IPMAT Indore 2021 QA PYQs

Q1:ipmat indore 2021QA › FactorisationMediumSA · TITA

The number of positive integers that divide

(1890) \cdot (130) \cdot (170)

and are not divisible by

45

is ________

Enter your answer to attempt

The Setup: We need the total number of positive integer divisors for a massive product, filtering out the ones divisible by 45. Think of it like running a script to ban any accounts in your Discord server that have a specific toxic role. Step 1: Combine the numbers and find the prime factorization of the main boss,

N=(1890)(130)(170)

.

1890 = 2 \times 3^3 \times 5 \times 7

130 = 2 \times 5 \times 13

170 = 2 \times 5 \times 17

Multiply them together to get the ultimate form:

N = 2^3 \times 3^3 \times 5^3 \times 7^1 \times 13^1 \times 17^1

Step 2: Calculate the total number of positive integers that divide

N

. Add 1 to each exponent and multiply:

\text{Total Divisors} = (3+1)(3+1)(3+1)(1+1)(1+1)(1+1)

\text{Total} = 4 \times 4 \times 4 \times 2 \times 2 \times 2 = 512

Step 3: Figure out how many of these are divisible by 45. Since

45=3^2 \times 5^1

, any divisor that is a multiple of 45 must have at least

3^2

and

5^1

locked into its prime factorization. We lock in those required factors, leaving us with a remaining pool of choices: * Powers of 3 left:

3-2=1

(so 2 choices:

3^2, 3^3

) * Powers of 5 left:

3-1=2

(so 3 choices:

5^1, 5^2, 5^3

) * The rest (2, 7, 13, 17) keep all their original choices (4, 2, 2, 2).

\text{Banned Divisors} = 4 \times 2 \times 3 \times 2 \times 2 \times 2 = 192

Step 4: Subtract the banned divisors from the total pool.

\text{Clean Divisors} = 512 - 192 = 320

Final Answer: 320

Q2:ipmat indore 2021QA › Progression & SeriesMediumSA · TITA

The sum up to

10

terms of the series

1 \cdot 3 + 5 \cdot 7 + 9 \cdot 11 + ...

is

Enter your answer to attempt

The Setup: We need the sum of the first 10 terms of the series

1\cdot3+5\cdot7+9\cdot11+...

. We are dealing with products of Arithmetic Progressions (APs). We need to unmask the

n

-th term so we can just blast it with summation formulas and speedrun the calculation. Step 1: Find the pattern for the first part of each term (1, 5, 9...). This is an AP with a common difference of 4.

a_n = 1 + (n-1)4 = 4n - 3

Step 2: Find the pattern for the second part of each term (3, 7, 11...). This is another AP, also with a common difference of 4.

b_n = 3 + (n-1)4 = 4n - 1

Step 3: Construct the general term

T_n

by multiplying them, and expand it into a quadratic polynomial:

T_n = (4n - 3)(4n - 1) = 16n^2 - 16n + 3

Step 4: Deploy the summation formulas for the first 10 terms.

\sum_{n=1}^{10} T_n = 16 \sum n^2 - 16 \sum n + \sum 3

Using the standard formulas

\sum n^2 = \frac{n(n+1)(2n+1)}{6}

and

\sum n = \frac{n(n+1)}{2}

for

n=10

: *

16 \times \left(\frac{10 \times 11 \times 21}{6}\right) = 16 \times 385 = 6160

*

-16 \times \left(\frac{10 \times 11}{2}\right) = -16 \times 55 = -880

*

10 \times 3 = 30

Step 5: Crunch the final numbers.

6160 - 880 + 30 = 5310

Final Answer: 5310

Q3:ipmat indore 2021QA › Progression & SeriesMediumSA · TITA

It is given that the sequence {

x_n

} satisfies

x_1 = 0, x_{n+1} = x_n + 1 + 2\sqrt(1+x_n)

for

n = 1,2,...

