Attempt PYPs as full mocks →
Past Year Questions › IPMAT Indore › 2021 › MCQ

IPMAT Indore 2021 — MCQ

All 20 MCQ previous year questions (PYQs) from the IPMAT Indore 2021 past year paper, with answers and full solutions.

Free SolutionsNo Login20 Questions
Q1:ipmat indore 2021QA › FunctionsMediumMCQ · MCQ
Suppose that a real-valued function f(x)f(x) of real numbers satisfies f(x+xy)=f(x)+f(xyf(x + xy) = f(x) + f(xy) for all real x,y,x, y, and that f(2020)=1f(2020) = 1. Compute f(2021)f(2021).
  • A20212020\frac{2021}{2020}
  • B20202019\frac{2020}{2019}
  • C1
  • D20202021\frac{2020}{2021}
Pick an option to attempt
The Setup: We are dealing with functional equations, which is essentially algebra's version of reverse-engineering a game engine. We need to find the core behavior of f(x)f(x) by plugging in strategic dummy variables until the underlying algorithm reveals itself. Step 1: Test the waters by setting y=0y=0 to find the baseline. f(x+0)=f(x)+f(0)  ⟹  f(x)=f(x)+f(0)  ⟹  f(0)=0f(x+0)=f(x)+f(0) \implies f(x)=f(x)+f(0) \implies f(0)=0 Step 2: Unlock the additive property. For any x≠0x \neq 0 and any real zz, just pick y=z/xy=z/x - then xy=zxy=z, and the given equation morphs into: f(x+z)=f(x)+f(z)f(x+z)=f(x)+f(z) Together with Step 1 (which covers x=0x=0), ff is additive across the reals. Step 3: Stack that property. Applying it repeatedly gives f(nt)=nf(t)f(nt)=nf(t) for every positive integer nn. That is the *only* thing we need - so we never have to assume ff is linear or even continuous, which would be an unearned leap. Step 4: Use the given stat f(2020)=1f(2020)=1, taking t=1t=1 and n=2020n=2020. 2020f(1)=f(2020)=1  ⟹  f(1)=120202020f(1)=f(2020)=1 \implies f(1)=\frac{1}{2020} Step 5: Calculate the final target with n=2021n=2021. f(2021)=2021f(1)=20212020f(2021)=2021f(1)=\frac{2021}{2020} Final Answer: 20212020\frac{2021}{2020}
Q2:ipmat indore 2021QA › LogarithmsEasyMCQ · MCQ
Suppose that log⁡2[log⁡3(log⁡4a)]=log⁡3[log⁡4(log⁡2b)]=log⁡4[log⁡2(log⁡3c)]=0\log_2[\log_3 (\log_4a)] = \log_3 [\log_4 (\log_2b)] = \log_4 [\log_2 (\log_3c)] = 0 then the value of a+b+ca + b + c is
  • A105
  • B71
  • C89
  • D37
Pick an option to attempt
The Setup: This is a logarithm inception puzzle. We just need to peel it back layer by layer like an onion. The golden rule here: if log⁡x(Y)=0\log_x(Y)=0, then Y=x0=1Y=x^0=1. Step 1: Crack the code for aa. log⁡2[log⁡3(log⁡4a)]=0  ⟹  log⁡3(log⁡4a)=20=1\log_2[\log_3(\log_4a)]=0 \implies \log_3(\log_4a)=2^0=1 log⁡4a=31=3  ⟹  a=43=64\log_4a=3^1=3 \implies a=4^3=64 Step 2: Crack the code for bb. log⁡3[log⁡4(log⁡2b)]=0  ⟹  log⁡4(log⁡2b)=30=1\log_3[\log_4(\log_2b)]=0 \implies \log_4(\log_2b)=3^0=1 log⁡2b=41=4  ⟹  b=24=16\log_2b=4^1=4 \implies b=2^4=16 Step 3: Crack the code for cc. log⁡4[log⁡2(log⁡3c)]=0  ⟹  log⁡2(log⁡3c)=40=1\log_4[\log_2(\log_3c)]=0 \implies \log_2(\log_3c)=4^0=1 log⁡3c=21=2  ⟹  c=32=9\log_3c=2^1=2 \implies c=3^2=9 Step 4: Add the squad together to find a+b+ca+b+c. a+b+c=64+16+9=89a+b+c=64+16+9=89 Final Answer: 89
Q3:ipmat indore 2021QA › Progression & SeriesMediumMCQ · MCQ
Let SnS_n be sum of the first nn terms of an A.P. If S5=S9S_5 = S_9, what is the ratio of a3:a5a_3 : a_5
  • A9:59:5
  • B5:95:9
  • C3:53:5
  • D5:35:3
Pick an option to attempt
