IPMAT Indore 2019 SA PYQs

Q1:ipmat indore 2019QA › PolygonsEasySA · TITA

The sum of the interior angles of a convex

n

-sided polygon is less than

2019^\circ

. The maximum possible value of

n

is

Enter your answer to attempt

The Setup: A pure geometry fundamental. The sum of the interior angles of any convex

n

-sided polygon is the GOAT formula

(n-2)\times 180^\circ

. Set up the inequality, solve, and take the largest integer that survives. Step 1: Set up the inequality. The sum is strictly less than

2019^\circ

:

(n-2)\times 180 < 2019

Step 2: Isolate the variable. Divide both sides by 180 (positive, so the inequality direction is safe):

n-2<\frac{2019}{180}

Step 3: Crunch the division. Since

180\times 11=1980

and

180\times 12=2160

, the quotient sits between 11 and 12:

\frac{2019}{180}=11.216\ldots \implies n<13.216\ldots

Step 4: Lock in the maximum integer. A polygon has a whole number of sides, so the largest admissible

n

is 13. Step 5: Verify both sides of the boundary. A bound is only trustworthy if the next value up actually fails: *

n=13

: sum

=(13-2)\times 180=1980^\circ

, and

1980<2019

. Valid. *

n=14

: sum

=(14-2)\times 180=2160^\circ

, and

2160>2019

. Fails. So 13 works and 14 does not, which pins the maximum exactly. Final Answer: 13

Q2:ipmat indore 2019QA › LogarithmsMediumSA · TITA

Suppose that a, b, and c are real numbers greater than 1. Then the value of

\dfrac{1}{1+\log_{a^2 b} \frac{c}{a}} + \dfrac{1}{1+\log_{b^2 c} \frac{a}{b}} + \dfrac{1}{1+\log_{c^2 a} \frac{b}{c}}

is

Enter your answer to attempt

The Setup: Log base inception. The trick is to convert that awkward

1

in each denominator into a logarithm sharing the same base, merge it with the term beside it, and then flip each fraction with the reciprocal rule. **Step 0: Note why

a,b,c>1

matters.** A logarithm needs a positive base other than 1. Because every one of

a,b,c

exceeds 1, each base

a^2b

,

b^2c

,

c^2a

and the eventual base

abc

is greater than 1, so every logarithm here is defined and the reciprocal rule is legal. That condition in the stem is doing real work, not decoration. Step 1: Convert the 1s. Using

1=\log_x x

with the base already present:

1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(a^2b)+\log_{a^2b}\left(\frac{c}{a}\right)

Step 2: Merge the logs. By the product rule

\log x+\log y=\log(xy)

, and noting that one factor of

a

cancels against the

\frac{1}{a}

:

a^2b\times\frac{c}{a}=a^{2-1}bc=abc \implies 1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(abc)

The same collapse happens in the other two, each landing on the identical argument

abc

:

b^2c\times\frac{a}{b}=abc, \qquad c^2a\times\frac{b}{c}=abc

So the whole expression becomes:

\frac{1}{\log_{a^2b}(abc)}+\frac{1}{\log_{b^2c}(abc)}+\frac{1}{\log_{c^2a}(abc)}

Step 3: Flip the base. Apply the reciprocal rule

\frac{1}{\log_x y}=\log_y x

, which swaps base and argument:

\log_{abc}(a^2b)+\log_{abc}(b^2c)+\log_{abc}(c^2a)

Step 4: The final fusion. They now share a base, so the sum becomes a single log of the product:

\log_{abc}\left(a^2b\cdot b^2c\cdot c^2a\right)=\log_{abc}\left(a^3b^3c^3\right)=\log_{abc}\left((abc)^3\right)

Pull the exponent out front:

3\log_{abc}(abc)=3\times 1=3

Note the answer is a bare constant - it does not depend on

a

,

b

or

c

at all, which is the elegance the question is built around. Final Answer: 3

Q3:ipmat indore 2019QA › FunctionsHardSA · TITA

A real-valued function

f

satisfies the relation

f(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6y

, for all real numbers

x

and

y

, then the value of

f(8)

is

Enter your answer to attempt

The Setup: Functional equations are the final boss of algebra. The standard strat is to substitute a convenient value to collapse one variable, and here

x=0

does something better than simplify - it *proves* the function is linear rather than making us assume it. **Step 1: Substitute

x=0

.**

f(0)f(y)=f(3)+3f(y)-3f(y)+6y \implies f(0)f(y)=f(3)+6y

Write

c=f(0)

, so that

c\,f(y)=6y+f(3)

. **Step 2: Rule out

c=0

before dividing.** If

c

were 0, the left side would be 0 for every

y

, forcing

6y+f(3)=0

for all real

y

- impossible, since the left side varies with

y

and the right is fixed. So

c\neq 0

, and we may divide:

f(y)=\frac{6}{c}y+\frac{f(3)}{c}

This is a genuine deduction, not a guess:

f

must be linear. Write

f(x)=kx+c

, noting

f(0)=c

is consistent and that

k=\frac{6}{c}

, i.e.

ck=6

. Step 3: Substitute the linear form into the original relation.