Then

x_{31}

is _______

Enter your answer to attempt

The Setup: We are faced with a recursive sequence:

x_{n+1}=x_n+1+2\sqrt{1+x_n}

with a starting state of

x_1=0

. It looks cursed, but it's actually just a perfect square in disguise. We just need to transform the variables to reveal the hidden meta. Step 1: Add 1 to both sides to complete the transformation.

x_{n+1} + 1 = (x_n + 1) + 2\sqrt{1+x_n} + 1

Step 2: Substitute

y_n=x_n+1

to clean up the UI.

y_{n+1} = y_n + 2\sqrt{y_n} + 1

Notice that the right side is a literal perfect square:

(\sqrt{y_n} + 1)^2

.

y_{n+1} = (\sqrt{y_n} + 1)^2

Step 3: Take the square root of both sides.

\sqrt{y_{n+1}} = \sqrt{y_n} + 1

This means the sequence

\sqrt{y_n}

is just a basic Arithmetic Progression increasing by 1 every step. Since

x_1=0

, our base case is

y_1=1

, which means

\sqrt{y_1}=1

. Step 4: Find the

n

-th term of this AP.

\sqrt{y_n} = 1 + (n-1)(1) = n

Step 5: Reverse the transformation to find

x_n

.

y_n = n^2 \implies x_n + 1 = n^2 \implies x_n = n^2 - 1

Step 6: Calculate the 31st term to finish the level.

x_{31} = 31^2 - 1 = 961 - 1 = 960

Final Answer: 960

Q4:ipmat indore 2021QA › Permutation & CombinationEasySA · TITA

There are 5 parallel lines on the plane. On the same plane, there are

n

other lines that are perpendicular to the 5 parallel lines. If the number of distinct rectangles formed by these lines is 360, what is the value of

n

?

Enter your answer to attempt

The Setup: We are stepping into the Combinatorics arena. There are 5 parallel lines and

n

perpendicular lines. To build a rectangle, you need a squad of exactly 4 lines: 2 horizontal and 2 vertical. It's like picking your duo for Fortnite-you need exactly two from each distinct set to lock in the shape. Step 1: Calculate the number of pairs we can pull from the 5 parallel lines. Choosing 2 out of 5 is basic combinations:

\binom{5}{2} = \frac{5 \times 4}{2} = 10

Step 2: Set up the equation for the perpendicular lines. We need 2 lines from the

n

squad, which gives us

\binom{n}{2}

. The total number of distinct rectangles is 360.

10 \times \binom{n}{2} = 360

Step 3: Solve for

n

. Divide both sides by 10 to strip it down:

\binom{n}{2} = 36

\frac{n(n-1)}{2} = 36

n(n-1) = 72

Step 4: Find the consecutive integers that multiply to 72. Since

9 \times 8 = 72

, we know

n=9

. No cap, it's that clean. Final Answer: 9

Q5:ipmat indore 2021QA › Time & WorkMediumSA · TITA

There are two taps, T1 and T2, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T1 is opened keeping T2 closed, the water tank (initially full) becomes empty in half an hour. If both T1 and T2 are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T2 is opened while T1 is closed is

Enter your answer to attempt

The Setup: This is a classic 'Time and Work' rate problem, just flipped to deal with emptying a tank instead of filling it. T1 clears the tank in 30 minutes solo, and T1 and T2 together clear it in 20 minutes. We treat the full tank as 1 complete unit of work and need to isolate the standalone speed of tap T2. Step 1: Establish T1's solo drain rate.

\text{Rate of T1} = \frac{1}{30} \text{ tanks per minute}

Step 2: Look at their combined synergy.

\text{Rate of T1} + \text{Rate of T2} = \frac{1}{20} \text{ tanks per minute}

Step 3: Subtract T1's rate from the combined rate to expose T2's true solo speed.

\text{Rate of T2} = \frac{1}{20} - \frac{1}{30}

Step 4: Find a common denominator (which is 60) to crunch the fractions:

\frac{3}{60} - \frac{2}{60} = \frac{1}{60}

This means T2 clears

\frac{1}{60}

of the tank every minute. Step 5: Flip the rate to find the total time. If T2 does

\frac{1}{60}

of the job per minute, it takes exactly 60 minutes to finish the whole thing solo. Final Answer: 60

Q6:ipmat indore 2021QA › Mean, Median & ModeMediumSA · TITA

A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is

Enter your answer to attempt

The Setup: This is a weighted average and total sum problem. 30 students are taking 5 courses, with each exam out of 200 marks. Averages are just totals spread out evenly. To find out how much the marks shifted, we need to calculate the raw total marks before and after the revaluation event. Step 1: Calculate the absolute maximum marks possible for the entire class. 30 students taking 5 courses means there are 150 individual exams.

