IPMAT Indore 2023 QA PYQs

Q1:ipmat indore 2023QA › Time, Speed & DistanceHardSA · TITA

Vinita drives a car which has four gears. The speed of the car in the fourth gear is five times its speed in the first gear. The car takes twice the time to travel a certain distance in the second gear as compared to the third gear. In a 100 km journey, if Vinita travels equal distances in each of the gears, she takes 585 minutes to complete the journey. Instead, if the distances covered in the first, second, third, and fourth gears are 4 km, 4 km, 32 km, and 60 km, respectively, then the total time taken, in minutes, to complete the journey, will be

Enter your answer to attempt

The Setup: We are given relative speed and time constraints across four car gears to complete a uniform distance, which we must use to evaluate the time taken for a mixed-distance journey. Step 1: Define the speed variables and their relations. Let the speeds in the four gears be

v_1, v_2, v_3,

and

v_4

. We are given

v_4 = 5v_1

. We are also told the time in the 2nd gear is twice the time in the 3rd gear for a constant distance:

\frac{d}{v_2} = 2\left(\frac{d}{v_3}\right) \implies v_3 = 2v_2

. Step 2: Express the times for the

100\text{ km}

uniform journey. The

100\text{ km}

journey consists of equal distances in each gear, meaning

25\text{ km}

per gear. Let

x = \frac{1}{v_1}

and

y = \frac{1}{v_2}

. This means

\frac{1}{v_4} = \frac{x}{5}

and

\frac{1}{v_3} = \frac{y}{2}

. The total time equation is:

25\left(\frac{1}{v_1}\right) + 25\left(\frac{1}{v_2}\right) + 25\left(\frac{1}{v_3}\right) + 25\left(\frac{1}{v_4}\right) = 585

25x + 25y + 25\left(\frac{y}{2}\right) + 25\left(\frac{x}{5}\right) = 585

25x + 5x + 25y + 12.5y = 585

30x + 37.5y = 585

Step 3: Simplify the baseline equation. Divide the entire equation by

7.5

:

4x + 5y = 78

Step 4: Calculate the time for the newly requested journey. The new distances are

4\text{ km}

(1st),

4\text{ km}

(2nd),

32\text{ km}

(3rd), and

60\text{ km}

(4th).

\text{Total Time} = 4\left(\frac{1}{v_1}\right) + 4\left(\frac{1}{v_2}\right) + 32\left(\frac{1}{v_3}\right) + 60\left(\frac{1}{v_4}\right)

Substitute

x

and

y

:

\text{Total Time} = 4x + 4y + 32\left(\frac{y}{2}\right) + 60\left(\frac{x}{5}\right)

\text{Total Time} = 4x + 4y + 16y + 12x

\text{Total Time} = 16x + 20y

Factor out a

4

:

\text{Total Time} = 4(4x + 5y)

Step 5: Substitute the baseline value to find the final time.

\text{Total Time} = 4(78) = 312\text{ minutes}

Final Answer: 312

Q2:ipmat indore 2023QA › RemainderEasySA · TITA

The remainder when

1! + 2! + 3! + ... + 95!

is divided by

15

is

Enter your answer to attempt

The Setup: We need to find the remainder when a factorial summation sequence (

1!+2!+3!+...+95!

) is divided by

15

. Step 1: Analyze the divisibility of factorials by

15

. Since

15 = 3 \times 5

, any factorial

n!

where

n \ge 5

will strictly contain both

3

and

5

as prime factors. Therefore, for all

n \ge 5

,

n! \equiv 0 \pmod{15}

. Step 2: Evaluate the terms that are not divisible by

15

. The only terms in the sequence that do not contain

5!

as a factor are the first four terms:

1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

Step 3: Calculate the final remainder. The entire sequence modulo

15

reduces simply to

33 \pmod{15}

.

33 = 2 \times 15 + 3

The remainder is

3

. Final Answer: 3

Q3:ipmat indore 2023QA › Ratio, Proportion & VariationMediumSA · TITA

Let

a, b, c, d

be positive integers such that

a + b + c + d = 2023

. If

a: b = 2:5

and

c:d=5:2

, then the maximum possible value of

a + c

is

Enter your answer to attempt

The Setup: Given a sum equation

a+b+c+d=2023

for positive integers and specified ratio pairs, we must maximize the sum

a+c

. Step 1: Express the variables using their ratios. Given

a:b=2:5

, let

a = 2x

and

b = 5x

for some positive integer

x

. Given

c:d=5:2

, let

c = 5y

and

d = 2y

for some positive integer

y

. Step 2: Formulate the sum constraint. Substitute the variables into the total sum equation:

2x + 5x + 5y + 2y = 2023

7x + 7y = 2023

x + y = 289

Step 3: Optimize the target function. We want to maximize

a + c

, which translates to maximizing

2x + 5y

. To maximize the value of

2x + 5y

given the constraint

x + y = 289

, we must make

y

(which carries the larger coefficient

5

) as large as possible. Since

a,b,c,d

are strictly positive integers, the scaling factors

x

and

y

must be at least

1

. The maximum valid integer value for

y

occurs when

x

is minimized at

1

.

1 + y = 289 \implies y = 288

Step 4: Calculate the maximum value.

\text{Max}(a+c) = 2(1) + 5(288) = 2 + 1440 = 1442

Final Answer: 1442

Q4:ipmat indore 2023QA › Linear EquationsEasySA · TITA

In an election with only two contesting candidates, 15% of the voters did not turn up to vote and 50 voters cast invalid votes. It is known that 44% of all the voters in the voting list voted for the winner. If the winner got 200 votes more than the other candidate, then the number of voters in the voting list is

Enter your answer to attempt

The Setup: We are asked to determine the total number of voters on a list given percentage-based turnout and fractional vote shares in a two-candidate election. Step 1: Define the election parameters. Let

V

be the total number of eligible voters on the list. Voters who didn't turn up =

0.15V

. Total votes cast =

0.85V

. Valid votes cast =

0.85V - 50

. Step 2: Represent the candidates' vote shares. The winner got exactly

44\%

of *all voters on the voting list* (not just valid votes).

W = 0.44V

The other candidate (the loser) received the remaining valid votes:

L = \text{Valid Votes} - W = (0.85V - 50) - 0.44V = 0.41V - 50

Step 3: Apply the winning margin constraint to solve for

V

. The winner won by exactly

200

votes.

W - L = 200

0.44V - (0.41V - 50) = 200

0.03V + 50 = 200

0.03V = 150

V = \frac{150}{0.03} = 5000

Final Answer: 5000

Q5:ipmat indore 2023QA › FunctionsMediumSA · TITA

If

f(1) = 1

and

f(n) = 3n - f(n - 1)

for all integers

n > 1

, then the value of

f(2023)

is

Enter your answer to attempt

The Setup: A recursive sequence is defined as

f(n)=3n-f(n-1)

with base case

f(1)=1

. We must find the value of

f(2023)

. Step 1: Generate the first few terms to identify a pattern.

f(1) = 1

f(2) = 3(2) - 1 = 5

f(3) = 3(3) - 5 = 4

f(4) = 3(4) - 4 = 8

f(5) = 3(5) - 8 = 7

f(6) = 3(6) - 7 = 11

Step 2: Formulate a generalized relationship. Let's find the difference between every second term (

f(n)

and

f(n-2)

):

f(n) = 3n - f(n-1)

Substitute the recursive definition for

f(n-1)

:

f(n) = 3n - (3(n-1) - f(n-2))

f(n) = 3n - 3n + 3 + f(n-2)

f(n) = f(n-2) + 3

This confirms that the odd-indexed terms and even-indexed terms form independent Arithmetic Progressions, each with a common difference of

3

. Step 3: Evaluate

f(2023)

. Since

2023

is odd, we use the sequence of odd-indexed terms:

1, 4, 7, 10, \dots

The generic formula for the odd index sequence

f(2k-1)

is an AP where the first term is

1

and the difference is

3

. Let

n = 2023

. The number of steps (or the term position in the separated odd sequence) is

\frac{n-1}{2}

.