The Setup: We are working with Arithmetic Progressions. The standard sum formula is Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}(2a+(n-1)d). We need to build an algebraic equation equating the two given sums to find the direct relationship between the first term (aa) and the common difference (dd). Step 1: Expand both sides using the sum formula. S5=52(2a+4d)=5(a+2d)=5a+10dS_5=\frac{5}{2}(2a+4d)=5(a+2d)=5a+10d S9=92(2a+8d)=9(a+4d)=9a+36dS_9=\frac{9}{2}(2a+8d)=9(a+4d)=9a+36d Step 2: Set them equal and isolate aa in terms of dd. 5a+10d=9a+36d5a+10d=9a+36d −4a=26d  ⟹  2a=−13d  ⟹  a=−6.5d-4a=26d \implies 2a=-13d \implies a=-6.5d Step 3: Find the expressions for the specific requested terms, a3a_3 and a5a_5. a3=a+2d=−6.5d+2d=−4.5da_3=a+2d=-6.5d+2d=-4.5d a5=a+4d=−6.5d+4d=−2.5da_5=a+4d=-6.5d+4d=-2.5d Step 4: Calculate the ratio. The negative signs and the dd variables cancel out cleanly. Ratio=−4.5d−2.5d=4.52.5=4525=95\text{Ratio}=\frac{-4.5d}{-2.5d}=\frac{4.5}{2.5}=\frac{45}{25}=\frac{9}{5} Final Answer: 9:59:5
Q4:ipmat indore 2021QA › Matrices & DeterminantsMediumMCQ · MCQ
If A,BA, B and A+BA + B are non singular matrices and AB=BAAB = BA then 2A−B−A(A+B)−1A+B(A+B)−1B2A - B - A(A + B)^{-1}A + B(A + B)^{-1} B equals
  • AA
  • BB
  • CA + B
  • DI
Pick an option to attempt
The Setup: Matrix algebra usually blocks you from treating matrix multiplication like standard variables because AB≠BAAB \neq BA. However, the prompt gives us the AB=BAAB=BA buff. This means AA and BB commute. Even better, since A(A+B)=A2+AB=A2+BA=(A+B)AA(A+B)=A^2+AB=A^2+BA=(A+B)A, matrix AA also commutes with (A+B)(A+B) and its inverse (A+B)−1(A+B)^{-1}. Step 1: Group the complicated inverse terms together to see what we're working with. Because AA commutes with (A+B)−1(A+B)^{-1}, we can rewrite A(A+B)−1AA(A+B)^{-1}A as A2(A+B)−1A^2(A+B)^{-1}. The identical argument runs for BB (it commutes with A+BA+B for exactly the same reason), so B(A+B)−1BB(A+B)^{-1}B becomes B2(A+B)−1B^2(A+B)^{-1}. −A2(A+B)−1+B2(A+B)−1=(B2−A2)(A+B)−1-A^2(A+B)^{-1}+B^2(A+B)^{-1}=(B^2-A^2)(A+B)^{-1} Step 2: Factor the difference of squares. Because AB=BAAB=BA, the cross terms cancel out perfectly, allowing standard factorization. B2−A2=(B−A)(B+A)B^2-A^2=(B-A)(B+A) Step 3: Substitute the factored form back in. Notice that matrix addition is always commutative, so (B+A)=(A+B)(B+A)=(A+B). (B−A)(A+B)(A+B)−1(B-A)(A+B)(A+B)^{-1} Since (A+B)(A+B) multiplied by its inverse is the Identity matrix (II), this entire chunk simplifies to just (B−A)(B-A). Step 4: Plug this beautifully simplified chunk back into the main equation. 2A−B+(B−A)2A-B+(B-A) 2A−B+B−A=A2A-B+B-A=A Final Answer: A
Q5:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ
If the angles A,B,CA, B, C of a triangle are in arithmetic progression such that sin⁡(2A+B)=1/2\sin(2A + B) = 1/2 then sin⁡(B+2C)\sin(B + 2C) is equal to
  • A−12\frac{-1}{2}
  • B12\frac{1}{2}
  • C−12\frac{-1}{\sqrt{2}}
  • D32\frac{3}{\sqrt{2}}
Pick an option to attempt
The Setup: Geometry meets Trigonometry. The angles of a triangle always sum to 180∘180^\circ. If they are in an Arithmetic Progression, the middle angle acts as the perfect average, completely locking in its value from the start. Step 1: Find angle BB. Since AA, BB, CC are in AP: A+C=2BA+C=2B Substitute this into the triangle angle sum theorem: A+B+C=180∘  ⟹  2B+B=180∘  ⟹  3B=180∘  ⟹  B=60∘A+B+C=180^\circ \implies 2B+B=180^\circ \implies 3B=180^\circ \implies B=60^\circ Step 2: Use the sine condition to find angle AA. We know sin⁡(2A+B)=1/2\sin(2A+B)=1/2. Since A+C=120∘A+C=120^\circ, angle AA is stuck below 120∘120^\circ, so 2A+60∘2A+60^\circ never climbs past 300∘300^\circ - that leaves exactly two candidates, 30∘30^\circ and 150∘150^\circ. * If 2A+60∘=30∘  ⟹  2A=−30∘2A+60^\circ=30^\circ \implies 2A=-30^\circ (Angles in a triangle must be positive, so this is a wipe). * If 2A+60∘=150∘  ⟹  2A=90∘  ⟹  A=45∘2A+60^\circ=150^\circ \implies 2A=90^\circ \implies A=45^\circ. Step 3: Find angle CC. C=180∘−(A+B)=180∘−(45∘+60∘)=75∘C=180^\circ-(A+B)=180^\circ-(45^\circ+60^\circ)=75^\circ Step 4: Plug the stats into the final equation sin⁡(B+2C)\sin(B+2C). B+2C=60∘+2(75∘)=60∘+150∘=210∘B+2C=60^\circ+2(75^\circ)=60^\circ+150^\circ=210^\circ sin⁡(210∘)=sin⁡(180∘+30∘)=−sin⁡(30∘)=−12\sin(210^\circ)=\sin(180^\circ+30^\circ)=-\sin(30^\circ)=-\frac{1}{2} Final Answer: −12\frac{-1}{2}