150 \times 200 = 30000 \text{ total possible marks}

Step 2: Find the raw total marks before the revaluation. The squad averaged an 80% hit rate.

0.80 \times 30000 = 24000 \text{ marks}

Step 3: Find the raw total marks after the revaluation bump. The new average slightly leveled up to 80.02%.

0.8002 \times 30000 = 24006 \text{ marks}

Step 4: Find the net increase. The total pool of marks literally went up by exactly 6 (

24006 - 24000

). Step 5: Allocate the points. We know two students got revaluations and neither lost points. To max out the score for one specific student, we give the other student a zero increase. This lets one student hog all 6 newly discovered points. Final Answer: 6

Q7:ipmat indore 2021QA › Straight LinesHardSA · TITA

If one of the lines given by the equation

2x^2 + axy + 3y^2 = 0

coincides with one of those given by

2x^2 + bxy - 3y^2 = 0

and the other lines represented by them are perpendicular then

a^2+b^2 =

Enter your answer to attempt

The Setup: We're dealing with homogenous equations of degree 2, which represent pairs of straight lines passing through the origin. The equations are

2x^2+axy+3y^2=0

and

2x^2+bxy-3y^2=0

. One line coincides, and the other lines are perpendicular. This is pure coordinate geometry boss-fight level. Let the slope of the shared line be

m

, and the perpendicular ones be

m_1

and

m_2

. Step 1: Translate the given equations to quadratic form in terms of slope (

y=mx

). For the first pair, dividing by

x^2

gives:

3m^2 + am + 2 = 0

The product of the roots (slopes) is

m \cdot m_1 = \frac{2}{3} \implies m_1 = \frac{2}{3m}

. Step 2: Repeat for the second pair.

-3m^2 + bm + 2 = 0

The product of slopes here is

m \cdot m_2 = -\frac{2}{3} \implies m_2 = -\frac{2}{3m}

. Step 3: Use the perpendicularity condition for the non-coinciding lines. Their slopes multiply to

-1

:

m_1 \cdot m_2 = -1

\left(\frac{2}{3m}\right) \left(-\frac{2}{3m}\right) = -1 \implies -\frac{4}{9m^2} = -1 \implies m^2 = \frac{4}{9}

Step 4: Set up the algebra to find

(a^2+b^2)

. We know

m

is a root for both equations, so both evaluate to 0:

3m^2 + am + 2 = 0

(Eq. 1)

-3m^2 + bm + 2 = 0

(Eq. 2) Add them together:

(a+b)m + 4 = 0 \implies a+b = -\frac{4}{m}

Subtract Eq. 2 from Eq. 1:

6m^2 + (a-b)m = 0 \implies a-b = -6m

Step 5: Calculate the final target,

a^2 + b^2

. Use the identity

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

. Square the sums:

(a+b)^2 = \frac{16}{m^2} = \frac{16}{4/9} = 36

(a-b)^2 = 36m^2 = 36(\frac{4}{9}) = 16

2(a^2+b^2) = 36 + 16 = 52 \implies a^2+b^2 = 26

Final Answer: 26

Q8:ipmat indore 2021QA › FunctionsMediumSA · TITA

If a function

f(a) = max (a, 0)

then the smallest integer value of

x

for which the equation

f(x - 3) + 2f(x + 1) = 8

holds true is _______.

Enter your answer to attempt

The Setup: This is a piecewise function check. The function

f(a)=\max(a,0)

basically acts like a bouncer: if

a

is negative, it gets zeroed out. We need to find the smallest integer

x

where

f(x-3)+2f(x+1)=8

. Step 1: Test the lowest tier (where both inputs are negative). If

x \le -1

, then both

x-3

and

x+1

are negative or zero. The bouncer zeroes them both:

0 + 2(0) = 0 \neq 8

That's a hard pass. Step 2: Level up to the middle tier (

-1 < x \le 3

). Here,

x+1

is strictly positive (gets VIP access), but

x-3

is still negative or zero (gets zeroed). The equation becomes:

0 + 2(x+1) = 8

x + 1 = 4

x = 3

This works perfectly and fits exactly at the upper boundary of our assumed domain. Step 3: Double-check the top tier just to be safe (

x>3

). Here, both inputs are strictly positive:

(x-3) + 2(x+1) = 8

3x - 1 = 8

3x = 9 \implies x = 3

Since

x=3

is the only valid root across the whole domain, that's our main character. Final Answer: 3

Q9:ipmat indore 2021QA › Set TheoryEasySA · TITA

In a class, 60% and 68% of students passed their Physics and Mathematics examinations respectively. Then at least ________ percentage of students passed both their Physics and Mathematics examinations.