\text{LHS}=(kx+c)(ky+c)=k^2xy+kcx+kcy+c^2

\text{RHS}=\left[k(2xy+3)+c\right]+3\left[k(x+y)+c\right]-3\left[ky+c\right]+6y=2kxy+3kx+6y+3k+c

The

3ky

and

-3ky

cancel, which is why no

y

-term survives on the right beyond the

6y

. Step 4: Match all four coefficients - not just one.

Term	LHS	RHS	Equation
$xy$	$k^2$	$2k$	$k^2=2k$
$x$	$kc$	$3k$	$kc=3k$
$y$	$kc$	$6$	$kc=6$
constant	$c^2$	$3k+c$	$c^2=3k+c$

From

k^2=2k

we get

k=0

or

k=2

;

k=0

contradicts

ck=6

, so

k=2

. Then

kc=6

gives

c=3

. Check the two remaining rows hold rather than assuming they do:

kc=6=3k

✓, and

c^2=9=3(2)+3

✓. All four agree, so:

f(x)=2x+3

Step 5: Execute.

f(8)=2(8)+3=19

Final Answer: 19

Q4:ipmat indore 2019QA › Matrices & DeterminantsHardSA · TITA

Let

A, B, C

be three

4 \times 4

matrices such that

det \ A = 5, det \ B = -3

, and

det \ C = \frac{1}{2}

. Then the

det

2AB^{-1}C^3B^T

is

Enter your answer to attempt

The Setup: A determinant-properties speedrun. Four rules do all the work:

\det(kM)=k^n\det M

for an

n\times n

matrix,

\det(XY)=\det X\det Y

,

\det(X^{-1})=\frac{1}{\det X}

, and

\det(X^T)=\det X

. Step 1: Extract the scalar. The matrices are

4\times 4

, so

n=4

and the scalar 2 comes out to the fourth power - not the first, which is the classic slip here:

\det\left(2AB^{-1}C^3B^T\right)=2^4\det\left(AB^{-1}C^3B^T\right)=16\det\left(AB^{-1}C^3B^T\right)

Step 2: Shatter the product. The determinant is multiplicative, so it distributes across the four factors:

16\times\det A\times\det\left(B^{-1}\right)\times\det\left(C^3\right)\times\det\left(B^T\right)

Step 3: Substitute the properties.

\det(B^{-1})=\frac{1}{\det B}

,

\det(C^3)=(\det C)^3

, and

\det(B^T)=\det B

:

16\times\det A\times\frac{1}{\det B}\times(\det C)^3\times\det B

Step 4: Cancel and calculate. The transpose contributes

\det B

and the inverse contributes

\frac{1}{\det B}

, so they annihilate each other:

16\times\det A\times(\det C)^3=16\times 5\times\left(\frac{1}{2}\right)^3=80\times\frac{1}{8}=10

**Step 5: Notice what

\det B=-3

was for.** Its *value* never reaches the answer - it cancels. But its being non-zero is exactly what guarantees

B

is invertible, so

B^{-1}

exists and the expression is meaningful at all. It is a load-bearing hypothesis disguised as a red herring. Final Answer: 10

Q5:ipmat indore 2019QA › Matrices & DeterminantsMediumSA · TITA

If

A

is a

3 \times 3

non-zero matrix such that

A^2 = 0

then the determinant of

(I + A)^{50} - 50A

is equal to

Enter your answer to attempt

The Setup: A binomial expansion on matrices. The condition

A^2=0

(a nilpotent matrix) is a kill switch: every power of

A

from the second onward vanishes, truncating a 51-term expansion into two terms. Step 0: Check the binomial theorem is even allowed. For matrices,

(X+Y)^n

expands binomially **only if

X

and

Y

commute** - otherwise the cross terms will not collect. Here

X=I

, and the identity commutes with every matrix, so

IA=AI=A

and the expansion is valid. Skipping this check is how the same move goes wrong on a harder problem. Step 1: Expand.

(I+A)^{50}=I^{50}+\binom{50}{1}I^{49}A+\binom{50}{2}I^{48}A^2+\cdots+A^{50}

Step 2: Trigger the kill switch. From

A^2=0

it follows that

A^3=A\cdot A^2=A\cdot 0=0

, and inductively every higher power is zero too. So every term from the

A^2

one onward dies, and

I^m=I

:

(I+A)^{50}=I+50A

Step 3: Simplify the expression.