Q6:ipmat indore 2021QA › Unit DigitMediumMCQ · MCQ
The unit digit in (743)85−(525)37+(987)96(743)^{85} - (525)^{37} + (987)^{96} is ________
  • A9
  • B3
  • C1
  • D5
Pick an option to attempt
The Setup: Unit digit mechanics operate on a strict modulo-10 cyclicity. We don't care about the massive numbers; we only care about the final digit of the base and the remainder of the exponent when divided by its cycle length. Step 1: Analyze the first term: 74385743^{85}. The unit digit depends entirely on 3853^{85}. The powers of 33 cycle in a 4-step loop (33, 99, 77, 11). Divide the exponent 8585 by 44 to find the remainder. 85=4×21+185=4\times21+1 Since the remainder is 11, it hits the first step of the cycle. The unit digit is 31=33^1=3. Step 2: Analyze the second term: 52537525^{37}. This is a freebie. Any positive integer power of a number ending in 55 will always end in 55. The unit digit is 55. Step 3: Analyze the third term: 98796987^{96}. The unit digit depends on 7967^{96}. The powers of 77 also cycle in a 4-step loop (77, 99, 33, 11). Divide 9696 by 44. 96=4×24+096=4\times24+0 Since it divides perfectly (remainder 00), it lands on the final step of the cycle. The unit digit is 11. Step 4: Combine the unit digits. 3−5+1=−13-5+1=-1 Since we can't output a negative unit digit in a standard integer, we simulate borrowing a 1010 from the next placeholder column over (just like elementary school subtraction). 10−1=910-1=9 Final Answer: 9
Q7:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ
The set of all real value of pp for which the equation 3sin⁡2x+12cos⁡x−3=p3 \sin^2x + 12 \cos x - 3 = p has at least one solution is
  • A[−12,12][-12, 12]
  • B[−12,9][-12, 9]
  • C[−15,9][-15, 9]
  • D[−15,12][-15, 12]
Pick an option to attempt
The Setup: We are converting a trig equation into a standard quadratic to find its absolute range (the min and max values of pp). Since the equation involves both sine and cosine, we use the identity sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x to unify the variables. Step 1: Substitute and simplify the expression. Let f(x)=pf(x)=p. 3(1−cos⁡2x)+12cos⁡x−3=p3(1-\cos^2x)+12\cos x-3=p 3−3cos⁡2x+12cos⁡x−3=p3-3\cos^2x+12\cos x-3=p −3cos⁡2x+12cos⁡x=p-3\cos^2x+12\cos x=p Step 2: Set up a boundary replacement. Let t=cos⁡xt=\cos x. Because of the constraints of the cosine function, tt is strictly bounded: t∈[−1,1]t\in[-1,1]. Our new function is g(t)=−3t2+12tg(t)=-3t^2+12t. We need the absolute min and max of this quadratic on the locked interval [−1,1][-1,1]. Step 3: Find the vertex of the parabola. For a quadratic at2+bt+cat^2+bt+c, the vertex occurs at t=−b/(2a)t=-b/(2a). t=−122(−3)=−12−6=2t=\frac{-12}{2(-3)}=\frac{-12}{-6}=2 The vertex sits at t=2t=2, which is completely outside our playable zone of [−1,1][-1,1]. Because it's a downward-opening parabola (negative aa), it is strictly increasing across our entire [−1,1][-1,1] interval. Step 4: Test the boundaries to find the range. * If t=−1t=-1 (Minimum): g(−1)=−3(−1)2+12(−1)=−3−12=−15g(-1)=-3(-1)^2+12(-1)=-3-12=-15. * If t=1t=1 (Maximum): g(1)=−3(1)2+12(1)=−3+12=9g(1)=-3(1)^2+12(1)=-3+12=9. The function seamlessly spans every value between these two extremes. Final Answer: [−15,9][-15, 9]
Q8:ipmat indore 2021QA › QuadrilateralsMediumMCQ · MCQ
ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If triangles AOB and COD have areas 4 and 9 respectively, then the minimum area that ABCD can have is
