IPMAT Indore 2019 QA 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

Q21:ipmat indore 2019QA › Simple & Compound InterestEasyMCQ · MCQ

If the compound interest earned on a certain sum for 2 years is twice the amount of simple interest for 2 years, then the rate of interest per annum is _______ percent

A200%
B2%
C4%
D400%

Pick an option to attempt

The Setup: Let's speedrun this finance glitch. Compound interest beats simple interest only because of interest earned *on* the first year's interest - so demanding that CI be exactly twice SI is a very aggressive ask, and the rate it forces is correspondingly extreme. Step 1: Define the base formulas. Let

P

be the principal and

r

the annual rate as a decimal.

SI=P\cdot r\cdot 2=2Pr, \qquad CI=P(1+r)^2-P

Step 2: Equate and expand. The condition is

CI=2\times SI

:

P(1+r)^2-P=2(2Pr)

Divide through by

P

(a principal of zero is meaningless) and expand:

1+2r+r^2-1=4r \implies r^2+2r=4r

Step 3: Solve for the rate.

r^2-2r=0 \implies r(r-2)=0

The root

r=0

would mean no interest at all, making

CI=SI=0

- technically satisfying the equation but describing no loan. Discard it, leaving

r=2

as a decimal, i.e.

200\%

. Step 4: Sanity-check, because 200% looks absurd. Take

P=100

at

r=200\%

: * Simple interest:

100\times 2\times 2=400

* Compound:

100(1+2)^2-100=900-100=800

And

800=2\times 400

exactly. The answer really is 200% - the reason it feels wrong is that doubling CI relative to SI is a far harsher demand than it sounds, and only a triple-per-year growth factor achieves it. Option 2 is the trap: solve correctly, get

r=2

, and then read that 2 as a percentage instead of as the decimal it is. Final Answer: 200%

Q22:ipmat indore 2019QA › FactorisationEasyMCQ · MCQ

The maximum value of the natural number

n

for which

21^n

divides

50!

is

A6
B7
C8
D9

Pick an option to attempt

The Setup: Factorials scale up like a boss-level mob. To count how many times

21

divides

50!

, break 21 into primes and count each prime's supply separately - whichever runs out first is the bottleneck. Step 1: Prime factorise the base.

21=3\times 7

Each copy of 21 consumes one 3 and one 7, so

21^n

needs

n

of each. Step 2: Count the supply of each prime with Legendre's formula. The exponent of a prime

p

in

n!

is

\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor

.

E_7(50!)=\left\lfloor\frac{50}{7}\right\rfloor+\left\lfloor\frac{50}{49}\right\rfloor=7+1=8

E_3(50!)=\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{9}\right\rfloor+\left\lfloor\frac{50}{27}\right\rfloor=16+5+1=22

Step 3: Take the minimum. The larger prime really is the scarcer one, and that is a theorem rather than a guess: for primes

p<q

we have

p^i<q^i

, so

\left\lfloor\frac{n}{p^i}\right\rfloor\geq\left\lfloor\frac{n}{q^i}\right\rfloor

term by term, and summing gives

E_p(n!)\geq E_q(n!)

. Still, computing both counts is worth the ten seconds - it turns a general principle into the concrete numbers the answer needs:

n=\min\left(E_3,\ E_7\right)=\min(22,\ 8)=8

With 22 threes available and only 8 sevens, the sevens are exhausted first - we can assemble 8 copies of 21, and the 14 spare threes are useless without partners. Final Answer: 8

Q23:ipmat indore 2019QA › RemainderMediumMCQ · MCQ

The remainder when

(29^{29})^{29}

is divided by

9

is

A1
B2
C3
D4

Pick an option to attempt

The Setup: Modular arithmetic is overpowered for shrinking absurd numbers. Rather than computing a skyscraper of digits, reduce the base mod 9 and exploit the fact that its powers cycle. Step 1: Read the exponent tower correctly. This is the first trap:

\left(29^{29}\right)^{29}

has brackets, so the exponents multiply:

\left(29^{29}\right)^{29}=29^{29\times 29}=29^{841}

This is emphatically not

29^{\left(29^{29}\right)}

, which would be an astronomically larger number. When the tower is bracketed from the bottom, multiply. Step 2: Simplify the base modulo 9. Since

29=27+2

and

27

is a multiple of 9:

29\equiv 2\pmod 9

So the task reduces to finding

2^{841}\bmod 9

. Step 3: Find the cycle. List powers of 2 modulo 9 until they repeat:

2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1

Hitting 1 at the sixth power means the cycle length is exactly 6, consistent with

\phi(9)=6

. Step 4: Reduce the exponent modulo the cycle length.

841=6\times 140+1 \implies 841\equiv 1\pmod 6

So

2^{841}

sits at the same position in the cycle as

2^1

:

2^{841}\equiv 2^1\equiv 2\pmod 9

Final Answer: 2

Q24:ipmat indore 2019QA › Divisibility RulesEasyMCQ · MCQ

Placing which of the following two digits at the right end of

4530

makes the resultant six digit number divisible by

6, 7

and

9

:

A96
B78
C42
D54

Pick an option to attempt

The Setup: Rather than running three separate divisibility side quests, fuse them into one by taking the LCM - a number divisible by 6, 7 and 9 is exactly a number divisible by their least common multiple. Step 1: Find the master divisor. Factorising:

6=2\times 3

,

7=7

,

9=3^2

. The LCM takes the highest power of each prime:

\text{LCM}(6,7,9)=2\times 3^2\times 7=126

Note it is not

6\times 7\times 9=378

- 6 and 9 share a factor of 3, so the product over-counts. Step 2: Build the number. Appending two digits to 4530 gives

453000+x

, where

x

runs from 00 to 99. Step 3: Find the gap. Divide the base by 126:

453000=126\times 3595+30

The base overshoots a multiple of 126 by 30, so to reach the next one we must add:

126-30=96

That gives

453096=126\times 3596

. And since the following candidate would be

96+126=222

- far beyond two digits - 96 is the only possible answer in range, which is worth noting rather than stopping at the first hit. Step 4: Confirm the distractors fail. All three wrong options clear the divisibility by 6, so that test alone separates nothing - 78 and 42 both fall at the 7, and 54 falls at the 9:

Appended	Number	by 6	by 7	by 9
96	453096	yes	yes	yes
78	453078	yes	no	yes
42	453042	yes	no	yes
54	453054	yes	yes	no

Final Answer: 96

Q25:ipmat indore 2019QA › Set TheoryEasyMCQ · MCQ

In a school

70\%

of the boys like cricket and

50\%

like football. If

x\%

like both Cricket and Football, then

A $20 \leq x \leq 50$
B $x \leq 20$
C $x \geq 50$
D $10 \leq x \leq 70$

Pick an option to attempt

The Setup: Two overlapping sets, and we need the floor and ceiling on the overlap

x

. Both bounds come from hard constraints, not from assumptions about the school. Step 1: The ceiling. An intersection can never exceed either of the sets forming it. Only 50% like football, and everyone who likes both is among them:

x\leq\min(70,\ 50)=50

Step 2: The floor - derived, not assumed. The union of the two groups cannot exceed the whole school, so

n(C\cup F)\leq 100

. By inclusion-exclusion:

n(C\cup F)=70+50-x=120-x

Combining the two:

120-x\leq 100 \implies x\geq 20

This is worth stating as an inequality rather than as *suppose everyone likes at least one sport*. That supposition is only the extreme case; the real reason

x

cannot drop below 20 is that 70 and 50 together already claim 120 percentage points of a 100-point school, so at least 20 points must be double-counted. The overlap is forced by arithmetic, whether or not anyone likes neither sport. Step 3: Combine. Both bounds are attainable -

x=50

when every football fan also likes cricket, and

x=20

when the two groups spread across the whole school with nobody left out:

20\leq x\leq 50

Final Answer:

20 \leq x \leq 50

Q26:ipmat indore 2019QA › Set TheoryMediumMCQ · MCQ

In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A7
B8
C10
D12

Pick an option to attempt

The Setup: A triple-set problem where the union is handed to us - every one of the 65 students likes at least one sport - so inclusion-exclusion has exactly one unknown and can simply be solved. Step 1: The master formula. With

C

,

F

,

H

for cricket, football and hockey:

n(C\cup F\cup H)=n(C)+n(F)+n(H)-n(C\cap F)-n(F\cap H)-n(C\cap H)+n(C\cap F\cap H)

Step 2: Substitute. Since every student likes at least one sport, the union is the full class of 65. Let

y=n(C\cap H)

:

65=40+25+20-10-8-y+5

Step 3: Solve.

