IPMAT Indore 2024 QA PYQs

Q1:ipmat indore 2024QA › LogarithmsMediumSA · TITA

If

4^{\log_2{x}} - 4x + 9^{\log_3{y}} - 16y + 68 = 0

, then

y - x

equals:

Enter your answer to attempt

The Setup: We are given the equation

4^{\log_{2}x}-4x+9^{\log_{3}y}-16y+68=0

. We need to evaluate the expression

y-x

. Step 1: Simplify the logarithmic terms using the base-change exponent identity

a^{\log_a b} = b

. For the first term:

4^{\log_{2}x} = (2^2)^{\log_{2}x} = 2^{2\log_{2}x} = 2^{\log_{2}(x^2)} = x^2

For the third term:

9^{\log_{3}y} = (3^2)^{\log_{3}y} = 3^{2\log_{3}y} = 3^{\log_{3}(y^2)} = y^2

Step 2: Substitute the simplified terms back into the algebraic equation.

x^2 - 4x + y^2 - 16y + 68 = 0

Step 3: Complete the square for both the

x

and

y

variables. Isolate the respective variables and add the required constants:

(x^2 - 4x + 4) + (y^2 - 16y + 64) = 0

(x - 2)^2 + (y - 8)^2 = 0

Note that

4 + 64 = 68

, which perfectly balances the original constant. Step 4: Solve for

x

and

y

. The sum of two real squares equals zero if and only if each independent square evaluates to zero.

x - 2 = 0 \Rightarrow x = 2

y - 8 = 0 \Rightarrow y = 8

Both values are strictly positive, satisfying the logarithmic domain restrictions. Step 5: Calculate the final target expression

y - x

.

8 - 2 = 6

Final Answer: 6

Q2:ipmat indore 2024QA › Ratio, Proportion & VariationEasySA · TITA

A fruit seller has oranges, apples, and bananas in the ratio

3:6:7

. If the number of oranges is a multiple of both 5 and 6, then the minimum number of fruits the seller has is:

Enter your answer to attempt

The Setup: A fruit seller has oranges, apples, and bananas in a ratio of

3:6:7

. The quantity of oranges is a multiple of both

5

and

6

, and we must find the minimum total number of fruits. Step 1: Define the quantities using a common scaling factor. Let

k

be a positive integer. Oranges =

3k

Apples =

6k

Bananas =

7k

Total Fruits =

3k + 6k + 7k = 16k

Step 2: Apply the divisibility constraint to the number of oranges. The quantity of oranges (

3k

) must be a multiple of both

5

and

6

. Calculate the Least Common Multiple (LCM) of

5

and

6

:

\text{LCM}(5, 6) = 30

Therefore,

3k

must be a multiple of

30

. Step 3: Determine the minimum valid scaling factor

k

. Let

m

be a positive integer such that:

3k = 30m \Rightarrow k = 10m

To minimize the total number of fruits, we must minimize

k

, which occurs when

m = 1

. Thus,

k = 10

. Step 4: Calculate the total number of fruits using the minimal scale factor.

\text{Total} = 16(10) = 160

Final Answer: 160

Q3:ipmat indore 2024QA › PolynomialsMediumSA · TITA

The number of real solutions of the equation

(x^2 - 15x + 55)^{x^2-5x+6} = 1

is:

Enter your answer to attempt

The Setup: We must find the number of real solutions for the exponential equation

(x^{2}-15x+55)^{x^{2}-5x+6}=1

. Step 1: Evaluate Condition 1 where the exponent is

0

and the base is non-zero.

x^2 - 5x + 6 = 0

(x - 2)(x - 3) = 0

This yields potential solutions

x = 2

and

x = 3

. We must verify the base is non-zero for these values: For

x = 2

, base =

2^2 - 15(2) + 55 = 29 \neq 0

. (Valid) For

x = 3

, base =

3^2 - 15(3) + 55 = 19 \neq 0

. (Valid) Step 2: Evaluate Condition 2 where the base is exactly

1

.

x^2 - 15x + 55 = 1

x^2 - 15x + 54 = 0

(x - 6)(x - 9) = 0

This yields solutions

x = 6

and

x = 9

. Both are valid for any real exponent. Step 3: Evaluate Condition 3 where the base is

-1

and the exponent is an even integer.

x^2 - 15x + 55 = -1

x^2 - 15x + 56 = 0

(x - 7)(x - 8) = 0

This yields potential solutions

x = 7

and

x = 8

. We must verify the exponent is even: For

x = 7

, exponent =

7^2 - 5(7) + 6 = 49 - 35 + 6 = 20

. (Even

\Rightarrow

Valid) For

x = 8

, exponent =

8^2 - 5(8) + 6 = 64 - 40 + 6 = 30

. (Even

\Rightarrow

Valid) Step 4: Aggregate all valid real solutions. The complete set of solutions is

\{2, 3, 6, 9, 7, 8\}

. Counting these unique values yields

6

distinct solutions. Final Answer: 6

Q4:ipmat indore 2024QA › Simple & Compound InterestEasySA · TITA

Person A borrows Rs. 4000 from another person B for a duration of 4 years. He borrows a portion of it at 3% simple interest per annum, while the rest at 4% simple interest per annum. If B gets Rs. 520 as total interest, then the amount A borrowed at 3% per annum in Rs. is:

Enter your answer to attempt

The Setup: Person A borrows Rs.

4000

for

4

years, splitting the principal between a

3\%

simple interest rate and a

4\%

simple interest rate. The total interest earned is Rs.

520

, and we must find the amount borrowed at

3\%

. Step 1: Define the variables for the split principal. Let

x

represent the principal amount borrowed at the

3\%

rate. Let

4000 - x

represent the remaining principal borrowed at the

4\%

rate. Step 2: Calculate the annualized interest yield. The total interest accrued over

4

years is

520

. Because it is simple interest, the annual interest is constant:

\text{Annual Interest} = \frac{520}{4} = 130

Step 3: Construct the linear equation for the annual interest.

0.03x + 0.04(4000 - x) = 130

Step 4: Solve for

x

. Distribute the terms and isolate the variable:

0.03x + 160 - 0.04x = 130

-0.01x = -30

x = 3000

Final Answer: 3000

Q5:ipmat indore 2024QA › TrianglesEasySA · TITA

The number of triangles with integer sides and with perimeter 15 is:

Enter your answer to attempt

The Setup: We need to calculate the total number of triangles that can be formed with integer sides and a fixed perimeter of

15

. Step 1: Define the basic parameters and constraints. Let the integer sides of the triangle be

a

,

b

, and

c

, strictly ordered such that

a \le b \le c

. The perimeter condition dictates:

a + b + c = 15

Step 2: Apply the triangle inequality theorem to bound the longest side

c

. The sum of the two shorter sides must strictly exceed the longest side:

a + b > c

Substitute

a + b = 15 - c

into the inequality:

15 - c > c \Rightarrow 2c < 15 \Rightarrow c \le 7.5

Since

c

is an integer, the maximum valid dimension for

c

is

7

. Furthermore,

c

is the maximum side, so it must be at least the average length of the perimeter:

c \ge \frac{15}{3} = 5

Thus,

c \in \{5, 6, 7\}

. Step 3: Systematically evaluate integer pairs

(a, b)

for each possible value of

c

, maintaining

a \le b \le c

. Case 1:

c = 5

Requires

a + b = 10

. The only integer pair satisfying

a \le b \le 5

is

(5, 5)

. (Yields

1

triangle) Case 2:

c = 6

Requires

a + b = 9

. The pairs satisfying

a \le b \le 6

are

(3, 6)

and

(4, 5)

. (Yields

2

triangles) Case 3:

c = 7

Requires

a + b = 8

. The pairs satisfying

a \le b \le 7

are

(1, 7)

,

(2, 6)

,

(3, 5)

, and

(4, 4)

. (Yields

4

triangles) Step 4: Aggregate the valid triangle formations.

\text{Total Triangles} = 1 + 2 + 4 = 7

Final Answer: 7

Q6:ipmat indore 2024QA › Matrices & DeterminantsMediumSA · TITA

If

A = \begin{bmatrix} x_1 & x_2 & 7 \\ y_1 & y_2 & y_3 \\ z_1 & 8 & 3 \end{bmatrix}

is a matrix such that the sum of all three elements along any row, column or diagonal are equal to each other, then the value of determinant of A is:

Enter your answer to attempt

The Setup: A

3 \times 3

matrix is given with elements

[x_1, x_2, 7]

,

[y_1, y_2, y_3]

, and

[z_1, 8, 3]

such that the sums across any row, column, or diagonal are identical. We must find its determinant. Step 1: Determine the magic sum

S

and the center element. Let

S

equal the constant sum of any row, column, or diagonal. Analyze the bottom row:

z_1 + 8 + 3 = S \Rightarrow z_1 = S - 11

Analyze the right-to-left diagonal:

7 + y_2 + z_1 = S

Substitute

z_1

into the diagonal equation:

7 + y_2 + (S - 11) = S \Rightarrow y_2 - 4 = 0 \Rightarrow y_2 = 4

In a

3 \times 3

magic square, the central element is always exactly

\frac{1}{3}

of the magic sum.