\text{Term Index} = \frac{2023 - 1}{2} = 1011

f(2023) = \text{Base} + (\text{Term Index}) \times \text{Difference}

f(2023) = 1 + 1011 \times 3

f(2023) = 1 + 3033 = 3034

Final Answer: 3034

Q6:ipmat indore 2023QA › CirclesHardSA · TITA

In the xy-plane let

A = (- 2, 0), B = (2, 0)

. Define the set S as the collection of all points C on the circle

x ^ 2 + y ^ 2 = 4

such that the area of the triangle ABC is an integer. The number of points in the set S is

Enter your answer to attempt

The Setup: Points

A=(-2,0)

and

B=(2,0)

lie on the

x

-axis. We must find all points

C

on the circle

x^2+y^2=4

such that the area of

\triangle ABC

is an integer. Step 1: Define the area of the triangle. The base of the triangle is the segment

AB

connecting

(-2,0)

and

(2,0)

. Its length is exactly

4

. Let point

C

have coordinates

(x, y)

. The height of the triangle relative to base

AB

is the absolute value of the

y

-coordinate,

|y|

.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times |y| = 2|y|

Step 2: Establish integer constraints for the area. We require the Area (

2|y|

) to be a positive integer

k

(a valid geometric triangle must have an area strictly greater than 0).

2|y| = k \implies |y| = \frac{k}{2}

Since

C

lies on the circle

x^2+y^2=4

, the maximum possible value for

|y|

is the radius,

2

.

0 < |y| \le 2 \implies 0 < \frac{k}{2} \le 2 \implies k \in \{1, 2, 3, 4\}

Step 3: Map the valid area states back to coordinate points. For each valid

k

, we substitute

|y| = k/2

back into the circle equation

x^2 = 4 - y^2

to count the distinct points: * If

k = 1

:

|y| = 0.5 \implies y = \pm 0.5

.

x^2 = 4 - 0.25 = 3.75

. This yields

2

values for

x

per

y

, giving

4

points. * If

k = 2

:

|y| = 1 \implies y = \pm 1

.

x^2 = 4 - 1 = 3

. This gives

4

points. * If

k = 3

:

|y| = 1.5 \implies y = \pm 1.5

.

x^2 = 4 - 2.25 = 1.75

. This gives

4

points. * If

k = 4

:

|y| = 2 \implies y = \pm 2

.

x^2 = 4 - 4 = 0

. Here

x = 0

uniquely, giving exactly

1

point for

y=2

and

1

point for

y=-2

, totaling

2

points. Step 4: Sum the valid points.

\text{Total Points} = 4 + 4 + 4 + 2 = 14

Final Answer: 14

Q7:ipmat indore 2023QA › Binomial TheoremMediumSA · TITA

If three consecutive coefficients in the expansion of

(x+y)^n

are in the ratio

1:9:63

, then the value of

n

is

Enter your answer to attempt

The Setup: We are given three consecutive coefficients in the binomial expansion of

(x+y)^n

strictly in the ratio

1:9:63

. We must determine the power

n

. Step 1: Express the coefficients algebraically. Let the three consecutive coefficients be

\binom{n}{r-1}

,

\binom{n}{r}

, and

\binom{n}{r+1}

. We are given two ratio equations:

\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{9}{1} = 9

\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{63}{9} = 7

Step 2: Apply the standard binomial coefficient ratio formula. The ratio

\frac{\binom{n}{k}}{\binom{n}{k-1}}

simplifies universally to

\frac{n - k + 1}{k}

. Apply this to our first equation (

k = r

):

\frac{n - r + 1}{r} = 9 \implies n - r + 1 = 9r \implies n = 10r - 1

Apply this to our second equation (

k = r + 1

):

\frac{n - (r+1) + 1}{r+1} = 7 \implies \frac{n - r}{r + 1} = 7 \implies n - r = 7r + 7 \implies n = 8r + 7

Step 3: Solve the linear system for

r

and

n

. Equate the two expressions for

n

:

10r - 1 = 8r + 7

2r = 8 \implies r = 4

Substitute

r

back into either equation to find

n

:

n = 10(4) - 1 = 39

Final Answer: 39

Q8:ipmat indore 2023QA › Permutation & CombinationHardSA · TITA

In a chess tournament, there are four groups, each containing an equal number of players. Each player plays 1. against every other player belonging to one's own group exactly once; 2. against each player belonging to one of the remaining three groups exactly twice; 3. against each player belonging to one of the remaining two groups exactly three times; and 4. against each player belonging to the remaining group exactly four times. If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is

Enter your answer to attempt

The Setup: A chess tournament consists of four equal groups. Players play matches inside their group and against the other three groups at varying frequencies (

1\times, 2\times, 3\times, 4\times

). We must find the minimum group size to surpass

1000

total matches. Step 1: Analyze the match distribution logic. Let each of the

4

groups have

k

players. Total players =

4k

. Every group must be assigned an inter-group match frequency of

2, 3,

and

4

against the other three groups. This is perfectly modeled by weighting the edges of a

K_4

graph with values

\{2, 3, 4\}

such that every vertex touches exactly one of each weight. Such a configuration requires three distinct perfect matchings on

K_4

, meaning the edge weights between the

6

pairs of groups are exactly: two

2

s, two

3

s, and two

4

s. Step 2: Calculate total inter-group matches. Between any two groups of size

k

playing each other

w

times, the number of matches is

k \times k \times w = w k^2

. Summing over the 6 pairs of groups:

\text{Inter-Group Matches} = k^2(2 + 2 + 3 + 3 + 4 + 4) = 18k^2

Step 3: Calculate total intra-group matches. Each player plays everyone in their own group exactly once. For one group, the matches are

\binom{k}{2} = \frac{k(k-1)}{2}

. Since there are

4

groups:

\text{Intra-Group Matches} = 4 \times \frac{k(k-1)}{2} = 2k(k-1) = 2k^2 - 2k

Step 4: Formulate the total matches inequality.

\text{Total Matches} = 18k^2 + 2k^2 - 2k = 20k^2 - 2k

We are given that there are more than

1000

matches:

20k^2 - 2k > 1000 \implies 10k^2 - k > 500

Step 5: Test integer values for

k

to find the minimum. * If

k = 7

:

10(49) - 7 = 490 - 7 = 483

(Fails,

483 \ngtr 500

) * If

k = 8

:

10(64) - 8 = 640 - 8 = 632

(Succeeds,

632 > 500

) The minimum possible number of players in each group is

8

. Final Answer: 8

Q9:ipmat indore 2023QA › LogarithmsEasySA · TITA

The product of the roots of the equation

\log_{2} 2^{(\log_{2}x)^{2}} -5 \log_{2}x+6=0

is

Enter your answer to attempt

The Setup: We need to find the product of the roots for the logarithmic equation

log_{2}2^{(log_{2}x)^{2}}-5~log_{2}x+6=0

. Step 1: Simplify the first term using logarithm properties. The term

\log_2(2^y)

universally simplifies to

y

. Thus,

\log_2(2^{(\log_2 x)^2})

simplifies directly to

(\log_2 x)^2

. The equation becomes:

(\log_2 x)^2 - 5\log_2 x + 6 = 0

Step 2: Apply a variable substitution. Let

t = \log_2 x

. The equation is a standard quadratic:

t^2 - 5t + 6 = 0

Step 3: Solve for

t

and reverse the substitution. Factor the quadratic:

(t - 2)(t - 3) = 0 \implies t = 2 \text{ or } t = 3

Reverse the substitution to solve for the roots in terms of

x

: If

t = 2 \implies \log_2 x = 2 \implies x = 2^2 = 4

If

t = 3 \implies \log_2 x = 3 \implies x = 2^3 = 8

Step 4: Calculate the product of the roots.

\text{Product} = 4 \times 8 = 32

Final Answer: 32

Q10:ipmat indore 2023QA › Permutation & CombinationMediumSA · TITA

The total number of positive integer solutions of

21 \leq a + b + c \leq 25

is

Enter your answer to attempt

The Setup: We must find the total number of positive integer solutions to the bounded inequality

21 \le a+b+c \le 25

. Step 1: Normalize the variables to utilize Stars and Bars. Since

a, b, c

must be strictly positive integers (

\ge 1

), let:

a = x + 1

b = y + 1

c = z + 1

where

x, y, z

are non-negative integers (

\ge 0

). Substitute these into the inequality:

21 \le (x+1) + (y+1) + (z+1) \le 25

21 \le x + y + z + 3 \le 25

18 \le x + y + z \le 22

Step 2: Apply the combinatorics formula for exact sums. The number of non-negative integer solutions to

x+y+z = n

is given by

\binom{n+3-1}{3-1} = \binom{n+2}{2}

. We need to sum this over

n \in \{18, 19, 20, 21, 22\}

:

\text{Solutions} = \binom{20}{2} + \binom{21}{2} + \binom{22}{2} + \binom{23}{2} + \binom{24}{2}

Step 3: Evaluate the sum. Calculate each combination

\binom{k}{2} = \frac{k(k-1)}{2}

:

\binom{20}{2} = \frac{20 \times 19}{2} = 190

\binom{21}{2} = \frac{21 \times 20}{2} = 210

\binom{22}{2} = \frac{22 \times 21}{2} = 231

\binom{23}{2} = \frac{23 \times 22}{2} = 253

\binom{24}{2} = \frac{24 \times 23}{2} = 276

\text{Total} = 190 + 210 + 231 + 253 + 276 = 1160

Final Answer: 1160

Q11:ipmat indore 2023QA › ModulusMediumSA · TITA

The length of the line segment joining the two intersection points of the curves

y = 4970 - |x|

and

y = x ^ 2

is

Enter your answer to attempt

The Setup: We must find the geometric length of the line segment connecting the two intersection points of the absolute value function

y = 4970 - |x|

and the parabola

y = x^2

. Step 1: Set the equations equal to find intersection points.

x^2 = 4970 - |x|

Rearrange to form a quadratic in terms of

|x|

, recognizing that

x^2 = |x|^2

:

|x|^2 + |x| - 4970 = 0

Step 2: Factor the quadratic equation. We need two integers that multiply to

-4970

and add to

1

. Note that

4970 = 10 \times 497 = 10 \times 7 \times 71 = 70 \times 71

.

