IPMAT Indore 2020 QA PYQs

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 __________.

Enter your answer to attempt

The Setup: We are dealing with the foundational division algorithm:

Dividend=(Divisor\times Quotient)+Remainder

. The absolute golden rule of division is that the remainder must be strictly non-negative and less than the divisor (

0\leq 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\times 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)=24

10q-q^2=24

q^2-10q+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\times 6)+4

Dividend=30+4=34

Final Answer: 34

Q2:ipmat indore 2020QA › CirclesMediumSA · TITA

The shortest distance from the point

(-4,3)

to the circle

(x^2 + y^2 = 1)

is __________.

Enter your answer to attempt

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

x^2+y^2=1

. This perfectly matches the standard form

(x-h)^2+(y-k)^2=r^2

. Thus, the center is the origin

(0,0)

and the radius is

r=\sqrt{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=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d=\sqrt{(-4-0)^2+(3-0)^2}

d=\sqrt{(-4)^2+(3)^2}

d=\sqrt{16+9}=\sqrt{25}=5

Step 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

Q3:ipmat indore 2020QA › LogarithmsMediumSA · TITA

The value of

(0.04^{log_{\sqrt{5}}(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...)})

is __________.

Enter your answer to attempt

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:

\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

This is an infinite GP where the first term

a=\frac{1}{4}

and the common ratio

r=\frac{1}{2}

. Since

|r|<1

the series converges, so the sum to infinity

S=\frac{a}{1-r}

applies:

S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}

Step 2: Reconstruct the base and the logarithm. The original expression's base is

0.04

, which translates to the fraction

\frac{4}{100}=\frac{1}{25}

. We can write this as an exponent:

5^{-2}

. The logarithm's base is

\sqrt{5}

, which is

5^{\frac{1}{2}}

. So the exponent is

\log_{5^{1/2}}\left(\frac{1}{2}\right)

. Step 3: Simplify the exponent using log properties. We use the rule

\log_{b^n}(x)=\frac{1}{n}\log_b(x)

:

\log_{5^{1/2}}\left(\frac{1}{2}\right)=\frac{1}{\frac{1}{2}}\log_5\left(\frac{1}{2}\right)=2\log_5\left(\frac{1}{2}\right)

Bring the multiplier inside as a power:

\log_5\left(\left(\frac{1}{2}\right)^2\right)=\log_5\left(\frac{1}{4}\right)

. Step 4: Combine the final expression. Substitute the cleaned-up base and exponent back together:

\left(5^{-2}\right)^{\log_5\left(\frac{1}{4}\right)}=5^{-2\log_5\left(\frac{1}{4}\right)}

Move the

-2

inside the log as a power:

5^{\log_5\left(\left(\frac{1}{4}\right)^{-2}\right)}

Using the identity

b^{\log_b(x)}=x

, the

5

and

\log_5

cancel out. We are left with:

\left(\frac{1}{4}\right)^{-2}=4^2=16

Final Answer: 16

Q4:ipmat indore 2020QA › Matrices & DeterminantsHardSA · TITA

Suppose

\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0

, where

\mathrm{a}, \mathrm{b}

and

\mathrm{c}

are distinct real numbers. If

{a}=3

, then the value of

abc

is:

Enter your answer to attempt

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

\left(a^3-1,\ b^3-1,\ c^3-1\right)

into two determinants:

\Delta=\left|\begin{matrix}a&a^2&a^3\\b&b^2&b^3\\c&c^2&c^3\end{matrix}\right|-\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=0

Step 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:

abc\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|

Step 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):

\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=-\left|\begin{matrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{matrix}\right|

Now swap Column 2 with Column 1 (sign flips back):

=+\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|

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

\Delta=0

now looks like this:

abc(V)-V=0 \implies 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\neq 0

. Therefore the other multiplier must be zero:

abc-1=0\implies abc=1

Final 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|\geq |(x+2)-(x-5)|=7

with equality for every

x

in the interval

[-2,5]

. Step 3: Bound the middle term. Obviously

|x-3|\geq 0

, with equality only at

x=3

. Adding the two bounds gives

f(x)\geq 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

Q6:ipmat indore 2020QA › Ratio, Proportion & VariationEasySA · TITA

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 __________.

Enter your answer to attempt

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=112-4x

Step 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 \implies k(2x+3y)=86

Step 3: Substitute, then kill every wrong branch. Plug the isolated

y

from Step 1 into the Step 2 equation:

k(2x+3(112-4x))=86

k(2x+336-12x)=86 \implies k(336-10x)=86

So

336-10x=\frac{86}{k}

, which forces

k

to be a positive factor of

86

. The factors are

1, 2, 43, 86

, giving

\frac{86}{k}\in\{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 \implies 10x=250 \implies 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\times 25=100

. Final Answer: 100

Q7:ipmat indore 2020QA › MiscellaneousEasySA · TITA

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 __________.