S = 3(y_2) = 12

Step 2: Populate the remaining elements of matrix

A

using

S = 12

. From Step 1,

z_1 = 12 - 11 = 1

. Left-to-Right Diagonal:

x_1 + y_2 + 3 = 12 \Rightarrow x_1 + 4 + 3 = 12 \Rightarrow x_1 = 5

Top Row:

5 + x_2 + 7 = 12 \Rightarrow x_2 = 0

Left Column:

5 + y_1 + 1 = 12 \Rightarrow y_1 = 6

Right Column:

7 + y_3 + 3 = 12 \Rightarrow y_3 = 2

Step 3: Construct the populated matrix

A

.

A = \begin{bmatrix} 5 & 0 & 7 \\ 6 & 4 & 2 \\ 1 & 8 & 3 \end{bmatrix}

Step 4: Calculate the determinant

|A|

by expanding along the top row.

|A| = 5((4)(3) - (8)(2)) - 0 + 7((6)(8) - (1)(4))

|A| = 5(12 - 16) + 7(48 - 4)

|A| = 5(-4) + 7(44) = -20 + 308 = 288

Final Answer: 288

Q7:ipmat indore 2024QA › FactorisationEasySA · TITA

The number of factors of 1800 that are multiple of 6 is:

Enter your answer to attempt

The Setup: We are asked to determine the number of factors of the integer

1800

that are also multiples of

6

. Step 1: Extract the prime factorization of

1800

.

1800 = 18 \times 100 = (2 \times 3^2) \times (2^2 \times 5^2)

1800 = 2^3 \times 3^2 \times 5^2

Any generic factor of

1800

takes the structure

2^a \times 3^b \times 5^c

, where constraints are

0 \le a \le 3

,

0 \le b \le 2

, and

0 \le c \le 2

. Step 2: Apply the multiple-of-

6

constraint to the exponents. Because

6 = 2^1 \times 3^1

, any factor that is a multiple of

6

must include at least one

2

and at least one

3

in its prime factorization. The restricted exponent ranges become:

a \in \{1, 2, 3\}

(yielding

3

valid choices)

b \in \{1, 2\}

(yielding

2

valid choices)

c \in \{0, 1, 2\}

(yielding

3

valid choices, as

5

is unconstrained) Step 3: Calculate the combinatorics of the restricted factor set. Multiply the independent choices together:

\text{Total Factors} = 3 \times 2 \times 3 = 18

Final Answer: 18

Q8:ipmat indore 2024QA › Mean, Median & ModeEasySA · TITA

The following table shows the number of employees and their median age in eight companies located in a district.

Company	Number of employees	Median age
A	32	24
B	28	30
C	43	39
D	39	45
E	35	49
F	29	54
G	23	59
H	16	63

It is known that the age of all employees are integers. It is known that the age of every employee in A is strictly less than the age of every employee in B, the age of every employee in B is strictly less than the age of every employee in C, ..., the age of every employee in G is strictly less than the age of every employee in H. The median age of employees across the eight companies is:

Enter your answer to attempt

The Setup: We are looking for the global median of a massive, perfectly sorted dataset. Since we know the strict hierarchical order of the companies (

A < B < C \dots

), the entire server population is already sorted in ascending order. We just need to locate exactly which company houses the middle employee and extract their stats. It is an absolute 'Where's Waldo?' situation, but with array indices. Step 1: Calculate the total server population. We sum up all employees across the companies to find our

N

:

32 + 28 + 43 + 39 + 35 + 29 + 23 + 16 = 245

Step 2: Find the index of the global median. Since the total

N

is odd (

245

), the median is simply the exact middle value in the sorted list.

\frac{245 + 1}{2} = 123

We need to find the exact target coordinates for the

123

rd employee overall. Step 3: Track the cumulative frequencies to locate the target's spawn zone. * Company A:

32

employees (Cumulative:

32

) * Company B:

28

employees (Cumulative:

32 + 28 = 60

) * Company C:

43

employees (Cumulative:

60 + 43 = 103

) * Company D:

39

employees (Cumulative:

103 + 39 = 142

) Since

103 < 123 \le 142

, the

123

rd employee lives right inside Company D's roster. Step 4: Pinpoint the exact age of this specific NPC. The

123

rd employee overall is exactly the

20

th employee within Company D (since

123 - 103 = 20

). Company D has exactly

39

employees. Let's find Company D's local median index:

\frac{39 + 1}{2} = 20

The

20

th employee *is* the exact median of Company D! Since the table explicitly states the median age of Company D is

45

, the

20

th employee's age is hard-locked at

45

. Final Answer: 45

Q9:ipmat indore 2024QA › TrianglesEasySA · TITA

Let

\triangle ABC

be a triangle right-angled at

B

with

AB = BC = 18

. The area of the largest rectangle that can be inscribed in this triangle and has

B

as one of the vertices is:

Enter your answer to attempt

The Setup: A right-angled triangle has vertex

B

at the right angle, with legs

AB = BC = 18

. We need to find the area of the largest inscribed rectangle sharing vertex

B

. Step 1: Establish the coordinate system. Place the right-angled vertex

B

at the origin

(0,0)

. Given leg lengths

AB = BC = 18

, set vertex

A

at

(0,18)

and vertex

C

at

(18,0)

. The hypotenuse

AC

forms a line passing through

(0,18)

and

(18,0)

. Its linear equation is:

x + y = 18 \Rightarrow y = 18 - x

Step 2: Define the area function of the inscribed rectangle. A rectangle sharing vertex

B

at

(0,0)

and bounded by the triangle will have its opposing vertex

(x, y)

strictly on the hypotenuse

AC

. The Area (

A

) is the product of its length and width:

A_{\text{rect}} = x \times y = x(18 - x) = 18x - x^2

Step 3: Maximize the quadratic area function. The function is a downward-opening parabola. We find its maximum via differentiation (or the vertex formula

x = \frac{-b}{2a}

):

\frac{d A_{\text{rect}}}{dx} = 18 - 2x = 0 \Rightarrow x = 9

Step 4: Calculate the maximum area footprint. Substitute

x = 9

into the bounding equation to find

y

:

y = 18 - 9 = 9

\text{Max Area} = 9 \times 9 = 81

Final Answer: 81

Q10:ipmat indore 2024QA › ModulusMediumSA · TITA

The number of pairs

(x, y)

of integers satisfying the inequality

|x - 5| + |y - 5| \leq 6

is:

Enter your answer to attempt

The Setup: We must find the number of integer coordinate pairs

(x,y)

that satisfy the absolute value inequality

|x-5|+|y-5| \le 6

. Step 1: Translate the bounded region to center it at the origin. Let

u = x - 5

and

v = y - 5

. Since

x

and

y

are elements of

\mathbb{Z}

,

u

and

v

must also be elements of

\mathbb{Z}

. Substitute into the inequality:

|u| + |v| \le 6

This bounded region forms a solid square rotated

45^{\circ}

on the Cartesian plane. Step 2: Apply the lattice point summation formula for Manhattan boundaries. The exact number of integer coordinate pairs

(u, v)

satisfying

|u| + |v| \le k

for an integer

k \ge 0

is governed by the discrete sequence formula

2k(k+1) + 1

. Step 3: Evaluate the formula for the target boundary distance

k = 6

.

\text{Total Pairs} = 2(6)(6+1) + 1

\text{Total Pairs} = 2(6)(7) + 1 = 84 + 1 = 85

Because the translation mapping

(x,y) \rightarrow (u,v)

is a bijective

1:1

map, the count remains identical for the original uncentered inequality. Final Answer: 85

Q11:ipmat indore 2024QA › Mean, Median & ModeEasySA · TITA

The following table shows the number of employees and their median age in eight companies located in a district.