(I+A)^{50}-50A=(I+50A)-50A=I

The

50A

subtraction is built into the question precisely to cancel the surviving term, leaving the identity. Step 4: Take the determinant. The determinant of the identity matrix is 1 in any dimension:

\det(I)=1

Note the answer never depended on *which* nilpotent

A

was chosen, only on

A^2=0

- the 'non-zero' in the stem just stops the problem being trivial. Final Answer: 1

Q6:ipmat indore 2019QA › Ratio, Proportion & VariationEasySA · TITA

Three friends divided some apples in the ratio

3 : 5 : 7

. After consuming 16 apples they found that the remaining number of apples with them was equal to the largest number of apples received by one of them at the beginning. The total number of apples these friends initially had was

Enter your answer to attempt

The Setup: A linear equation dressed as a ratio problem. Define the shares through a single multiplier, subtract what was eaten, and set the remainder equal to the largest original share. Step 1: Define the shares. Let the ratio multiplier be

x

, so the three friends received

3x

,

5x

and

7x

apples.

\text{Total at the start}=3x+5x+7x=15x

The largest individual share is

7x

. Step 2: Construct the equation. After 16 apples are eaten,

15x-16

remain, and the stem says this equals the largest starting share:

15x-16=7x

Step 3: Solve for the multiplier.

15x-7x=16 \implies 8x=16 \implies x=2

Step 4: Answer the question that was asked. The prompt wants the initial total, not the multiplier and not one person's share:

\text{Total}=15x=15(2)=30

Step 5: Verify against the story. With

x=2

the shares are 6, 10 and 14, totalling 30. Eat 16 and

30-16=14

remain - exactly the largest starting share of 14. The story checks out. Final Answer: 30

Q7:ipmat indore 2019QA › Profit & LossEasySA · TITA

A shopkeeper reduces the price of a pen by 25% as a result of which the sales quantity increased by 20%. If the revenue made by the shopkeeper decreases by x% then x is

Enter your answer to attempt

The Setup: Revenue is

\text{Price}\times\text{Quantity}

, so a percentage change in each multiplies rather than adds. Anyone who subtracts

25\%-20\%=5\%

has fallen for the trap this question exists to set. Step 1: Define the baseline. Let the original price be

P

and quantity

Q

, so the original revenue is

R_1=PQ

. Step 2: Apply the percentage shifts as multipliers. * Price falls 25%: new price

=(1-0.25)P=0.75P

* Quantity rises 20%: new quantity

=(1+0.20)Q=1.20Q

Step 3: Calculate the new revenue.

R_2=(0.75P)(1.20Q)=(0.75\times 1.20)PQ=0.90\,PQ

Step 4: Convert to a percentage drop. Revenue went from

1.00PQ

to

0.90PQ

, a fall of

0.10PQ

:

\frac{R_1-R_2}{R_1}\times 100=\frac{0.10PQ}{PQ}\times 100=10\%

So

x=10

. Sanity check with real numbers: at

P=100

,

Q=100

, revenue goes from 10,000 to

75\times 120=9{,}000

- down 1,000, which is 10%. Final Answer: 10

Q8:ipmat indore 2019QA › Minima & MaximaHardSA · TITA

For all real values of

x

,

\dfrac{3x^2 - 6x + 12}{x^2 + 2x + 4}

lies between

1

and

k

, and does not take any value above

k

. Then

k

equals:

Enter your answer to attempt

The Setup: To find the range of a rational function, set it equal to

y

, rearrange into a quadratic in

x

, and force the discriminant to be non-negative - because

x

is required to be real. The values of

y

that survive are exactly the range. Step 0: Confirm the function is defined everywhere. The denominator

x^2+2x+4

has discriminant

2^2-4(1)(4)=-12<0

, so it never hits zero and the expression is defined for all real

x

. Without this, chunks of the range argument could be vacuous. **Step 1: Create the quadratic in

x

.**

y=\frac{3x^2-6x+12}{x^2+2x+4} \implies y(x^2+2x+4)=3x^2-6x+12

Collecting terms:

x^2(y-3)+x(2y+6)+(4y-12)=0

**Step 2: Handle the

y=3

case separately - the discriminant method assumes a genuine quadratic.** If

y=3

the

x^2

coefficient vanishes and this is not a quadratic at all; it degenerates to

12x+0=0

, giving

x=0

. So

y=3

is attained, at

x=0

, and indeed

\frac{12}{4}=3

. It sits inside the range we are about to derive, so the final answer is unaffected - but the case has to be dismissed explicitly, not ignored. **Step 3: Enforce the discriminant for

y\neq 3

.** Real

x

requires

b^2-4ac\geq 0

. Noticing that

4y-12=4(y-3)

keeps this tidy:

(2y+6)^2-4(y-3)\cdot 4(y-3)\geq 0 \implies (2y+6)^2-16(y-3)^2\geq 0

Since

(2y+6)^2=4(y+3)^2

, divide throughout by 4:

(y+3)^2-4(y-3)^2\geq 0

Step 4: Difference of squares. With

A=y+3

and

B=2(y-3)

:

\left[(y+3)-2(y-3)\right]\left[(y+3)+2(y-3)\right]\geq 0

(-y+9)(3y-3)\geq 0 \implies 3(9-y)(y-1)\geq 0

Dividing by 3 and negating the first bracket flips the inequality:

(y-9)(y-1)\leq 0 \implies 1\leq y\leq 9

Step 5: Confirm both endpoints are actually reached. An inequality gives bounds; it does not by itself prove they are attained, and the stem's wording requires that the function really does climb to

k

. *

y=9

: the quadratic becomes

6x^2+24x+24=0

, i.e.

(x+2)^2=0

, so

x=-2

. Check:

\frac{12+12+12}{4-4+4}=\frac{36}{4}=9

. *

y=1

: it becomes

-2x^2+8x-8=0

, i.e.

(x-2)^2=0

, so

x=2

. Check:

\frac{12-12+12}{4+4+4}=\frac{12}{12}=1

. Both are hit exactly once, at the two points where the discriminant is zero. The maximum value is therefore

k=9

. Final Answer: 9

Q9:ipmat indore 2019QA › Straight LinesEasySA · TITA

The maximum distance between the point

(-5, 0)

and a point on the circle

x^2 + y^2 = 4

is

Enter your answer to attempt

The Setup: A coordinate geometry freebie. The farthest point of a circle from an external point lies on the line through the point and the centre, on the far side - so the maximum distance is the distance to the centre plus the radius. Step 1: Deconstruct the circle. From

x^2+y^2=4

, the centre is the origin

C(0,0)

and the radius is

r=\sqrt{4}=2

. Step 2: Distance from the point to the centre. Both

P(-5,0)

and

C(0,0)

lie on the x-axis, so the distance is just the gap in x-coordinates:

d=|-5-0|=5

Step 3: Check the point is outside. Since

d=5>r=2

, the point

P

lies outside the circle. (For a point *inside*, the maximum would still be

d+r

, but the minimum formula flips - worth knowing which case you are in.) Step 4: Add the radius. Travel from

P

through the centre and out to the opposite edge:

\text{Max distance}=d+r=5+2=7

Concretely, the farthest point is

(2,0)

, and the distance from

(-5,0)

to

(2,0)

is indeed 7 - while the nearest point

(-2,0)

sits only 3 away. Final Answer: 7

Q10:ipmat indore 2019QA › LogarithmsHardSA · TITA

If

x, y, z

are positive real numbers such that

x^{12} = y^{16} = z^{24}

and the three quantities

3 \log_y x, 4 \log_z y, n \log_x z

are in arithmetic progression, then the value of

n

is

Enter your answer to attempt

The Setup: A crossover between logarithms and arithmetic progressions. The chained equality

x^{12}=y^{16}=z^{24}

locks all three variables together, so we can rewrite everything in terms of a single variable and the logarithms collapse into plain numbers. **Step 0: Note

z\neq 1

.** If

z

were 1 then

x=y=z=1

and every logarithm here would have base 1, which is undefined. The variables being positive reals satisfying this chain forces a genuine base, so all three terms make sense. **Step 1: Express everything through

z

.** * From

x^{12}=z^{24}

, take the 12th root:

x=z^{2}

. * From

y^{16}=z^{24}

, take the 16th root:

y=z^{24/16}=z^{3/2}

. **Step 2: Evaluate the first term,

3\log_y x

.** Using

\log_{b^m}(b^n)=\frac{n}{m}

:

\log_y x=\log_{z^{3/2}}\left(z^{2}\right)=\frac{2}{3/2}=2\times\frac{2}{3}=\frac{4}{3} \implies 3\log_y x=3\times\frac{4}{3}=4

**Step 3: Evaluate the second term,

4\log_z y

.**

\log_z y=\log_z\left(z^{3/2}\right)=\frac{3}{2} \implies 4\log_z y=4\times\frac{3}{2}=6

**Step 4: Evaluate the third term,

n\log_x z

.**

\log_x z=\log_{z^{2}}(z)=\frac{1}{2} \implies n\log_x z=\frac{n}{2}

Step 5: Impose the AP condition. The three terms are

4

,

6

and

\frac{n}{2}

. In an arithmetic progression consecutive gaps are equal, so the middle term is the average of its neighbours - equivalently, the common difference