Enter your answer to attempt

The Setup: We are dealing with Set Theory here, specifically overlapping sets. 60% passed Physics and 68% passed Mathematics. We want to find the absolute floor (the minimum overlap) of students who survived both exams. Think of it like trying to find the minimum number of mutuals two influencers *must* share if they both have a massive following in the same city. Step 1: Define the sets and the absolute total. Let

n(P)=60

and

n(M)=68

. The maximum possible size of the entire union (everyone who passed at least one subject) literally cannot exceed 100%. So:

n(P \cup M) \le 100

Step 2: Hit them with the standard Set Theory formula:

n(P \cup M) = n(P) + n(M) - n(P \cap M)

Step 3: Plug in the known stats and the boundary we just established:

60 + 68 - n(P \cap M) \le 100

128 - n(P \cap M) \le 100

Step 4: Solve the inequality to find the minimum intersection threshold:

128 - 100 \le n(P \cap M)

n(P \cap M) \ge 28

The absolute bare minimum overlap is 28%. Period. Final Answer: 28

Q10:ipmat indore 2021QA › FunctionsMediumMCQ · MCQ

Suppose that a real-valued function

f(x)

of real numbers satisfies

f(x + xy) = f(x) + f(xy

) for all real

x, y,

and that

f(2020) = 1

. Compute

f(2021)

.

A $\frac{2021}{2020}$
B $\frac{2020}{2019}$
C1
D $\frac{2020}{2021}$

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)

by plugging in strategic dummy variables until the underlying algorithm reveals itself. Step 1: Test the waters by setting

y=0

to find the baseline.

f(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 \neq 0

and any real

z

, just pick

y=z/x

- then

xy=z

, and the given equation morphs into:

f(x+z)=f(x)+f(z)

Together with Step 1 (which covers

x=0

),

f

is additive across the reals. Step 3: Stack that property. Applying it repeatedly gives

f(nt)=nf(t)

for every positive integer

n

. That is the *only* thing we need - so we never have to assume

f

is linear or even continuous, which would be an unearned leap. Step 4: Use the given stat

f(2020)=1

, taking

t=1

and

n=2020

.

2020f(1)=f(2020)=1 \implies f(1)=\frac{1}{2020}

Step 5: Calculate the final target with

n=2021

.

f(2021)=2021f(1)=\frac{2021}{2020}

Final Answer:

\frac{2021}{2020}

Q11: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

then the value of

a + 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

, then

Y=x^0=1

. Step 1: Crack the code for

a

.

\log_2[\log_3(\log_4a)]=0 \implies \log_3(\log_4a)=2^0=1

\log_4a=3^1=3 \implies a=4^3=64

Step 2: Crack the code for

b

.

\log_3[\log_4(\log_2b)]=0 \implies \log_4(\log_2b)=3^0=1

\log_2b=4^1=4 \implies b=2^4=16

Step 3: Crack the code for

c

.

\log_4[\log_2(\log_3c)]=0 \implies \log_2(\log_3c)=4^0=1

\log_3c=2^1=2 \implies c=3^2=9

Step 4: Add the squad together to find

a+b+c

.

a+b+c=64+16+9=89

Final Answer: 89

Q12:ipmat indore 2021QA › Progression & SeriesMediumMCQ · MCQ

Let

S_n

be sum of the first

n

terms of an A.P. If

S_5 = S_9

, what is the ratio of

a_3 : a_5

A $9:5$
B $5:9$
C $3:5$
D $5:3$

Pick an option to attempt

The Setup: We are working with Arithmetic Progressions. The standard sum formula is

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 (

a

) and the common difference (

d

). Step 1: Expand both sides using the sum formula.

S_5=\frac{5}{2}(2a+4d)=5(a+2d)=5a+10d

S_9=\frac{9}{2}(2a+8d)=9(a+4d)=9a+36d

Step 2: Set them equal and isolate

a

in terms of

d

.