Company	Number of employees	Median age
A	32	24
B	28	30
C	43	39
D	39	45
E	35	49
F	29	54
G	23	59
H	16	63

It is known that the age of all employees are integers. It is known that the age of every employee in A is strictly less than the age of every employee in B, the age of every employee in B is strictly less than the age of every employee in C, ..., the age of every employee in G is strictly less than the age of every employee in H. In company F, the lowest possible sum of the ages of all employees is:

Enter your answer to attempt

The Setup: This is a pure min-max puzzle. We need to minimize the total sum of an array while anchored by a fixed median and bounded by a strict lower limit from the previous dataset (Company E). We are basically min-maxing a character build for the absolute lowest possible stats to clear this challenge. Step 1: Analyze Company F's required median. Company F has

29

employees. The median index is:

\frac{29 + 1}{2} = 15

So, the

15

th employee's age must be exactly

54

(from the table). To minimize the total sum, all employees from index

15

to

29

should be exactly

54

years old. Going any higher is an automatic L for our minimum sum objective. Step 2: Find the absolute minimum age for the first

14

employees in F. Because of the strict inequality rule, the youngest person in F must be strictly older than the oldest person in E.

f_1 > e_{35}

We need to shrink E's oldest age as much as possible to give F a lower floor. Company E has

35

employees with a median age of

49

. The median is the

18

th employee. We can set E's entire upper half to exactly

49

:

e_{18} = e_{19} = \dots = e_{35} = 49

So, the maximum age in E can be successfully nerfed down to

49

. Step 3: Set the lower half of F's ages. Since

f_1 > e_{35}

, and

e_{35} = 49

, the lowest possible valid integer age for any employee in Company F is

50

. We generously assign this bare minimum age to all employees below F's median:

f_1 = f_2 = \dots = f_{14} = 50

Step 4: Calculate the final minimized sum for Company F. We have

14

employees at age

50

, and

15

employees (the median and everyone above) at age

54

.

\text{Sum} = (14 \times 50) + (15 \times 54)

\text{Sum} = 700 + 810 = 1510

Final Answer: 1510

Q12:ipmat indore 2024QA › Set TheoryEasySA · TITA

In a group of 150 students, 52 like tea, 48 like juice and 62 like coffee. If each student in the group likes at least one among tea, juice and coffee, then the maximum number of students that like more than one drink is:

Enter your answer to attempt

The Setup: Out of

150

students,

52

like tea,

48

like juice, and

62

like coffee. Since everyone likes at least one drink, we must maximize the number of students who like more than one drink. Step 1: Define the cardinalities of the sets. Let the subsets of students be

T

(Tea),

J

(Juice), and

C

(Coffee).

|T| = 52

,

|J| = 48

, and

|C| = 62

. The sum of independent choices is

52 + 48 + 62 = 162

. Because every student likes at least one drink, the total union encapsulates the whole group:

|T \cup J \cup C| = 150

. Step 2: Construct the overlapping set union equation. Let

x

denote the exact count of students liking precisely two drinks. Let

y

denote the exact count of students liking precisely three drinks. The Inclusion-Exclusion Principle formula for exact counts states:

|T \cup J \cup C| = (|T| + |J| + |C|) - x - 2y

150 = 162 - x - 2y \Rightarrow x + 2y = 12

Step 3: Formulate and execute the optimization constraint. We must maximize the parameter of students liking *more than one* drink, which equates mathematically to maximizing the sum

(x + y)

. Rewrite

(x + y)

by isolating

x

in our established equation (

x = 12 - 2y

):

x + y = (12 - 2y) + y = 12 - y

To maximize the function

(12 - y)

, we must apply the minimum valid boundary for

y

. Since cardinalities must be non-negative integers, the minimum for

y

is

0

. Step 4: Evaluate the maximum value. If

y = 0

, then

x = 12

.

\text{Max}(x + y) = 12 + 0 = 12

Final Answer: 12

Q13:ipmat indore 2024QA › Profit & LossEasySA · TITA

The price of a chocolate is increased by x% and then reduced by x%. The new price is 96.76% of the original price. Then x is:

Enter your answer to attempt

The Setup: A chocolate's price undergoes a sequential

x\%

increase followed by an

x\%

decrease. The resulting price is

96.76\%

of the original price, and we must determine the value of

x

. Step 1: Construct the sequential price multiplier function. Let

P

represent the baseline price. An

x\%

increase applies a multiplier of

(1 + \frac{x}{100})

. An

x\%

decrease applies a subsequent multiplier of

(1 - \frac{x}{100})

.

\text{New Price} = P \times \left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{100}\right)

Step 2: Simplify via the difference of squares identity.

\text{New Price} = P \left(1 - \frac{x^2}{10000}\right)

Step 3: Equate to the provided net proportional change. The final state is

96.76\%

(or

0.9676

) of

P

.

P \left(1 - \frac{x^2}{10000}\right) = 0.9676 P

Divide out

P

from both sides since initial price is arbitrary:

1 - \frac{x^2}{10000} = 0.9676

Step 4: Solve for the absolute rate

x

.

\frac{x^2}{10000} = 1 - 0.9676 = 0.0324

x^2 = 324

Because a percentage rate scaling magnitude must be positive, take the principal square root:

x = 18

Final Answer: 18

Q14:ipmat indore 2024QA › FunctionsMediumSA · TITA

Let

f

and

g

be two functions defined by

f(x) = |x + |x||

and

g(x) = \frac{1}{x}

for

x \neq 0

. If

f(a) + g(f(a)) = \frac{13}{6}

for some real

a

, then the maximum possible value of

f(g(a))

is:

Enter your answer to attempt

The Setup: We are given the functions

f(x)=|x+|x||

and

g(x)=1/x

for non-zero

x

. Based on the constraint

f(a)+g(f(a))=13/6

, we must find the maximum possible value of

f(g(a))

. Step 1: Analyze the piecewise domains of

f(x) = |x + |x||

. If

x \le 0

, then

|x| = -x \Rightarrow f(x) = |x - x| = 0

. If

x > 0

, then

|x| = x \Rightarrow f(x) = |x + x| = 2x

. Step 2: Ascertain the domain of

a

using the given composite equation. The given condition is

f(a) + g(f(a)) = \frac{13}{6}

. The function

g(x) = \frac{1}{x}

is fundamentally undefined at

0

. Consequently,

f(a)

cannot equal

0

. By our piecewise analysis, if

f(a) \neq 0

, then

a

must be strictly positive (

a > 0

), locking

f(a) = 2a

. Step 3: Formulate and solve the rational equation for

a

. Substitute

f(a) = 2a

into the equation:

2a + g(2a) = \frac{13}{6}

2a + \frac{1}{2a} = \frac{13}{6}

Let

u = 2a

to clarify the quadratic structure:

u + \frac{1}{u} = \frac{13}{6}

Multiply entirely by

6u

:

6u^2 + 6 = 13u \Rightarrow 6u^2 - 13u + 6 = 0

Factor the resulting quadratic:

(2u - 3)(3u - 2) = 0

Thus,

u = \frac{3}{2}

or

u = \frac{2}{3}

. Because

u = 2a

, we trace back to two valid positive candidates for

a

:

a = \frac{3}{4} \quad \text{or} \quad a = \frac{1}{3}

Step 4: Evaluate the maximization query for

f(g(a))

. Since

a > 0

,

g(a) = \frac{1}{a} > 0

. Any positive input to

f

triggers the

2x

piecewise condition.

f(g(a)) = f\left(\frac{1}{a}\right) = 2\left(\frac{1}{a}\right) = \frac{2}{a}

Evaluate against both derived candidates of

a

: If

a = \frac{3}{4} \Rightarrow \frac{2}{3/4} = \frac{8}{3} \approx 2.67

If

a = \frac{1}{3} \Rightarrow \frac{2}{1/3} = 6

The strict mathematical maximum is

6

. Final Answer: 6

Q15:ipmat indore 2024QA › Progression & SeriesHardMCQ · MCQ

The terms of a geometric progression are real and positive. If the

p

-th term of the progression is

q

and the

q

-th term is

p

, then the logarithm of the first term is

A $\dfrac{(1-q)\log(p)-(1-p)\log(q)}{p-q}$
B $\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}$
C $\dfrac{(1-q)\log(p)+(1-p)\log(q)}{p-q}$
D $\dfrac{(1-q)\log(q)+(1-p)\log(p)}{p-q}$

Pick an option to attempt

The Setup: The terms of a geometric progression are real and positive, with the

p

-th term equal to

q

and the

q

-th term equal to

p

. We need to find the logarithm of the first term. Step 1: Define the sequence algebraically. Let the first term be

a

and the common ratio be

r

. The general term formula gives:

T_p = a r^{p-1} = q

T_q = a r^{q-1} = p

Step 2: Apply logarithms to both equations. Taking the log of both sides yields a system of linear equations:

\log(a) + (p-1)\log(r) = \log(q)

\log(a) + (q-1)\log(r) = \log(p)

Step 3: Eliminate

\log(r)

to solve for

\log(a)

. Subtract the second equation from the first:

(p-q)\log(r) = \log(q) - \log(p) \implies \log(r) = \frac{\log(q) - \log(p)}{p-q}

Substitute

\log(r)

back into the first equation:

\log(a) = \log(q) - (p-1)\left(\frac{\log(q) - \log(p)}{p-q}\right)

Step 4: Simplify the expression. Find a common denominator:

\log(a) = \frac{(p-q)\log(q) - (p-1)(\log(q) - \log(p))}{p-q}

\log(a) = \frac{p\log(q) - q\log(q) - p\log(q) + \log(q) + p\log(p) - \log(p)}{p-q}

\log(a) = \frac{(1-q)\log(q) - (1-p)\log(p)}{p-q}

Final Answer:

\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}

Q16:ipmat indore 2024QA › CirclesEasyMCQ · MCQ

If the shortest distance of a given point to a given circle is

4 \, \text{cm}

and the longest distance is

9 \, \text{cm}

, then the radius of the circle is

A $1.5 \, \text{cm}$ or $13 \, \text{cm}$
B $2.5 \, \text{cm}$
C $3.5 \, \text{cm}$
D $2.5 \, \text{cm}$ or $6.5 \, \text{cm}$

Pick an option to attempt

The Setup: The shortest distance from a given point to a given circle is

4\text{ cm}

and the longest distance is

9\text{ cm}

. We need to find the radius of the circle. Step 1: Establish the distance relations based on the point's location. Let the circle have center

O

and radius

r

. Let the point be

P

. The shortest and longest distances from a point to a circle lie along the straight line passing through the point and the center of the circle. Step 2: Evaluate Case 1 (Point

P

is outside the circle). Shortest distance:

OP - r = 4

Longest distance:

OP + r = 9

Subtract the two equations to isolate

r

:

(OP + r) - (OP - r) = 9 - 4

2r = 5 \implies r = 2.5\text{ cm}

Step 3: Evaluate Case 2 (Point

P

is inside the circle). Shortest distance:

r - OP = 4

Longest distance:

r + OP = 9

Add the two equations to isolate

r

:

(r - OP) + (r + OP) = 4 + 9

2r = 13 \implies r = 6.5\text{ cm}

Final Answer:

2.5\text{ cm}

or

6.5\text{ cm}

Q17:ipmat indore 2024QA › ModulusEasyMCQ · MCQ

If

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

and

ax - 3ay = 1

, then the value of

a

is

A $\frac{1}{5}$
B $\frac{1}{2}$
C $\frac{1}{7}$
D2

Pick an option to attempt

The Setup: We are given the equations

|x+1|+(y+2)^{2}=0

and

ax-3ay=1

. We must determine the value of

a

. Step 1: Analyze the non-negative components of the first equation. The absolute value function

|x+1|

and the squared term

(y+2)^2

must both be strictly greater than or equal to zero for all real numbers. For their sum to exactly equal zero, each individual term must independently be zero:

|x+1| = 0 \implies x = -1

(y+2)^2 = 0 \implies y = -2

Step 2: Substitute the coordinates into the second linear equation. Substitute

x = -1

and

y = -2

into

ax - 3ay = 1

:

a(-1) - 3a(-2) = 1

-a + 6a = 1

5a = 1 \implies a = \frac{1}{5}

Final Answer:

\frac{1}{5}

Q18:ipmat indore 2024QA › ModulusEasyMCQ · MCQ

The number of real solutions of the equation

x^2 - 10|x| - 56 = 0

is

A1
B4
C3
D2

Pick an option to attempt

The Setup: We need to find the number of real solutions for the absolute value quadratic equation

x^{2}-10|x|-56=0

. Step 1: Use substitution to simplify the quadratic. Recognize that for any real number,

x^2 = |x|^2

. Rewrite the equation in terms of

|x|

:

|x|^2 - 10|x| - 56 = 0

Step 2: Factor the quadratic equation. We look for two numbers that multiply to

-56

and add to

-10

. These are

-14

and

+4

:

(|x| - 14)(|x| + 4) = 0

Step 3: Solve for

|x|

and determine real values for

x

. This yields two potential cases for

|x|

: Case 1:

|x| - 14 = 0 \implies |x| = 14 \implies x = 14 \text{ or } x = -14

Case 2:

|x| + 4 = 0 \implies |x| = -4

Because the absolute value of a real number cannot be negative, Case 2 produces no real solutions. This leaves exactly

2

valid real solutions from Case 1. Final Answer: 2

Q19:ipmat indore 2024QA › IndicesEasyMCQ · MCQ

The greatest number among

2^{300}

,

3^{200}

,

4^{100}

,

2^{100} + 3^{100}

is

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

Pick an option to attempt

The Setup: We need to identify the greatest number among the expressions

2^{300}

,

3^{200}

,

4^{100}

, and

2^{100}+3^{100}

. Step 1: Normalize the exponents to a common power for direct comparison. We can rewrite each single-term expression by factoring out a power of

100

in the exponent:

2^{300} = (2^3)^{100} = 8^{100}

3^{200} = (3^2)^{100} = 9^{100}

4^{100} = 4^{100}

Step 2: Evaluate the additive term against the largest single base. Compare the addition term

2^{100} + 3^{100}

to the largest normalized term

9^{100}

: Since

2^{100} < 3^{100}

, their sum satisfies

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

. Clearly,

2 \times 3^{100}

is vastly smaller than

9^{100}

(which is

3^{200}

). Step 3: Conclude the greatest term. Among the expressions

8^{100}

,

9^{100}

, and

4^{100}

, the term with the largest base is

9^{100}

, which corresponds back to

3^{200}

. Final Answer:

3^{200}

Q20:ipmat indore 2024QA › Progression & SeriesMediumMCQ · MCQ

The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of

n

for which the sum of its first

n

terms is closest to 100, is

A4
B5
C7
D6

Pick an option to attempt

The Setup: An infinite geometric progression has a sum of

80

, and the sum of its first two terms is

35

. We need to find the number of terms

n

for which the partial sum is closest to

100

. Step 1: Establish equations for the GP. Let the first term be

a

and the common ratio be

r

.

S_\infty = \frac{a}{1-r} = 80 \implies a = 80(1-r)

S_2 = a + ar = a(1+r) = 35

Step 2: Solve for the common ratio

r

and first term

a

. Substitute the expression for

a

into the second equation:

80(1-r)(1+r) = 35

80(1-r^2) = 35 \implies 1-r^2 = \frac{35}{80} = \frac{7}{16}

r^2 = \frac{9}{16} \implies r = \pm\frac{3}{4}

If

r = \frac{3}{4}

, then

a = 80(1 - 3/4) = 20

. The partial sum for positive terms converging to

80

will never exceed

80

or approach

100

. If

r = -\frac{3}{4}

, then

a = 80(1 + 3/4) = 140

. This alternating series can exceed

80

and approach

100

. Step 3: Formulate the partial sum

S_n

for

r = -3/4

.

S_n = \frac{140(1 - (-3/4)^n)}{1 - (-3/4)} = \frac{140}{7/4}(1 - (-3/4)^n) = 80(1 - (-3/4)^n)

Step 4: Test values of

n

to find the sum closest to

100

. For the sum to exceed

80

,

n

must be odd (making

(-3/4)^n

negative). Test odd integers: If

n=3

:

S_3 = 80(1 - (-27/64)) \approx 113.75

(Distance to 100 =

13.75

) If

n=5

:

S_5 = 80(1 - (-243/1024)) \approx 98.98

(Distance to 100 =

1.02

) If

n=7

:

S_7 = 80(1 - (-2187/16384)) \approx 90.68

(Distance to 100 =

9.32

) The value

S_5

yields the closest proximity to

100

. Final Answer: 5

Q21:ipmat indore 2024QA › Permutation & CombinationHardMCQ · MCQ

Let

n

be the number of ways in which 20 identical balloons can be distributed among 5 girls and 3 boys such that everyone gets at least one balloon and no girl gets fewer balloons than a boy does. Then

A $9000 \leq n < 10000$
B $8000 \leq n < 9000$
C $7000 \leq n < 8000$
D $6000 \leq n < 7000$

Pick an option to attempt

This question was cancelled by the exam board. No option was accepted as correct and all candidates were awarded full marks. The Setup: There are

20

identical balloons to be distributed among

5

girls and

3

boys such that everyone receives at least one balloon. A condition specifies that no girl receives fewer balloons than any boy does. Step 1: Translate the logical constraints mathematically. Let

G_i

be the balloons received by girl

i

and

B_j

be the balloons received by boy

j

. The condition implies

\min(G_i) \ge \max(B_j)

. Let

M = \max(B_j)

. Since

B_j \ge 1

,

M \ge 1

. The boys must consume at least

M + 2

balloons (one gets

M

, the others at least

1

). The girls must consume at least

5M

balloons. Total bounded constraint:

5M + M + 2 \le 20 \implies 6M \le 18 \implies M \le 3

. Step 2: Analyze valid configurations partitioned by

M

. **Case 1:

M = 1

.** All

3

boys get

1

balloon (

1

way). The

5

girls distribute the remaining

17

balloons (

G_i \ge 1

). Using stars and bars:

\binom{17-1}{5-1} = \binom{16}{4} = 1820

. **Case 2:

M = 2

.** Boy combinations with a max of

2

:

(2,2,2)

[1 way],

(2,2,1)

[3 ways],

(2,1,1)

[3 ways]. Sum(B)=6, remaining 14 to girls (

G_i \ge 2

). Ways

= 1 \times 70 = 70

. Sum(B)=5, remaining 15 to girls (

G_i \ge 2

). Ways

= 3 \times 126 = 378

. Sum(B)=4, remaining 16 to girls (

G_i \ge 2

). Ways

= 3 \times 210 = 630

. Total for Case 2

= 70 + 378 + 630 = 1078

. **Case 3:

M = 3

.** Boys must sum to

\le 5

since girls need at least

5 \times 3 = 15

. The only boy configuration is

(3,1,1)

[3 ways]. Sum(B)=5, remaining 15 to girls (

G_i \ge 3

). Ways

= 3 \times 1 = 3

. Step 3: Sum the cases. Total exact valid distributions

= 1820 + 1078 + 3 = 2901

. Note: All four options give ranges from

6000

upward, so none of them contains

2901

. This question was cancelled by the exam board and all candidates were awarded full marks. The value

2901

is what the stated constraints actually yield. Final Answer: 2901

Q22:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

Let

a = \dfrac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_{7} 25 (\log_7 8 - \log_7 4)}

. Then the value of

5^a

is

A8
B $\frac{5}{2}$
C5
D $\frac{7}{2}$

Pick an option to attempt

The Setup: We are provided a complex logarithmic expression for a constant

a

, and we need to evaluate

5^a

. Step 1: Simplify the numerator and denominator using logarithmic properties. Expression:

a = \frac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}

Using the quotient rule

\log_x(y) - \log_x(z) = \log_x(y/z)

: Numerator:

(\log_7 4)(\log_7 5 - \log_7 2) = (2\log_7 2)(\log_7 5 - \log_7 2)

Denominator:

(\log_7 25)(\log_7 8 - \log_7 4) = (2\log_7 5)(\log_7 (8/4)) = (2\log_7 5)(\log_7 2)

Step 2: Expand terms to locate cancellations.

a = \frac{2\log_7 2 (\log_7 5 - \log_7 2)}{2\log_7 5 \log_7 2}

Cancel the common factor

2\log_7 2

:

a = \frac{\log_7 5 - \log_7 2}{\log_7 5}

Step 3: Separate the fraction and change bases.

a = 1 - \frac{\log_7 2}{\log_7 5}

Using the change of base formula:

a = 1 - \log_5 2

Step 4: Evaluate

5^a

.

5^a = 5^{1 - \log_5 2} = \frac{5^1}{5^{\log_5 2}} = \frac{5}{2}

Final Answer:

\frac{5}{2}

Q23:ipmat indore 2024QA › InequalitiesEasyMCQ · MCQ

The smallest possible number of students in a class if the girls in the class are less than 50% but more than 48% is

A27
B100
C200
D25

Pick an option to attempt

The Setup: We need to find the smallest possible total number of students in a class, given the percentage of girls is strictly between

48\%

and

50\%

. Step 1: Formulate the inequality constraint. Let

G

be the integer number of girls and

N

be the total integer number of students.

48\% < \frac{G}{N} < 50\%

\frac{12}{25} < \frac{G}{N} < \frac{1}{2}

Step 2: Test values for

N

starting from the lower bound constraints. For the fraction to sit strictly between

0.48

and

0.50

,

N

must be minimally larger than

25

. If

N = 25

:

\frac{12}{25} = 0.48

(Fails the strict greater-than inequality). If

N = 26

: The maximum integer

G

strictly less than half of

26

is

12

.

\frac{12}{26} \approx 0.461

(Fails, as it is less than

0.48

). If

N = 27

: The maximum integer

G

strictly less than half of

27

is

13

.

\frac{13}{27} \approx 0.48148

(Valid). Step 3: Conclude the minimum total. Since

0.48 < 0.48148 < 0.50

, the minimum total class size

N

that supports a valid integer amount of girls is

27

. Final Answer: 27

Q24:ipmat indore 2024QA › TrianglesHardMCQ · MCQ

The side AB of a triangle ABC is c. The median BD is of length k. If

\angle BDA = \theta

and

\theta < 90^\circ

, then the area of triangle ABC is

A $\dfrac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$
B $\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$
C $\dfrac{k^2 \cos 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$
D $\dfrac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$

Pick an option to attempt

The Setup: In triangle

ABC

, the side

AB

is

c

, the median

BD

has length

k

, and the angle

\angle BDA = \theta

where

\theta < 90^\circ

. We need to find the area of the entire triangle

ABC

. Step 1: Establish the area formula using the median. A median divides a triangle into two smaller triangles of identical area.

\text{Area}(ABC) = 2 \times \text{Area}(ABD)

Using the base

AD

and the altitude from

B

to

AD

(which is

k\sin\theta

):

\text{Area}(ABD) = \frac{1}{2} \times AD \times (k\sin\theta)

\text{Area}(ABC) = 2 \left( \frac{1}{2} \times AD \times k\sin\theta \right) = AD \cdot k\sin\theta

Step 2: Use the Cosine Rule to express

AD

. In

\triangle ABD

, apply the Law of Cosines:

c^2 = AD^2 + k^2 - 2(AD)(k)\cos\theta

Rearrange into a quadratic equation in terms of

AD

:

AD^2 - (2k\cos\theta)AD + (k^2 - c^2) = 0

Solve for

AD

via the quadratic formula (taking the positive valid geometric root):

AD = \frac{2k\cos\theta + \sqrt{4k^2\cos^2\theta - 4(k^2 - c^2)}}{2}

AD = k\cos\theta + \sqrt{c^2 - k^2(1 - \cos^2\theta)} = k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta}

Step 3: Substitute

AD

back into the Area equation.

\text{Area}(ABC) = \left( k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta} \right) k\sin\theta

\text{Area}(ABC) = k^2\sin\theta\cos\theta + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}

Using the double-angle identity

2\sin\theta\cos\theta = \sin(2\theta)

:

\text{Area}(ABC) = \frac{k^2\sin 2\theta}{2} + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}

Final Answer:

\dfrac{k^{2}\sin 2\theta}{2}+k\sin\theta\sqrt{c^{2}-k^{2}\sin^{2}\theta}

Q25:ipmat indore 2024QA › TrianglesMediumMCQ · MCQ

Let

\triangle ABC

be a triangle with

AB = AC

and

D

be a point on

BC

such that

\angle BAD = 30^\circ

. If

E

is a point on

AC

such that

AD = AE

, then

\angle CDE

equals

A $60^\circ$
B $30^\circ$
C $10^\circ$
D $15^\circ$

Pick an option to attempt

The Setup: In a triangle

ABC

with

AB=AC

,

D

is a point on

BC

creating

\angle BAD=30^\circ

, and

E

is a point on

AC

creating

AD=AE

. We need to determine the angle

\angle CDE

. Step 1: Assign variables to the base angles of the primary triangle. Let

\angle B = \angle C = \alpha

(since

AB = AC

). The vertex angle of the large triangle is

\angle BAC = 180^\circ - 2\alpha

. The remaining upper angle inside is

\angle DAC = \angle BAC - \angle BAD = (180^\circ - 2\alpha) - 30^\circ = 150^\circ - 2\alpha

. Step 2: Determine the base angles of the internal isosceles triangle. In

\triangle ADE

, since

AD = AE

, the base angles are equal:

\angle ADE = \angle AED = \frac{180^\circ - \angle DAC}{2}

\angle ADE = \frac{180^\circ - (150^\circ - 2\alpha)}{2} = \frac{30^\circ + 2\alpha}{2} = 15^\circ + \alpha

Step 3: Use the exterior angle theorem to map

\angle CDE

. Look at

\triangle ABD

. The exterior angle at

D

is

\angle ADC

, equaling the sum of the remote interior angles:

\angle ADC = \angle B + \angle BAD = \alpha + 30^\circ

Geometrically, angle

\angle ADC

is comprised of two adjacent components:

\angle ADC = \angle ADE + \angle CDE

Substitute the known expressions:

\alpha + 30^\circ = (15^\circ + \alpha) + \angle CDE

\angle CDE = 30^\circ - 15^\circ = 15^\circ

Final Answer:

15^{\circ}

Q26:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

If

\log_4 x = a

and

\log_{25} x = b

, then

\log_x 10

is

A $\dfrac{a + b}{2}$
B $\dfrac{a - b}{2ab}$
C $\dfrac{a + b}{2ab}$
D $\dfrac{a + b}{2(a - b)}$

Pick an option to attempt

The Setup: We are provided the logarithmic equations

\log_{4}x=a

and

\log_{25}x=b

. We need to determine the value of

\log_{x} 10

expressed in terms of variables

a

and

b

. Step 1: Change the base of the given logarithms to

x

. Using the base-change inversion rule

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

:

\log_4 x = a \implies \log_x 4 = \frac{1}{a}

\log_{25} x = b \implies \log_x 25 = \frac{1}{b}

Step 2: Simplify the bases to prime numbers. Using the logarithm power rule

\log_x(y^c) = c \log_x(y)

:

\log_x(2^2) = \frac{1}{a} \implies 2 \log_x 2 = \frac{1}{a} \implies \log_x 2 = \frac{1}{2a}

\log_x(5^2) = \frac{1}{b} \implies 2 \log_x 5 = \frac{1}{b} \implies \log_x 5 = \frac{1}{2b}

Step 3: Calculate

\log_x 10

. Since

10 = 2 \times 5

, we can separate the target logarithm using the product rule:

\log_x 10 = \log_x(2 \times 5) = \log_x 2 + \log_x 5

Step 4: Substitute the derived fractions and combine.