6-4=2

carries forward:

\frac{n}{2}=6+2=8 \implies n=16

Check the progression reads

4,\,6,\,8

- a clean AP with common difference 2. Final Answer: 16

Q11:ipmat indore 2019QA › TrigonometryHardSA · TITA

The number of pairs

(x, y)

satisfying the equation

\sin x + \sin y = \sin(x + y)

and

|x| + |y| = 1

is

Enter your answer to attempt

The Setup: A trigonometric boss phase. Factor the first equation with sum-to-product identities to expose its root conditions, then run those through the absolute-value geometry of

|x|+|y|=1

to see which survive. Step 1: Expand and factor. Sum-to-product on the left, double angle on the right:

2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)

Bring everything to one side and pull out the common factor:

2\sin\left(\frac{x+y}{2}\right)\left[\cos\left(\frac{x-y}{2}\right)-\cos\left(\frac{x+y}{2}\right)\right]=0

Apply

\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)

to the bracket, with

A=\frac{x-y}{2}

and

B=\frac{x+y}{2}

, so that

\frac{A+B}{2}=\frac{x}{2}

and

\frac{B-A}{2}=\frac{y}{2}

:

4\sin\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\sin\left(\frac{x+y}{2}\right)=0

Step 2: Identify the root scenarios. A product is zero only if a factor is, giving three parallel timelines:

\sin\left(\frac{x}{2}\right)=0\implies x=2n\pi, \qquad \sin\left(\frac{y}{2}\right)=0\implies y=2n\pi, \qquad \sin\left(\frac{x+y}{2}\right)=0\implies x+y=2n\pi

Step 3: Apply the domain constraint. On

|x|+|y|=1

we have

|x|\leq 1

,

|y|\leq 1

, and by the triangle inequality

|x+y|\leq|x|+|y|=1

. Since

2\pi\approx 6.28

is far outside that, **only

n=0

survives** in all three families, collapsing them to

x=0

,

y=0

, or

x+y=0

. Step 4: Hunt the coordinates. Intersect each line with the diamond

|x|+|y|=1

: *

x=0 \implies |y|=1

: gives

(0,1)

and

(0,-1)

*

y=0 \implies |x|=1

: gives

(1,0)

and

(-1,0)

*

x=-y \implies 2|x|=1

: gives

\left(\tfrac{1}{2},-\tfrac{1}{2}\right)

and

\left(-\tfrac{1}{2},\tfrac{1}{2}\right)

Step 5: Check for double-counting before adding. The only point that could belong to two of these lines is

(0,0)

, where

x=0

and

y=0

meet - but it fails

|x|+|y|=1

, so it never enters the list. The six points above are pairwise distinct, and the count is genuinely

2+2+2

. Final Answer: 6

Q12:ipmat indore 2019QA › CirclesHardSA · TITA

The circle

x^2 + y^2 - 6x - 10y + k = 0

does not touch or intersect the coordinate axes. If the point

(1, 4)

does not lie outside the circle, and the range of

k

is

(a, b]

, then

a + b

is

Enter your answer to attempt

The Setup: A circle with a floating parameter

k

. We must force it to sit clear of both axes while keeping the point

(1,4)

inside or on its boundary, then read off the two ends of the resulting interval. Step 1: Extract the circle's stats. Complete the square:

(x-3)^2-9+(y-5)^2-25+k=0 \implies (x-3)^2+(y-5)^2=34-k

Centre

C(3,5)

and radius

r=\sqrt{34-k}

. Step 0 (implicit, but state it): the circle must exist. A real circle needs

r^2>0

, i.e.

34-k>0

, so

k<34

. Keep this in the background - it turns out to be implied by the stricter bound below, but a parameter range derived without it can silently include values where there is no circle at all. Step 2: Keep it clear of the axes. The centre sits in the first quadrant, so the distance to the y-axis is its x-coordinate 3, and to the x-axis its y-coordinate 5. Avoiding both strictly (neither touching nor intersecting) requires:

r<3 \implies 34-k<9 \implies k>25

r<5 \implies 34-k<25 \implies k>9

The first is the stricter of the two, and it absorbs the second, so the binding condition is

k>25

- an open bound, because touching is disallowed. Step 3: Trap the point. 'does not lie outside' means on or inside the boundary, so the distance from

(1,4)

to the centre is at most

r

:

d^2=(1-3)^2+(4-5)^2=4+1=5 \implies 5\leq 34-k \implies k\leq 29

This bound is closed, since lying exactly on the circle is permitted. Step 4: Merge and read off. Combining,