65=85-18+5-y=72-y \implies y=72-65=7

Step 4: Verify the Venn diagram is actually consistent. A value that satisfies the equation can still be nonsense if it forces a negative region, so break the class into its seven disjoint parts. Subtracting the triple overlap of 5 from each pair:

Region	Count
All three	5
Cricket & football only	$10-5=5$
Football & hockey only	$8-5=3$
Cricket & hockey only	$7-5=2$
Cricket only	$40-5-5-2=28$
Football only	$25-5-5-3=12$
Hockey only	$20-5-3-2=10$

Every region is non-negative, and they total

5+5+3+2+28+12+10=65

- exactly the class size. The answer is not just algebraically valid but genuinely realisable. Final Answer: 7

Q27:ipmat indore 2019QA › InequalitiesHardMCQ · MCQ

If

x \in (a, b)

satisfies the inequality

\dfrac{x - 3}{x^2 + 3x + 2} \geq 1

, then the largest possible value of

b - a

is

A3
B1
C2
DNo real values of x satisfies the inequality

Pick an option to attempt

The Setup: A rational inequality. We cannot cross-multiply, because the denominator's sign is unknown and multiplying by a negative would flip the inequality. Everything goes to one side instead. Step 1: Reframe. Move the 1 across and combine over a common denominator:

\frac{x-3}{x^2+3x+2}-1\geq 0 \implies \frac{x-3-\left(x^2+3x+2\right)}{x^2+3x+2}\geq 0 \implies \frac{-x^2-2x-5}{x^2+3x+2}\geq 0

Step 2: Kill the numerator's sign ambiguity. Multiply by

-1

and reverse the inequality:

\frac{x^2+2x+5}{x^2+3x+2}\leq 0

Check the numerator's discriminant:

\Delta=2^2-4(1)(5)=4-20=-16<0

With

\Delta<0

and a positive leading coefficient,

x^2+2x+5

is **strictly positive for every real

x

- it never touches zero. So the numerator can never make the fraction zero or negative on its own. Step 3: Force the denominator negative.** A positive over something, being

\leq 0

, requires the denominator to be strictly negative (zero is not allowed - the expression would be undefined):

(x+1)(x+2)<0

An upward parabola sits below zero strictly between its roots:

x\in(-2,-1)

Note the interval is open because the endpoints

-2

and

-1

make the denominator zero, where the original expression is undefined. Separately worth seeing: the numerator

x^2+2x+5

never vanishes, so the fraction is never exactly zero - meaning equality in the original

\geq 1

is never attained, and the condition behaves throughout as a strict

>1

. Step 4: Compute the width. With

a=-2

and

b=-1

:

b-a=-1-(-2)=1

Option 4 - claiming no real

x

works - is the trap for anyone who mishandles the sign flip in Step 2 and concludes the fraction can never be negative. Final Answer: 1

Q28:ipmat indore 2019QA › IdentitiesHardMCQ · MCQ

If

a, b, c

are real numbers and

a^2 + b^2 + c^2 = 1

, then the set of values

ab + bc + ca

can take is:

A[-1,2]
B[- $\frac{1}{2}$ , 2]
C[-1,1]
D[- $\frac{1}{2}$ , 1]

Pick an option to attempt

The Setup: We need the exact range of

ab+bc+ca

on the unit sphere

a^2+b^2+c^2=1

. Both bounds fall out of the fact that a real square is never negative - but a bound is only the answer if it is actually reached, so we must exhibit a witness for each. Step 1: The lower bound. Expand a perfect square, which cannot be negative:

(a+b+c)^2\geq 0 \implies a^2+b^2+c^2+2(ab+bc+ca)\geq 0

Substituting the constraint:

1+2(ab+bc+ca)\geq 0 \implies ab+bc+ca\geq -\frac{1}{2}

Step 2: The upper bound. Use the sum of pairwise squared differences, also non-negative:

(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2\left(a^2+b^2+c^2\right)-2(ab+bc+ca)\geq 0

2(1)-2(ab+bc+ca)\geq 0 \implies ab+bc+ca\leq 1

Step 3: Show both ends are attained. Inequalities alone would allow a smaller range, so produce explicit points on the sphere: * Upper: take

a=b=c=\frac{1}{\sqrt{3}}

. Then

a^2+b^2+c^2=3\cdot\frac{1}{3}=1

✓, and

ab+bc+ca=3\cdot\frac{1}{3}=1

. Equality holds because

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

exactly when all three are equal. * Lower: take

a=\frac{1}{\sqrt{2}}

,

b=-\frac{1}{\sqrt{2}}

,

c=0

. Then

a^2+b^2+c^2=\frac{1}{2}+\frac{1}{2}+0=1

✓, and

ab+bc+ca=-\frac{1}{2}+0+0=-\frac{1}{2}

. Equality holds because

a+b+c=0

. Step 4: Conclude. The expression is continuous on a connected sphere, so it sweeps every value between the two attained extremes - the set is the closed interval, endpoints included:

-\frac{1}{2}\leq ab+bc+ca\leq 1

Final Answer: [-

\frac{1}{2}

, 1]

Q29:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{2} \frac{3x - 1}{2 - x} < 1

holds true for

A $x \in \left( \frac{1}{3}, 1 \right)$
B $x \in \left( \frac{1}{3}, 2 \right)$
C $x \in \left( 0, \frac{1}{3} \right) \cup \left( 1, 2 \right)$
D $x \in \left( -\infty, 1 \right)$

Pick an option to attempt

The Setup: Before touching the inequality we must respect the logarithm's domain. A log of a non-positive number does not exist, so any

x

that breaks the argument is disqualified before the inequality is even considered - and skipping this step is exactly how the wrong option gets picked here. Step 1: Determine the domain. The argument must be strictly positive:

\frac{3x-1}{2-x}>0

The critical points are

x=\frac{1}{3}

(numerator zero) and

x=2

(denominator zero). Testing the three regions: for

x<\frac{1}{3}

the numerator is negative and denominator positive, giving a negative; for

\frac{1}{3}<x<2

both are positive; for

x>2

the numerator is positive and denominator negative. So:

\text{Domain}=\left(\frac{1}{3},\ 2\right)

Step 2: Unpack the inequality. The base is

2>1

, so the logarithm is increasing and the inequality direction is preserved when exponentiating:

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

Move everything to one side rather than cross-multiplying, since

2-x

changes sign:

\frac{3x-1}{2-x}-2<0 \implies \frac{3x-1-2(2-x)}{2-x}<0 \implies \frac{5x-5}{2-x}<0

Step 3: Solve. Factor and tidy the sign:

\frac{5(x-1)}{-(x-2)}<0 \implies \frac{x-1}{x-2}>0

A quotient is positive when both parts share a sign, giving

x<1

or

x>2

. Step 4: Intersect with the domain - this is where the question is won or lost.

\left[(-\infty,1)\cup(2,\infty)\right]\cap\left(\frac{1}{3},2\right)=\left(\frac{1}{3},\ 1\right)

The branch

x>2

solves the algebra perfectly but lies entirely outside the domain, where the logarithm does not exist at all. Discarding it is not optional bookkeeping: carry it through and the answer set is simply wrong, no matter how clean the algebra behind it looked. Final Answer:

x \in \left( \frac{1}{3}, 1 \right)

Q30:ipmat indore 2019QA › InequalitiesMediumMCQ · MCQ

The set of values of

x

which satisfy the inequality

0.7^{(2x^2 - 3x + 4)} < 0.343

is

A $\left( \frac{1}{2}, 1 \right)$
B $\left( \frac{1}{2}, \infty \right)$
C $\left( -\infty, \frac{1}{2} \right)$
D $\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)$

Pick an option to attempt

The Setup: An exponential inequality with a trap built into the base. Because

0.7<1

, the function

0.7^t

is decreasing, so comparing exponents reverses the inequality - an UNO reverse card most people forget to play. Step 1: Match the bases. Since

7^3=343

, scaling the decimal places gives:

0.343=(0.7)^3 \implies 0.7^{\left(2x^2-3x+4\right)}<0.7^3

Step 2: Drop the bases and flip. For a base in

(0,1)

, a smaller value corresponds to a larger exponent:

2x^2-3x+4>3

Step 3: Solve the quadratic inequality.