\log_x 10 = \frac{1}{2a} + \frac{1}{2b}

Find a common denominator to combine the rational expressions:

\log_x 10 = \frac{b + a}{2ab} = \frac{a+b}{2ab}

Final Answer:

\dfrac{a+b}{2ab}

Q27:ipmat indore 2024QA › Permutation & CombinationHardMCQ · MCQ

If 5 boys and 3 girls sit randomly around a circular table, the probability that there will be at least one boy sitting between any two girls is

A $\frac{1}{7}$
B $\frac{2}{7}$
C $\frac{3}{5}$
D $\frac{1}{4}$

Pick an option to attempt

The Setup: There are

5

boys and

3

girls sitting randomly around a circular table. We must find the probability that there is at least one boy sitting between any two girls. Step 1: Calculate the total number of unrestricted circular arrangements. For

n

distinct individuals sitting in a circle, the number of unique arrangements is

(n - 1)!

. Total people

= 5 + 3 = 8

.

\text{Total Arrangements} = (8 - 1)! = 7! = 5040

Step 2: Calculate the number of restricted arrangements (girls separated). First, arrange the

5

boys in a circle.

\text{Ways to seat boys} = (5 - 1)! = 4! = 24

Seating the

5

boys creates exactly

5

gaps between them. To ensure no two girls sit adjacent to each other, we place the

3

girls into these distinct gaps. Choose

3

gaps out of

5

, and arrange the girls:

\text{Ways to seat girls} = \binom{5}{3} \times 3! = 10 \times 6 = 60

Total valid arrangements

= 24 \times 60 = 1440

. Step 3: Calculate the probability.

\text{Probability} = \frac{1440}{5040} = \frac{144}{504} = \frac{12}{42} = \frac{2}{7}

Final Answer:

\frac{2}{7}

Q28:ipmat indore 2024QA › Linear EquationsHardMCQ · MCQ

A fruit seller had a certain number of apples, bananas, and oranges at the start of the day. The number of bananas was 10 more than the number of apples, and the total number of bananas and apples was a multiple of 11. She was able to sell 70% of the apples, 60% of bananas, and 50% of oranges during the day. If she was able to sell 55% of the fruits she had at the start of the day, then the minimum number of oranges she had at the start of the day was

A190
B210
C180
D220

Pick an option to attempt

The Setup: A fruit seller has apples, bananas, and oranges, where the number of bananas is

10

more than the number of apples, and the sum of apples and bananas is a multiple of

11

. She sells

70\%

of apples,

60\%

of bananas, and

50\%

of oranges, representing

55\%

of her total starting inventory. We need to find the minimum initial number of oranges. Step 1: Construct the total inventory linear equation. The total fruit sold is mapped to the percentages:

0.7A + 0.6B + 0.5O = 0.55(A + B + O)

Expand and group similar terms:

0.15A + 0.05B = 0.05O

Divide entirely by

0.05

:

3A + B = O

Step 2: Substitute the banana relation to express

O

strictly in terms of

A

. We are given that

B = A + 10

.

O = 3A + (A + 10) = 4A + 10

To minimize

O

, we must find the absolute minimum integer value for

A

. Step 3: Apply integer constraints. For the seller to sell

70\%

of apples and

60\%

of bananas as integer whole fruits,

A

must be a multiple of

10

, and

B

must be a multiple of

5

. Let

A = 10m

for some positive integer

m

. Then

B = 10m + 10

. We are given that

(A + B)

must be a multiple of

11

.

A + B = 10m + (10m + 10) = 20m + 10

Set this equal to

11k

:

20m + 10 = 11k \implies 9m + 10 = 11(k - m)

Let

p = k - m

:

9m = 11p - 10

Test positive integers for

m

sequentially: If

m = 1 \implies 9 = 11p - 10 \implies 11p = 19

(No integer

p

) If

m = 5 \implies 45 = 11p - 10 \implies 11p = 55 \implies p = 5

(Valid) Step 4: Calculate the final values. Using the minimum multiplier

m = 5

:

A = 10(5) = 50

O = 4(50) + 10 = 210

Final Answer: 210

Q29:ipmat indore 2024QA › Time, Speed & DistanceMediumMCQ · MCQ

A boat goes 96 km upstream in 8 hours and covers the same distance moving downstream in 6 hours. On the next day it starts from point A, goes downstream for 1 hour, then upstream for 1 hour, and repeats this for four more times, that is, 5 upstream and 5 downstream journeys. Then the boat would be

A22.5 km downstream of A
B20 km downstream of A
C15 km downstream of A
D12.5 km downstream of A

Pick an option to attempt

The Setup: A boat travels

96\text{ km}

upstream in

8

hours and the same distance downstream in

6

hours. The next day, it alternates

1

hour downstream and

1

hour upstream from point

A

for

5

full cycles. We must find its final location. Step 1: Calculate the upstream and downstream speeds.

\text{Upstream Speed } (U) = \frac{96\text{ km}}{8\text{ hours}} = 12\text{ km/h}

\text{Downstream Speed } (D) = \frac{96\text{ km}}{6\text{ hours}} = 16\text{ km/h}

Step 2: Calculate the net displacement per cycle. Each cycle consists of

1

hour moving downstream followed immediately by

1

hour moving upstream. Distance traveled downstream in

1

hr

= 16 \times 1 = 16\text{ km}

. Distance traveled upstream in

1

hr

= 12 \times 1 = 12\text{ km}

. Net displacement per cycle

= 16 - 12 = 4\text{ km}

(in the downstream direction). Step 3: Calculate the total displacement over all cycles. The boat completes

5

identical cycles (5 upstream and 5 downstream journeys in total).

\text{Total Displacement} = 5 \times 4\text{ km} = 20\text{ km}

The boat finishes exactly

20\text{ km}

downstream of its starting point

A

. Final Answer: 20 km downstream of A

Q30:ipmat indore 2024QA › Permutation & CombinationMediumMCQ · MCQ

The number of solutions of the equation

x_1 + x_2 + x_3 + x_4 = 50

, where

x_1, x_2, x_3, x_4

are integers with

x_1 \geq 1, x_2 \geq 2, x_3 \geq 0, x_4 \geq 0

is

A20200
B19200
C19600
D18400

Pick an option to attempt

The Setup: We are asked to find the number of integer solutions to a linear equation subject to specific lower-bound constraints. We will use a variable substitution technique to normalize the lower bounds to zero, allowing the application of the stars and bars combinatorial method. Step 1: Define the equation and initial constraints. Equation:

x_1 + x_2 + x_3 + x_4 = 50

Constraints:

x_1 \ge 1, x_2 \ge 2, x_3 \ge 0, x_4 \ge 0

Step 2: Normalize the variables to zero-bounded equivalents. Let

y_1 = x_1 - 1

, which ensures

y_1 \ge 0

. Let

y_2 = x_2 - 2

, which ensures

y_2 \ge 0

. Let

y_3 = x_3

, keeping

y_3 \ge 0

. Let

y_4 = x_4

, keeping

y_4 \ge 0

. Step 3: Substitute the normalized variables back into the original equation.

(y_1 + 1) + (y_2 + 2) + y_3 + y_4 = 50

y_1 + y_2 + y_3 + y_4 + 3 = 50

y_1 + y_2 + y_3 + y_4 = 47

Step 4: Apply the stars and bars formula. The number of non-negative integer solutions to

y_1 + y_2 + \dots + y_k = n

is given by

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

. Here,

n = 47

and

k = 4

:

\text{Solutions} = \binom{47 + 4 - 1}{4 - 1} = \binom{50}{3}

Step 5: Evaluate the binomial coefficient.