Enter your answer to attempt

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\times 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\times 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 __________.

Enter your answer to attempt

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\cap P\cap 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\leq n(M\cap P), \qquad x\leq n(P\cap C), \qquad x\leq n(M\cap C)

The problem caps all three pairwise intersections at 20. Therefore

x\leq 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=20

Step 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.

\text{Laps to finish}=\frac{3000}{300}=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:

\text{Slower runner}=\frac{2}{3}\times 10=\frac{20}{3}\text{ laps}

Keep this as an exact fraction -

\frac{20}{3}

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:

\text{Lead}=10-\frac{20}{3}=\frac{30-20}{3}=\frac{10}{3}\text{ 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

\frac{10}{3}=3\frac{1}{3}

is strictly greater than

3

but short of

4

, the third overtake happens before the finish line and a fourth never does.

\text{Number of passes}=\left\lfloor \frac{10}{3}\right\rfloor=3

Final Answer: 3

Q10:ipmat indore 2020QA › Permutation & CombinationHardSA · TITA

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 __________.

Enter your answer to attempt

The Setup: This combinatorics problem mixes indistinguishable (identical) objects with distinct ones. The standard combination formula

\binom{n}{r}

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):

\binom{9}{4}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}=126

ways * Case 2 (1 identical, 3 distinct):

1\times \binom{9}{3}=\frac{9\times 8\times 7}{3\times 2\times 1}=84

ways * Case 3 (2 identical, 2 distinct):

1\times \binom{9}{2}=\frac{9\times 8}{2\times 1}=36

ways * Case 4 (3 identical, 1 distinct):

1\times \binom{9}{1}=9

ways * Case 5 (4 identical, 0 distinct):

1\times \binom{9}{0}=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=256

Step 4: Cross-check with a symmetry argument. That total is

\binom{9}{0}+\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\binom{9}{4}

- the lower half of row 9 of Pascal's triangle. Since

\binom{9}{k}=\binom{9}{9-k}

, the ten entries of that row pair off perfectly into two equal halves, so this sum is exactly half the row total:

\frac{2^9}{2}=\frac{512}{2}=256

Two independent routes, same number, so the count is confirmed. Final Answer: 256

Q11:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

The probability that a randomly chosen factor of

10^{19}

is a multiple of

10^{15}

is

A $\frac{1}{25}$
B $\frac{1}{12}$
C $\frac{1}{20}$
D $\frac{1}{16}$

Pick an option to attempt

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of

10^{19}

(the sample space) and the total number of factors that carry at least

10^{15}

in their DNA (the winning outcomes). Step 1: Break down the base stats. Prime factorize the main number:

10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}

The formula for the total number of factors of

p^a\times q^b

is

(a+1)(b+1)

, since each exponent is chosen freely:

\text{Total factors}=(19+1)(19+1)=20\times 20=400

Step 2: Isolate the target condition. A factor of

10^{19}

has the shape

2^x\times 5^y

with

0\leq x,y\leq 19

. For it to be a multiple of

10^{15}=2^{15}\times 5^{15}

, it must carry at least fifteen of each prime, so

x\geq 15

and

y\geq 15

. Combined with the ceiling of 19, the valid range for both

x

and

y

is

\{15,16,17,18,19\}

. Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

\text{Valid factors}=5\times 5=25

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

P=\frac{25}{400}=\frac{1}{16}

Final Answer:

\frac{1}{16}

Q12:ipmat indore 2020QA › TrianglesHardMCQ · MCQ

The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is

A28
B29
C31
D33

Pick an option to attempt

The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins. Step 1: Set up the side lengths. Let the three consecutive integer sides be

x-1

,

x

and

x+1

, with

x-1\geq 1

. For any triangle the two smaller sides must beat the largest:

(x-1)+x>x+1\implies x>2

. Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is

x+1

, so the single condition to check is

a^2+b^2>c^2

:

(x-1)^2+x^2>(x+1)^2

x^2-2x+1+x^2>x^2+2x+1

x^2>4x

Since

x

is positive we may divide by

x

, giving

x>4

. As

x

is an integer, the lowest valid value is

x=5

(the smallest qualifying triangle being

4,5,6

). Note this condition is strictly tighter than Step 1's

x>2

, so the triangle inequality is absorbed and adds nothing further. Step 3: Apply the perimeter cap. The perimeter is at most 100:

(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots

Since

x

is an integer, the maximum is

x=33

, giving sides

32,33,34

and a perimeter of

99

. Step 4: Count the surviving squad. The valid range for

x

runs from 5 to 33 inclusive:

33-5+1=29

Final Answer: 29

Q13:ipmat indore 2020QA › Straight LinesMediumMCQ · MCQ

The equation of the straight line passing through the point

M (-5,4)

, such that the portion of it between the axes is divided by the point

M

into two equal halves, is

A $10y - 8x = 80$
B $8y + 10x = 80$
C $10y + 8x = 80$
D $8y + 10x + 80 = 0$

Pick an option to attempt

The Setup: This is a coordinate geometry speedrun using intercept form. The line crosses the x-axis at some point

A(a,0)

and the y-axis at

B(0,b)

. Saying

M

divides the portion between the axes into two equal halves is just a fancy way of saying

M

is the midpoint of segment

AB

. Step 1: Apply the midpoint formula. The midpoint of

A(a,0)

and

B(0,b)

is

\left(\frac{a}{2},\frac{b}{2}\right)

, and we are told that point is

M(-5,4)

:

\frac{a}{2}=-5\implies a=-10

\frac{b}{2}=4\implies b=8

Step 2: Construct the line equation. Use the two-intercept form

\frac{x}{a}+\frac{y}{b}=1

:

\frac{x}{-10}+\frac{y}{8}=1

Step 3: Clear the fractions in one shot. Multiply every term by

80

(a common multiple of 10 and 8 chosen to land directly on the answer's scale):

80\cdot\frac{x}{-10}+80\cdot\frac{y}{8}=80 \implies -8x+10y=80

Reordering gives

10y-8x=80

. Step 4: Verify against all three known points. A midpoint slip is easy to make, so test the line on the two intercepts and

M

itself: *

A(-10,0)

:

10(0)-8(-10)=80

✓ *

B(0,8)

:

10(8)-8(0)=80

✓ *

M(-5,4)

:

10(4)-8(-5)=40+40=80

✓ All three sit on the line, and

M

is genuinely halfway between

A

and

B

. Running the other three options through

M(-5,4)

fails every time, so the match is unique. Final Answer:

10y - 8x = 80

Q14:ipmat indore 2020QA › TrigonometryHardMCQ · MCQ

The value of

\cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right) + \cos^2\left(\frac{5\pi}{8}\right) + \cos^2\left(\frac{7\pi}{8}\right)

is

A $1$
B $\frac{3}{2}$
C $2$
D $\frac{9}{4}$

Pick an option to attempt

The Setup: This trig expression looks intimidating, but it has zero plot armour once you spot the symmetry in the angles. We use the supplementary angle identity to fold the expression in half, then the complementary identity to finish it off. Step 1: Fold the back half. Notice that

\frac{7\pi}{8}=\pi-\frac{\pi}{8}

, and

\cos(\pi-\theta)=-\cos\theta

, so

\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right)

. Squaring kills the minus sign:

\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right)

Identically,

\frac{5\pi}{8}=\pi-\frac{3\pi}{8}

, so

\cos^2\left(\frac{5\pi}{8}\right)=\cos^2\left(\frac{3\pi}{8}\right)

. Step 2: Simplify the sum. The four terms collapse into two identical pairs:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]

Step 3: Exploit the complementary angles. Notice that

\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}

. The two angles are complementary, and

\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta

, so:

\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{8}\right)

Step 4: Use the legendary identity. Substitute sine into the folded expression:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right]

By the Pythagorean identity

\cos^2\theta+\sin^2\theta=1

, the bracket collapses to 1:

2\times 1=2

Final Answer:

2

Q15:ipmat indore 2020QA › Progression & SeriesHardMCQ · MCQ

If

\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots

up to

\infty = \frac{\pi^2}{6}

, then the value of

\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots

up to

\infty

is

A $\frac{\pi^2}{8}$
B $\frac{\pi^2}{16}$
C $\frac{\pi^2}{12}$
D $\frac{\pi^2}{36}$

Pick an option to attempt

The Setup: We are manipulating the famous Basel problem. We are handed the sum of *all* inverse squares and asked for the sum of only the *odd* ones. The trick is to split the total into its even and odd halves, then notice the even half is a scaled copy of the whole thing. Step 1: Define the series. Let the total be

S=\frac{\pi^2}{6}

. Every positive integer is either odd or even, so the total splits cleanly with nothing counted twice and nothing missed:

S=\text{Sum(Odds)}+\text{Sum(Evens)}

Every term here is positive and the series converges, so splitting and rearranging it like this is legitimate rather than a formal trick. Step 2: Factor the even series. Write out the even terms and pull out the common factor. Each even number is

2n

, so its square is

4n^2

:

\text{Sum(Evens)}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right)

