IPMAT Indore 2024 MCQ PYQs

Q1: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}

Q2: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}

Q3: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}

Q4: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

Q5: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}

Q6: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

Q7: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

Q8: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}

Q9: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

Q10: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}

Q11: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}

Q12: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}

Q13: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}

Q14: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

Q15: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

Q16: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

Q17: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

Q18: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}

Q19: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}

Q20: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)

Q21: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

Q22: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}

Q23: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

Q24: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

Q25: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

Q26:ipmat indore 2024LRDI › ArrangementsHardMCQ · MCQ

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage

The constituency in which B got lower number of votes compared to A and C is

AS3
BS4
CS2
DS1

Pick an option to attempt

The Setup: This is a logical reasoning Data Interpretation problem based on an election matrix. We must deduce the exact vote distribution for parties A, B, and C across 5 constituencies using their total vote counts, win conditions, and numerical sequence constraints. Step 1: Analyze Party C's winning constraints. Every constituency has exactly

20

voters and a clear winner (no ties for first place). To guarantee a win without tying, a party must secure a minimum of

8

votes (since the remaining

12

could be split

6-6

; a score of

7

allows a

7-7-6

tie). Party C won *only* S2 and S3. Thus,

c_2 \ge 8

and

c_3 \ge 8

. Party C's total votes across all constituencies is

16

. Therefore, C must have obtained exactly

8

votes in S2,

8

votes in S3, and

0

votes in S1, S4, and S5. Step 2: Analyze Party B's winning constraints. Party A won only S1, and C won only S2 and S3. By elimination, Party B must be the winner of S4 and S5. In S4 and S5, C has

0

votes, meaning A and B split the

20

votes entirely. For B to win clearly, B must secure more than half the votes:

b_4 \ge 11

and

b_5 \ge 11

. Step 3: Apply Party B's sequence constraints. B's votes across S1 to S5 are distinct natural numbers in increasing order:

b_1 < b_2 < b_3 < b_4 < b_5

. Since

b_4 \ge 11

, and

b_5

must be strictly greater than

b_4

,

b_5 \ge 12

. B's total votes equal

35

. To leave enough votes for the first three constituencies, we must minimize

b_4

and

b_5

. Let

b_4 = 11

and

b_5 = 12

. The remaining votes for the first three constituencies are:

b_1 + b_2 + b_3 = 35 - 11 - 12 = 12

. Step 4: Determine Party B's exact sequence. In S2 and S3, C wins with

8

votes. Therefore, A and B must each have fewer than

8

votes (

b_2 \le 7, b_3 \le 7, a_2 \le 7, a_3 \le 7

). Since

a_i + b_i = 20 - 8 = 12

in these constituencies, the only valid integer pairs for

(a_i, b_i)

bounded by

7

are

(7,5), (6,6),

and

(5,7)

. Thus, B's votes in S2 and S3 must be chosen from the set

\{5, 6, 7\}

. Maintaining the strictly increasing sequence

b_2 < b_3

, we test combinations to satisfy

b_1 + b_2 + b_3 = 12

: If

(b_2, b_3) = (5, 6)

, then

b_1 = 12 - 11 = 1

. (Valid natural number) If

(b_2, b_3) = (5, 7)

, then

b_1 = 12 - 12 = 0

. (Invalid, natural numbers begin at

1

) Therefore, B's exact vote sequence across S1-S5 is

1, 5, 6, 11, 12

. Step 5: Calculate Party A's vote distribution. Using the formula

a_i = 20 - b_i - c_i

: S1:

a_1 = 20 - 1 - 0 = 19

(A wins) S2:

a_2 = 20 - 5 - 8 = 7

(C wins) S3:

a_3 = 20 - 6 - 8 = 6

(C wins) S4:

a_4 = 20 - 11 - 0 = 9

(B wins) S5:

a_5 = 20 - 12 - 0 = 8

(B wins) Checking the total:

19 + 7 + 6 + 9 + 8 = 49

, which perfectly matches A's given total. Step 6: Answer the specific prompt. We must find the constituency where B got fewer votes than both A and C. In S2, B has

5

votes, A has

7

votes, and C has

8

votes.