5a+10d=9a+36d

-4a=26d \implies 2a=-13d \implies a=-6.5d

Step 3: Find the expressions for the specific requested terms,

a_3

and

a_5

.

a_3=a+2d=-6.5d+2d=-4.5d

a_5=a+4d=-6.5d+4d=-2.5d

Step 4: Calculate the ratio. The negative signs and the

d

variables cancel out cleanly.

\text{Ratio}=\frac{-4.5d}{-2.5d}=\frac{4.5}{2.5}=\frac{45}{25}=\frac{9}{5}

Final Answer:

9:5

Q13:ipmat indore 2021QA › Matrices & DeterminantsMediumMCQ · MCQ

If

A, B

and

A + B

are non singular matrices and

AB = BA

then

2A - 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 \neq BA

. However, the prompt gives us the

AB=BA

buff. This means

A

and

B

commute. Even better, since

A(A+B)=A^2+AB=A^2+BA=(A+B)A

, matrix

A

also commutes with

(A+B)

and its inverse

(A+B)^{-1}

. Step 1: Group the complicated inverse terms together to see what we're working with. Because

A

commutes with

(A+B)^{-1}

, we can rewrite

A(A+B)^{-1}A

as

A^2(A+B)^{-1}

. The identical argument runs for

B

(it commutes with

A+B

for exactly the same reason), so

B(A+B)^{-1}B

becomes

B^2(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=BA

, the cross terms cancel out perfectly, allowing standard factorization.

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)(A+B)^{-1}

Since

(A+B)

multiplied by its inverse is the Identity matrix (

I

), this entire chunk simplifies to just

(B-A)

. Step 4: Plug this beautifully simplified chunk back into the main equation.

2A-B+(B-A)

2A-B+B-A=A

Final Answer: A

Q14:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ

If the angles

A, B, C

of a triangle are in arithmetic progression such that

\sin(2A + B) = 1/2

then

\sin(B + 2C)

is equal to

A $\frac{-1}{2}$
B $\frac{1}{2}$
C $\frac{-1}{\sqrt{2}}$
D $\frac{3}{\sqrt{2}}$

Pick an option to attempt

The Setup: Geometry meets Trigonometry. The angles of a triangle always sum to

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

B

. Since

A

,

B

,

C

are in AP:

A+C=2B

Substitute this into the triangle angle sum theorem:

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

A

. We know

\sin(2A+B)=1/2

. Since

A+C=120^\circ

, angle

A

is stuck below

120^\circ

, so

2A+60^\circ

never climbs past

300^\circ

- that leaves exactly two candidates,

30^\circ

and

150^\circ

. * If

2A+60^\circ=30^\circ \implies 2A=-30^\circ

(Angles in a triangle must be positive, so this is a wipe). * If

2A+60^\circ=150^\circ \implies 2A=90^\circ \implies A=45^\circ

. Step 3: Find angle

C

.

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)

.

B+2C=60^\circ+2(75^\circ)=60^\circ+150^\circ=210^\circ

\sin(210^\circ)=\sin(180^\circ+30^\circ)=-\sin(30^\circ)=-\frac{1}{2}

Final Answer:

\frac{-1}{2}

Q15:ipmat indore 2021QA › Unit DigitMediumMCQ · MCQ

The unit digit in

(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:

743^{85}

. The unit digit depends entirely on

3^{85}

. The powers of

3

cycle in a 4-step loop (

3

,

9

,

7

,

1

). Divide the exponent

85

by

4

to find the remainder.

85=4\times21+1

Since the remainder is

1

, it hits the first step of the cycle. The unit digit is

3^1=3

. Step 2: Analyze the second term:

525^{37}

. This is a freebie. Any positive integer power of a number ending in

5

will always end in

5

. The unit digit is

5

. Step 3: Analyze the third term:

987^{96}

. The unit digit depends on

7^{96}

. The powers of

7

also cycle in a 4-step loop (

7

,

9

,

3

,

1

). Divide

96

by

4

.

96=4\times24+0

Since it divides perfectly (remainder

0

), it lands on the final step of the cycle. The unit digit is

1

. Step 4: Combine the unit digits.

3-5+1=-1

Since we can't output a negative unit digit in a standard integer, we simulate borrowing a

10

from the next placeholder column over (just like elementary school subtraction).