The bracket is literally the original total

S

, so

\text{Sum(Evens)}=\frac{S}{4}

. This self-similarity is the whole engine of the problem. Step 3: Solve for the odds. Substitute back into the master equation:

S=\text{Sum(Odds)}+\frac{S}{4} \implies \text{Sum(Odds)}=S-\frac{S}{4}=\frac{3S}{4}

Step 4: Final substitution. Plug in the known value of

S

:

\text{Sum(Odds)}=\frac{3}{4}\times \frac{\pi^2}{6}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}

Final Answer:

\frac{\pi^2}{8}

Q16:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is

A $\frac{4}{9}$
B $\frac{5}{9}$
C $\frac{4}{15}$
D $\frac{2}{15}$

Pick an option to attempt

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between

\frac{4}{5}

and the answer is the entire point of the question. Step 1: Map the base probabilities. * Probability of rolling a five:

P(A)=\frac{1}{6}

* Probability of rolling anything else:

P(\text{not }A)=\frac{5}{6}

* Man tells the truth:

P(T)=\frac{4}{5}

* Man lies:

P(L)=\frac{1}{5}

Step 2: Map the conditional claims. If it genuinely is a five, he reports five precisely when he is being truthful:

P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}

. If it is not a five, he can only report five by lying, so this branch is taken as

P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}

- the exam's convention is that a lie about the target face lands on that face. *Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be

\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}

and the answer would come out as

\frac{4}{5}

. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended. Step 3: Apply Bayes' Theorem.

P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}

Step 4: Crunch the numbers.

\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}

\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}

P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}

Sanity check on the direction:

\frac{4}{9}<\frac{1}{2}

, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying. Final Answer:

\frac{4}{9}

Q17:ipmat indore 2020QA › LogarithmsEasyMCQ · MCQ

If

\log_5(\log_8(x^2 - 1)) = 0

, then a possible value of

x

is

A2 $\sqrt{2}$
B $\sqrt{2}$
C2
D3

Pick an option to attempt

The Setup: This is a logarithm inception. We unpack it from the outside in using the fundamental definition

\log_b(A)=C\implies A=b^C

, then check the candidates against the option list. Step 1: Destroy the outer log.

\log_5\left(\log_8(x^2-1)\right)=0

Convert from log form to exponential form:

\log_8(x^2-1)=5^0=1

Step 2: Destroy the inner log. Apply the same rule to what remains:

x^2-1=8^1=8

Step 3: Solve the quadratic.

x^2=9 \implies x=\pm 3

Step 4: Pick the value that is actually on the menu. Both

x=3

and

x=-3

satisfy the equation - the expression only ever sees

x^2

, so the sign is irrelevant, and

x^2-1=8>0

keeps every logarithm well defined. This is exactly why the stem asks for *a possible* value rather than *the* value. Of the two, only

3

appears among the options, so it is the answer. The other three options all fail, and one of them is a genuine trap: *

x=2\sqrt{2}

:

x^2-1=7

, and

\log_8 7\approx 0.936

, so the outer log is not 0. *

x=\sqrt{2}

:

x^2-1=1

, so

\log_8 1=0

- and

\log_5 0

is undefined, not zero. Mistaking one for the other is the intended snare. *

x=2

:

x^2-1=3

, and

\log_8 3\approx 0.528

, so the outer log is not 0. Final Answer: 3

Q18:ipmat indore 2020QA › InequalitiesMediumMCQ · MCQ

Consider the following statements: (i) When

(0 < x < 1)

, then

(\frac{1}{1+x} < 1 - x + x^2)

(ii) When

(0 < x < 1)

, then

(\frac{1}{1+x} > 1 - x + x^2)

(iii) When

(-1 < x < 0)

, then

(\frac{1}{1+x} < 1 - x + x^2)

(iv) When

(-1 < x < 0)

, then

(\frac{1}{1+x} > 1 - x + x^2)

Then the correct statements are:

A(i) and (ii)
B(ii) and (iv)
C(i) and (iv)
D(ii) and (iii)

Pick an option to attempt

The Setup: Cross-multiplying an inequality whose denominator changes sign is how this question eats people alive. We dodge that entirely with the sum of cubes identity

a^3+b^3=(a+b)(a^2-ab+b^2)

, taking

a=1

and

b=x

, which turns the whole comparison into a single fraction whose sign we can read off. Step 0: Note the structure. Statements (i) and (ii) are exact opposites on the same interval, as are (iii) and (iv). So exactly one of each pair must be true, and the answer is guaranteed to be one statement from each pair - that alone kills option 1 ((i) and (ii) cannot both hold) before any algebra. Step 1: Exploit the identity. With

a=1

,

b=x

:

1+x^3=(1+x)(1-x+x^2)