(|x| + 71)(|x| - 70) = 0

Step 3: Solve for

x

and locate the coordinates. Since absolute value

|x|

must be non-negative,

|x| = -71

is rejected.

|x| = 70 \implies x = 70 \text{ or } x = -70

Because the equations are perfectly symmetric across the y-axis, the segment connecting the two points is perfectly horizontal. Step 4: Calculate the distance. The length of a horizontal line segment is simply the absolute difference between the

x

-coordinates.

\text{Length} = |70 - (-70)| = 140

Final Answer: 140

Q12:ipmat indore 2023QA › Progression & SeriesHardSA · TITA

If

f(n)= 1 + 2 + 3 +\cdots+(n+1)

and

g(n)= \sum_{k=1}^{k=n} \dfrac{1}{f(k)}

, then the least value of

n

for which

g(n)

exceeds the value

\dfrac{99}{100}

is:

Enter your answer to attempt

The Setup: Given an arithmetic sum base function

f(n)

, we have a reciprocal series

g(n)

. We must find the least integer

n

that pushes

g(n)

above the threshold of

99/100

. Step 1: Simplify the base function

f(n)

. The function is the sum of consecutive integers:

f(n) = 1 + 2 + 3 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}

Step 2: Construct the series

g(n)

and apply partial fractions.

g(n) = \sum_{k=1}^n \frac{1}{f(k)} = \sum_{k=1}^n \frac{2}{(k+1)(k+2)}

Split the fraction using partial fraction decomposition:

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

Step 3: Evaluate the telescoping sum. When summing from

k=1

to

n

, all intermediate terms cancel out:

g(n) = 2 \left[ \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \right]

g(n) = 2 \left( \frac{1}{2} - \frac{1}{n+2} \right) = 1 - \frac{2}{n+2}

Step 4: Apply the threshold inequality. We need

g(n) > \frac{99}{100}

:

1 - \frac{2}{n+2} > 0.99

0.01 > \frac{2}{n+2}

\frac{1}{100} > \frac{2}{n+2} \implies n+2 > 200 \implies n > 198

Since

n

must be an integer, the least value fulfilling this is

199

. Final Answer: 199

Q13:ipmat indore 2023QA › RemainderEasySA · TITA

The polynomial

4x ^ {10} - x ^ 9 + 3x ^ {11} - 5x ^ 7 + c x ^ 6 + 2x ^ 5 - x ^ 4 + x ^ 3 - 4x ^ 2 + 6x - 2

when divided by

x - 1

leaves a remainder

2.

Then the value of

c + 6

is

Enter your answer to attempt

The Setup: A polynomial divided by

(x-1)

leaves a remainder of

2

. We apply the Polynomial Remainder Theorem to find a missing coefficient

c

and then compute

c+6

. Step 1: Apply the Remainder Theorem. If a polynomial

P(x)

is divided by

(x - a)

, the remainder is exactly

P(a)

. Here, dividing by

(x - 1)

gives a remainder of

2

, meaning

P(1) = 2

. Step 2: Evaluate

P(1)

algebraically. Substitute

x = 1

into the polynomial. For

x=1

, all powers of

x

resolve to

1

, leaving just the sum of the coefficients.

P(1) = 4(1) - 1(1) + 3(1) - 5(1) + c(1) + 2(1) - 1(1) + 1(1) - 4(1) + 6(1) - 2

Group the numerical terms:

P(1) = 4 - 1 + 3 - 5 + 2 - 1 + 1 - 4 + 6 - 2 + c

P(1) = 3 + c

Step 3: Equate to the remainder and solve.

c + 3 = 2 \implies c = -1

Step 4: Calculate the final expression. We need the value of

c + 6

.

-1 + 6 = 5

Final Answer: 5

Q14:ipmat indore 2023QA › Simple & Compound InterestMediumSA · TITA

Assume it is the beginning of the year today. Ankita will earn INR 10,000 at the end of the year, which she plans to invest in a bank deposit immediately at a fixed simple interest of 0.5% per annum. Her yearly income will increase by INR 10,000 every year, and the fixed simple interest offered by the bank on new deposits will also increase by 0.5% per annum every year. If Ankita continues to invest all her yearly income in new bank deposits at the end of each year, the total interest earned by her, in INR, in five years from today will be

Enter your answer to attempt

The Setup: We evaluate sequential deposits placed at the end of each year with growing principals and scaling simple interest rates over a bounded 5-year timeline from 'today' (Start of Year 1). Step 1: Chart the timeline and investment parameters. Five years from today implies the timeline ends exactly at the conclusion of Year 5 (T=5). Simple Interest Formula:

\text{Interest} = P \times R \times T

. * Deposit 1 (End of Yr 1, T=1): Earns

10,000

. Rate =

0.5\% = 0.005

. Time invested =

4

years (T=1 to T=5). * Deposit 2 (End of Yr 2, T=2): Income rises by 10k, so earns

20,000

. Rate rises by 0.5%, so =

1.0\% = 0.01

. Time invested =

3

years. * Deposit 3 (End of Yr 3, T=3): Earns

30,000

. Rate =

1.5\% = 0.015

. Time invested =

2

years. * Deposit 4 (End of Yr 4, T=4): Earns

40,000

. Rate =

2.0\% = 0.02

. Time invested =

1

year. * Deposit 5 (End of Yr 5, T=5): Earns

50,000

. Rate =

2.5\% = 0.025

. Time invested =

0

years (cashed exactly as deposited). Step 2: Calculate interest for each independent deposit. * Interest 1:

10000 \times 0.005 \times 4 = 200

* Interest 2:

20000 \times 0.010 \times 3 = 600

* Interest 3:

30000 \times 0.015 \times 2 = 900

* Interest 4:

40000 \times 0.020 \times 1 = 800

* Interest 5:

50000 \times 0.025 \times 0 = 0

Step 3: Sum the total interest.

\text{Total Interest} = 200 + 600 + 900 + 800 + 0 = 2500

Final Answer: 2500

Q15:ipmat indore 2023QA › Time & WorkMediumSA · TITA

Amisha can complete a particular task in twenty days. After working for four days she fell sick for four days and resumed the work on the ninth day but with half of her original work rate. She completed the task in another twelve days with the help of a co-worker who joined her from the ninth day. The number of days required for the co-worker to complete the task alone would be

Enter your answer to attempt

The Setup: A worker finishes a partial task at shifting rates and enlists a co-worker to complete the remaining fraction. By mapping their combined work capacity against the remaining work, we can isolate the co-worker's individual rate. Step 1: Establish Amisha's work rates. Amisha's standard rate allows her to finish in

20

days, so her normal rate is

A = \frac{1}{20}

tasks/day. When she resumes, she works at half her original rate, making her new rate

A' = \frac{1}{40}

tasks/day. Step 2: Calculate work completed before the break. Amisha works normally for

4

days.

\text{Work Done} = 4 \times \left(\frac{1}{20}\right) = \frac{4}{20} = \frac{1}{5}

\text{Remaining Work} = 1 - \frac{1}{5} = \frac{4}{5}

Step 3: Formulate the combined work phase. She is sick for

4

days, so Day 9 begins the new phase. From Day 9, she and her co-worker work together for

12

days to finish the job. Let the co-worker's rate be

C

tasks/day.