10-1=9

Final Answer: 9

Q16:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ

The set of all real value of

p

for which the equation

3 \sin^2x + 12 \cos x - 3 = p

has at least one solution is

A $[-12, 12]$
B $[-12, 9]$
C $[-15, 9]$
D $[-15, 12]$

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

p

). Since the equation involves both sine and cosine, we use the identity

\sin^2x=1-\cos^2x

to unify the variables. Step 1: Substitute and simplify the expression. Let

f(x)=p

.

3(1-\cos^2x)+12\cos x-3=p

3-3\cos^2x+12\cos x-3=p

-3\cos^2x+12\cos x=p

Step 2: Set up a boundary replacement. Let

t=\cos x

. Because of the constraints of the cosine function,

t

is strictly bounded:

t\in[-1,1]

. Our new function is

g(t)=-3t^2+12t

. We need the absolute min and max of this quadratic on the locked interval

[-1,1]

. Step 3: Find the vertex of the parabola. For a quadratic

at^2+bt+c

, the vertex occurs at

t=-b/(2a)

.

t=\frac{-12}{2(-3)}=\frac{-12}{-6}=2

The vertex sits at

t=2

, which is completely outside our playable zone of

[-1,1]

. Because it's a downward-opening parabola (negative

a

), it is strictly increasing across our entire

[-1,1]

interval. Step 4: Test the boundaries to find the range. * If

t=-1

(Minimum):

g(-1)=-3(-1)^2+12(-1)=-3-12=-15

. * If

t=1

(Maximum):

g(1)=-3(1)^2+12(1)=-3+12=9

. The function seamlessly spans every value between these two extremes. Final Answer:

[-15, 9]

Q17: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

O

, the products of the areas of opposite triangles are perfectly equal. Let the areas be

A_1

,

A_2

,

A_3

,

A_4

circling the intersection. We are given

A_1=4

and

A_3=9

. We need to optimize

A_2

and

A_4

to minimize the total sum. Step 1: Apply the cross-product area rule.

A_1\times A_3=A_2\times A_4

4\times9=A_2\times A_4 \implies A_2\times A_4=36

Step 2: Set up the total area equation.

\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 \ge 2\sqrt{xy}

.

A_2+A_4 \ge 2\sqrt{A_2\times A_4}

A_2+A_4 \ge 2\sqrt{36}=2(6)=12

The absolute minimum sum these two triangles can mathematically contribute is

12

(which occurs when

A_2=A_4=6

). Step 4: Add the locked constants back to find the overall minimum area.

\text{Min Total Area}=13+12=25

Final Answer: 25

Q18: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+d

. We want to maximize the ratio

\frac{1000a+100b+10c+d}{a+b+c+d}

. Step 1: Use an algebraic reduction to isolate the variables. Let

S=a+b+c+d

. We can rewrite the numerator in terms of

S

.

1000a+100b+10c+d=1000(S-b-c-d)+100b+10c+d

=1000S-1000b-1000c-1000d+100b+10c+d

=1000S-900b-990c-999d

Step 2: Reconstruct the ratio with the new expression.

\text{Ratio}=\frac{1000S-900b-990c-999d}{S}

\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

b

,

c

, and

d

are digits (meaning they are non-negative integers

\ge 0

), the absolute smallest value for the subtracted term is

0

. This occurs when we hard-lock the nerfed variables to zero:

b=0

,

c=0

,

d=0

. Step 4: Calculate the ratio with the optimized stats.

\text{Ratio}=1000-0=1000

(For example, if the number is

9000

, the sum of digits is

9

.

9000/9=1000

. It mathematically cannot get higher than this). Final Answer: 1000

Q19:ipmat indore 2021QA › Straight LinesMediumMCQ · MCQ

Consider the polynomials

f(x) = ax^2 + bx + c

, where

a > 0, b, c

are real,

g(x) = -2x

. If

f(x)

cuts the x-axis at

(-2, 0)

and

g(x)

passes through

(a, b)

, then the minimum value of

f(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)

entirely in terms of the variable

a

. 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)

coordinate to crack the value of

b

. The line

g(x)

passes through

(a,b)

, so we literally just plug those coordinates in.