  • A26
  • B25
  • C21
  • D16
Pick an option to attempt
The Setup: This relies on a sick geometric property. For *any* quadrilateral where diagonals intersect at OO, the products of the areas of opposite triangles are perfectly equal. Let the areas be A1A_1, A2A_2, A3A_3, A4A_4 circling the intersection. We are given A1=4A_1=4 and A3=9A_3=9. We need to optimize A2A_2 and A4A_4 to minimize the total sum. Step 1: Apply the cross-product area rule. A1×A3=A2×A4A_1\times A_3=A_2\times A_4 4×9=A2×A4  ⟹  A2×A4=364\times9=A_2\times A_4 \implies A_2\times A_4=36 Step 2: Set up the total area equation. Total Area=A1+A2+A3+A4=4+A2+9+A4=13+A2+A4\text{Total Area}=A_1+A_2+A_3+A_4=4+A_2+9+A_4=13+A_2+A_4 Step 3: Deploy the AM-GM (Arithmetic Mean-Geometric Mean) inequality to minimize the variable part. AM-GM states that for any non-negative numbers, their sum is bounded by their product: x+y≥2xyx+y \ge 2\sqrt{xy}. A2+A4≥2A2×A4A_2+A_4 \ge 2\sqrt{A_2\times A_4} A2+A4≥236=2(6)=12A_2+A_4 \ge 2\sqrt{36}=2(6)=12 The absolute minimum sum these two triangles can mathematically contribute is 1212 (which occurs when A2=A4=6A_2=A_4=6). Step 4: Add the locked constants back to find the overall minimum area. Min Total Area=13+12=25\text{Min Total Area}=13+12=25 Final Answer: 25
Q9:ipmat indore 2021QA › Ratio, Proportion & VariationEasyMCQ · MCQ
The highest possible value of the ratio of a four-digit number and the sum of its four digits is
  • A1000
  • B277.75
  • C900.1
  • D999
Pick an option to attempt
The Setup: We want to min-max a fraction mathematically. Let the four-digit number be formatted as 1000a+100b+10c+d1000a+100b+10c+d. We want to maximize the ratio 1000a+100b+10c+da+b+c+d\frac{1000a+100b+10c+d}{a+b+c+d}. Step 1: Use an algebraic reduction to isolate the variables. Let S=a+b+c+dS=a+b+c+d. We can rewrite the numerator in terms of SS. 1000a+100b+10c+d=1000(S−b−c−d)+100b+10c+d1000a+100b+10c+d=1000(S-b-c-d)+100b+10c+d =1000S−1000b−1000c−1000d+100b+10c+d=1000S-1000b-1000c-1000d+100b+10c+d =1000S−900b−990c−999d=1000S-900b-990c-999d Step 2: Reconstruct the ratio with the new expression. Ratio=1000S−900b−990c−999dS\text{Ratio}=\frac{1000S-900b-990c-999d}{S} Ratio=1000−900b+990c+999dS\text{Ratio}=1000-\frac{900b+990c+999d}{S} Step 3: Optimize the equation. To maximize the overall ratio, we must completely minimize the fraction being subtracted. Since bb, cc, and dd are digits (meaning they are non-negative integers ≥0\ge 0), the absolute smallest value for the subtracted term is 00. This occurs when we hard-lock the nerfed variables to zero: b=0b=0, c=0c=0, d=0d=0. Step 4: Calculate the ratio with the optimized stats. Ratio=1000−0=1000\text{Ratio}=1000-0=1000 (For example, if the number is 90009000, the sum of digits is 99. 9000/9=10009000/9=1000. It mathematically cannot get higher than this). Final Answer: 1000
Q10:ipmat indore 2021QA › Straight LinesMediumMCQ · MCQ
Consider the polynomials f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where a>0,b,ca > 0, b, c are real, g(x)=−2xg(x) = -2x. If f(x)f(x) cuts the x-axis at (−2,0)(-2, 0) and g(x)g(x) passes through (a,b)(a, b), then the minimum value of f(x)+9a+1f(x) + 9a + 1 is
  • A0
  • B1
  • C2
  • D3
Pick an option to attempt
