle Venn diagram problem. We have a total population of 80 students across Mathematics (M), Physics (P) and Chemistry (C). We need to maximise the triple intersection (students who passed all three) without violating the 'at most 20' limits placed on the pairwise overlaps.
Step 1: Define the variables and totals. Let x=n(M∩P∩C) be the number of students who passed all three. We want to push x as high as possible.
The individual subject totals are n(M)=50, n(P)=30 and n(C)=40.
Step 2: Establish the ceiling. By the definition of intersection, anyone who passed all three also passed each pair, so the triple intersection is a subset of every pairwise intersection:
x≤n(M∩P),x≤n(P∩C),x≤n(M∩C)
The problem caps all three pairwise intersections at 20. Therefore x≤20 - this is a hard upper bound, no construction can beat it.
Step 3: Build a scenario that hits the ceiling. An upper bound is useless until we show it is reachable. Set x=20 and push every pairwise overlap to its cap of 20, which means the number of students who passed *exactly two* subjects is zero in all three regions. The *single-subject-only* groups then fall out:
* Passed M only: 50−20=30
* Passed P only: 30−20=10
* Passed C only: 40−20=20Step 4: Verify the headcount closes. Total students accounted for = (M only) + (P only) + (C only) + (all three):
30+10+20+20=80
This lands exactly on the 80 students who appeared, with no region going negative and every pairwise intersection sitting at 20 rather than above it. The scenario is structurally sound, so the ceiling of 20 is genuinely achievable and is therefore the maximum.
Final Answer: 20
Q9:ipmat indore 2020QA › Time, Speed & DistanceMediumSA · TITA
Two friends run a 3-kilometer race along a circular course of length 300 meters. If their speeds are in the ratio 3:2, the number of times the winner passes the other is __________.
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The Setup: This is a relative speed problem on a circular track. When runners start together and run the same way, the faster one *passes* the slower one every time he gains exactly one full lap. So the whole question reduces to: how many whole laps of lead does the winner build up before the race ends?
Step 1: Determine the total race parameters. The track is 300 metres long and the race is 3 kilometres, which equals 3000 metres.
Laps to finish=3003000=10
The winner finishes the instant he completes his 10th lap, and that is when the clock stops for both runners.
Step 2: Leverage the speed ratio. The speeds are in the ratio 3:2. Both runners are on the track for the same duration, and at constant speed distance is directly proportional to speed, so their distances are in that same 3:2 ratio.
Step 3: Locate the slower runner at the finish. When the winner has covered 10 laps:
Slower runner=32×10=320 laps
Keep this as an exact fraction - 320 laps, not a rounded 6.67 - because the final count depends on which side of a whole number the lead lands.
Step 4: Count the overtakes. The lead the winner builds over the full race is the difference in laps covered:
Lead=10−320=330−20=310 laps
The winner overtakes once for each whole lap of lead he gains, i.e. as the lead sweeps past 1, 2 and 3. Since 310=331 is strictly greater than 3 but short of 4, the third overtake happens before the finish line and a fourth never does.
Number of passes=⌊310⌋=3Final Answer: 3
Out of 13 objects, 4 are indistinguishable and the rest are distinct. The number of ways we can choose 4 objects out of 13 objects is __________.
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The Setup: This combinatorics problem mixes indistinguishable (identical) objects with distinct ones. The standard combination formula (rn) only works for choosing distinct items. When dealing with indistinguishable objects, choosing one of them gives the same exact outcome as choosing another, so there is only 1 way to select any given quantity of them. We must break this down into cases based on how many identical items we draft into the final squad of 4.
Step 1: Catalog the inventory. Total pool =13 objects.
Indistinguishable objects =4. Distinct objects =13−4=9.
We need to form a team of exactly 4 objects.
Step 2: Calculate combinations case-by-case. We iterate through every possible quantity of indistinguishable objects, from 0 up to 4 (we cannot take more than the 4 that exist).
* Case 1 (0 identical, 4 distinct):(49)=4×3×2×19×8×7×6=126 ways
* Case 2 (1 identical, 3 distinct):1×(39)=3×2×19×8×7=84 ways
* Case 3 (2 identical, 2 distinct):1×(29)=2×19×8=36 ways
* Case 4 (3 identical, 1 distinct):1×(19)=9 ways
* Case 5 (4 identical, 0 distinct):1×(09)=1 way
Step 3: Aggregate the total possibilities. The cases are mutually exclusive (they differ in how many identical objects were taken), so we add them:
126+84+36+9+1=256Step 4: Cross-check with a symmetry argument. That total is (09)+(19)+(29)+(39)+(49) - the lower half of row 9 of Pascal's triangle. Since (k9)=(9−k9), the ten entries of that row pair off perfectly into two equal halves, so this sum is exactly half the row total:
229=2512=256
Two independent routes, same number, so the count is confirmed.
Final Answer: 256