b=-2(a) \implies b=-2a

Step 2: Use the x-intercept to crack

c

. The function

f(x)

hits the axis at

x=-2

, meaning

f(-2)=0

.

a(-2)^2+b(-2)+c=0

4a-2b+c=0

Substitute

b=-2a

into this equation:

4a-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)=ax^2-2ax-8a=a(x^2-2x-8)

For an upward-facing parabola (

a>0

), the minimum value occurs exactly at its vertex

x=-b_{new}/(2a_{new})

.

x=\frac{-(-2a)}{2(a)}=\frac{2a}{2a}=1

Step 4: Calculate the minimum y-value of

f(x)

by plugging in

x=1

.

f(1)=a(1^2)-2a(1)-8a=a-2a-8a=-9a

Step 5: Evaluate the final target expression

f(x)+9a+1

at its minimum point.

(-9a)+9a+1=1

Final Answer: 1

Q20: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\%

. The people speaking zero languages (the 'None' group) make up the remainder:

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,

\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.

\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:

\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\%+\text{Target}

52\%=30\%+\text{Target} \implies \text{Target}=22\%

Final Answer: 22%

Q21: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

V_1

and

V_2

.

8V_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

V_1

.

V_1+V_2=140

V_1+\frac{4}{3}V_1=140 \implies \frac{7}{3}V_1=140 \implies V_1=60 \text{ km/hr}

Since

V_1

is

60

,

V_2=140-60=80 \text{ km/hr}

. Step 4: Set up the alternate timeline. Train 2 now starts 5 hours late. Let

t

be the total hours Train 1 travels until it gets caught. Train 2 will have traveled

t-5

hours. Equate the new distances:

60t=80(t-5)

60t=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

Q22: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

5

odd digits available (

1

,

3

,

5

,

7

,

9

). We need to select

3

of them and arrange them into the

3

odd slots. Number of ways to fill odd places =

P(5,3)=5\times4\times3=60

ways. Step 2: Assess the remaining free agents. We have

5

even digits (

0

,

2

,

4

,

6

,

8

) and

2

leftover odd digits that didn't get drafted into the odd slots. That gives us a pool of

7

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

2

digits from the pool of

7

and arrange them. Number of ways to fill even places =

P(7,2)=7\times6=42

ways. Step 4: Multiply the independent events to find the total combinations.

\text{Total Numbers}=60\times42=2520

Final Answer: 2520

Q23: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

2

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

2

points out of

10

gives us:

\binom{10}{2}=\frac{10\times9}{2}=45 \text{ possible pairs}

Step 2: Calculate the overlapping spam. The

5

collinear points are generating redundant lines. If we picked pairs exclusively from these

5

points, we'd get:

\binom{5}{2}=\frac{5\times4}{2}=10 \text{ redundant pairs}

Step 3: Clean up the geometry. We subtract all

10

redundant pairs from our total because they don't form unique lines. However, we still have to add

1

back in to represent the single, massive straight line that all

5

of those points are sitting on.

\text{Total Distinct Lines}=45-10+1=36

Final Answer: 36

Q24:ipmat indore 2021QA › Straight LinesMediumMCQ · MCQ

The x-intercept of the line that passes through the intersection of the lines

x + 2y = 4

and

2x + 3y = 6

, and is perpendicular to the line

3x - 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=4

and

2x+3y=6

. Multiply the first equation by

2

:

2x+4y=8

. Subtract the second equation from it:

(2x+4y)-(2x+3y)=8-6 \implies y=2

Plug

y=2

back into the first equation:

x+2(2)=4 \implies x+4=4 \implies x=0

The lines intersect at

(0,2)

. This is our anchor point. Step 2: Steal the slope from the reference line. The reference is

3x-y=2

, which rearranges to

y=3x-2

. The slope (

m

) is

3

. Step 3: Calculate the perpendicular slope. Perpendicular lines have negative reciprocal slopes.

m_{\perp}=-\frac{1}{3}

Step 4: Build the final line equation using the point-slope form with our anchor point

(0,2)

and new slope

-1/3

.

y-2=-\frac{1}{3}(x-0)

3y-6=-x \implies x+3y=6

Step 5: Find the x-intercept. The x-intercept always occurs exactly when

y=0

.

x+3(0)=6 \implies x=6

Final Answer: 6

IPMAT Indore 2021 — QA

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