The Setup: We need to trace the clues to build the quadratic equation f(x)f(x) entirely in terms of the variable aa. Once we have the pure function, we can calculate its vertex to find its lowest possible floor, and then plug that into the final target expression. Step 1: Use the g(x)g(x) coordinate to crack the value of bb. The line g(x)g(x) passes through (a,b)(a,b), so we literally just plug those coordinates in. b=−2(a)  ⟹  b=−2ab=-2(a) \implies b=-2a Step 2: Use the x-intercept to crack cc. The function f(x)f(x) hits the axis at x=−2x=-2, meaning f(−2)=0f(-2)=0. a(−2)2+b(−2)+c=0a(-2)^2+b(-2)+c=0 4a−2b+c=04a-2b+c=0 Substitute b=−2ab=-2a into this equation: 4a−2(−2a)+c=0  ⟹  4a+4a+c=0  ⟹  8a+c=0  ⟹  c=−8a4a-2(-2a)+c=0 \implies 4a+4a+c=0 \implies 8a+c=0 \implies c=-8a Step 3: Construct the final quadratic and find its minimum. f(x)=ax2−2ax−8a=a(x2−2x−8)f(x)=ax^2-2ax-8a=a(x^2-2x-8) For an upward-facing parabola (a>0a>0), the minimum value occurs exactly at its vertex x=−bnew/(2anew)x=-b_{new}/(2a_{new}). x=−(−2a)2(a)=2a2a=1x=\frac{-(-2a)}{2(a)}=\frac{2a}{2a}=1 Step 4: Calculate the minimum y-value of f(x)f(x) by plugging in x=1x=1. f(1)=a(12)−2a(1)−8a=a−2a−8a=−9af(1)=a(1^2)-2a(1)-8a=a-2a-8a=-9a Step 5: Evaluate the final target expression f(x)+9a+1f(x)+9a+1 at its minimum point. (−9a)+9a+1=1(-9a)+9a+1=1 Final Answer: 1
Q11:ipmat indore 2021QA › Set TheoryMediumMCQ · MCQ
In a city, 50% of the population can speak in exactly one language among Hindi, English and Tamil, while 40% of the population can speak in at least two of these three languages. Moreover, the number of people who cannot speak in any of these three languages is twice the number of people who can speak in all these three languages. If 52% of the population can speak in Hindi and 25% of the population can speak exactly in one language among English and Tamil, then the percentage of the population who can speak in Hindi and in exactly one more language among English and Tamil is
  • A22%
  • B25%
  • C30%
  • D38%
Pick an option to attempt
The Setup: This is a 3-set Venn Diagram problem. Think of the three languages as mutual circles in a social network. We're going to break down the total population (100%) into specific overlapping zones to find the targeted demographic. Step 1: Establish the baseline stats. We know 50% speak exactly one language, and 40% speak at least two. That means the total active user base speaking *any* of the three languages is 50%+40%=90%50\%+40\%=90\%. The people speaking zero languages (the 'None' group) make up the remainder: 100%−90%=10%100\%-90\%=10\%. Step 2: Isolate the top-tier group (All 3). The prompt states the 'None' group (10%) is exactly twice the size of the 'All 3' group. So, All 3=10%/2=5%\text{All } 3=10\%/2=5\%. Step 3: Break down the 'At least two' category. This group (40%) includes people speaking exactly two languages AND people speaking all three. Exactly Two=40%−5%=35%\text{Exactly Two}=40\%-5\%=35\% Step 4: Find the solo Hindi speakers. The 'Exactly One' group is 50% total. We are told 25% speak exactly one language among English and Tamil (meaning Only E + Only T = 25%). Therefore, the Only Hindi group is the remainder: Only Hindi=50%−25%=25%\text{Only Hindi}=50\%-25\%=25\% Step 5: Solve for the target. We know 52% of the population speaks Hindi in total. The total Hindi circle is made up of: Only Hindi (25%) + All 3 (5%) + Hindi & Exactly One More (which is our target). 52%=25%+5%+Target52\%=25\%+5\%+\text{Target} 52%=30%+Target  ⟹  Target=22%52\%=30\%+\text{Target} \implies \text{Target}=22\% Final Answer: 22%
Q12:ipmat indore 2021QA › Time, Speed & DistanceEasyMCQ · MCQ
A train left point A at 12 noon. Two hours later, another train started from point A in the same direction. It overtook the first train at 8 PM. It is known that the sum of the speeds of the two trains is 140 km/hr. Then, at what time would the second train overtake the first train, if instead the second train had started from point A in the same direction 5 hours after the first train? Assume that both the trains travel at constant speeds.