2x^2-3x+1>0

Factor by splitting the middle term:

2x^2-2x-x+1>0 \implies 2x(x-1)-1(x-1)>0 \implies (2x-1)(x-1)>0

The roots are

x=\frac{1}{2}

and

x=1

. The leading coefficient is positive, so the parabola opens upward and is positive outside its roots and negative between them. Step 4: Write the interval.

x<\frac{1}{2} \quad\text{or}\quad x>1

Option 1, the interval

\left(\frac{1}{2},1\right)

, is precisely what you get by not flipping the sign in Step 2: that error yields

2x^2-3x+1<0

, whose solution is the region *between* the roots rather than outside them. The whole question turns on that single step. Final Answer:

\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)

Q31:ipmat indore 2019QA › CirclesHardMCQ · MCQ

A chord is drawn inside a circle, such that the length of the chord is equal to the radius of the circle. Now, two circles are drawn, one on each side of the chord, each touching the chord at its midpoint and the original circle. Let k be the ratio of the areas of the bigger inscribed circle and the smaller inscribed circle, then k equals

A $2 + \sqrt{3}$
B $1 + \sqrt{2}$
C $7 + 4\sqrt{3}$
D $97 + 56\sqrt{3}$

Pick an option to attempt

The Setup: Pure equilateral-triangle energy. With the chord equal to the radius, joining the centre to the chord's endpoints makes an equilateral triangle, which hands us the centre-to-chord distance. From there the two inscribed circles are pinned down exactly. Step 1: Find the distance to the chord. Take

R=1

to keep the arithmetic light. A chord of length 1 subtends an equilateral triangle with the centre, and the perpendicular distance from centre to chord is that triangle's height:

d=\sqrt{R^2-\left(\frac{R}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}

Step 2: Pin down each inscribed circle - this is the load-bearing step. Put the centre

O

at the origin with the chord horizontal at height

d

, so its midpoint is

M=\left(0,\tfrac{\sqrt{3}}{2}\right)

and the perpendicular through

M

is the

y

-axis. A circle touching the chord **at

M

** must have its centre on that perpendicular, at distance

r

from

M

. * Above the chord (away from

O

): centre

\left(0,d+r\right)

. Internal tangency to the big circle means the centres are

R-r

apart, so

d+r=R-r

, giving

2r=R-d

- its **diameter is

R-d

**. * Below the chord (past

O

): centre

\left(0,d-r\right)

with

r>d

, so the distance to

O

is

r-d

, and internal tangency gives

r-d=R-r

, i.e.

2r=R+d

- its **diameter is

R+d

**. So the two circles have as diameters exactly the two pieces into which

M

cuts the vertical diameter of the original circle, running from

M

up to

(0,1)

and from

M

down to

(0,-1)

:

\text{smaller}=1-\frac{\sqrt{3}}{2}, \qquad \text{bigger}=1+\frac{\sqrt{3}}{2}

Step 3: Take the ratio of areas. Areas scale as the square of any linear measure, so the area ratio is the square of the diameter ratio:

k=\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right)^2=\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)^2

Rationalise the inner fraction by multiplying top and bottom by

2+\sqrt{3}

, noting

(2-\sqrt{3})(2+\sqrt{3})=4-3=1

:

k=\left(\frac{(2+\sqrt{3})^2}{1}\right)^2=(2+\sqrt{3})^4

Step 4: Expand in two stages.

(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}

k=(7+4\sqrt{3})^2=49+56\sqrt{3}+48=97+56\sqrt{3}

Watch the option list:

7+4\sqrt{3}

is sitting there as option 3, and it is precisely the ratio of the diameters, not the areas. Stopping one squaring too early lands exactly on it. Final Answer:

97 + 56\sqrt{3}

Q32:ipmat indore 2019QA › QuadrilateralsMediumMCQ · MCQ

Points

P

,

Q

,

R

, and

S

are taken on sides

AB

,

BC

,

CD

, and

DA

of square

ABCD

respectively, so that

\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{n}

. Then the ratio of the area of

PQRS

to the area of

ABCD

is

A $1 : (1 + n)$
B $1 : n$
C $1 + n^2 : (1 + n)^2$
D $(1 + n) : (1 + n^2)$

Pick an option to attempt

The Setup: A tilted quadrilateral inside a square. Rather than chase the inner side length directly, compute the four corner triangles and subtract them from the whole - the symmetry makes all four identical. Step 1: Define the segments. Let the big square have side

x

. The ratio

\frac{AP}{PB}=\frac{1}{n}

splits each side into

1+n

equal parts, one part then

n

parts:

AP=BQ=CR=DS=\frac{x}{n+1}, \qquad PB=QC=RD=SA=\frac{nx}{n+1}

Step 2: Compute one corner triangle, then argue the rest. Triangle

PBQ

sits in corner

B

, right-angled there, with legs

PB

and

BQ

:

[\,PBQ\,]=\frac{1}{2}\cdot\frac{nx}{n+1}\cdot\frac{x}{n+1}=\frac{nx^2}{2(n+1)^2}

The other three are congruent, not merely similar: at corner

C

the legs are

QC=\frac{nx}{n+1}

and

CR=\frac{x}{n+1}

- the same two lengths, and likewise at

D

and

A

. The whole configuration maps to itself under a

90^\circ

rotation, so:

\text{Total corner area}=4\times\frac{nx^2}{2(n+1)^2}=\frac{2nx^2}{(n+1)^2}

Step 3: Subtract.

[\,PQRS\,]=x^2-\frac{2nx^2}{(n+1)^2}=x^2\left(\frac{(n+1)^2-2n}{(n+1)^2}\right)=x^2\left(\frac{n^2+2n+1-2n}{(n+1)^2}\right)=x^2\cdot\frac{n^2+1}{(n+1)^2}

Step 4: Form the ratio. Dividing by

[\,ABCD\,]=x^2

, the side length cancels entirely - the answer depends only on

n

:

\frac{[\,PQRS\,]}{[\,ABCD\,]}=\frac{1+n^2}{(1+n)^2}

Sanity check at

n=1

, where every point is a midpoint: the ratio becomes

\frac{2}{4}=\frac{1}{2}

, the well-known result that joining the midpoints of a square halves its area. Final Answer:

1 + n^2 : (1 + n)^2

Q33:ipmat indore 2019QA › CirclesHardMCQ · MCQ

On a circular path of radius 6 m a boy starts from a point

A

on the circumference and walks along a chord

AB

of length 3 m. He then walks along another chord

BC

of length 2 m to reach point

C

. The point

B

lies on the minor arc

AC

. The distance between point

C

from point

A

is

A $\frac{\sqrt{15} + \sqrt{35}}{2}$ m
B $8$ m
C $\sqrt{13}$ m
D $6$ m

Pick an option to attempt

The Setup: Two chords laid end to end inside a circle. Each chord subtends a central angle we can extract with the cosine rule, and because

B

lies on the minor arc

AC

, those two angles simply add to give the angle subtending

AC

. Step 1: Find each central angle. Let

O

be the centre,

R=6

. Triangle

OAB

has sides

6,6,3

, so with

\angle AOB=\alpha

:

\cos\alpha=\frac{6^2+6^2-3^2}{2(6)(6)}=\frac{72-9}{72}=\frac{63}{72}=\frac{7}{8}

Triangle

OBC

has sides

6,6,2

, so with

\angle BOC=\beta

:

\cos\beta=\frac{6^2+6^2-2^2}{2(6)(6)}=\frac{72-4}{72}=\frac{68}{72}=\frac{17}{18}

Step 2: Add the angles. Since

B

lies on the minor arc

AC

, the two central angles sit side by side without overlapping, so

\angle AOC=\alpha+\beta

. Both are central angles in

(0,\pi)

, where sine is positive, so we may take the positive root:

\sin\alpha=\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{8}, \qquad \sin\beta=\sqrt{1-\frac{289}{324}}=\frac{\sqrt{35}}{18}

Now the cosine addition formula:

\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{7}{8}\cdot\frac{17}{18}-\frac{\sqrt{15}}{8}\cdot\frac{\sqrt{35}}{18}=\frac{119-\sqrt{525}}{144}=\frac{119-5\sqrt{21}}{144}

using

\sqrt{525}=\sqrt{25\times 21}=5\sqrt{21}

. **Step 3: Cosine rule once more, on triangle

OAC

.**

AC^2=6^2+6^2-2(6)(6)\cos(\alpha+\beta)=72-72\cdot\frac{119-5\sqrt{21}}{144}=72-\frac{119-5\sqrt{21}}{2}

AC^2=\frac{144-119+5\sqrt{21}}{2}=\frac{25+5\sqrt{21}}{2}=\frac{50+10\sqrt{21}}{4}