5 < 7

and

5 < 8

. Final Answer: S2

Q27:ipmat indore 2024LRDI › ArrangementsHardMCQ · MCQ

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage

The number of votes obtained by B in S2 is

A6
B7
C5
D4

Pick an option to attempt

The Setup: We must identify the specific number of votes obtained by Party B in constituency S2, using the comprehensive election matrix derived from the logical constraints. Step 1: Reference the derived election matrix. As established through the total vote counts and sequence constraints: Party C's votes across S1-S5:

0, 8, 8, 0, 0

Party B's votes across S1-S5:

1, 5, 6, 11, 12

Party A's votes across S1-S5:

19, 7, 6, 9, 8

Step 2: Isolate the requested data point. We look at Party B's vote sequence (

b_1, b_2, b_3, b_4, b_5

) and identify the value for S2 (

b_2

).

b_2 = 5

. Final Answer: 5

Q28:ipmat indore 2024LRDI › ArrangementsHardMCQ · MCQ

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage

The number of votes obtained by A in S5 is

A6
B9
C8
D7

Pick an option to attempt

The Setup: We must identify the specific number of votes obtained by Party A in constituency S5, utilizing the completed election matrix. Step 1: Reference the derived election matrix. The calculated vote distribution for Party A across the five constituencies (S1 through S5) is

19, 7, 6, 9,

and

8

. Step 2: Isolate the requested data point. We evaluate Party A's sequence (

a_1, a_2, a_3, a_4, a_5

) and extract the specific value corresponding to S5 (

a_5

).

a_5 = 8

. Final Answer: 8

Q29:ipmat indore 2024LRDI › ArrangementsHardMCQ · MCQ

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage

Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency

AS4
BS2
CS5
DS3

Pick an option to attempt

The Setup: We need to compare Party A's vote counts across all five constituencies to determine which constituency yielded their lowest performance. Step 1: Retrieve Party A's vote distribution. From our derived election matrix, the votes obtained by Party A in constituencies S1, S2, S3, S4, and S5 are respectively:

19, 7, 6, 9, 8

Step 2: Identify the minimum value. Comparing the integers in the set

\{19, 7, 6, 9, 8\}

, the lowest number is

6

. Step 3: Map the minimum value back to its constituency. The vote count of

6

corresponds to constituency S3. Final Answer: S3

Q30:ipmat indore 2024LRDI › ArrangementsHardMCQ · MCQ

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage

Assume that A and C had formed an alliance and any voter who voted for either A or C would have voted for this alliance. Then the number of seats this alliance would have won is

A4
B2
C3
D5

Pick an option to attempt

The Setup: We must model a hypothetical political alliance between Party A and Party C, combining their votes in each constituency to see how many total seats the new alliance would win against Party B. Step 1: Calculate the alliance's combined votes per constituency. We sum the individual votes of A and C for each constituency (

a_i + c_i

): S1:

19 + 0 = 19

S2:

7 + 8 = 15

S3:

6 + 8 = 14

S4:

9 + 0 = 9

S5:

8 + 0 = 8

Step 2: Compare the alliance's votes against Party B's votes. To win a seat, the alliance's combined votes must exceed Party B's votes in that constituency. S1: Alliance (

19

) vs Party B (

1

)

\Rightarrow

Alliance wins. S2: Alliance (

15

) vs Party B (

5

)

\Rightarrow

Alliance wins. S3: Alliance (

14

) vs Party B (

6

)

\Rightarrow

Alliance wins. S4: Alliance (

9

) vs Party B (

11

)

\Rightarrow

Party B wins. S5: Alliance (

8

) vs Party B (

12

)

\Rightarrow

Party B wins. Step 3: Total the seats won by the alliance. The alliance successfully wins constituencies S1, S2, and S3, totaling

3

seats. Final Answer: 3

IPMAT Indore 2024 — MCQ

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