25<k\leq 29

, i.e. the interval

(25,29]

- open at the left, closed at the right, exactly matching the

(a,b]

shape the question announces. That the two bounds come out with the right open/closed pairing is a good sign the conditions have been read correctly.

a=25,\quad b=29 \implies a+b=54

Final Answer: 54

Q13:ipmat indore 2019QA › Matrices & DeterminantsEasySA · TITA

If a

3 \times 3

matrix is filled with +1's and -1's such that the sum of each row and column of the matrix is 1, then the absolute value of its determinant is

Enter your answer to attempt

The Setup: A matrix built under severe constraints. Three entries drawn from

\{+1,-1\}

can only sum to 1 as

(+1)+(+1)+(-1)

- no other combination works, since three

+1

s give 3 and anything with two

-1

s gives

-1

. So **every row and every column holds exactly two

+1

s and one

-1

. Step 1: See the structure.** Exactly one

-1

per row and exactly one per column means the positions of the

-1

s form a permutation of the three columns across the three rows. There are therefore exactly

3!=6

matrices satisfying the constraints - the problem is far more rigid than it first looks. Step 2: Compute the prototype. Take the permutation that puts the

-1

s on the main diagonal:

M=\begin{pmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{pmatrix}

Row sums:

-1+1+1=1

. Column sums: likewise 1. A valid build. Expanding along the first row:

|M|=-1(1-1)-1(-1-1)+1(1-(-1))=-1(0)-1(-2)+1(2)=0+2+2=4

Step 3: Show the value is forced, not lucky. Every other valid matrix comes from this one by permuting rows (equivalently, by relabelling which column carries each row's

-1

). A row swap flips the sign of a determinant and nothing else, so across the whole family the determinant is

\pm 4

and its absolute value is invariant at 4. Since

3!=6

is a small family, this is checkable exhaustively rather than by assertion: all six matrices give

|\det|=4

, three with

\det=+4

and three with

\det=-4

, matching the even and odd permutations. That is exactly why the question asks for the *absolute* value - the determinant itself is not determined by the constraints, but its magnitude is. Final Answer: 4

Q14:ipmat indore 2019QA › Set TheoryHardSA · TITA

Let the set

P= \{2,3,4,..., 25\}

. For each

k \in P

, define

Q(k)= \{x \in P

such that

x > k

and

k

divides

x\}

. Then the number of elements in the set

P - U_{k=2}^{25} Q(k)

is

Enter your answer to attempt

The Setup: A Sieve of Eratosthenes in disguise. The whole difficulty is decoding the set notation into plain English; once translated, the counting is immediate. **Step 1: Decode

Q(k)

.** For a given

k

, the set

Q(k)

collects every member of

P

that is strictly greater than

k

and divisible by

k

- that is, the **proper multiples of

k

** lying inside

P

. Step 2: Decode the union. The union

\bigcup_{k=2}^{25}Q(k)

therefore holds every

x\in P

admitting some divisor

k

with

2\leq k<x

. Such an

x

is precisely a composite number. Both directions are worth confirming rather than assuming: * If

x

lands in the union, it has a divisor strictly between 1 and itself, so

x

is composite. * If

x\in P

is composite, it has a prime factor

p\leq\sqrt{x}\leq\sqrt{25}=5

. That

p

satisfies

2\leq p<x

and lies in

P

, so

x

genuinely gets caught. (Checking the largest case:

x=25

is caught by

k=5

.) So the union is exactly the composites in

P

- nothing more, nothing less. Step 3: Decode the subtraction. Removing the union from

P

deletes every composite, leaving the numbers with no divisor other than 1 and themselves: the primes. Note

P

starts at 2, so there is no 1 to worry about - 1 is neither prime nor composite and would have muddied the count had it been included. Step 4: Count the survivors. The primes in

\{2,3,\ldots,25\}

are:

2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23

That is a squad of 9. Final Answer: 9

Q15:ipmat indore 2019QA › SolidsMediumSA · TITA

The number of whole metallic tiles that can be produced by melting and recasting a circular metallic plate, if each of the tiles has a shape of a right-angled isosceles triangle and the circular plate has a radius equal in length to the longest side of the tile (Assume that the tiles and plate are of uniform thickness, and there is no loss of material in the melting and recasting process) is

Enter your answer to attempt

The Setup: Conservation of material dressed up as a 3D melting problem. Because the thickness is uniform and nothing is lost, the volume ratio equals the area ratio, and the whole thing reduces to plane geometry. Step 1: Total source material. With the plate's radius

R

:

\text{Plate area}=\pi R^2

Step 2: Area of one tile. The tile is a right-angled isosceles triangle, whose longest side is the hypotenuse, and the stem sets that hypotenuse equal to