\text{Combined Rate} = A' + C = \frac{1}{40} + C

\text{Work done together} = 12 \times \left(\frac{1}{40} + C\right)

Set this equal to the remaining work:

12 \left(\frac{1}{40} + C\right) = \frac{4}{5}

Step 4: Solve for

C

and convert to days. Divide both sides by

12

:

\frac{1}{40} + C = \frac{4}{60} = \frac{1}{15}

C = \frac{1}{15} - \frac{1}{40}

Find a common denominator (

120

):

C = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24}

Since the co-worker's rate is

\frac{1}{24}

of the task per day, it would take them exactly

24

days to complete the task alone. Final Answer: 24

Q16:ipmat indore 2023QA › Unit DigitEasyMCQ · MCQ

Let

p

be a positive integer such that the unit digit of

p ^ 3

is

4

. What are the possible unit digits of

(p + 3) ^ 3

A3
B1, 7, 9
C4, 7
D1, 3, 7

Pick an option to attempt

The Setup: We are given a positive integer

p

such that the unit digit of

p^3

is exactly

4

. We need to determine the possible unit digits of the expression

(p+3)^3

. Step 1: Determine the unit digit of

p

. We evaluate the cubes of all base digits from

0

to

9

to see which produces a unit digit of

4

:

0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 9^3 = 729

. The only digit whose cube ends in

4

is

4

itself. Therefore, the unit digit of

p

must be

4

. Step 2: Calculate the unit digit of

(p+3)^3

. Since

p \equiv 4 \pmod{10}

, we substitute this into the target expression:

p + 3 \equiv 4 + 3 \equiv 7 \pmod{10}

Now, cube this resulting unit digit:

7^3 = 343

The unit digit of

343

is

3

. Thus, the only possible unit digit for

(p+3)^3

is

3

. Final Answer: 3

Q17:ipmat indore 2023QA › Integral SolutionsMediumMCQ · MCQ

Let

[x]

denote the greatest integer not exceeding

x

and {

x

}

= x - [x]

If

n

is a natural number, then the sum of all values of

x

satisfying the equation

2[x] = x + n

{

x

} is

A $\dfrac{n(n+2)}{2}$
B $\dfrac{3}{2}$
C $\dfrac{n(n+1)}{2}$
D $n$

Pick an option to attempt

The Setup: We are given the equation

2[x] = x + n\{x\}

, where

n

is a natural number,

[x]

is the greatest integer function, and

\{x\}

is the fractional part. We must find the sum of all values of

x

that satisfy this equation. Step 1: Express

x

in terms of its integer and fractional parts. By definition, any real number

x

can be written as

x = [x] + \{x\}

. Substitute this into the given equation:

2[x] = ([x] + \{x\}) + n\{x\}

[x] = (n+1)\{x\}

Step 2: Isolate the fractional part and apply its bounds.

\{x\} = \frac{[x]}{n+1}

By the definition of the fractional part function,

0 \le \{x\} < 1

.

0 \le \frac{[x]}{n+1} < 1

Since

n

is a natural number (

n \ge 1

),

n+1

is strictly positive. Multiply the inequality by

n+1

:

0 \le [x] < n+1

Because

[x]

must be an integer, the possible values for

[x]

are the integers

0, 1, 2, \dots, n

. Step 3: Formulate

x

and calculate the sum. Substitute the expression for

\{x\}

back into

x = [x] + \{x\}

:

x = [x] + \frac{[x]}{n+1} = [x]\left(1 + \frac{1}{n+1}\right) = [x]\left(\frac{n+2}{n+1}\right)

To find the sum of all valid

x

, we sum this expression over all possible values of

[x]

from

0

to

n

:

\text{Sum} = \sum_{k=0}^{n} k\left(\frac{n+2}{n+1}\right) = \left(\frac{n+2}{n+1}\right) \sum_{k=0}^{n} k

Using the sum of the first

n

integers formula

\sum_{k=0}^n k = \frac{n(n+1)}{2}

:

\text{Sum} = \left(\frac{n+2}{n+1}\right) \left(\frac{n(n+1)}{2}\right) = \frac{n(n+2)}{2}

Final Answer:

\frac{n(n+2)}{2}

Q18:ipmat indore 2023QA › InequalitiesEasyMCQ · MCQ

The set of all real values of x satisfying the inequality

\dfrac{x^2(x+1)}{(x-1)(2x+1)^3} > 0

is

A $(-\infty, -1)$ U $(-\dfrac{1}{2}, 0)$ U $(1, +\infty)$
B $(-1, -\dfrac{1}{2})$ U $(1,+\infty)$
C $(-1, 0)$ U $(1,+\infty)$
D $(-1, -\dfrac{1}{2})$ U $(0,+\infty)$

Pick an option to attempt

The Setup: We must find the set of all real values of

x

that satisfy the rational inequality

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

. Step 1: Identify the critical points. Set the numerator and denominator factors to zero to find the critical points where the expression changes sign or becomes undefined: Numerator:

x = 0

(multiplicity 2, even),

x = -1

(multiplicity 1, odd). Denominator:

x = 1

(multiplicity 1, odd),

x = -1/2

(multiplicity 3, odd). The critical points in ascending order are:

-1, -1/2, 0, 1

. Step 2: Evaluate the sign intervals using a test-point method. Because the factor

x^2

is always non-negative and touches

0

at

x=0

, the sign does not change across

x=0

. We test the regions partitioned by the odd-multiplicity critical points (

-1, -1/2, 1

): * **Region

x > 1

:** Test

x = 2

. Expression is

(+)(+)/(+)(+) > 0

. (Valid) * **Region

-1/2 < x < 1

(excluding

0

):** Since sign alternates at

x=1

, this region is negative. (Invalid) * **Region

-1 < x < -1/2

:** Since sign alternates at

x=-1/2

, this region is positive. (Valid) * **Region

x < -1

:** Since sign alternates at

x=-1

, this region is negative. (Invalid) Step 3: Combine the valid intervals. The strict inequality requires the expression to be strictly greater than

0

, so the critical points themselves are not included. The valid intervals are

(-1, -1/2)

and

(1, +\infty)

.

\text{Set} = \left(-1, -\frac{1}{2}\right) \cup (1, +\infty)

Final Answer:

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

Q19:ipmat indore 2023QA › Profit & LossEasyMCQ · MCQ

A goldsmith bought a large solid golden ball at INR 1,000,000 and melted it to make a certain number of solid spherical beads such that the radius of each bead was one-fifth of the radius of the original ball. Assume that the cost of making golden beads is negligible. If the goldsmith sold all the beads at 20% discount on the listed price and made a total profit of 20%, then the listed price or each golden bead, in INR, was

A48000
B12000
C9600
D24000

Pick an option to attempt

The Setup: A solid golden ball costing INR

1,000,000

is melted into smaller spherical beads. The radius of each bead is one-fifth of the original ball's radius. The beads are sold at a

20\%

discount for a total overall profit of

20\%

. We must find the listed price per bead. Step 1: Calculate the number of beads produced. Let the radius of the original ball be

R

and the radius of a bead be

r

. We are given

r = \frac{R}{5}

. The volume of a sphere is proportional to the cube of its radius.

\text{Number of beads} = \frac{\text{Volume of original ball}}{\text{Volume of one bead}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \left(\frac{R}{r}\right)^3 = (5)^3 = 125

So,

125

identical beads are produced. Step 2: Calculate the required revenue per bead. The total cost is INR

1,000,000

. The total profit made is

20\%

, so the total revenue generated is

120\%

of the cost:

\text{Total Revenue} = 1,000,000 \times 1.20 = 1,200,000

Since this revenue comes from selling

125

beads, the selling price (

SP

) per bead is:

SP = \frac{1,200,000}{125} = 9600

Step 3: Calculate the listed price. The selling price is derived after a

20\%

discount is applied to the listed price (

L

).

0.80 \times L = 9600

L = \frac{9600}{0.80} = 12000

Final Answer: 12000

Q20:ipmat indore 2023QA › CirclesMediumMCQ · MCQ

Which of the following straight lines are both tangent to the circle

x ^ 2 + y ^ 2 - 6x + 4y - 12 = 0

?

