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IPMAT Indore 2023 — SA

All 15 SA previous year questions (PYQs) from the IPMAT Indore 2023 past year paper, with answers and full solutions.

Free SolutionsNo Login15 Questions

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

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

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

f(1) = 1

and

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

for all integers

n > 1

, then the value of

f(2023)

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

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

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

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

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

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

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

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

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

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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

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