Step 4: Recognise the perfect square. The surd untangles because

50+10\sqrt{21}

is a square:

\left(\sqrt{15}+\sqrt{35}\right)^2=15+35+2\sqrt{15\cdot 35}=50+2\sqrt{525}=50+10\sqrt{21}

AC=\frac{\sqrt{15}+\sqrt{35}}{2}\approx 4.89\text{ m}

A quick plausibility check:

AC

must be less than

AB+BC=5

(triangle inequality) and more than each chord, and

4.89

sits exactly there. The distractors fail this outright - 8 m and 6 m both exceed the straight-line distance

AB+BC=5

, which is impossible. Final Answer:

\frac{\sqrt{15} + \sqrt{35}}{2}

m

Q34:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

The area enclosed by the curve

2|x| + 3|y| = 6

is

A12 square units
B3 square units
C4 square units
D24 square units

Pick an option to attempt

The Setup: Absolute values mirror the graph across both axes, so this curve is a quadrilateral centred at the origin. Find the intercepts, confirm the shape, and the area follows from one formula. Step 1: Find the intercepts. Setting

y=0

:

2|x|=6 \implies |x|=3 \implies x=\pm 3

Setting

x=0

:

3|y|=6 \implies |y|=2 \implies y=\pm 2

Step 2: Identify the shape. In each quadrant the equation is linear - for instance

2x+3y=6

when both are positive - so the curve is four straight segments joining

(3,0)

,

(0,2)

,

(-3,0)

and

(0,-2)

. Each side runs between an

x

-intercept and a

y

-intercept, so every side has length

\sqrt{3^2+2^2}=\sqrt{13}

: all four equal, making it a genuine rhombus with perpendicular diagonals along the axes. Step 3: Measure the diagonals.

d_1=3-(-3)=6, \qquad d_2=2-(-2)=4

Step 4: Compute the area. For a rhombus (indeed any quadrilateral with perpendicular diagonals), the area is half the product of the diagonals:

\text{Area}=\frac{1}{2}d_1d_2=\frac{1}{2}(6)(4)=12

Equivalently, the first-quadrant piece is a right triangle of legs 3 and 2, area

3

, and symmetry gives four of them:

4\times 3=12

. Two routes, same number. Final Answer: 12 square units

Q35:ipmat indore 2019QA › Straight LinesHardMCQ · MCQ

Two points on a ground are 1 m apart. If a cow moves in the field in such a way that its distance from the two points is always in ratio

3: 2

then

Athe cow moves in a straight line
Bthe cow moves in a circle
Cthe cow moves in a parabola
Dthe cow moves in a hyperbola

Pick an option to attempt

The Setup: A locus question. When a point moves so that the ratio of its distances to two fixed points stays constant, the path is completely determined by that constant - this is the Circle of Apollonius. Step 1: State the condition. Let the fixed points be

A

and

B

with

AB=1

m, and let

P

be the cow. The condition is:

\frac{PA}{PB}=\frac{3}{2}

Step 2: Apply the Apollonius result. For

\frac{PA}{PB}=k

with

k>0

: * If

k=1

the locus is the perpendicular bisector of

AB

- a straight line, since equal distances means equidistant. * If

k\neq 1

the locus is a circle. Here

k=\frac{3}{2}\neq 1

, so the cow traces a circle. Step 3: See it concretely rather than just quoting the theorem. Put

A

at the origin and

B

at

(1,0)

, and let

P=(x,y)

. Then

PA^2=\tfrac{9}{4}PB^2

gives:

4\left(x^2+y^2\right)=9\left((x-1)^2+y^2\right) \implies 5x^2+5y^2-18x+9=0

x^2+y^2-\frac{18}{5}x+\frac{9}{5}=0 \implies \left(x-\frac{9}{5}\right)^2+y^2=\left(\frac{6}{5}\right)^2

That is unmistakably a circle - centre

\left(1.8,\,0\right)

, radius

1.2

m. The two points where it crosses the line

AB

are the internal and external division points of

AB

in the ratio

3:2

, namely

x=0.6

and

x=3

, which is exactly the diameter

[0.6,\,3]

of width

2.4

. Note the circle is not centred on

AB

's midpoint and does not pass through

A

or

B

- a useful check that this really is Apollonius and not the perpendicular bisector. Final Answer: the cow moves in a circle

Q36:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

Given that

\cos x + \cos y = 1

, the range of

\sin x - \sin y

is

A $[-1, 1]$
B $[-2, 2]$
C $[0, \sqrt{3}]$
D $[-\sqrt{3}, \sqrt{3}]$

Pick an option to attempt

The Setup: One equation constrains the pair

(x,y)

, and we want the range of a second expression. Squaring both and adding is the standard move, because the Pythagorean identity then collapses most of the terms. Step 1: Name the target. Let

S=\sin x-\sin y

, given

\cos x+\cos y=1

. Step 2: Square both and add.

(\cos x+\cos y)^2=1 \implies \cos^2x+2\cos x\cos y+\cos^2y=1

(\sin x-\sin y)^2=S^2 \implies \sin^2x-2\sin x\sin y+\sin^2y=S^2

Adding, and grouping

\sin^2+\cos^2=1

for each of

x

and

y

:

1+1+2\left(\cos x\cos y-\sin x\sin y\right)=1+S^2

The bracket is exactly the expansion of

\cos(x+y)

, so:

2+2\cos(x+y)=1+S^2 \implies S^2=1+2\cos(x+y)

**Step 3: Bound

\cos(x+y)

using the constraint.** Sum-to-product on the given equation, writing

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

and

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

:

2\cos u\cos v=1 \implies \cos u\cos v=\frac{1}{2}

Since

|\cos v|\leq 1

, we need

|\cos u|\geq\frac{1}{2}

, hence

\cos^2u\geq\frac{1}{4}

. The double-angle formula then gives:

\cos(x+y)=2\cos^2u-1 \geq 2\cdot\frac{1}{4}-1=-\frac{1}{2}

and of course

\cos(x+y)\leq 1

always. Substituting into

S^2=1+2\cos(x+y)

:

S^2\in[\,1-1,\ 1+2\,]=[0,\ 3]

Step 4: Confirm both extremes are actually reachable. Bounds are worthless unless attained, so produce explicit angles satisfying

\cos x+\cos y=1

: *

x=\frac{\pi}{3},\ y=-\frac{\pi}{3}

: then

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

✓ and

S=\frac{\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}

, the maximum. *

x=y=\frac{\pi}{3}

: then

\cos x+\cos y=1

✓ and

S=0

, the minimum of

|S|

. Swapping

x

and

y

negates

S

, so

-\sqrt{3}

is reached too, and since

S

varies continuously it sweeps everything between:

S\in\left[-\sqrt{3},\ \sqrt{3}\right]

Final Answer:

[-\sqrt{3}, \sqrt{3}]

Q37:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

If

sin \theta + cos \theta = m

, then

sin^6 \theta + cos^6 \theta

equals

A $\frac{3(m^2+1)}{4}$
B $\frac{3(m^2-1)}{4}$
C $1-\frac{3(m^2-1)}{4}$
D $1-\frac{3(m^2-1)^2}{4}$

Pick an option to attempt

The Setup: An algebraic identity wearing a trigonometry costume. We must climb from first powers to sixth powers, and the bridge is the single quantity

\sin\theta\cos\theta

, which squaring the given equation hands us immediately. Step 1: Extract the cross-term. Square both sides:

(\sin\theta+\cos\theta)^2=m^2 \implies \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=m^2

Since

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

:

1+2\sin\theta\cos\theta=m^2 \implies \sin\theta\cos\theta=\frac{m^2-1}{2}

Step 2: Reduce the sixth powers to that one quantity. Use the sum of cubes

a^3+b^3=(a+b)\left(a^2-ab+b^2\right)

with

a=\sin^2\theta

and

b=\cos^2\theta

:

\sin^6\theta+\cos^6\theta=\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta\right)

The first bracket is 1. For the second, use the identity

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

on

\sin^4\theta+\cos^4\theta

, which turns

a^2-ab+b^2

into

(a+b)^2-3ab

:

\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta=\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\cos^2\theta=1-3\left(\sin\theta\cos\theta\right)^2

Step 3: Substitute. Note the cross-term gets squared, so the bracket from Step 1 is squared whole:

\sin^6\theta+\cos^6\theta=1-3\left(\frac{m^2-1}{2}\right)^2=1-\frac{3\left(m^2-1\right)^2}{4}