A $4x+3y+19=0, 4x+3y-31=0$
B $4x+3y-19=0, 4x+3y+31=0$
C $4x+3y-19=0, 4x+3y-31=0$
D $4x+3y+19=0, 4x+3y+31=0$

Pick an option to attempt

The Setup: We are asked to identify which pair of straight lines are both tangents to the circle defined by

x^2+y^2-6x+4y-12=0

. Step 1: Determine the center and radius of the circle. Complete the square for both

x

and

y

to convert the equation into standard circle form

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

:

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4

(x - 3)^2 + (y + 2)^2 = 25

The center of the circle is

C = (3, -2)

and the radius is

r = \sqrt{25} = 5

. Step 2: Apply the tangent distance formula. For a line

Ax + By + C = 0

to be tangent to a circle, the perpendicular distance from the circle's center to the line must exactly equal the radius

r

. Looking at the options, all candidate lines have the form

4x + 3y + c = 0

. Set the perpendicular distance from

(3, -2)

to the line equal to

5

:

\frac{|4(3) + 3(-2) + c|}{\sqrt{4^2 + 3^2}} = 5

\frac{|12 - 6 + c|}{\sqrt{25}} = 5

\frac{|6 + c|}{5} = 5

|6 + c| = 25

Step 3: Solve for the constant

c

. This absolute value equation yields two valid cases for

c

: Case 1:

6 + c = 25 \implies c = 19

Case 2:

6 + c = -25 \implies c = -31

The two tangent lines are

4x + 3y + 19 = 0

and

4x + 3y - 31 = 0

. Final Answer:

4x+3y+19=0,4x+3y-31=0

Q21:ipmat indore 2023QA › Matrices & DeterminantsMediumMCQ · MCQ

If

A = \begin{bmatrix} 1 & 2 \newline 3 & a \end{bmatrix}

where

a

is a real number and det

(A ^ 3 - 3A ^ 2 - 5A) = 0

then one of the values of

a

can be

A4
B6
C1
D5

Pick an option to attempt

The Setup: We are given a

2 \times 2

matrix

A

with a variable

a

. We must find a valid value for

a

given that

\det(A^3-3A^2-5A)=0

. Step 1: Factor the matrix polynomial.

A^3 - 3A^2 - 5A = A(A^2 - 3A - 5I)

Using the determinant product rule

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

, the condition becomes:

\det(A) \cdot \det(A^2 - 3A - 5I) = 0

This means that either

\det(A) = 0

or

\det(A^2 - 3A - 5I) = 0

. Step 2: Test the first determinant condition. Calculate the determinant of matrix

A

:

A = \begin{bmatrix} 1 & 2 \\ 3 & a \end{bmatrix}

\det(A) = (1)(a) - (2)(3) = a - 6

If

\det(A) = 0

, then

a - 6 = 0 \implies a = 6

. Checking the multiple-choice options provided in the prompt,

6

is a listed option. Final Answer: 6

Q22:ipmat indore 2023QA › Time, Speed & DistanceEasyMCQ · MCQ

A helicopter flies along the sides of a square field of side length 100 kms. The first side is covered at a speed of 100 kmph, and for each subsequent side the speed is increased by 100 kmph till it covers all the sides. The average speed of the helicopter is

A250 kmph
B184 kmph
C192 kmph
D200 kmph

Pick an option to attempt

The Setup: A helicopter flies around the perimeter of a

100\text{ km} \times 100\text{ km}

square field. Its speed starts at

100\text{ kmph}

for the first side and increases by

100\text{ kmph}

for each subsequent side. We must calculate its average speed for the entire trip. Step 1: Determine the distance and speed for each leg. Since the field is a square with side length

100\text{ km}

, the helicopter flies four distinct legs of

100\text{ km}

each. * Side 1: Distance =

100\text{ km}

, Speed =

100\text{ kmph}

* Side 2: Distance =

100\text{ km}

, Speed =

200\text{ kmph}

* Side 3: Distance =

100\text{ km}

, Speed =

300\text{ kmph}

* Side 4: Distance =

100\text{ km}

, Speed =

400\text{ kmph}

Step 2: Calculate the time taken for each leg. Using Time = Distance / Speed: *

T_1 = \frac{100}{100} = 1\text{ hour}

*

T_2 = \frac{100}{200} = \frac{1}{2}\text{ hour}

*

T_3 = \frac{100}{300} = \frac{1}{3}\text{ hour}

*

T_4 = \frac{100}{400} = \frac{1}{4}\text{ hour}

Step 3: Calculate the average speed.

\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

\text{Total Distance} = 4 \times 100 = 400\text{ km}

\text{Total Time} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12 + 6 + 4 + 3}{12} = \frac{25}{12}\text{ hours}

\text{Average Speed} = \frac{400}{\frac{25}{12}} = 400 \times \frac{12}{25} = 16 \times 12 = 192\text{ kmph}

Final Answer: 192 kmph

Q23:ipmat indore 2023QA › Ratio, Proportion & VariationEasyMCQ · MCQ

\dfrac{(a + b)}{(b + c)} = \dfrac{(c + d)}{(d + a)}

which of the following statements is always true?

A $a + b + c + d = 0$
B $a=c,$ or $a+b+c+d = 0$
C $a=c$
D $a=c,$ and $b=d$

Pick an option to attempt

The Setup: We are given the algebraic equality

\frac{a+b}{b+c} = \frac{c+d}{d+a}

. We need to evaluate which conditional statement must necessarily follow. Step 1: Cross-multiply to clear the fractions.

(a+b)(a+d) = (c+d)(b+c)

Step 2: Expand both sides of the equation.

a^2 + ad + ab + bd = bc + c^2 + bd + cd

Step 3: Simplify and factor the expression. Subtract the right side from the left side. Notice that the

+bd

term exists on both sides and cancels out:

a^2 + ad + ab - bc - c^2 - cd = 0

Group the terms strategically to factor by grouping. Pair the squares, and group the remaining terms by common factors

b

and

d

:

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

Apply the difference of squares identity to the first term:

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

Factor out the common binomial term

(a - c)

:

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

Step 4: Evaluate the Zero Product Property. For this product to be zero, at least one of the factors must be zero: Factor 1:

a - c = 0 \implies a = c

Factor 2:

a + b + c + d = 0

Thus, it must always be true that

a=c

or

a+b+c+d=0

. Final Answer:

a=c,

or

a+b+c+d=0

Q24:ipmat indore 2023QA › PolynomialsMediumMCQ · MCQ

If the harmonic mean of the roots of the equation

(5 + \sqrt{2}) x ^ 2 - bx + 8 + 2\sqrt{5} = 0

is

4

then the value of

b

is

A $2$
B $4-\sqrt{5}$
C $3$
D $4+\sqrt{5}$

Pick an option to attempt

The Setup: We are given a quadratic equation

(5+\sqrt{2})x^2 - bx + 8+2\sqrt{5} = 0

whose roots have a harmonic mean of

4

. We must find the value of

b

. Step 1: Establish the Harmonic Mean formula for the roots. Let the roots of the quadratic equation be

\alpha

and

\beta

. The Harmonic Mean (HM) of two numbers is defined as:

HM = \frac{2\alpha\beta}{\alpha+\beta}

We are given that

HM = 4

. Step 2: Extract the sum and product of the roots from the quadratic. Using Vieta's formulas for

Ax^2 + Bx + C = 0

:

\text{Sum of roots } (\alpha+\beta) = -\frac{B}{A} = \frac{b}{5+\sqrt{2}}

\text{Product of roots } (\alpha\beta) = \frac{C}{A} = \frac{8+2\sqrt{5}}{5+\sqrt{2}}

Step 3: Substitute Vieta's results into the Harmonic Mean equation.

4 = \frac{2 \left( \frac{8+2\sqrt{5}}{5+\sqrt{2}} \right)}{\left( \frac{b}{5+\sqrt{2}} \right)}

Notice that the complex denominator

(5+\sqrt{2})

neatly cancels out from both the numerator and the denominator of the large fraction:

4 = \frac{2(8+2\sqrt{5})}{b}

4 = \frac{16+4\sqrt{5}}{b}

Step 4: Solve for

b

.

4b = 16 + 4\sqrt{5}

Divide both sides by

4

:

b = 4 + \sqrt{5}

Final Answer:

4+\sqrt{5}

Q25:ipmat indore 2023QA › Simple & Compound InterestEasyMCQ · MCQ

If the difference between compound interest and simple interest for a certain amount of money invested for

3

years at an annual interest rate of

10\%

is INR

527

, then the amount invested in INR is

A17000
B15000
C1500
D170000

Pick an option to attempt

The Setup: The difference between Compound Interest (CI) and Simple Interest (SI) on a principal amount invested for

3

years at a

10\%

annual rate is INR

527

. We need to determine the original principal amount. Step 1: State the 3-year CI and SI difference formula. For a principal

P

invested for exactly

3

years at an annual interest rate

R

(expressed as a percentage), the difference

D

between CI and SI is given by the standard derived formula:

D = P \left(\frac{R}{100}\right)^2 \left(\frac{R}{100} + 3\right)

Step 2: Substitute the known values into the formula. We are given

D = 527

and

R = 10

.

527 = P \left(\frac{10}{100}\right)^2 \left(\frac{10}{100} + 3\right)

527 = P (0.1)^2 (0.1 + 3)

527 = P (0.01)(3.1)

527 = 0.031 P

Step 3: Solve for the Principal

P

.