  • A3 AM the next day
  • B4 AM the next day
  • C8 AM the next day
  • D11 PM the same day
Pick an option to attempt
The Setup: This is a Time, Speed, and Distance chase sequence. We have a slow train (Train 1) and a fast train (Train 2) catching up. First, we need to mathematically unmask their individual speeds using the 8 PM catch-up event. Step 1: Calculate travel times for the first scenario. Train 1 left at 12 noon and got caught at 8 PM, meaning it traveled for 8 hours. Train 2 left 2 hours later (2 PM) and caught up at 8 PM, meaning it traveled for 6 hours. Step 2: Equate their distances. Since Train 2 caught Train 1, they covered the exact same distance. Let their speeds be V1V_1 and V2V_2. 8V1=6V2  ⟹  4V1=3V2  ⟹  V2=43V18V_1=6V_2 \implies 4V_1=3V_2 \implies V_2=\frac{4}{3}V_1 Step 3: Use the combined speed stat to solve for V1V_1. V1+V2=140V_1+V_2=140 V1+43V1=140  ⟹  73V1=140  ⟹  V1=60 km/hrV_1+\frac{4}{3}V_1=140 \implies \frac{7}{3}V_1=140 \implies V_1=60 \text{ km/hr} Since V1V_1 is 6060, V2=140−60=80 km/hrV_2=140-60=80 \text{ km/hr}. Step 4: Set up the alternate timeline. Train 2 now starts 5 hours late. Let tt be the total hours Train 1 travels until it gets caught. Train 2 will have traveled t−5t-5 hours. Equate the new distances: 60t=80(t−5)60t=80(t-5) 60t=80t−400  ⟹  20t=400  ⟹  t=20 hours60t=80t-400 \implies 20t=400 \implies t=20 \text{ hours} Step 5: Map the time back to the clock. Train 1 started at 12 noon. 20 hours later lands us exactly at 8 AM the next day. Final Answer: 8 AM the next day
Q13:ipmat indore 2021QA › Permutation & CombinationEasyMCQ · MCQ
The number of 5-digit numbers consisting of distinct digits that can be formed such that only odd digits occur at odd places is
  • A5250
  • B6240
  • C2520
  • D3360
Pick an option to attempt
The Setup: Permutations and Combinations puzzle with strict placement rules. We are building a 5-digit number (Positions 1, 2, 3, 4, 5). The odd positions (1st, 3rd, 5th) are strictly reserved for odd digits. Step 1: Draft the odd squad. We have exactly 55 odd digits available (11, 33, 55, 77, 99). We need to select 33 of them and arrange them into the 33 odd slots. Number of ways to fill odd places = P(5,3)=5×4×3=60P(5,3)=5\times4\times3=60 ways. Step 2: Assess the remaining free agents. We have 55 even digits (00, 22, 44, 66, 88) and 22 leftover odd digits that didn't get drafted into the odd slots. That gives us a pool of 77 distinct digits. Step 3: Draft for the even slots (2nd and 4th). The prompt says 'only odd digits occur at odd places', which means odd places are restricted, but even places are fully unrestricted. They can accept any remaining digit, including zero, because the 1st digit is already guaranteed to be a non-zero odd digit. We select 22 digits from the pool of 77 and arrange them. Number of ways to fill even places = P(7,2)=7×6=42P(7,2)=7\times6=42 ways. Step 4: Multiply the independent events to find the total combinations. Total Numbers=60×42=2520\text{Total Numbers}=60\times42=2520 Final Answer: 2520
Q14:ipmat indore 2021QA › Permutation & CombinationEasyMCQ · MCQ
There are 10 points in the plane, of which 5 points are collinear and no three among the remaining are collinear. Then the number of distinct straight lines that can be formed out of these 10 points is
  • A10
  • B25
  • C35
  • D36
Pick an option to attempt
The Setup: Geometric Combinatorics. To draw a straight line, you literally just need to connect 22 points. Normally, you'd calculate all possible pairs. But when points are collinear (sitting on the exact same line), connecting any two of them just redraws the exact same line. We have to filter out the duplicates. Step 1: Calculate the raw, unfiltered max number of lines. Choosing any 22 points out of 1010 gives us: (102)=10×92=45 possible pairs\binom{10}{2}=\frac{10\times9}{2}=45 \text{ possible pairs} Step 2: Calculate the overlapping spam. The 55 collinear points are generating redundant lines. If we picked pairs exclusively from these 55 points, we'd get: (52)=5×42=10 redundant pairs\binom{5}{2}=\frac{5\times4}{2}=10 \text{ redundant pairs} Step 3: Clean up the geometry. We subtract all 1010 redundant pairs from our total because they don't form unique lines. However, we still have to add 11 back in to represent the single, massive straight line that all 55 of those points are sitting on. Total Distinct Lines=45−10+1=36\text{Total Distinct Lines}=45-10+1=36 Final Answer: 36
Q15:ipmat indore 2021QA › Straight LinesMediumMCQ · MCQ
The x-intercept of the line that passes through the intersection of the lines x+2y=4x + 2y = 4 and 2x+3y=62x + 3y = 6, and is perpendicular to the line 3x−y=23x - y = 2 is
  • A2
  • B0.5
  • C4
  • D6
Pick an option to attempt