Option 3 drops that squaring and reads

1-\frac{3(m^2-1)}{4}

- the difference is one exponent, so check it before committing. Spot-check at

\theta=0

: then

m=\sin 0+\cos 0=1

, and the formula gives

1-\frac{3(1-1)^2}{4}=1

, matching

\sin^6 0+\cos^6 0=0+1=1

. Final Answer:

1-\frac{3(m^2-1)^2}{4}

Q38:ipmat indore 2019QA › Matrices & DeterminantsEasyMCQ · MCQ

If inverse of the matrix

\left[\begin{array}{cc}2 & -0.5 \\ -1 & x\end{array}\right]

is

\left[\begin{array}{ll}1 & 1 \\ 2 & 4 \end{array}\right]

, then the value of

x

is

A $0.5$
B $1$
C $2$
D $3$

Pick an option to attempt

The Setup: A matrix times its inverse is the identity, in either order. That single fact turns this into two-by-two multiplication with no determinant work at all. Step 1: Set up the product. With

A=\begin{bmatrix}2&-0.5\\-1&x\end{bmatrix}

and

A^{-1}=\begin{bmatrix}1&1\\2&4\end{bmatrix}

:

A^{-1}A=\begin{bmatrix}1&1\\2&4\end{bmatrix}\begin{bmatrix}2&-0.5\\-1&x\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

**Step 2: Pick the entry that isolates

x

.** The

(1,2)

entry pairs row 1 of

A^{-1}

with column 2 of

A

, and must equal the identity's 0:

(1)(-0.5)+(1)(x)=x-0.5=0 \implies x=0.5

Step 3: Verify all four entries, not just one. A single equation can be satisfied by a value that breaks the others, so substitute

x=0.5

and multiply out completely:

\begin{bmatrix}1&1\\2&4\end{bmatrix}\begin{bmatrix}2&-0.5\\-1&0.5\end{bmatrix}=\begin{bmatrix}2-1&-0.5+0.5\\4-4&-1+2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

All four land correctly. Checking the other order too, since an inverse must work from both sides:

\begin{bmatrix}2&-0.5\\-1&0.5\end{bmatrix}\begin{bmatrix}1&1\\2&4\end{bmatrix}=\begin{bmatrix}2-1&2-2\\-1+1&-1+2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

Step 4: A determinant cross-check. Since

\det(A)\det\left(A^{-1}\right)

must be 1:

\det A=2(0.5)-(-0.5)(-1)=1-0.5=0.5

and

\det A^{-1}=(1)(4)-(1)(2)=2

, and

0.5\times 2=1

✓. Every other option fails -

x=1,2,3

each leave the

(1,2)

entry non-zero. Final Answer:

0.5

Q39:ipmat indore 2019QA › FunctionsHardMCQ · MCQ

The function

f(x) = \dfrac{x^3 - 5x^2 - 8x}{3}

is

Apositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2}, +\infty)$
Bnegative and monotonically decreasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Cnegative and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and positive and monotonically increasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Dpositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and negative and monotonically decreasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$

Pick an option to attempt

The Setup: A two-part question: where the function is positive or negative, and where it is increasing or decreasing. The options pair those up, so both must be settled and then matched on the same intervals. Step 1: Find the roots - this fixes the sign. Factor out

x

:

f(x)=\frac{x\left(x^2-5x-8\right)}{3}=0

The roots are

x=0

together with the solutions of

x^2-5x-8=0

:

x=\frac{5\pm\sqrt{25+32}}{2}=\frac{5\pm\sqrt{57}}{2}

Write

\alpha=\frac{5-\sqrt{57}}{2}\approx -1.27

and

\beta=\frac{5+\sqrt{57}}{2}\approx 6.27

, so the three roots in order are

\alpha<0<\beta

. Being a cubic with positive leading coefficient,

f

alternates sign across them:

Interval	Sign of $f$
$x<\alpha$	negative
$\alpha<x<0$	positive
$0<x<\beta$	negative
$x>\beta$	positive

Step 2: Differentiate - this fixes the monotonicity.

f'(x)=\frac{3x^2-10x-8}{3}=\frac{(3x+2)(x-4)}{3}

The critical points are

x=-\frac{2}{3}

and

x=4

, and since the parabola opens upward,

f'>0

outside them:

f \text{ is increasing on } \left(-\infty,-\tfrac{2}{3}\right) \text{ and } (4,\infty), \text{ decreasing on } \left(-\tfrac{2}{3},4\right)

Step 3: Match the two pictures on the same intervals. The containments are what make the answer work, so check them numerically: * On

(-\infty,\alpha)

: here

\alpha\approx -1.27<-\frac{2}{3}\approx -0.67

, so this whole interval lies inside the increasing region

\left(-\infty,-\frac{2}{3}\right)

. From the table,

f

is negative there. So: negative and monotonically increasing. * On

(\beta,\infty)

: here

\beta\approx 6.27>4

, so this interval lies inside the increasing region

(4,\infty)

. From the table,

f

is positive there. So: positive and monotonically increasing. Both intervals are increasing, but they differ in sign - which is precisely what separates the correct option from the three that pair the signs the other way round or call one of them decreasing. Final Answer: negative and monotonically increasing for x

\in (-\infty, \frac{5-\sqrt{57}}{2})

and positive and monotonically increasing for x

\in (\frac{5+\sqrt{57}}{2},+\infty)

Q40:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

For

a > b > c > 0

, the minimum value of the function

f(x) = |x - a| + |x - b| + |x - c|

is

A $2a - b - c$
B $a + b - 2c$
C $a + b + c$
D $a - c$

Pick an option to attempt

The Setup: The sum of distances from a moving point

x

to three fixed points

c<b<a

on a number line. The median minimises it - but rather than quote that rule, we can prove it in two lines and get the exact minimum for free. Step 1: Bound the outer pair. By the triangle inequality, for any real

x

:

|x-a|+|x-c|\geq |(x-c)-(x-a)|=a-c

with equality precisely when

x

lies between

c

and

a

, i.e.

x\in[c,a]

. Step 2: Bound the middle term. Trivially

|x-b|\geq 0

, with equality only at

x=b

. Step 3: Add the bounds. For every real

x

:

f(x)=|x-a|+|x-b|+|x-c|\geq (a-c)+0=a-c

Step 4: Show the floor is reached. The two equality conditions must hold simultaneously, and they can:

x=b

satisfies the second exactly, and since

c<b<a

it also lies inside

[c,a]

, satisfying the first. Substituting

x=b

, and using

a>b

so

|b-a|=a-b

, and

b>c

so

|b-c|=b-c

:

f(b)=(a-b)+0+(b-c)=a-c

The

b

terms cancel, which is why the answer depends only on the outermost two points - the middle value

b

affects where the minimum occurs but not what it is. The bound is attained, so it is the true minimum rather than merely a lower estimate. Final Answer:

a - c

Q41:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

Let

\alpha, \beta

be the roots of

x^2 - x + p = 0

and

\gamma, \delta

be the roots of

x^2 - 4x + q = 0

where p and q are integers. If

\alpha, \beta, \gamma, \delta

are in geometric progression then

p + q

is

A-34
B30
C26
D-38

Pick an option to attempt

The Setup: Two quadratics crossed with a geometric progression. Assign standard GP terms to the four roots, pull sum and product relations out with Vieta, and let the integrality of

p

and

q

pick the branch. Step 1: Assign the GP terms. The stem lists the roots in the order

\alpha,\beta,\gamma,\delta

, so writing the progression as

a,ar,ar^2,ar^3

:

\alpha=a,\quad \beta=ar \quad\text{(roots of the first quadratic)}, \qquad \gamma=ar^2,\quad \delta=ar^3 \quad\text{(roots of the second)}

Step 2: Apply Vieta to the sums. For

x^2-x+p=0

the roots sum to 1; for

x^2-4x+q=0

they sum to 4:

a+ar=a(1+r)=1, \qquad ar^2+ar^3=ar^2(1+r)=4

**Step 3: Divide to isolate

r

.** The first equation says

a(1+r)=1

, which is non-zero, so dividing by it is safe:

\frac{ar^2(1+r)}{a(1+r)}=\frac{4}{1} \implies r^2=4 \implies r=2 \text{ or } r=-2