P = \frac{527}{0.031} = \frac{527,000}{31}

Divide

527

by

31

to simplify the fraction:

527 \div 31 = 17

P = 17,000

Final Answer: 17000

Q26:ipmat indore 2023QA › Progression & SeriesMediumMCQ · MCQ

Let

a_{1}, a_{2}, a_{3}

be three distinct real numbers in geometric progression. If the equations

a_{1} x ^ 2 + 2a_{2}x + a_{3} = 0

and

b_{1} x ^ 2 + 2b_{2}x + b_{3} = 0

have a common root, then which of the following is necessarily true?

A $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in geometric progression
B $b_{1}, b_{2}, b_{3}$ are in geometric progression
C $b_{1}, b_{2}, b_{3}$ are in arithmetic progression
D $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in arithmetic progression

Pick an option to attempt

The Setup: Given three distinct real numbers

a_1, a_2, a_3

in a geometric progression (GP), we are told the quadratics

a_1x^2+2a_2x+a_3=0

and

b_1x^2+2b_2x+b_3=0

share a common root. We must deduce the relationship between the ratios of their coefficients. Step 1: Analyze the first quadratic equation. Since

a_1, a_2, a_3

are in GP, they satisfy the property

a_2^2 = a_1a_3

. Check the discriminant (

D

) of the first quadratic equation

a_1x^2 + 2a_2x + a_3 = 0

:

D = (2a_2)^2 - 4(a_1)(a_3) = 4a_2^2 - 4a_1a_3

Substitute

a_1a_3

for

a_2^2

:

D = 4(a_1a_3) - 4a_1a_3 = 0

Since

D = 0

, the first equation has a single repeated real root. We find this root using the quadratic formula:

x = \frac{-2a_2}{2a_1} = -\frac{a_2}{a_1}

Step 2: Apply the common root constraint. Because the two equations share a common root, and the first equation only has *one* unique root, this root (

x = -a_2/a_1

) must exactly be the root of the second equation

b_1x^2+2b_2x+b_3=0

. Substitute

x = -a_2/a_1

into the second equation:

b_1\left(-\frac{a_2}{a_1}\right)^2 + 2b_2\left(-\frac{a_2}{a_1}\right) + b_3 = 0

b_1\frac{a_2^2}{a_1^2} - 2b_2\frac{a_2}{a_1} + b_3 = 0

Step 3: Simplify the relationship to match the options. Divide the entire equation by

a_3

:

b_1\frac{a_2^2}{a_1^2 a_3} - 2b_2\frac{a_2}{a_1 a_3} + \frac{b_3}{a_3} = 0

Use the GP identity

a_2^2 = a_1a_3

to simplify the denominators. For the first term:

\frac{a_2^2}{a_1^2 a_3} = \frac{a_1 a_3}{a_1^2 a_3} = \frac{1}{a_1}

For the second term:

\frac{a_2}{a_1 a_3} = \frac{a_2}{a_2^2} = \frac{1}{a_2}

Substitute these back:

\frac{b_1}{a_1} - 2\frac{b_2}{a_2} + \frac{b_3}{a_3} = 0

2\left(\frac{b_2}{a_2}\right) = \frac{b_1}{a_1} + \frac{b_3}{a_3}

This fits the exact definition of an Arithmetic Progression (AP) where the middle term is the arithmetic mean of the outer terms. Final Answer:

\frac{b_{1}}{a_{1}}.

\frac{b_{2}}{a_{2}},

\frac{b_{3}}{a_{3}}

are in arithmetic progression

Q27:ipmat indore 2023QA › TrianglesHardMCQ · MCQ

In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be

A $11:4\sqrt{10}$
B $12:11$
C $11:12$
D $4\sqrt{10}:11$

Pick an option to attempt

The Setup: In

\triangle ABC

,

D

is the midpoint of

BC

, and

AM

is the altitude on

BC

. With side ratios

AB:BC:CA = 2:4:3

, we must find the length ratio

BM:AD

. Step 1: Assign algebraic lengths to the sides. Let the sides be

c = AB = 2x

,

a = BC = 4x

, and

b = CA = 3x

. Since

D

is the midpoint of

BC

,

BD = \frac{BC}{2} = 2x

. Step 2: Calculate the length of

BM

. Use the Law of Cosines to find

\cos B

in

\triangle ABC

:

\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4x)^2 + (2x)^2 - (3x)^2}{2(4x)(2x)}

\cos B = \frac{16x^2 + 4x^2 - 9x^2}{16x^2} = \frac{11x^2}{16x^2} = \frac{11}{16}

In the right-angled

\triangle ABM

, the segment

BM

represents the adjacent side to angle

B

. Thus,

BM = c \cdot \cos B

:

BM = 2x \left(\frac{11}{16}\right) = \frac{11x}{8}

Step 3: Calculate the length of the median

AD

. Apply Apollonius's Theorem to the median

AD

:

AB^2 + AC^2 = 2(AD^2 + BD^2)

(2x)^2 + (3x)^2 = 2(AD^2 + (2x)^2)

4x^2 + 9x^2 = 2AD^2 + 8x^2

13x^2 - 8x^2 = 2AD^2

5x^2 = 2AD^2 \implies AD^2 = \frac{5x^2}{2} \implies AD = \frac{x\sqrt{5}}{\sqrt{2}} = \frac{x\sqrt{10}}{2}

Step 4: Compute the ratio

BM : AD

.

\text{Ratio} = \frac{\frac{11x}{8}}{\frac{x\sqrt{10}}{2}} = \frac{11}{8} \times \frac{2}{\sqrt{10}} = \frac{22}{8\sqrt{10}} = \frac{11}{4\sqrt{10}}

Final Answer:

11:4\sqrt{10}

Q28:ipmat indore 2023QA › LogarithmsMediumMCQ · MCQ

Let

a, b, c

be real numbers greater than 1, and

n

be a positive real number not equal to 1. If

log_n(log_2a) = 1; log_n(log_2b) = 2

and

log_n(log_2c) = 3

then which of the following is true?

A $(b-a)^n=(c-b)$
B $a^n+b^n=c^n$
C $a+b=c$
D $(a^n+b)^n=ac$

Pick an option to attempt

The Setup: We are given a system of nested logarithms:

\log_n(\log_2 a)=1

,

\log_n(\log_2 b)=2

, and

\log_n(\log_2 c)=3

. We must establish which given algebraic relationship holds true. Step 1: Convert the logarithmic equations into exponential form. Using the rule

\log_x(y) = z \implies y = x^z

: Equation 1:

\log_2 a = n^1 = n \implies a = 2^n

Equation 2:

\log_2 b = n^2 \implies b = 2^{n^2}

Equation 3:

\log_2 c = n^3 \implies c = 2^{n^3}

Step 2: Express variables in terms of each other. Notice that the exponent of

b

is the square of the exponent of

a

:

b = 2^{n^2} = (2^n)^n = a^n

Notice that the exponent of

c

relates to

b

:

c = 2^{n^3} = (2^{n^2})^n = b^n

We also know that

c

relates to

a

by

c = (2^n)^{n^2} = a^{n^2}

. Additionally, multiply

a

and

c

:

ac = 2^n \cdot 2^{n^3} = 2^{n + n^3}

Alternatively,

ac = a \cdot b^n

. Step 3: Evaluate the multiple-choice options. We test the specific option

(a^n+b)^n = ac

to see if it holds true. Substitute

b = a^n

into the left side:

\text{Left Side} = (b + b)^n = (2b)^n

Distribute the exponent:

(2b)^n = 2^n \cdot b^n

Now, substitute our initial mappings back in. We know

2^n = a

and

b^n = c

:

2^n \cdot b^n = a \cdot c

The left side mathematically perfectly matches the right side

ac

. Final Answer:

(a^{n}+b)^{n}=ac

Q29:ipmat indore 2023QA › Permutation & CombinationEasyMCQ · MCQ

Consider an

8 \times 8

chessboard. The number of ways 8 rooks can be placed on the board such that no two rooks are in the same row and no two are in the same column is

A7
B7!
C8
D8!

Pick an option to attempt

The Setup: We need to find the total number of ways to place

8

indistinguishable rooks on a standard

8 \times 8

chessboard such that no two rooks threaten each other (no two share the same row or column). Step 1: Place the rooks sequentially row by row. To ensure no two rooks share a row, exactly one rook must be placed in each of the 8 rows. * Row 1: The first rook can be placed in any of the

8

squares (columns) in the first row. * Row 2: The second rook must be placed in the second row, but it cannot share the column occupied by the first rook. This leaves

7

valid squares. * Row 3: The third rook cannot share a column with the first two rooks, leaving

6

valid squares. *

\dots

* Row 8: The final rook is forced into the single remaining unoccupied column. Step 2: Calculate total configurations. The total number of valid placements is the product of the independent choices for each row:

\text{Total Ways} = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 8!