On both intervals

x>-1

, so

1+x>0

- it is never zero, and we may safely divide by it:

1-x+x^2=\frac{1+x^3}{1+x}

Step 2: Find the delta. Subtract the two quantities being compared and put them over the common denominator:

(1-x+x^2)-\frac{1}{1+x}=\frac{1+x^3}{1+x}-\frac{1}{1+x}=\frac{x^3}{1+x}

Everything now rides on the sign of

\frac{x^3}{1+x}

, and since

1+x>0

throughout, that sign is simply the sign of

x^3

- which is the sign of

x

. **Step 3: Evaluate Case 1,

0<x<1

.** The numerator

x^3

is positive and the denominator

1+x

is positive, so the delta is positive:

\frac{x^3}{1+x}>0 \implies 1-x+x^2>\frac{1}{1+x}

Flipping that around to the form the paper uses gives

\frac{1}{1+x}<1-x+x^2

, which is exactly Statement (i) - and therefore (ii) is false. *Spot check at

x=\frac{1}{2}

:*

\frac{1}{1+x}=\frac{2}{3}

and

1-x+x^2=\frac{3}{4}

, and indeed

\frac{2}{3}<\frac{3}{4}

, with the gap

\frac{1}{12}

matching

\frac{x^3}{1+x}=\frac{1/8}{3/2}=\frac{1}{12}

. **Step 4: Evaluate Case 2,

-1<x<0

.** Here the numerator

x^3

is negative (a negative number cubed stays negative), while the denominator

1+x

is still positive - for instance

1+(-0.5)=0.5

. A negative over a positive is negative:

\frac{x^3}{1+x}<0 \implies 1-x+x^2<\frac{1}{1+x}

Flipping it round gives

\frac{1}{1+x}>1-x+x^2

, which is exactly Statement (iv) - and therefore (iii) is false. *Spot check at

x=-\frac{1}{2}

:*

\frac{1}{1+x}=2

and

1-x+x^2=\frac{7}{4}

, and indeed

2>\frac{7}{4}

, with the gap

-\frac{1}{4}

matching

\frac{x^3}{1+x}=\frac{-1/8}{1/2}=-\frac{1}{4}

. Step 5: Assemble. The true statements are (i) and (iv), one from each contradictory pair as Step 0 predicted. Final Answer: (i) and (iv)

Q19:ipmat indore 2020QA › Mixture & AlligationMediumMCQ · MCQ

Fifty litres of a mixture of milk and water contains 30 percent of water. This mixture is added to eighty litres of another mixture of milk and water that contains 20 percent of water. Then, how many litres of water should be added to the resulting mixture to obtain a final mixture that contains 25 percent of water?

A1
B2
C3
D4

Pick an option to attempt

The Setup: Mixture problems just require tracking the absolute volumes of the components like inventory. We find the exact litres of water and milk we currently hold, merge them, then solve for the extra water needed to hit the target 25%. The key discipline is that adding pure water changes the water column *and* the total, but leaves the milk column frozen. Step 1: Audit Mixture 1. Total

=50

L, water is 30%.

\text{Water}=0.30\times 50=15\text{ L}, \qquad \text{Milk}=50-15=35\text{ L}

Step 2: Audit Mixture 2. Total

=80

L, water is 20%.

\text{Water}=0.20\times 80=16\text{ L}, \qquad \text{Milk}=80-16=64\text{ L}

Step 3: Merge the inventory. Tip both into the same barrel:

\text{Total volume}=50+80=130\text{ L}, \quad \text{Water}=15+16=31\text{ L}, \quad \text{Milk}=35+64=99\text{ L}

Step 4: Calculate the required additive. Let

w

litres of pure water go in. Both the water and the total grow by

w

, and a 25% water mix means water is

\frac{1}{4}

of the total:

\frac{31+w}{130+w}=\frac{1}{4}

Cross-multiply and solve:

4(31+w)=130+w \implies 124+4w=130+w \implies 3w=6 \implies w=2

Step 5: Verify from the milk side. The milk column never moved, so it is an independent check. Final total

=130+2=132

L, of which milk is still 99 L:

\frac{99}{132}=\frac{3}{4}=75\% \text{ milk}, \text{ hence } 25\% \text{ water}

The two routes agree, so

w=2

is right. Final Answer: 2

Q20:ipmat indore 2020QA › Time & WorkHardMCQ · MCQ

Three workers working together need 1 hour to construct a wall. The first worker, working alone, can construct the wall twice as fast as the third worker, and can complete the task an hour sooner than the second worker. Then, the average of time taken in hours by three workers, when working alone, to construct the wall is

A $\dfrac{\sqrt{33} + 4}{3}$
B $\dfrac{\sqrt{33} + 5}{3}$
C $\dfrac{\sqrt{33} + 6}{3}$
D $\dfrac{\sqrt{33} + 7}{3}$