R

. Letting each of the two equal legs be

x

, Pythagoras gives:

x^2+x^2=R^2 \implies 2x^2=R^2 \implies x^2=\frac{R^2}{2}

The legs are perpendicular, so they serve directly as base and height:

\text{Tile area}=\frac{1}{2}\cdot x\cdot x=\frac{x^2}{2}=\frac{1}{2}\cdot\frac{R^2}{2}=\frac{R^2}{4}

Step 3: Divide. Note

R

cancels completely - the answer does not depend on the plate's size, only on the stated relationship between the two:

n=\frac{\pi R^2}{\frac{R^2}{4}}=4\pi

Step 4: Filter for whole tiles. With

\pi\approx 3.14159

:

n\approx 4\times 3.14159=12.566

The stem asks for whole tiles, and you cannot cast a fraction of one, so the count floors to 12 - with material worth about

0.57

of a tile left unused. Rounding to 13 here is the trap: that would require metal that does not exist. Final Answer: 12

Q16:ipmat indore 2019QA › Integral SolutionsHardSA · TITA

If

|x| <100

and

|y| <100

, then the number of integer solutions of

(x, y)

satisfying the equation

4x + 7y = 3

is

Enter your answer to attempt

The Setup: A linear Diophantine equation. Find one integer solution, generate the full family from it, then clamp that family with the two bounds and count what survives. Step 1: Secure a base solution. Reduce

4x+7y=3

modulo 4 to strip out

x

:

7y\equiv 3\pmod 4 \implies 3y\equiv 3\pmod 4 \implies y\equiv 1\pmod 4

Testing

y=1

:

4x+7=3\implies 4x=-4\implies x=-1

. The anchor is

(x_0,y_0)=(-1,1)

. Step 2: Generate the whole family. For

ax+by=c

with

d=\gcd(a,b)

, the general solution is

x=x_0+\frac{b}{d}k

,

y=y_0-\frac{a}{d}k

. Here

\gcd(4,7)=1

, which also confirms solutions exist at all (a solution requires

d\mid c

, and

1\mid 3

):

x=-1+7k, \qquad y=1-4k, \qquad k\in\mathbb{Z}

Step 3: Clamp each variable. For

x

:

-100<-1+7k<100 \implies -99<7k<101 \implies -14.14\ldots<k<14.42\ldots \implies -14\leq k\leq 14

For

y

, note the coefficient of

k

is negative, so dividing by

-4

reverses both inequality signs - the classic place to lose this question:

-100<1-4k<100 \implies -101<-4k<99 \implies -24.75<k<25.25 \implies -24\leq k\leq 25

Step 4: Intersect and count. Both conditions must hold at once, and

[-14,14]

is strictly the tighter of the two, so it is the binding constraint:

14-(-14)+1=29

Each

k

yields exactly one

(x,y)

, and distinct

k

give distinct

x

, so there is no double counting. Spot-check the extremes:

k=14

gives

(97,-55)

and

k=-14

gives

(-99,57)

- both satisfy

4x+7y=3

and sit inside the bounds, while

k=15

would push

x

to 104 and out of range. Final Answer: 29

Q17:ipmat indore 2019QA › Mean, Median & ModeMediumSA · TITA

The average of five distinct integers is 110 and the smallest number among them is 100. The maximum possible value of the largest integer is

Enter your answer to attempt

The Setup: A zero-sum min-maxing problem. To push the largest integer to its ceiling, starve every other member of the set down to its smallest legal value - the total is fixed, so every point saved elsewhere is a point gained at the top. Step 1: Fix the total. An average of 110 across five integers means:

\text{Sum}=5\times 110=550

Step 2: Pin the floor. Write the sorted values as

x_1<x_2<x_3<x_4<x_5

. The stem says the smallest is 100, so

x_1=100

exactly - not merely at least 100. Step 3: Minimise the supporting cast. The integers must be distinct, so the three middle values cannot repeat 100 or each other. The smallest they can legally be is the run of consecutive integers immediately above:

x_2=101,\quad x_3=102,\quad x_4=103

Step 4: Solve for the ceiling.

100+101+102+103=406 \implies x_5=550-406=144

Step 5: Verify the set is legal. The five values are

100,101,102,103,144

- all distinct integers, summing to

550

so the average is exactly 110, with 100 genuinely the smallest and

144>103

so the ordering holds. Nothing was forced past a constraint to get here, which is what makes 144 achievable rather than just an upper estimate. Final Answer: 144

Q18:ipmat indore 2019QA › Progression & SeriesMediumSA · TITA

Assume that all positive integers are written down consecutively from left to right as in 1234567891011...... The 6389th digit in this sequence is

Enter your answer to attempt

The Setup: Walk through the concatenated string block by block. Count how many digits the 1-, 2- and 3-digit numbers consume, subtract that from the target position, then divide cleanly into the 4-digit block. Step 1: Count the early blocks.