Final Answer: 8!

Q30:ipmat indore 2023QA › Conic SectionsEasyMCQ · MCQ

The equation

x^2 + y^2 - 2x - 4y +5 = 0

represents

Aa pair of straight lines
Ba circle
Can ellipse
Da point

Pick an option to attempt

The Setup: We are given the conic equation

x^2+y^2-2x-4y+5=0

and asked to classify the specific geometric shape it represents. Step 1: Reformat the equation using completing the square. Group the

x

terms and

y

terms:

(x^2 - 2x) + (y^2 - 4y) = -5

Complete the square for

x

by adding

(-2/2)^2 = 1

: Complete the square for

y

by adding

(-4/2)^2 = 4

: Balance the equation by adding these to the right side as well:

(x^2 - 2x + 1) + (y^2 - 4y + 4) = -5 + 1 + 4

Step 2: Simplify and classify the equation.

(x - 1)^2 + (y - 2)^2 = 0

This is the standard form of a circle

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

. Here, the radius squared is exactly

0

(

r = 0

). A circle with a radius of

0

mathematically collapses into a single coordinate point located at its center,

(1, 2)

. Final Answer: a point

Q31:ipmat indore 2023QA › Progression & SeriesHardMCQ · MCQ

A person standing at the centre of an open ground first walks 32 meters towards the east, takes a right turn and walks 16 meters, takes another right turn and walks 8 meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?

A $32$ meters
B $\frac{32}{\sqrt{5}}$ meters
C $64$ meters
D $\frac{64}{\sqrt{5}}$ meters

Pick an option to attempt

The Setup: A person maps out a path taking consecutive

90^\circ

right turns, with each leg of the journey halving in distance (

32\text{m}, 16\text{m}, 8\text{m} \dots

). We must find their net displacement from the origin after infinite turns. Step 1: Establish a coordinate system. Let the starting point be the origin

(0,0)

. * Leg 1:

32\text{m}

East (Positive

x

-direction). * Leg 2:

16\text{m}

South (Negative

y

-direction). * Leg 3:

8\text{m}

West (Negative

x

-direction). * Leg 4:

4\text{m}

North (Positive

y

-direction). * Leg 5:

2\text{m}

East (Positive

x

-direction). This pattern continues infinitely. Step 2: Calculate the net displacement on the

x

-axis. The

x

-coordinate sequence alternates direction every two steps:

32

(East),

-8

(West),

2

(East),

-0.5

(West)

\dots

This forms an infinite geometric progression where the first term

a = 32

and the common ratio

r = -\frac{1}{4}

.

X_{\text{final}} = \frac{a}{1 - r} = \frac{32}{1 - (-1/4)} = \frac{32}{5/4} = 32 \times \frac{4}{5} = \frac{128}{5}

Step 3: Calculate the net displacement on the

y

-axis. The

y

-coordinate sequence begins on the second step:

-16

(South),

4

(North),

-1

(South)

\dots

This forms an infinite geometric progression where

a = -16

and

r = -\frac{1}{4}

. (We'll use magnitude for distance).

Y_{\text{final}} = \frac{16}{1 - (-1/4)} = \frac{16}{5/4} = 16 \times \frac{4}{5} = \frac{64}{5}

Step 4: Calculate the final straight-line distance. Use the Pythagorean theorem:

D = \sqrt{X_{\text{final}}^2 + Y_{\text{final}}^2}

.

D = \sqrt{\left(\frac{128}{5}\right)^2 + \left(\frac{64}{5}\right)^2} = \frac{64}{5} \sqrt{2^2 + 1^2} = \frac{64\sqrt{5}}{5}

Simplify by converting the denominator:

D = \frac{64\sqrt{5}}{(\sqrt{5}\cdot\sqrt{5})} = \frac{64}{\sqrt{5}}

Final Answer:

\frac{64}{\sqrt{5}}

meters

Q32:ipmat indore 2023QA › LogarithmsHardMCQ · MCQ

If

\log_{(cos x)}(sin x) + \log_{(sin x)}(cos x) = 2,

then the value of

x

is

A $n\pi + \dfrac{\pi}{4}, n$ is an integer
B $2n\pi + \dfrac{\pi}{4}, n$ is an integer
C $\dfrac{n\pi}{4}, n$ is an integer
D $\dfrac{n\pi}{4} + \dfrac{\pi}{4}, n$ is an integer

Pick an option to attempt

The Setup: We must solve for

x

given the trigonometric logarithmic equation

\log_{\cos x}(\sin x) + \log_{\sin x}(\cos x) = 2

. Step 1: Use substitution to solve the algebra. By the base-change inversion property of logarithms,

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

. Let

t = \log_{\cos x}(\sin x)

. The equation becomes:

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

Multiply by

t

to form a quadratic:

t^2 - 2t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1

Step 2: Re-substitute to find the trigonometric relation.

\log_{\cos x}(\sin x) = 1 \implies \cos^1 x = \sin x \implies \sin x = \cos x

Dividing by

\cos x

yields

\tan x = 1

. Step 3: Evaluate the domain restrictions for logarithms. For a logarithm

\log_b a

to be defined, the base

b > 0, b \neq 1

, and the argument

a > 0

. Thus, we strictly require

\sin x > 0

,

\sin x \neq 1

,

\cos x > 0

, and

\cos x \neq 1

. This strictly restricts valid solutions for

x

exclusively to the first quadrant of the unit circle. Step 4: Find the general solution for

x

. The first-quadrant angle where

\tan x = 1

is

\frac{\pi}{4}

. To represent all coterminal first-quadrant angles, we add full

360^\circ

rotations, which is

2n\pi

, where

n

is an integer.

x = 2n\pi + \frac{\pi}{4}

Final Answer:

2n\pi+\frac{\pi}{4},

n is an integer

Q33:ipmat indore 2023QA › Permutation & CombinationMediumMCQ · MCQ

A rabbit is sitting at the base of a staircase which has 10 steps. It proceeds to the top of the staircase by climbing either one step at a time or two steps at a time. The number of ways it can reach the top is

A144
B89
C34
D55

Pick an option to attempt

The Setup: A rabbit climbs a

10

-step staircase, taking either one or two steps at a time. We must find the total number of unique ways to reach the top. Step 1: Establish the recursive relationship. Let

W_n

be the number of ways to reach the

n

-th step. To reach step

n

, the rabbit must have either taken a single step from step

n-1

, or a double step from step

n-2

. Therefore, the total ways to reach step

n

is the sum of the ways to reach the previous two steps:

W_n = W_{n-1} + W_{n-2}

. This forms the Fibonacci sequence. Step 2: Define the base cases. * To reach Step 1 (

W_1

): Exactly

1

way (one 1-step). * To reach Step 2 (

W_2

): Exactly

2

ways (two 1-steps, or one 2-step). Step 3: Compute the sequence up to step 10. *

W_3 = W_2 + W_1 = 2 + 1 = 3

*

W_4 = 3 + 2 = 5

*

W_5 = 5 + 3 = 8

*

W_6 = 8 + 5 = 13

*

W_7 = 13 + 8 = 21

*

W_8 = 21 + 13 = 34

*

W_9 = 34 + 21 = 55

*

W_{10} = 55 + 34 = 89

Final Answer: 89

Q34:ipmat indore 2023QA › ProbabilityMediumMCQ · MCQ

The probability that a randomly chosen positive divisor of

10 ^ {2023}

is an integer multiple of

10 ^ {2001}

is

A $\dfrac{22}{2023}$
B $\dfrac{23^2}{{2024}^2}$
C $\dfrac{22}{2024}$
D $\dfrac{22^2}{{2023}^2}$

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen positive divisor of

10^{2023}

is also an integer multiple of

10^{2001}

. Step 1: Determine the total number of divisors (the sample space). First, find the prime factorization of the base number:

10^{2023} = (2 \times 5)^{2023} = 2^{2023} \times 5^{2023}

The formula for the total number of divisors of

p^a \cdot q^b

is

(a+1)(b+1)

.

\text{Total Divisors} = (2023 + 1)(2023 + 1) = 2024^2

Step 2: Determine the number of valid target divisors. A divisor

D

is a multiple of

10^{2001}

(which is

2^{2001} \times 5^{2001}

) if its prime factorization

D = 2^x \times 5^y

meets the constraints: * For base 2:

2001 \le x \le 2023

* For base 5:

2001 \le y \le 2023

Calculate the number of integer choices for the exponents

x

and

y

: Choices for

x = 2023 - 2001 + 1 = 23

. Choices for

y = 2023 - 2001 + 1 = 23

.

\text{Number of valid multiples} = 23 \times 23 = 23^2

Step 3: Calculate the probability.