Pick an option to attempt

The Setup: This is a rates problem, and the golden rule is that *times do not add, rates do*. We convert *hours to build the wall* into *walls per hour*, sum the three rates to 1 wall/hour, and only convert back to time at the very end. Step 1: Define the rates. Let

r_1,r_2,r_3

be the workers' rates in walls per hour, and

t_i=\frac{1}{r_i}

their solo times. All three together take 1 hour, so their rates sum to one whole wall per hour:

r_1+r_2+r_3=\frac{1}{1}=1

Worker 1 is twice as fast as Worker 3, so

r_1=2r_3

, i.e.

r_3=\frac{r_1}{2}

(equivalently

t_3=2t_1

). Step 2: Link Workers 1 and 2. Worker 1 finishes an hour sooner than Worker 2, and that condition is about *times*, so it must be written in times first:

t_1=t_2-1 \implies \frac{1}{r_1}=\frac{1}{r_2}-1 \implies \frac{1}{r_2}=\frac{1}{r_1}+1=\frac{1+r_1}{r_1}

Flipping gives

r_2=\frac{r_1}{1+r_1}

. Step 3: Solve the master rate equation. Substitute

r_2

and

r_3

into the sum:

r_1+\frac{r_1}{1+r_1}+\frac{r_1}{2}=1

Multiply through by

2(1+r_1)

to clear denominators:

2r_1(1+r_1)+2r_1+r_1(1+r_1)=2(1+r_1)

2r_1+2r_1^2+2r_1+r_1+r_1^2=2+2r_1

3r_1^2+5r_1=2+2r_1 \implies 3r_1^2+3r_1-2=0

By the quadratic formula:

r_1=\frac{-3\pm\sqrt{9-4(3)(-2)}}{6}=\frac{-3\pm\sqrt{33}}{6}

A rate cannot be negative, and

\sqrt{33}\approx 5.745

, so the minus root is discarded and

r_1=\frac{\sqrt{33}-3}{6}

. Step 4: Convert back to times. Rationalising the denominator each time:

t_1=\frac{1}{r_1}=\frac{6}{\sqrt{33}-3}=\frac{6(\sqrt{33}+3)}{33-9}=\frac{6(\sqrt{33}+3)}{24}=\frac{\sqrt{33}+3}{4}

t_2=t_1+1=\frac{\sqrt{33}+3}{4}+\frac{4}{4}=\frac{\sqrt{33}+7}{4}, \qquad t_3=2t_1=\frac{2\sqrt{33}+6}{4}

Step 5: Average them. Note the question asks for the average of the *times*, not of the rates - averaging rates instead is the single most common way to lose this mark.

t_1+t_2+t_3=\frac{(\sqrt{33}+3)+(\sqrt{33}+7)+(2\sqrt{33}+6)}{4}=\frac{4\sqrt{33}+16}{4}=\sqrt{33}+4

\text{Average}=\frac{\sqrt{33}+4}{3}

Step 6: Sanity check the numbers. Numerically

t_1\approx 2.19

,

t_2\approx 3.19

,

t_3\approx 4.37

hours. Every one exceeds 1 hour, as it must - a lone worker cannot beat all three together. Their gap

t_2-t_1

is exactly 1,

t_3

is exactly double

t_1

, and

\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}=1

to twelve decimal places. All three given conditions hold. Final Answer:

\dfrac{\sqrt{33} + 4}{3}

Q21:ipmat indore 2020QA › Set TheoryMediumMCQ · MCQ

In a class, students are assigned roll numbers from 1 to 140. All students with even roll numbers opted for cricket, all those whose roll numbers are divisible by 5 opted for football, and all those whose roll numbers are divisible by 3 opted for basketball. The number of students who did not opt for any of the three sports is

A102
B38
C98
D42

Pick an option to attempt

The Setup: This is Inclusion-Exclusion. We count the students playing at least one sport (the union of the three sets), then subtract from 140 to get the NPCs playing nothing. Note the sets overlap heavily - a roll number like 30 is in all three - which is exactly why we cannot simply add. Step 1: Calculate the single sets. Counting multiples of

k

up to 140 is

\left\lfloor\frac{140}{k}\right\rfloor

.

|A|\ (\text{cricket, }2)=\left\lfloor\frac{140}{2}\right\rfloor=70, \quad |B|\ (\text{football, }5)=\left\lfloor\frac{140}{5}\right\rfloor=28, \quad |C|\ (\text{basketball, }3)=\left\lfloor\frac{140}{3}\right\rfloor=46