Step 4: Let integrality choose the branch. Both roots of

r^2=4

satisfy the *sum* equations, so the deciding constraint is that

p

and

q

are integers: *

r=2

: then

a(1+2)=1

gives

a=\frac{1}{3}

, and the roots are

\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}

. This yields

p=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}

and

q=\frac{4}{3}\cdot\frac{8}{3}=\frac{32}{9}

- neither an integer, so this branch is rejected. *

r=-2

: then

a(1-2)=1

gives

-a=1

, so

a=-1

, and the roots are

-1,\ 2,\ -4,\ 8

. **Step 5: Compute

p

and

q

from the products.**

p=\alpha\beta=(-1)(2)=-2, \qquad q=\gamma\delta=(-4)(8)=-32

p+q=-2+(-32)=-34

Step 6: Verify the whole construction. The four roots

-1,2,-4,8

do form a GP with common ratio

-2

throughout. And they genuinely solve their quadratics:

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

has roots

-1

and

2

✓, while

x^2-4x-32=(x+4)(x-8)

has roots

-4

and

8

✓. A negative ratio is what makes the signs alternate, which is exactly why the integer branch is the one most people discard by reflex. Final Answer: -34

Q42:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

If

(1 + x - 2x^2)^6 = A_0 + \sum_{r=1}^{12} A_r x^r

, then the value of

A_2 + A_4 + A_6 + \cdots + A_{12}

is

A31
B32
C30
D29

Pick an option to attempt

The Setup: A coefficient-extraction trick. Expanding a sixth power of a trinomial by hand is a losing game; evaluating the polynomial at

x=\pm 1

collects the even-indexed coefficients in one move. **Step 1: Evaluate at

x=1

and

x=-1

.** Write

P(x)=(1+x-2x^2)^6=A_0+A_1x+A_2x^2+\cdots+A_{12}x^{12}

.

P(1)=\left(1+1-2\right)^6=0^6=0

P(-1)=\left(1-1-2\right)^6=(-2)^6=64

Note

P(1)=0

because

x=1

is a root of

1+x-2x^2

- the trinomial factors as

(1-x)(1+2x)

, so raising it to any power keeps that root. Step 2: Add them to kill the odd terms. In

P(-1)

every odd power picks up a minus sign while every even power is unchanged, so adding cancels the odd coefficients and doubles the even ones:

P(1)+P(-1)=2\left(A_0+A_2+A_4+\cdots+A_{12}\right)

0+64=2\left(A_0+A_2+\cdots+A_{12}\right) \implies A_0+A_2+\cdots+A_{12}=32

**Step 3: Strip out

A_0

.** The question starts the sum at

A_2

, so the constant term must be removed. Get it by evaluating at

x=0

, where every term with an

x

vanishes:

A_0=P(0)=(1+0-0)^6=1

A_2+A_4+\cdots+A_{12}=32-1=31

That last subtraction is the whole difficulty of the question - option 2 is 32, sitting there for anyone who stops one step early. Final Answer: 31

Q43:ipmat indore 2019QA › Progression & SeriesEasyMCQ · MCQ

The number of terms common to both the arithmetic progressions

2, 5, 8, 11, ..., 179

and

3, 5, 7, 9, ..., 101

is

A17
B16
C19
D15

Pick an option to attempt

The Setup: Where two arithmetic progressions overlap, the shared terms form a third AP whose common difference is the LCM of the two originals. Find its first term, its step, and where it has to stop. Step 1: Identify the two APs.

\text{AP 1}: 2,5,8,11,\ldots,179 \quad (d_1=3), \qquad \text{AP 2}: 3,5,7,9,\ldots,101 \quad (d_2=2)

Step 2: Build the common AP. Scanning the openings, AP 1 gives

2,5,8,\ldots

and AP 2 gives

3,5,7,\ldots

- the first shared value is

5

. A number common to both must step by a multiple of 3 *and* of 2, so the common difference is:

\text{LCM}(3,2)=6

giving the sequence

5,11,17,23,\ldots

Step 3: Find where it must stop. A common term has to belong to both lists, so it cannot exceed either final term:

\text{upper limit}=\min(179,\ 101)=101

Step 4: Count the terms.

5+(n-1)6\leq 101 \implies (n-1)6\leq 96 \implies n-1\leq 16 \implies n\leq 17

Step 5: Check the endpoint is genuinely common. The count is only right if the 17th term,

5+16\times 6=101

, really appears in both lists - and it does:

101=2+33\times 3

sits in AP 1, and

101=3+49\times 2

is the last term of AP 2. So the bound is tight rather than an overshoot, and the count is exactly 17. Final Answer: 17

Q44:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

From a pack of 52 cards, we draw one by one, without replacement. If

f(n)

is the probability that an Ace will appear at the

n^{\text{th}}

turn, then

A $f(2) = \frac{1}{13} > f(3)$
B $\frac{1}{13} > f(2) > f(3)$
C $f(3) > f(2) = \frac{1}{13}$
D $f(2) = f(3) = \frac{1}{13}$

Pick an option to attempt

The Setup: This is built to bait a long conditional-probability chain - first draw, second draw given the first, and so on. Step back instead: with an unconditional question about a shuffled deck, every position is alike. Step 1: The symmetry argument. Deal all 52 cards face down in a row. Before anything is revealed, each of the 52 cards is equally likely to be sitting in any given position, so position

n

is no different from position 1. Formally, the number of orderings with an Ace at position

n

is the same for every

n

, so:

f(n)=\frac{\text{number of Aces}}{\text{number of cards}}=\frac{4}{52}=\frac{1}{13} \quad\text{for every } n=1,2,\ldots,52

Step 2: Why the chain reasoning does not contradict this. It is tempting to think the second draw *must* differ from the first, because if the first card was an Ace only 3 Aces remain. But we are not told what the first card was. Averaging over both possibilities restores the balance - take

n=2

:

f(2)=\underbrace{\frac{4}{52}\cdot\frac{3}{51}}_{\text{first was an Ace}}+\underbrace{\frac{48}{52}\cdot\frac{4}{51}}_{\text{first was not}}=\frac{12+192}{2652}=\frac{204}{2652}=\frac{1}{13}

The long route lands on precisely the same value the symmetry argument gives in one line. Conditioning on unknown information changes nothing. Step 3: Conclude. Since the probability is constant across all turns:

f(2)=f(3)=\frac{1}{13}

Options 1, 2 and 3 all assert that

f(2)

and

f(3)

differ, which is exactly the trap of imagining the deck depletes in a way you can observe. Final Answer:

f(2) = f(3) = \frac{1}{13}

Q45:ipmat indore 2019QA › ProbabilityMediumMCQ · MCQ

A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{10}$

Pick an option to attempt

The Setup: Conditional probability with the answer already partly spoiled - we know the sum is 15. So the sample space shrinks to just those outcomes, and we count how many of them open with a 4. Step 1: Enumerate the outcomes summing to 15. Three dice, maximum 18, so 15 is near the top and the possibilities are few. Group by which values appear, then count orderings:

Multiset	Orderings	Count
$\{3,6,6\}$	one repeat, so $\frac{3!}{2!}$	3
$\{4,5,6\}$	all distinct, so $3!$	6
$\{5,5,5\}$	all identical	1

\text{Total}=3+6+1=10

Step 2: Count those starting with 4. A 4 can only appear in the

\{4,5,6\}

group - neither

\{3,6,6\}

nor

\{5,5,5\}

contains a 4 at all. Of its six orderings, those beginning with 4 are

(4,5,6)

and

(4,6,5)

:

\text{Favourable}=2

Step 3: Divide. Every one of the 10 outcomes is equally likely (each is a specific ordered triple with probability

\frac{1}{216}

), so the conditional probability is a plain ratio of counts:

P\left(\text{first is }4\ \middle|\ \text{sum is }15\right)=\frac{2}{10}=\frac{1}{5}

Final Answer:

\frac{1}{5}

Q46:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

A4
B4.667
C4.333
D3.667

Pick an option to attempt

The Setup: A sampling-bias trap. The *families* are equally numerous, but the poll picks a person, and larger families put more people into the pool. Anyone who averages the three family sizes directly walks into it. Step 1: Build the population. Let each size have

N

families:

\text{people in 2-member families}=2N, \qquad 4N, \qquad 6N \implies \text{total}=12N

Note the families split

1:1:1

but the people split

2:4:6

, i.e.

1:2:3

. That mismatch is the entire question. Step 2: Probability a randomly chosen person comes from each group.