\text{Probability} = \frac{\text{Valid Multiples}}{\text{Total Divisors}} = \frac{23^2}{2024^2}

Final Answer:

\frac{23^{2}}{2024^{2}}

Q35:ipmat indore 2023QA › Set TheoryEasyMCQ · MCQ

In a group of 120 students, 80 students are from the Science stream and the rest are from the Commerce stream. It is known that 70 students support Mumbai Indians in the Indian Premier League; all the other students support Chennai Super Kings. The number of Science students who are supporters of Mumbai Indians is

AExactly 20
BBetween 15 and 25
C30 or more
DBetween 20 and 25

Pick an option to attempt

The Setup: A group of

120

students comprises

80

Science and

40

Commerce students. Exactly

70

students support Mumbai Indians (MI), while the rest support Chennai Super Kings (CSK). We must establish the possible range of Science students supporting MI. Step 1: Establish the basic parameters. Total Students =

120

Science Students =

80

Commerce Students =

120 - 80 = 40

Total MI Supporters =

70

Step 2: Calculate the maximum bound for Science MI supporters. Let

x

be the number of Science students who support MI. The maximum possible value for

x

occurs if as many MI supporters as possible are drawn from the Science pool. Since there are only

70

MI supporters total, and

80

Science students exist, it is mathematically possible for all

70

MI supporters to be Science students.

\text{Max}(x) = 70

Step 3: Calculate the minimum bound for Science MI supporters. The minimum possible value for

x

occurs if as many MI supporters as possible are drawn from the Commerce pool. There are only

40

Commerce students. If all

40

of them support MI, the remaining MI supporters must be Science students.

\text{Min}(x) = 70 (\text{Total MI}) - 40 (\text{Commerce MI}) = 30

Step 4: Conclude the range. The number of Science students supporting MI must sit exactly in the inclusive range

30 \le x \le 70

. Looking at the options, the constraint '30 or more' is the only logically correct bounding descriptor. Final Answer: 30 or more

Q36:ipmat indore 2023QA › Divisibility RulesEasyMCQ · MCQ

If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4?

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

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen

3

-digit number is divisible by *neither*

3

nor

4

. Step 1: Calculate the size of the sample space. The 3-digit numbers run inclusively from

100

to

999

.

\text{Total 3-digit numbers} = 999 - 100 + 1 = 900

Step 2: Use the Inclusion-Exclusion principle to find the complement. We first find the number of integers divisible by

3

OR

4

:

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

. * Divisible by 3: The sequence is

102, 105 \dots 999

.

\text{Count} = \frac{999 - 102}{3} + 1 = 297/3 + 1 = 300

* Divisible by 4: The sequence is

100, 104 \dots 996

.

\text{Count} = \frac{996 - 100}{4} + 1 = 896/4 + 1 = 225

* Divisible by 12 (Intersection): The sequence is

108, 120 \dots 996

.

\text{Count} = \frac{996 - 108}{12} + 1 = 888/12 + 1 = 75

\text{Divisible by 3 or 4} = 300 + 225 - 75 = 450

Step 3: Calculate the target probability. The target numbers are those *not* divisible by

3

or

4

.

\text{Target count} = \text{Total} - (\text{Divisible by 3 or 4}) = 900 - 450 = 450

\text{Probability} = \frac{450}{900} = \frac{1}{2}

Final Answer:

\frac{1}{2}

Q37:ipmat indore 2023QA › ProbabilityEasyMCQ · MCQ

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.8 is

A5
B6
C3
D7

Pick an option to attempt

The Setup: We must find the minimum number of independent coin tosses required for the probability of getting at least one head to exceed

0.8

. Step 1: Formulate the probability equation. The probability of getting *at least one* head is the complement of getting *zero* heads (all tails).

P(\text{at least 1 H}) = 1 - P(\text{all T})

For a fair coin tossed

n

times, the probability of getting all tails is

\left(\frac{1}{2}\right)^n

.

P(\text{at least 1 H}) = 1 - \left(\frac{1}{2}\right)^n

Step 2: Set up and solve the inequality. We require this probability to strictly exceed

0.8

.

1 - \left(\frac{1}{2}\right)^n > 0.8

1 - 0.8 > \left(\frac{1}{2}\right)^n

0.2 > \left(\frac{1}{2}\right)^n

Convert the decimal to a fraction (

0.2 = \frac{1}{5}

):

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

Inverting the fractions flips the inequality:

2^n > 5

Step 3: Test integer values for

n

. * If

n = 1

:

2^1 = 2 \ngtr 5

* If

n = 2

:

2^2 = 4 \ngtr 5

* If

n = 3

:

2^3 = 8 > 5

(Valid) The minimum integer number of tosses required is

3

. Final Answer: 3

Q38:ipmat indore 2023QA › TrigonometryMediumMCQ · MCQ

If

cos \alpha

+

cos \beta

= 1 then the maximum value of

sin \alpha - sin \beta

is

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

Pick an option to attempt

The Setup: Given the constraint

\cos\alpha + \cos\beta = 1

, we must find the maximum value of

\sin\alpha - \sin\beta

. Step 1: Rewrite both expressions with the sum-to-product identities. Let

u = \frac{\alpha+\beta}{2}

and

v = \frac{\alpha-\beta}{2}

. The constraint and the target expression become:

\cos\alpha + \cos\beta = 2\cos u \cos v = 1

\sin\alpha - \sin\beta = 2\cos u \sin v

Both share the same factor

2\cos u

, which is what makes the constraint substitutable. Step 2: Eliminate

u

using the constraint. The constraint gives

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

directly (note

\cos v \neq 0

). Substitute this into the target:

\sin\alpha - \sin\beta = \left(\frac{1}{\cos v}\right)\sin v = \tan v

The whole problem has collapsed to maximising

\tan v

. Step 3: Find the admissible range of

v

.

\cos u

is a cosine, so it is bounded by

|\cos u| \le 1

:

\left|\frac{1}{2\cos v}\right| \le 1 \implies |\cos v| \ge \frac{1}{2}

On the branch where

\tan v

is positive and increasing, this restricts

v

to

0 \le v \le \frac{\pi}{3}

. Step 4: Maximise on that range.

\tan v

increases throughout

\left[0, \frac{\pi}{3}\right]

, so the maximum sits at the right endpoint:

\text{Maximum} = \tan\frac{\pi}{3} = \sqrt{3}

Step 5: Confirm the maximum is actually attained. At

v = \frac{\pi}{3}

the constraint forces

\cos u = \frac{1}{2\cos(\pi/3)} = 1

, so

u = 0

, giving

\alpha = \frac{\pi}{3}

and

\beta = -\frac{\pi}{3}

. Check both conditions:

\cos\frac{\pi}{3} + \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} + \frac{1}{2} = 1

\sin\frac{\pi}{3} - \sin\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}

Both hold, so

\sqrt{3}

is genuinely achieved and is the maximum. Final Answer:

\sqrt{3}

Q39:ipmat indore 2023QA › RemainderMediumMCQ · MCQ

A polynomial

P(x)

leaves a remainder

2

when divided by

(x - 1)

and a remainder

1

when divided by

(x - 2)

The remainder when

P(x)

is divided by

(x - 1)(x - 2)

is

A $3-x$
B $3$
C $x-3$
D $2$

Pick an option to attempt

The Setup: A polynomial

P(x)

leaves specific remainders when divided by two linear binomials. We must use the Polynomial Remainder Theorem to deduce the linear remainder when divided by their quadratic product. Step 1: Establish the given values using the Remainder Theorem. The Remainder Theorem states that dividing

P(x)

by

(x-a)

yields a remainder of

P(a)

. * Divided by

(x-1)

, remainder is

2 \implies P(1) = 2

. * Divided by

(x-2)

, remainder is

1 \implies P(2) = 1

. Step 2: Formulate the Division Algorithm equation. When

P(x)

is divided by a quadratic polynomial

(x-1)(x-2)

, the maximum possible degree of the remainder is linear. Let the remainder be

R(x) = ax + b

.

P(x) = Q(x)(x-1)(x-2) + (ax + b)

Step 3: Substitute the known

x

values to create a system of equations. Substitute

x = 1

:

P(1) = Q(1)(0)(-1) + (a(1) + b) \implies a + b = 2

Substitute

x = 2

:

P(2) = Q(2)(1)(0) + (a(2) + b) \implies 2a + b = 1

Step 4: Solve the linear system for

a

and

b

. Subtract the first equation from the second equation:

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

a = -1

Substitute

a

back into the first equation:

-1 + b = 2 \implies b = 3

The resulting linear remainder is

R(x) = -1x + 3 = 3 - x

. Final Answer:

3-x

IPMAT Indore 2023 — QA

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