Step 2: Calculate the pairwise intersections. A student in two sets has a roll number divisible by both divisors, so the correct counter is their LCM - not their product. Here

2,3,5

are pairwise coprime, so the LCM happens to equal the product, but that coincidence is what makes this step safe; with divisors like 2 and 4 the product would give a wrong count.

|A\cap B|\ (\text{LCM}=10)=\left\lfloor\frac{140}{10}\right\rfloor=14, \quad |A\cap C|\ (\text{LCM}=6)=\left\lfloor\frac{140}{6}\right\rfloor=23, \quad |B\cap C|\ (\text{LCM}=15)=\left\lfloor\frac{140}{15}\right\rfloor=9

Step 3: Calculate the triple intersection.

|A\cap B\cap C|\ (\text{LCM}=30)=\left\lfloor\frac{140}{30}\right\rfloor=4

Step 4: Execute the formula.

|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|

=70+28+46-14-23-9+4=144-46+4=102

Step 5: Take the complement. That 102 is the number who play at least one sport, so:

140-102=38

Watch the trap: 102 is sitting right there as option 1. It is the union, not the answer - the question asks for everyone *outside* it. Final Answer: 38

Q22:ipmat indore 2020QA › FunctionsEasyMCQ · MCQ

Given

f(x) = x^2 + \log_3 x

and

g(y) = 2y + f(y)

, then the value of

g(3)

equals

A16
B15
C25
D26

Pick an option to attempt

The Setup: Pure function evaluation. No theory required - just substitute from the outside in, respecting each definition exactly as written. Step 1: Set up the outer function. The definition is

g(y)=2y+f(y)

, so at

y=3

:

g(3)=2(3)+f(3)=6+f(3)

Note the variable name is a red herring:

f

is written in

x

and

g

in

y

, but a function does not care what its argument is called. Feeding 3 into

f

is the same operation either way. Step 2: Evaluate the nested function. Using

f(x)=x^2+\log_3 x

at

x=3

:

f(3)=3^2+\log_3 3

We have

3^2=9

, and any log of its own base is 1, since

\log_b b=1

means *what power of

b

gives

b

?* - namely 1. So

\log_3 3=1

and

x=3>0

keeps the log defined.

f(3)=9+1=10

Step 3: Recombine. Substitute back:

g(3)=6+10=16

Final Answer: 16

Q23:ipmat indore 2020QA › Matrices & DeterminantsHardMCQ · MCQ

A

2 \times 2

matrix is filled with four distinct integers randomly chosen from the set

\{1,2,3,4,5,6\}

. Then the probability that the matrix generated in such a way is singular is

A $\frac{2}{45}$
B $\frac{1}{45}$
C $\frac{4}{15}$
D $\frac{1}{15}$

Pick an option to attempt

The Setup: A

2\times 2

matrix

M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is singular when its determinant vanishes, i.e.

ad-bc=0

, so we need the main-diagonal product to equal the anti-diagonal product:

ad=bc

. The whole question reduces to finding which four numbers from

\{1,2,3,4,5,6\}

can be split into two disjoint pairs of equal product. Step 1: Calculate the total sample space. We pick 4 distinct numbers from 6 and place them in 4 labelled positions, so order matters - this is a permutation:

\text{Total outcomes}=P(6,4)=6\times 5\times 4\times 3=360

Step 2: Find every pair of pairs with equal products. Rather than assert it, list all 15 pairs and their products outright:

Pair	Product	Pair	Product	Pair	Product
(1,2)	2	(2,4)	8	(3,5)	15
(1,3)	3	(2,5)	10	(3,6)	18
(1,4)	4	(2,6)	12	(4,5)	20
(1,5)	5	(3,4)	12	(4,6)	24
(1,6)	6	(2,3)	6	(5,6)	30

Exactly two product values repeat, and in both cases the two pairs are disjoint (they share no element, which is required since all four entries must be distinct): * Product

6

:

\{1,6\}

and

\{2,3\}

* Product

12

:

\{2,6\}

and

\{3,4\}

Every other product in the table is unique, so no further singular matrices exist. Step 3: Count matrix arrangements. Take product 6. Put

\{1,6\}

on the main diagonal

(a,d)

and

\{2,3\}

on the anti-diagonal

(b,c)

: * Arrangements of

\{1,6\}

across

a,d

:

2!=2

* Arrangements of

\{2,3\}

across

b,c

:

2!=2

That gives

2\times 2=4

matrices. But the two pairs can also switch diagonals -

\{2,3\}

on the main,

\{1,6\}

on the anti - for another 4. So product 6 contributes

8

matrices, and by identical logic so does product 12. Step 4: Final probability.