Block	Count of numbers	Digits each	Digits used	Running total
1-9	9	1	9	9
10-99	90	2	180	189
100-999	900	3	2,700	2,889

So position 2,889 is the final digit of 999, and everything beyond is 4-digit territory. Step 2: Penetrate the 4-digit block.

6389-2889=3500 \text{ digits into the block}

Every number here uses exactly 4 digits, so:

\frac{3500}{4}=875 \text{ with remainder } 0

A remainder of zero is the case to read carefully: it means the target does not fall partway into a number but lands exactly on the last digit of the 875th four-digit number. (A remainder of 1, 2 or 3 would instead point at that position within the *next* number.) Step 3: Identify the number. The block begins at 1000, which is the 1st, so the 875th is:

1000+875-1=1874

Step 4: Extract the digit. With the remainder zero, we want the final digit of 1874, which is 4. Step 5: Confirm the position. The last digit of 1874 sits at position

2889+875\times 4=2889+3500=6389

- exactly the digit asked for. Final Answer: 4

Q19:ipmat indore 2019QA › MiscellaneousEasySA · TITA

The number of pairs of integers whose sums are equal to their products is

Enter your answer to attempt

The Setup: A famous factoring trick (Simon's Favourite Factoring Trick). Turn the sentence into an equation, add a constant to force a clean factorisation, then use the fact that the factors must be integers. Step 1: Set up the algebra. Let the two integers be

x

and

y

:

x+y=xy \implies xy-x-y=0

Step 2: Apply the cheat code. Add 1 to both sides, which makes the left side factor:

xy-x-y+1=1 \implies (x-1)(y-1)=1

Step 3: Use the integer constraint. Two integers multiply to 1. Over the integers, 1 has only two factorisations -

1\times 1

and

(-1)\times(-1)

- because any factor of absolute value 2 or more forces a product of absolute value 2 or more. (Over the reals there would be infinitely many; the integrality is what makes the problem finite.) *

x-1=1

and

y-1=1 \implies (x,y)=(2,2)

*

x-1=-1

and

y-1=-1 \implies (x,y)=(0,0)

Step 4: Count. The solutions are

(2,2)

and

(0,0)

- exactly 2. Both are symmetric in

x

and

y

, so the count is 2 whether the pairs are read as ordered or unordered, and the question is unambiguous either way. Verify each against the original wording - sum equals product:

2+2=4=2\times 2

✓ and

0+0=0=0\times 0

✓. Final Answer: 2

Q20:ipmat indore 2019QA › Divisibility RulesHardSA · TITA

You have been asked to select a positive integer N which is less than 1000, such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is

Enter your answer to attempt

The Setup: A three-set counting problem that would be messy through full Inclusion-Exclusion. The elegant route is to notice the sets split by parity, which makes the awkward intersections vanish outright. Step 1: Handle the even sets. Let

A

be the multiples of 4 and

B

the multiples of 6, over

1\leq N\leq 999

:

|A|=\left\lfloor\frac{999}{4}\right\rfloor=249, \qquad |B|=\left\lfloor\frac{999}{6}\right\rfloor=166

Their overlap is the multiples of

\text{LCM}(4,6)=12

- the LCM, not the product 24, since 4 and 6 share a factor of 2:

|A\cap B|=\left\lfloor\frac{999}{12}\right\rfloor=83

Step 2: Combine them.

|A\cup B|=249+166-83=332

Step 3: Count the odd set. Let

C

be the odd multiples of 9, i.e.

9k

with

k

odd and

9k\leq 999

, so

k\leq 111

. The admissible

k

are

1,3,5,\ldots,111

:

|C|=\frac{111+1}{2}=56

Step 4: Exploit disjointness. Every multiple of 4 is even and every multiple of 6 is even, so

A\cup B

contains only even numbers. Every element of

C

is

9\times(\text{odd})

, hence odd. A number cannot be both even and odd, so

(A\cup B)\cap C=\varnothing

and the three-set Inclusion-Exclusion collapses to plain addition:

\text{Total}=332+56=388

Step 5: A note on why the 'odd' restriction costs nothing. An *even* multiple of 9 is a multiple of 18, and since

6\mid 18

every such number already sits in

B

. So the answer would be unchanged had the question said *a multiple of 9* instead - the restriction to odd multiples is what makes the sets disjoint and the counting clean, not what makes the total smaller. Final Answer: 388

IPMAT Indore 2019 — SA

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