The Setup: Coordinate Geometry boss rush. We have to find an intersection point, calculate a perpendicular slope, build a brand new line equation, and finally snipe its x-intercept. Step 1: Find the crossover point of x+2y=4x+2y=4 and 2x+3y=62x+3y=6. Multiply the first equation by 22: 2x+4y=82x+4y=8. Subtract the second equation from it: (2x+4y)−(2x+3y)=8−6  ⟹  y=2(2x+4y)-(2x+3y)=8-6 \implies y=2 Plug y=2y=2 back into the first equation: x+2(2)=4  ⟹  x+4=4  ⟹  x=0x+2(2)=4 \implies x+4=4 \implies x=0 The lines intersect at (0,2)(0,2). This is our anchor point. Step 2: Steal the slope from the reference line. The reference is 3x−y=23x-y=2, which rearranges to y=3x−2y=3x-2. The slope (mm) is 33. Step 3: Calculate the perpendicular slope. Perpendicular lines have negative reciprocal slopes. m⊥=−13m_{\perp}=-\frac{1}{3} Step 4: Build the final line equation using the point-slope form with our anchor point (0,2)(0,2) and new slope −1/3-1/3. y−2=−13(x−0)y-2=-\frac{1}{3}(x-0) 3y−6=−x  ⟹  x+3y=63y-6=-x \implies x+3y=6 Step 5: Find the x-intercept. The x-intercept always occurs exactly when y=0y=0. x+3(0)=6  ⟹  x=6x+3(0)=6 \implies x=6 Final Answer: 6
Q16:ipmat indore 2021LRDI › TournamentsHardMCQ · MCQ
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
TeamsPlayedWinsLossesDrawsPoints
A508
B526
C525
D515
E51
F5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Total number of matches ending in draw is
  • A12
  • B4
  • C5
  • D6
Pick an option to attempt
The Setup: Six teams, every pair meets once, so (62)=15\binom{6}{2}=15 matches. Each match hands out exactly 2 points (2-0 for a decision, 1-1 for a draw), so the whole tournament is worth 15×2=3015 \times 2 = 30 points - that fixed total is the key that opens the table. For any team, played == wins ++ losses ++ draws =5=5 and points =2(wins)+draws=2(\text{wins}) + \text{draws}. Step 1: Decode A through D. Two equations per team, and both are already given (losses and points): * A - 00 losses, so w+d=5w+d=5; with 2w+d=82w+d=8, subtracting gives w=3,d=2w=3, d=2. * B - 22 losses, so w+d=3w+d=3; with 2w+d=62w+d=6, w=3,d=0w=3, d=0. * C - 22 losses, so w+d=3w+d=3; with 2w+d=52w+d=5, w=2,d=1w=2, d=1. * D - 11 loss, so w+d=4w+d=4; with 2w+d=52w+d=5, w=1,d=3w=1, d=3. Step 2: E is forced by the under-5 clause. E has 11 loss, so w+d=4w+d=4 and points =2w+d=w+4=2w+d=w+4. For E to stay under 5 points we need w<1w<1, so w=0w=0, d=4d=4, and E finishes on 4 points - a single win anywhere would break the clause. Step 3: F falls out of the 30-point total. F=30−(8+6+5+5+4)=30−28=2 pointsF = 30 - (8+6+5+5+4) = 30 - 28 = 2 \text{ points} Step 4: Split F's 2 points - and this is the step worth slowing down on. Two ways to score 2: 11 win, 00 draws, 44 losses; or 00 wins, 22 draws, 33 losses. Look at E to kill the first one. E must draw *four* of its five matches. B has 00 draws, so B is not one of E's partners. If F also had 00 draws, E's only available partners would be A, C and D - three teams for four draws, which is impossible. So F must carry draws, and **F is 00 wins, 22 draws, 33 losses. Step 5:** The completed table.
TeamsPlayedWinsLossesDrawsPoints
A53028
B53206
C52215
D51135
E50144
F50322
Step 6: The table plus the two given results actually pins every single match. E draws A, C, D and F and loses to B (Step 4). D's one loss is to C, so its three draws are A, E, F and it beats B. A never loses and draws only D and E, so it beats B, C and F. C's two losses are to A and B; its win over D and its draw with E are known, so it beats F. That leaves F losing to A, B and C. Master Matrix - each cell is the row team's result:
ABCDEF
A-WWDDW
BL-WLWW
CLL-WDW
DDWL-DD
EDLDD-D
FLLLDD-
Step 7: Count the draws. The Draws column totals 2+0+1+3+4+2=122+0+1+3+4+2=12, but every drawn match is counted twice - once for each team. So the number of matches that ended in a draw is: 122=6\frac{12}{2}=6 Final Answer: 6
Q17:ipmat indore 2021LRDI › TournamentsHardMCQ · MCQ
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
TeamsPlayedWinsLossesDrawsPoints
A508
B526
C525
D515
E51
F5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Which team has the highest number of draws
  • AA
  • BC
  • CD
  • DE
Pick an option to attempt
The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.
TeamsPlayedWinsLossesDrawsPoints
A53028
B53206
C52215
D51135
E50144
F50322
Step 1: Read straight down the Draws column: A has 22, B has 00, C has 11, D has 33, E has 44, F has 22. Step 2: E tops it with 44 draws, and it is not close - D is next on 33. This is no accident: E is the team squeezed by the under-5-points clause, which forced four of its five matches to be draws. Final Answer: E
Q18:ipmat indore 2021LRDI › TournamentsHardMCQ · MCQ
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
TeamsPlayedWinsLossesDrawsPoints
A508
B526
C525
D515
E51
F5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Total points Team F scored was
  • A0
  • B1
  • C2
  • D3
Pick an option to attempt
The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.