\text{Singular matrices}=8+8=16 \implies P=\frac{16}{360}=\frac{2}{45}

Final Answer:

\frac{2}{45}

Q24:ipmat indore 2020QA › Ratio, Proportion & VariationEasyMCQ · MCQ

Ashok started a business with a certain investment. After a few months, Bharat joined him investing half the amount of Ashok's initial investment. At the end of the first year, the total profit was divided between them in the ratio

3:1

. Bharat joined Ashok after

A2 months
B3 months
C4 months
D6 months

Pick an option to attempt

The Setup: In partnership accounting, profit is shared in proportion to each partner's *capital-months* - the money put in multiplied by how long it stayed in. Two partners with the same capital but different durations do not split evenly, and that is the whole mechanism here. Step 1: Define Ashok's stats. Let Ashok's investment be

2x

(choosing

2x

rather than

x

keeps Bharat's half a whole number). He started the business, so his money was in for the full 12 months:

\text{Ashok's capital-months}=2x\times 12=24x

Step 2: Define Bharat's stats. Bharat invested half of Ashok's amount, so his capital is

x

. Let it stay in for

t

months:

\text{Bharat's capital-months}=x\times t=xt

Step 3: Equate to the profit ratio. Profit splits

3:1

in the order the partners are named, Ashok to Bharat:

\frac{\text{Ashok's capital-months}}{\text{Bharat's capital-months}}=\frac{3}{1} \implies \frac{24x}{xt}=3

Step 4: Solve for the duration. Capital is positive, so

x

cancels safely:

\frac{24}{t}=3 \implies 3t=24 \implies t=8

Step 5: Convert duration into a joining date - do not stop at Step 4. The number 8 is how long Bharat's money was *invested*, but the question asks when he joined. In a 12-month year, being invested for the last 8 months means he sat out the first four:

12-8=4\text{ months}

Answering 8 here is the intended trap, and it is why the question is phrased 'Bharat joined Ashok after' rather than asking for his investment period. Final Answer: 4 months

Q25:ipmat indore 2020QA › Mean, Median & ModeHardMCQ · MCQ

The average marks of

6

students in a test is

64

. All the students got different marks, one of the students obtained

70

marks and all other students scored

40

or above. The maximum possible difference between the second highest and the second lowest marks is

A50
B54
C57
D58

Pick an option to attempt

The Setup: A min-maxing puzzle wearing a statistics costume. Sort the six marks as

S_1>S_2>S_3>S_4>S_5>S_6

, so the second highest is

S_2

and the second lowest is

S_5

. We want to maximise

S_2-S_5

subject to a fixed total, all marks distinct integers of at least 40, and one of them being exactly 70. Step 1: Set the baseline constraints. Six students averaging 64 gives a fixed budget:

\text{Total}=6\times 64=384

Step 2: Pin the floor.

S_6\geq 40

, and since the marks are distinct integers,

S_5\geq S_6+1\geq 41

. So

S_5=41

is the best we can ever do, achieved by putting

S_6=40

. Every point saved at the bottom is a point available at the top, so this choice never costs us anything. Step 3: Decide where the mandatory 70 goes. This is the real work, and it needs all cases checked rather than a lucky guess: * **70 as

S_1

or

S_2

:** then

S_2\leq 70

, so the difference is at most

70-41=29

. Dead. * **70 as

S_5

or

S_6

:** impossible. If

S_5=70

then

S_1,\dots,S_4

all exceed 70, so they total at least

71+72+73+74=290

; adding

70

and

S_6\geq 40

already gives

400>384

. Placing 70 even lower is worse still. * **70 as

S_4

:** then

S_3\geq 71

, and with

S_5=41,S_6=40

the top three share

384-70-41-40=233

. Minimising

S_3

at 71 leaves

S_1+S_2=162

, and

S_1>S_2

forces

S_2\leq 80

. Difference at most

80-41=39

. Still beaten. * **70 as

S_3

: the surviving case, examined next. Step 4: Work the winning case.** With 70 at

S_3

, the two marks below it must be distinct integers between 41 and 69, and pushing them as low as possible frees the most budget for the top:

S_4=42

,

S_5=41

,

S_6=40

. That leaves:

S_1+S_2=384-(70+42+41+40)=384-193=191

To make

S_2

as large as possible we want

S_1

as close above it as possible. Since 191 is odd, the tightest legal split of

S_1>S_2

is

96

and

95

:

S_1=96,\quad S_2=95

Step 5: Confirm the set is legal and compute. The marks are

96,95,70,42,41,40

- six distinct integers, all at least 40, one exactly 70, summing to

384

. Every condition holds.

S_2-S_5=95-41=54

Since Step 3 ruled out every other placement of 70 with a strictly smaller ceiling, 54 is the true maximum. Final Answer: 54

IPMAT Indore 2020 — QA

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