P(\text{size }2)=\frac{2N}{12N}=\frac{1}{6}, \qquad P(\text{size }4)=\frac{4N}{12N}=\frac{1}{3}, \qquad P(\text{size }6)=\frac{6N}{12N}=\frac{1}{2}

A person is three times as likely to be drawn from a 6-member family as from a 2-member one. Step 3: Take the expectation. Each respondent reports their own family's size:

E[X]=2\cdot\frac{1}{6}+4\cdot\frac{1}{3}+6\cdot\frac{1}{2}=\frac{1}{3}+\frac{4}{3}+3=\frac{5}{3}+3=\frac{14}{3}\approx 4.667

Step 4: Name the trap explicitly. Averaging the family sizes as though each family were one data point gives

\frac{2+4+6}{3}=4

- which is option 1. That is the average family size *per family*, a genuinely different quantity from the average size *reported by a random person*. The poll samples people, so the answer is the larger

4.667

. The sample of 1000 is a red herring: it only means the estimate is close to the true expectation, not that the size matters. Final Answer: 4.667

Q47:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The value of

(\log_{3} 30)^{-1} + (\log_{4} 900)^{-1} + (\log_{5} 30)^{-1}

is

A0.5
B30
C2
D1

Pick an option to attempt

The Setup: Inverted logarithms. The reciprocal rule

\frac{1}{\log_a b}=\log_b a

flips each term so they can be forced onto a shared base and merged. Step 1: Invert all three.

\left(\log_3 30\right)^{-1}=\log_{30}3, \qquad \left(\log_4 900\right)^{-1}=\log_{900}4, \qquad \left(\log_5 30\right)^{-1}=\log_{30}5

Step 2: Bring the middle term to base 30. The odd one out has base 900, but

900=30^2

. Using

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

:

\log_{900}4=\log_{30^2}4=\frac{1}{2}\log_{30}4=\log_{30}\left(4^{1/2}\right)=\log_{30}2

Both the base and the argument cooperate here - halving the coefficient turns

4

into

\sqrt{4}=2

, exactly the factor needed later. Step 3: Merge and evaluate. With a common base, a sum of logs is the log of the product:

\log_{30}3+\log_{30}2+\log_{30}5=\log_{30}(3\times 2\times 5)=\log_{30}30=1

The three arguments

3

,

2

and

5

are precisely the prime factors of 30, which is why the expression collapses so cleanly - the question is built backwards from

30=2\times 3\times 5

. Final Answer: 1

Q48:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{a}{f(x)} < \log_{a}{g(x)}

implies that

A $f(x) > g(x) > 0$ for $0 < a < 1$ and $g(x) > f(x) > 0$ for $a > 1$
B $g(x) > f(x) > 0$ for $0 < a < 1$ and $f(x) > g(x) > 0$ for $a > 1$
C $f(x) > g(x) > 0$ for $a > 0$
D $g(x) > f(x) > 0$ for $a > 0$

Pick an option to attempt

The Setup: A logarithmic inequality is governed entirely by its base. Above 1 the function increases and the inequality is preserved; between 0 and 1 it decreases and the inequality reverses. On top of that, both arguments must be strictly positive for the logarithms to exist at all. Step 1: The domain condition, which applies in both cases. A logarithm is only defined on positive arguments, so regardless of the base:

f(x)>0 \quad\text{and}\quad g(x)>0

This is why every chain in the answer ends in

>0

rather than stopping at a comparison between

f

and

g

. **Step 2: Case

a>1

.** Here

\log_a

is strictly increasing, so the order of the arguments matches the order of their logarithms:

\log_a f(x)<\log_a g(x) \implies f(x)<g(x)

Combining with the domain condition, and noting

g

exceeds

f

which is already positive:

g(x)>f(x)>0

**Step 3: Case

0<a<1

.** Here

\log_a

is strictly decreasing - a larger argument gives a smaller logarithm - so the inequality reverses:

\log_a f(x)<\log_a g(x) \implies f(x)>g(x)

Combining with positivity:

f(x)>g(x)>0

Step 4: Note that no single answer covers both. Options 3 and 4 each claim one chain holds 'for

a > 0

', i.e. for every admissible base at once. That cannot be right: the two cases give opposite orderings of

f

and

g

, so any statement quantified over all bases is false for half of them. Only a case-split answer survives. Final Answer:

f(x) > g(x) > 0

for

0 < a < 1

and

g(x) > f(x) > 0

for

a > 1

Q49:ipmat indore 2019QA › SolidsHardMCQ · MCQ

Three cubes with integer edge lengths are given. It is known that the sum of their surface areas is

564 \ \text{cm}^2

. Then the possible values of the sum of their volumes are

A $764 \ \text{cm}^3$ and $586 \ \text{cm}^3$
B $586 \ \text{cm}^3$ and $564 \ \text{cm}^3$
C $764 \ \text{cm}^3$ and $564 \ \text{cm}^3$
D $586 \ \text{cm}^3$ and $786 \ \text{cm}^3$

Pick an option to attempt

The Setup: A Diophantine equation dressed as geometry. Convert the surface-area condition into a sum of three squares, find every integer solution, then convert back to volumes. Step 1: Reduce the surface-area equation. A cube of edge

a

has surface area

6a^2

, so with edges

a,b,c

:

6a^2+6b^2+6c^2=564 \implies a^2+b^2+c^2=94

Step 2: Bound the search before starting it. Order the edges

a\leq b\leq c

, with each at least 1 (an edge of 0 is not a cube). Then

c

is the largest, so

3c^2\geq 94

and

c^2\leq 94

:

\frac{94}{3}\leq c^2\leq 94 \implies 31.3\leq c^2\leq 94 \implies 6\leq c\leq 9

That leaves exactly four cases to test, which is what makes the search genuinely exhaustive rather than a lucky hunt. Step 3: Test all four.

$c$	$c^2$	$a^2+b^2$ needed	Solutions with $a\leq b\leq c$
9	81	13	$4+9$ -> edges $\{2,3,9\}$
8	64	30	none ( $30=1+29,4+26,9+21,16+14,25+5$ - no square pairs)
7	49	45	$9+36$ -> edges $\{3,6,7\}$
6	36	58	none (needs both $\leq 36$ ; $58-36=22$ and $58-25=33$ are not squares)

So there are exactly two admissible triples, and no more. Step 4: Convert to volumes.

\{2,3,9\}: \quad 2^3+3^3+9^3=8+27+729=764

\{3,6,7\}: \quad 3^3+6^3+7^3=27+216+343=586

Both triples do give surface area

6(94)=564

✓, so both are genuine and the answer must list both values. Final Answer:

764 \ \text{cm}^3

and

586 \ \text{cm}^3

Q50:ipmat indore 2019QA › IndicesMediumMCQ · MCQ

Determine the greatest number among the following four numbers:

A $2^{300}$
B $3^{200}$
C $2^{100} + 3^{100}$
D $4^{100}$

Pick an option to attempt

The Setup: A power-scaling comparison. Computing these directly is hopeless, so rewrite every term with the same exponent - then only the bases need comparing. Step 1: Normalise to the exponent 100. Each exponent is a multiple of 100, so pull that out:

2^{300}=\left(2^3\right)^{100}=8^{100}, \qquad 3^{200}=\left(3^2\right)^{100}=9^{100}, \qquad 4^{100}=4^{100}

Step 2: Compare the three pure powers. With a common positive exponent,

t\mapsto t^{100}

is increasing on positive bases, so the ordering of the bases carries straight over:

9^{100}>8^{100}>4^{100} \implies 3^{200}>2^{300}>4^{100}

Step 3: Dispose of the sum. The remaining candidate

2^{100}+3^{100}

is not a pure power, so bound it. Since

2^{100}<3^{100}

:

2^{100}+3^{100}<3^{100}+3^{100}=2\cdot 3^{100}

Compare that generous overestimate against the leader:

\frac{3^{200}}{3^{100}}=3^{100} \quad\text{which is vastly larger than }2

so

2\cdot 3^{100}<3^{100}\cdot 3^{100}=3^{200}

. The sum loses even after being inflated, and the gap is not close -

3^{200}

exceeds it by a factor of roughly

\frac{3^{100}}{2}

. Step 4: Conclude. The full ordering is

3^{200}>2^{300}>4^{100}>2^{100}+3^{100}

, and the greatest is

3^{200}

. The lesson: adding two large powers is worth far less than multiplying the exponent.