TeamsPlayedWinsLossesDrawsPoints
A53028
B53206
C52215
D51135
E50144
F50322
Step 1: The tournament is worth a fixed 3030 points - 1515 matches, each handing out exactly 22 points whether it is decided or drawn. Step 2: Subtract everyone else. A, B, C, D and E take 8+6+5+5+4=288+6+5+5+4=28 between them, so F is left with: 30−28=2 points30-28=2 \text{ points} Step 3: Sanity-check it against F's row: 00 wins and 22 draws is 2(0)+2=22(0)+2=2 points. Consistent, and comfortably under the 5-point cap the question imposes on F. Final Answer: 2
Q19:ipmat indore 2021LRDI › TournamentsHardMCQ · MCQ
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
TeamsPlayedWinsLossesDrawsPoints
A508
B526
C525
D515
E51
F5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Which team was not defeated by team A
  • AB
  • BC
  • CD
  • DF
Pick an option to attempt
The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.
TeamsPlayedWinsLossesDrawsPoints
A53028
B53206
C52215
D51135
E50144
F50322
Step 1: A won 33 matches, so exactly two teams escaped it - and the options only offer B, C, D and F, so we need to find which of those four A did not beat. Step 2: Go to D's row: D has exactly one loss. The question stem already tells us *who* dealt it - team C defeated D. That single loss is therefore spoken for. Step 3: So D cannot have lost to A as well. A's match against D was one of A's two draws, which means D is a team A did not defeat. Step 4: Check the others quickly. B and C both sit on 22 losses; A is unbeaten and B's and C's losses are accounted for by A and each other (B beat C, so C's other loss is to A, and B's two losses are to A and D). F lost 33, including to A. D is the only option left standing. (E also escaped A, via a draw, but E is not on the menu.) Final Answer: D
Q20:ipmat indore 2021LRDI › TournamentsHardMCQ · MCQ
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
TeamsPlayedWinsLossesDrawsPoints
A508
B526
C525
D515
E51
F5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Team E was defeated by
  • ATeams A and B only
  • BOnly Team A
  • COnly Team B
  • DTeams A, B and D only
Pick an option to attempt
The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.
TeamsPlayedWinsLossesDrawsPoints
A53028
B53206
C52215
D51135
E50144
F50322
Step 1: E's row shows exactly one loss. So E was beaten by exactly one team - that alone wipes out 'Teams A and B only' and 'Teams A, B and D only', which both name two or more. Step 2: Now find the one. E drew its other four matches. B is the only team in the whole table with zero draws, so B cannot be one of E's four draw partners - the only remaining relationship between B and E is a result, and E does not win at all (00 wins). Step 3: Therefore B beat E, and since E has just the one loss, no one else did. A drew with E rather than beating it, which is exactly why 'Only Team A' fails. Final Answer: Only Team B

Other IPMAT Indore papers & sections

IPMAT Indore 2026 LRDI 8IPMAT Indore 2026 MCQ 30IPMAT Indore 2026 QA 37IPMAT Indore 2026 SA 15IPMAT Indore 2026 VARC 45IPMAT Indore 2025 LRDI 9IPMAT Indore 2025 MCQ 30IPMAT Indore 2025 QA 36IPMAT Indore 2025 SA 15IPMAT Indore 2025 VARC 45IPMAT Indore 2024 LRDI 6IPMAT Indore 2024 MCQ 30IPMAT Indore 2024 QA 39IPMAT Indore 2024 SA 15IPMAT Indore 2024 VARC 45IPMAT Indore 2023 LRDI 6IPMAT Indore 2023 MCQ 30IPMAT Indore 2023 QA 39IPMAT Indore 2023 SA 15IPMAT Indore 2023 VARC 45IPMAT Indore 2022 LRDI 5IPMAT Indore 2022 MCQ 30IPMAT Indore 2022 QA 40IPMAT Indore 2022 SA 15IPMAT Indore 2022 VARC 45IPMAT Indore 2021 LRDI 6IPMAT Indore 2021 MCQ 20IPMAT Indore 2021 QA 24IPMAT Indore 2021 SA 10IPMAT Indore 2021 VARC 30IPMAT Indore 2020 LRDI 5IPMAT Indore 2020 MCQ 20IPMAT Indore 2020 QA 25IPMAT Indore 2020 SA 10IPMAT Indore 2020 VARC 30IPMAT Indore 2019 LRDI 5IPMAT Indore 2019 MCQ 40IPMAT Indore 2019 QA 55IPMAT Indore 2019 SA 20IPMAT Indore 2019 VARC 40JIPMAT 2025 LR 33JIPMAT 2025 QA 33JIPMAT 2025 VARC 34