\binom{50}{3} = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} = 50 \times 49 \times 8 = 19600

Final Answer: 19600

Q31:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

The numbers

2^{2024}

and

5^{2024}

are expanded and their digits are written out consecutively on one page. The total number of digits written on the page is

A1987
B2025
C2065
D2000

Pick an option to attempt

The Setup: This problem requires calculating the combined number of digits of two large numbers with a shared exponent. We can determine the number of digits by sandwiching the values between sequential powers of

10

. Step 1: Define the digit counting function. Let

d_1

be the number of digits in

2^{2024}

. Thus,

10^{d_1-1} < 2^{2024} < 10^{d_1}

. Let

d_2

be the number of digits in

5^{2024}

. Thus,

10^{d_2-1} < 5^{2024} < 10^{d_2}

. Step 2: Multiply the two bounding inequalities.

10^{d_1-1} \times 10^{d_2-1} < 2^{2024} \times 5^{2024} < 10^{d_1} \times 10^{d_2}

10^{d_1+d_2-2} < (2 \times 5)^{2024} < 10^{d_1+d_2}

10^{d_1+d_2-2} < 10^{2024} < 10^{d_1+d_2}

Step 3: Evaluate the integer constraints on the exponents. For the exact integer

10^{2024}

to be strictly bounded between these two powers of

10

, the exponent

2024

must equal the lower bound exponent plus one:

2024 = (d_1 + d_2 - 2) + 1

2024 = d_1 + d_2 - 1

d_1 + d_2 = 2025

The total number of consecutive digits written on the page is

2025

. Final Answer: 2025

Q32:ipmat indore 2024QA › CirclesHardMCQ · MCQ

If

\theta

is the angle between the pair of tangents drawn from the point

(0,\frac{7}{2})

to the circle

x^2 + y^2 - 14x + 16y + 88 = 0

, then

\tan \theta

equals

A $\frac{4}{5}$
B $\frac{2}{5}$
C $\frac{3}{4}$
D $\frac{20}{21}$

Pick an option to attempt

The Setup: We are asked to find the tangent of the angle between two tangents drawn from an external point to a circle. We will use the geometric properties of right triangles formed by the tangents, the circle's radius, and the distance to the center. Step 1: Determine the circle's center and radius. The equation is

x^2 + y^2 - 14x + 16y + 88 = 0

. Complete the square for

x

and

y

:

(x^2 - 14x + 49) + (y^2 + 16y + 64) = -88 + 49 + 64

(x - 7)^2 + (y + 8)^2 = 25

Center

C = (7, -8)

and Radius

r = \sqrt{25} = 5

. Step 2: Calculate the distance from the external point to the center. Point

P = (0, 7/2)

.

d = \sqrt{(7 - 0)^2 + (-8 - 7/2)^2}

d = \sqrt{49 + (-23/2)^2} = \sqrt{49 + \frac{529}{4}}

d = \sqrt{\frac{196 + 529}{4}} = \sqrt{\frac{725}{4}} = \frac{5\sqrt{29}}{2}

Step 3: Evaluate the trigonometric ratio for half the angle. Let

\theta

be the full angle between the tangents. The line connecting

P

to

C

bisects this angle. Let the half-angle be

\alpha

.

\sin\alpha = \frac{\text{radius}}{\text{distance}} = \frac{5}{\frac{5\sqrt{29}}{2}} = \frac{2}{\sqrt{29}}

Using the Pythagorean theorem, the adjacent side (tangent length) is

\sqrt{(\sqrt{29})^2 - 2^2} = \sqrt{25} = 5

. Thus,

\tan\alpha = \frac{2}{5}

. Step 4: Calculate

\tan\theta

using the double-angle identity.

\tan\theta = \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha}

\tan\theta = \frac{2(2/5)}{1 - (2/5)^2} = \frac{4/5}{1 - 4/25} = \frac{4/5}{21/25}

\tan\theta = \frac{4}{5} \times \frac{25}{21} = \frac{20}{21}

Final Answer:

\frac{20}{21}

Q33:ipmat indore 2024QA › PolynomialsMediumMCQ · MCQ

The difference between the maximum real root and the minimum real root of the equation

(x^2 - 5)^4 + (x^2 - 7)^4 = 16

is

A $\sqrt{10}$
B $2\sqrt{5}$
C $\sqrt{7}$
D $2\sqrt{7}$

Pick an option to attempt

The Setup: We are given a polynomial equation of degree 8. By substituting a symmetric variable centered between the two binomial constants, we can reduce the degree and find the real roots to determine their range. Step 1: Apply a symmetric variable substitution. The terms are

(x^2-5)

and

(x^2-7)

. The midpoint of

-5

and

-7

is

-6

. Let

y = x^2 - 6

. Substitute into the equation:

(y + 1)^4 + (y - 1)^4 = 16

Step 2: Expand the binomials. Using the binomial theorem:

(y^4 + 4y^3 + 6y^2 + 4y + 1) + (y^4 - 4y^3 + 6y^2 - 4y + 1) = 16

The odd powers cancel out perfectly:

2y^4 + 12y^2 + 2 = 16

2y^4 + 12y^2 - 14 = 0

y^4 + 6y^2 - 7 = 0

Step 3: Factor the resulting quadratic in terms of

y^2

.

(y^2 + 7)(y^2 - 1) = 0

Because

x

must be real,

y = x^2 - 6

must be real, making

y^2

necessarily non-negative. Therefore,

y^2 = -7

has no real solutions. We proceed with

y^2 = 1

.

y = 1 \quad \text{or} \quad y = -1

Step 4: Translate back to

x

and evaluate the roots. If

y = 1 \implies x^2 - 6 = 1 \implies x^2 = 7 \implies x = \pm\sqrt{7}

. If

y = -1 \implies x^2 - 6 = -1 \implies x^2 = 5 \implies x = \pm\sqrt{5}

. The maximum real root is

\sqrt{7}

, and the minimum is

-\sqrt{7}

.

\text{Difference} = \sqrt{7} - (-\sqrt{7}) = 2\sqrt{7}

Final Answer:

2\sqrt{7}

Q34:ipmat indore 2024QA › TrigonometryEasyMCQ · MCQ

The angle of elevation of the top of a pole from a point A on the ground is 30. The angle of elevation changes to 45, after moving 20 meters towards the base of the pole. Then the height of the pole, in meters, is

A $15(\sqrt{5} + 1)$
B $20(\sqrt{3} + 1)$
C $30$
D $10(\sqrt{3} + 1)$

Pick an option to attempt

The Setup: This problem models two angles of elevation to the top of a pole from different horizontal distances. We use right-triangle trigonometry to build a system of linear equations mapping distance to height. Step 1: Establish the initial trigonometric relation. Let

h

be the height of the pole, and

x

be the initial horizontal distance from point

A

to the pole's base.

\tan(30^\circ) = \frac{h}{x} \implies \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h\sqrt{3}

Step 2: Establish the secondary trigonometric relation. After moving

20

meters closer, the new distance is

x - 20

.

\tan(45^\circ) = \frac{h}{x - 20}

Since

\tan(45^\circ) = 1

:

1 = \frac{h}{x - 20} \implies h = x - 20

Step 3: Substitute and solve for

h

. Replace

x

with

h\sqrt{3}

:

h = h\sqrt{3} - 20

20 = h\sqrt{3} - h = h(\sqrt{3} - 1)

h = \frac{20}{\sqrt{3} - 1}

Step 4: Rationalize the denominator. Multiply the numerator and denominator by the conjugate

(\sqrt{3} + 1)

:

h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1}

h = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)

Final Answer:

10(\sqrt{3}+1)

Q35:ipmat indore 2024QA › Permutation & CombinationEasyMCQ · MCQ

The number of values of

x

for which

C \binom {17-x}{3x+1}

is defined as an integer is

A6
B2
C4
D5

Pick an option to attempt

The Setup: We need to find the number of values of

x

for which the binomial coefficient

\binom{17-x}{3x+1}

is a valid integer. This requires testing integer constraints dictated by the mathematical definition of combinatorics. Step 1: State the rules for a valid binomial coefficient

\binom{n}{k}

. For the coefficient to evaluate to a defined integer, the parameters must satisfy: 1. Both

n

and

k

must be non-negative integers. Therefore,

x

must be an integer. 2.

n \ge 0

3.

k \ge 0

4.

n \ge k

Step 2: Apply the rules to the given expressions. Condition 2:

17 - x \ge 0 \implies x \le 17

Condition 3:

3x + 1 \ge 0 \implies x \ge -1/3

. Since

x

is an integer,

x \ge 0

. Condition 4:

17 - x \ge 3x + 1 \implies 16 \ge 4x \implies x \le 4

Step 3: Evaluate the bounded integer set. Combining the inequalities yields

0 \le x \le 4

. The possible integer candidates are

x \in \{0, 1, 2, 3, 4\}

. Let's verify each candidate: If

x = 0 \implies \binom{17}{1}

(Valid) If

x = 1 \implies \binom{16}{4}

(Valid) If

x = 2 \implies \binom{15}{7}

(Valid) If

x = 3 \implies \binom{14}{10}

(Valid) If

x = 4 \implies \binom{13}{13}

(Valid) All

5

integer values of

x

produce a valid evaluation. Final Answer: 5

Q36:ipmat indore 2024QA › TrianglesHardMCQ · MCQ

Let ABC be an equilateral triangle, with each side of length

k

. If a circle is drawn with diameter AB, then the area of the portion of the triangle lying inside the circle is

A $\left(3\sqrt{3} + \pi\right) \frac{k^2}{24}$
B $\left(3\sqrt{3} + \pi\right) \frac{k^2}{6}$
C $\left(3\sqrt{3} - \pi\right) \frac{k^2}{24}$
D $\left(3\sqrt{3} + \pi\right) \frac{k^2}{8}$

Pick an option to attempt

The Setup: An equilateral triangle is partially overlapped by a circle whose diameter is one of its sides. We must calculate the area of the geometric intersection of the two figures. Step 1: Establish the geometry of the intersection. Let the equilateral triangle be

ABC

with side length

k

. The circle has diameter

AB

, meaning its center

O

is the midpoint of

AB

, and its radius is

k/2

. The circle intersects side

AC

at a point

D

and side

BC

at a point

E

. According to Thales's Theorem, any angle inscribed in a semicircle is a right angle. Since

AB

is the diameter,

\angle ADB = 90^\circ

and

\angle AEB = 90^\circ

. Thus,

BD

and

AE

are altitudes of the equilateral triangle

ABC

. In an equilateral triangle, altitudes also act as medians, meaning

D

and

E

exactly bisect sides

AC

and

BC

. Step 2: Partition the target area. The portion of the triangle lying inside the circle is the shape bounded by segments

AD

,

EB

,

AB

, and the circular arc

DE

. We can decompose this bounded region by connecting the center

O

to points

D

and

E

. The area consists of three non-overlapping geometric shapes:

\triangle AOD

,

\triangle BOE

, and the circular sector

DOE

. Step 3: Calculate the component areas. Because

O

is the midpoint of

AB

and

D

is the midpoint of

AC

,

AD = k/2

and

AO = k/2

. Since

D

is on the circle,

OD = k/2

. Thus,

\triangle AOD

is equilateral with side length

k/2

.

\text{Area}(\triangle AOD) = \frac{\sqrt{3}}{4} \left(\frac{k}{2}\right)^2 = \frac{\sqrt{3}}{16} k^2

By identical symmetry,

\triangle BOE

is also equilateral with the same area. Because

\angle AOD = 60^\circ

and

\angle BOE = 60^\circ

, the central angle of the sector

DOE

is

180^\circ - 60^\circ - 60^\circ = 60^\circ

.

\text{Area}(\text{Sector } DOE) = \frac{60^\circ}{360^\circ} \times \pi \left(\frac{k}{2}\right)^2 = \frac{1}{6} \times \frac{\pi k^2}{4} = \frac{\pi}{24} k^2

Step 4: Sum the components to find total area.

\text{Total Area} = 2 \left( \frac{\sqrt{3}}{16} k^2 \right) + \frac{\pi}{24} k^2

\text{Total Area} = \frac{\sqrt{3}}{8} k^2 + \frac{\pi}{24} k^2 = \left( \frac{3\sqrt{3}}{24} + \frac{\pi}{24} \right) k^2 = \frac{3\sqrt{3} + \pi}{24} k^2

Final Answer:

\left(3\sqrt{3}+\pi\right)\dfrac{k^{2}}{24}

Q37:ipmat indore 2024QA › Simple & Compound InterestHardMCQ · MCQ

Sagarika divides her savings of

10000

rupees to invest across two schemes A and B. Scheme A offers an interest rate of

10\%

per annum, compounded half-yearly, while scheme B offers a simple interest rate of

12\%

per annum. If at the end of first year, the value of her investment in scheme B exceeds the value of her investment in scheme A by

2310

rupees, then the total interest, in rupees, earned by Sagarika during the first year of investment is

A1111
B1000
C1100
D1130

Pick an option to attempt

The Setup: A principal sum of

10000

rupees is split into two investment schemes with different interest protocols (Compound vs Simple). Setting up a linear equation evaluating their final amounts will isolate the initial split quantities. Step 1: Define the variables and growth formulas for 1 year. Let the investment in Scheme A be

x

. Let the investment in Scheme B be

10000 - x

. Scheme A (

10\%

p.a. compounded half-yearly): The rate per half-year period is

5\% = 0.05

, and there are

2

compounding periods.

\text{Amount}_A = x(1 + 0.05)^2 = x(1.1025) = 1.1025x

Scheme B (

12\%

p.a. simple interest):

\text{Amount}_B = (10000 - x)(1 + 0.12) = 1.12(10000 - x)

Step 2: Construct the equation based on the given constraint. The value of Scheme B exceeds Scheme A by

2310

at the end of the year.

\text{Amount}_B - \text{Amount}_A = 2310

1.12(10000 - x) - 1.1025x = 2310

Step 3: Solve for

x

.

11200 - 1.12x - 1.1025x = 2310

11200 - 2310 = 2.2225x

8890 = 2.2225x

Recognize that

2.2225 = \frac{22225}{10000} = \frac{889}{400}

.

x = 8890 \times \frac{400}{889} = 10 \times 400 = 4000

So,

4000

was invested in A, and

6000

was invested in B. Step 4: Calculate the total interest earned. Interest from A

= 1.1025(4000) - 4000 = 410

Interest from B

= 6000 \times 0.12 = 720

\text{Total Interest} = 410 + 720 = 1130

Final Answer: 1130

Q38:ipmat indore 2024QA › Set TheoryEasyMCQ · MCQ

In a survey of 500 people, it was found that 250 owned a 4-wheeler but not a 2-wheeler, 100 owned a 2-wheeler but not a 4-wheeler, and 100 owned neither a 4-wheeler nor a 2-wheeler. Then the number of people who owned both is

A75
B60
C100
D50

Pick an option to attempt

The Setup: This is a classic Set Theory problem solvable by mapping the given disjoint subsets to the universal set total. Step 1: Identify the disjoint groups. The total population sampled (Universal Set) is

500

. Group 1 (Only 4-wheeler)

= 250

Group 2 (Only 2-wheeler)

= 100

Group 3 (Neither)

= 100

Group 4 (Both)

=

Let this be

x

. Step 2: Formulate the union equation. Because these four groups represent mutually exclusive, completely exhaustive subsets of the surveyed population, their sum must equal the total.

\text{Total} = (\text{Only 4W}) + (\text{Only 2W}) + (\text{Both}) + (\text{Neither})

500 = 250 + 100 + x + 100

500 = 450 + x

x = 50

Final Answer: 50

Q39:ipmat indore 2024QA › Linear EquationsEasyMCQ · MCQ

For some non-zero real values of

a, b

and

c

, it is given that

\left|\frac{c}{a}\right|=4,\left|\frac{a}{b}\right|=\frac{1}{3}

and

\frac{b}{c}=-\frac{3}{4}

. If

a c>0

, then

\left(\frac{b+c}{a}\right)

equals

A1
B-1
C7
D-7

Pick an option to attempt

The Setup: We are given absolute value equations mapping ratios of variables, alongside a specific sign constraint

ac > 0

. We must synthesize these constraints to compute a combined fractional expression. Step 1: Establish the magnitudes of the ratios. Given

\left|\frac{c}{a}\right| = 4

, we know

\frac{c}{a} = \pm 4

. Given

\left|\frac{a}{b}\right| = \frac{1}{3}

, we can invert it to find the magnitude of its reciprocal:

\left|\frac{b}{a}\right| = 3

, meaning

\frac{b}{a} = \pm 3

. Step 2: Apply the sign constraints to determine exact ratio values. We are given the condition

ac > 0

. This implies that variables

a

and

c

share the exact same sign (both positive or both negative). Consequently, their ratio must be strictly positive. Thus,

\frac{c}{a} = 4

. Next, we determine

\frac{b}{a}

using the provided relation

\frac{b}{c} = -\frac{3}{4}

. By multiplying

\frac{b}{c}

by

\frac{c}{a}

, we isolate

\frac{b}{a}

:

\frac{b}{a} = \left(\frac{b}{c}\right) \times \left(\frac{c}{a}\right)

\frac{b}{a} = \left(-\frac{3}{4}\right) \times (4) = -3

Step 3: Calculate the target expression. We need to find the value of

\frac{b+c}{a}

. We can separate this fraction into our known ratios:

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

\frac{b+c}{a} = -3 + 4 = 1

Final Answer: 1

IPMAT Indore 2024 — QA

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