2^{100}+3^{100}

is barely bigger than its larger half, while squaring

3^{100}

multiplies it by itself. Final Answer:

3^{200}

Q51:ipmat indore 2019QA › TrianglesMediumMCQ · MCQ

The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices

(0, 0), (0, 31),

and

(31, 0)

is

A435
B465
C450
D464

Pick an option to attempt

The Setup: We are hunting for lattice points strictly inside the triangle. The boundary lines are the axes (

x=0, y=0

) and the hypotenuse (

x+y=31

). To stay *interior*, our coordinates must be strictly positive integers that sum to less than 31. Boundary points fail the vibe check and get instantly disqualified. Step 1: Set the Constraints We need integers

x > 0

and

y > 0

such that

x + y < 31

. This means the absolute max limit for the sum is

x + y \le 30

. Step 2: Sum the Rows If

x = 1

,

y

can range from

1

to

29

(giving us 29 valid points). If

x = 2

,

y

can range from

1

to

28

(28 points). This sequence cascades linearly all the way down to

x = 29

, where

y

can only be

1

(1 point). Step 3: Calculate the Total We just need to sum the first 29 positive integers. We plug this into the classic arithmetic series sum formula

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

.

S = \frac{29(30)}{2}

S = 29 \times 15 = 435

Final Answer: 435

Q52:ipmat indore 2019QA › Time, Speed & DistanceMediumMCQ · MCQ

Two small insects, which are

x

metres apart, take

u

minutes to pass each other when they are flying towards each other, and

v

minutes to meet each other when they are flying in the same direction. Then, the ratio of the speed of the slower insect to that of the faster insect is

A $\dfrac{u}{v}$
B $\dfrac{u}{v-u}$
C $\dfrac{v-u}{v+u}$
D $\dfrac{u}{v+u}$

Pick an option to attempt

The Setup: This is a relative speed speedrun. When they fly head-on, their speeds stack (addition). When it's a chase scene (same direction), their speeds counter each other (subtraction). We just set up the two relative speed equations and isolate the ratio like a standard system of equations. Step 1: Define the Speeds Let the faster insect's speed be

S_1

and the slower insect's speed be

S_2

. Flying towards each other (closing the gap

x

in time

u

):

S_1 + S_2 = \frac{x}{u}

Flying in the same direction (closing the gap

x

in time

v

):

S_1 - S_2 = \frac{x}{v}

Step 2: Isolate the Ratio Instead of solving for individual speeds and making it messy, we can just divide the two equations to instantly wipe the distance

x

from the board:

\frac{S_1 + S_2}{S_1 - S_2} = \frac{\frac{x}{u}}{\frac{x}{v}} = \frac{v}{u}

Step 3: Cross-Multiply and Solve

u(S_1 + S_2) = v(S_1 - S_2)

uS_1 + uS_2 = vS_1 - vS_2

Group the

S_1

terms on one side and

S_2

terms on the other:

uS_2 + vS_2 = vS_1 - uS_1

S_2(u + v) = S_1(v - u)

Now, isolate the ratio of the slower speed to the faster speed (

S_2/S_1

):

\frac{S_2}{S_1} = \frac{v - u}{v + u}

Final Answer:

\dfrac{v-u}{v+u}

Q53:ipmat indore 2019QA › Mixture & AlligationHardMCQ · MCQ

An alloy

P

has copper and zinc in the proportion of

5:2

(by weight), while another alloy

Q

has the same metals in the proportion of

3:4

(by weight). If these two alloys are mixed in the proportion of

a:b

(by weight), a new alloy

R

is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy

S

, formed by mixing the two alloys

P

and

Q

in the proportion of

b:a

(by weight) is

A $7:9$
B $9:7$
C $9:5$
D $5:9$

Pick an option to attempt

The Setup: We've got an alchemy minigame. First, we need to unlock the exact ratio (

a:b

) that balances the metals 1:1. Once we have those initial mix weights, we hit them with an Uno reverse card, swap the weights (

b:a

) for a new batch, and calculate the final drop rates for copper and zinc. **Step 1: Find the Initial Mix Ratio (

a:b

)** In alloy P, Copper is

5/7

and Zinc is

2/7

. In alloy Q, Copper is

3/7

and Zinc is

4/7

. Mix them in parts

a

and

b

to get equal Copper and Zinc:

\frac{5a}{7} + \frac{3b}{7} = \frac{2a}{7} + \frac{4b}{7}

Multiply by 7 to drop the denominators and balance the equation:

5a + 3b = 2a + 4b

3a = b \implies \frac{a}{b} = \frac{1}{3}

So the ratio

a:b

is strictly

1:3

. **Step 2: Form Alloy S (Ratio

b:a

)** Now we mix them using the reverse ratio: 3 parts of P and 1 part of Q. Calculate the total Copper:

\text{Cu} = 3\left(\frac{5}{7}\right) + 1\left(\frac{3}{7}\right) = \frac{15}{7} + \frac{3}{7} = \frac{18}{7}

Calculate the total Zinc:

\text{Zn} = 3\left(\frac{2}{7}\right) + 1\left(\frac{4}{7}\right) = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}

Step 3: Final Proportion The ratio of Cu to Zn in alloy S is:

\frac{18}{7} : \frac{10}{7}

Multiply by 7 and simplify by dividing by 2:

18 : 10 = 9 : 5

Final Answer:

9:5

Q54:ipmat indore 2019QA › Permutation & CombinationMediumMCQ · MCQ

How many different numbers can be formed by using only the digits 1 and 3 which are smaller than 3000000?

A64
B128
C190
D254

Pick an option to attempt

The Setup: This is a pure combinatorics flex. We are restricted to binary-style choices (only 1s and 3s) and capped by a ceiling of 3 million. We need to sum the possibilities for all digit lengths from 1 to 7 without busting the limit. Step 1: Count the Smaller Lengths (1 to 6 digits) For any

n

-digit number made of just two specific digits, there are exactly

2^n

combinations. Let's tally everything up to 6 digits (since all of them easily slide under

3{,}000{,}000

): * 1-digit:

2^1 = 2

* 2-digit:

2^2 = 4

* 3-digit:

2^3 = 8

* 4-digit:

2^4 = 16

* 5-digit:

2^5 = 32

* 6-digit:

2^6 = 64

Total combinations for lengths 1-6:

2+4+8+16+32+64 = 126

. Step 2: Handle the 7-Digit Boss Level To be a 7-digit number strictly smaller than 3,000,000, the first digit *must* be 1. (Starting with a 3 instantly busts the cap). First digit = locked at '1' (1 way). Remaining 6 digits = any combo of 1s and 3s (

2^6 = 64

ways). Valid 7-digit numbers =

1 \times 64 = 64

. Step 3: Final Sum Combine the two pools to get the total number of valid permutations:

\text{Total} = 126 + 64 = 190

Final Answer: 190

Q55:ipmat indore 2019QA › Progression & SeriesMediumMCQ · MCQ

There are numbers

a_1, a_2, a_3, \ldots, a_n

each of them being

+1

or

-1

. If it is known that

a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0

then

A $n$ is a multiple of 2 but not a multiple of 4
B $n$ is a multiple of 3
C $n$ can be any multiple of 4
DThe only possible value of $n$ is 4

Pick an option to attempt

The Setup: We are dealing with a cyclic sum of products that perfectly zeroes out. Because the terms can only be +1 or -1, for the sum to be zero, there must be a perfectly balanced meta-game: an equal number of positive and negative terms. Step 1: Define the Terms Let

x_1 = a_1 a_2

,

x_2 = a_2 a_3, \dots, x_n = a_n a_1

. Each

x_i

is strictly either

+1

or

-1

. The given equation translates to

\sum x_i = 0

. This requires an equal number of

+1

s and

-1

s. Thus,

n

must be an even number. Let

n = 2k

, meaning there are exactly

k

positive terms and

k

negative terms. Step 2: Multiply Everything Together Let's look at the product of all

x_i

terms using two different perspectives. Method A (using the base variables):

\prod x_i = (a_1 a_2)(a_2 a_3)\dots(a_n a_1) = a_1^2 a_2^2 \dots a_n^2

Since every

a_i \in \{+1, -1\}

, their squares are all definitively

1

. So the overall product is exactly

1

. Method B (using the +1 and -1 counts): We know there are exactly

k

negative terms (

-1

) in the mix.

\prod x_i = (+1)^k (-1)^k = (-1)^k

Step 3: Match the Logic For Method A and Method B to agree without breaking math,

(-1)^k

must equal

1

. This means

k

MUST be an even integer. Let

k = 2m

. Since our total terms

n = 2k

, we substitute to get

n = 2(2m) = 4m

. This confirms

n

must strictly be a multiple of 4. Final Answer:

n

can be any multiple of 4

IPMAT Indore 2019 — QA

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