IPMAT Indore 2022 MCQ PYQs

Q1:ipmat indore 2022QA › TrigonometryHardMCQ · MCQ

For

0\lt\theta\lt\frac{\pi}{4}

, let

a=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \cos \theta\right), b=\left((\cos \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right), c=\left((\sin \theta)^{\cos \theta}\right)\left(\log _{2} \cos \theta\right)

and

d=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right)

. Then, the median value in the sequence

a, b, c, d

is

A $\frac{a+b}{2}$
B $\frac{a+d}{2}$
C $\frac{b+c}{2}$
D $\frac{c+d}{2}$

Pick an option to attempt

The Setup: We are asked to order four logarithmic-exponential expressions to find their median. By analyzing the base and arguments within the interval

0 < \theta < \pi/4

, we can determine their strict mathematical ordering. Step 1: Analyze the components within the given domain. For

\theta \in (0, \pi/4)

, the sine and cosine functions are bounded such that:

0 < \sin\theta < \frac{1}{\sqrt{2}} < \cos\theta < 1

Because both

\sin\theta

and

\cos\theta

are strictly between

0

and

1

, their base-

2

logarithms are strictly negative:

\log_2(\sin\theta) < \log_2(\cos\theta) < 0

Step 2: Compare the components of expressions

a, b, c, d

. The definitions are:

a = (\sin\theta)^{\sin\theta} \log_2(\cos\theta)

b = (\cos\theta)^{\sin\theta} \log_2(\sin\theta)

c = (\sin\theta)^{\cos\theta} \log_2(\cos\theta)

d = (\sin\theta)^{\sin\theta} \log_2(\sin\theta)

Notice that all four overall expressions yield negative values (a positive exponential multiplied by a negative logarithm). Step 3: Establish the inequalities. First, compare

a

and

c

: Since

\sin\theta < \cos\theta

and the base

\sin\theta \in (0, 1)

, a larger exponent yields a smaller value:

(\sin\theta)^{\cos\theta} < (\sin\theta)^{\sin\theta}

Because

\log_2(\cos\theta)

is negative, multiplying both sides by it reverses the inequality:

(\sin\theta)^{\cos\theta} \log_2(\cos\theta) > (\sin\theta)^{\sin\theta} \log_2(\cos\theta) \implies c > a

Second, compare

b

and

d

: Because

\sin\theta < \cos\theta

, raising them to the positive power of

\sin\theta

preserves the inequality:

(\sin\theta)^{\sin\theta} < (\cos\theta)^{\sin\theta}

Because

\log_2(\sin\theta)

is negative, multiplying reverses the inequality:

(\sin\theta)^{\sin\theta} \log_2(\sin\theta) > (\cos\theta)^{\sin\theta} \log_2(\sin\theta) \implies d > b

Third, compare

a

and

d

: They share the identical positive exponential factor

(\sin\theta)^{\sin\theta}

. Since

\log_2(\sin\theta) < \log_2(\cos\theta)

, multiplying by the positive exponential preserves the inequality:

d < a

Step 4: Determine the median. Combining the inequalities gives the strict ascending order:

b < d < a < c

. For a set of 4 elements, the median is the arithmetic mean of the two central terms (

d

and

a

).

\text{Median} = \frac{a + d}{2}

Final Answer:

\frac{a+d}{2}

Q2:ipmat indore 2022QA › LogarithmsMediumMCQ · MCQ

The set of real values of

x

for which the inequality

\log _{27} 8 \leq \log _{3} x \lt 9^{\frac{1}{\log _{2} 3}}

holds is

A $[2,81)$
B $(2,27)$
C $[2,81]$
D $(2,27]$

Pick an option to attempt

The Setup: We are given a compound logarithmic inequality:

\log_{27}8 \le \log_3 x < 9^{1/\log_2 3}

. We must simplify the bounds to isolate

x

. Step 1: Simplify the lower bound.

\log_{27} 8 = \log_{3^3} 2^3

Using the exponent rule

\log_{b^k} a^p = \frac{p}{k} \log_b a

:

\log_{27} 8 = \frac{3}{3} \log_3 2 = \log_3 2

Step 2: Simplify the upper bound. Use the base-change inversion rule

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

:

9^{1/\log_2 3} = 9^{\log_3 2}

Rewrite the base

9

as

3^2

:

(3^2)^{\log_3 2} = 3^{2 \log_3 2} = 3^{\log_3(2^2)} = 3^{\log_3 4}

By the fundamental log identity

b^{\log_b k} = k

, this equals

4

. Step 3: Solve the resulting inequality for

x

. Substitute the simplified bounds back into the original inequality:

\log_3 2 \le \log_3 x < 4

Since the logarithmic base

3

is greater than

1

, the function is strictly increasing, meaning we can safely exponentiate all parts by base

3

without flipping signs:

3^{\log_3 2} \le 3^{\log_3 x} < 3^4

2 \le x < 81

Final Answer:

[2,81)

Q3:ipmat indore 2022QA › FunctionsMediumMCQ · MCQ

The set of all possible values of

f(x)

for which

(81)^{x}+(81)^{f(x)}=3

is

A( $0.25,3)$
B $(-\infty, 4)$
C $(-\infty, 0.25)$
D $(3,4)$

Pick an option to attempt

The Setup: Given the equation

81^x + 81^{f(x)} = 3

, we need to determine the complete set of possible values for the output of the function

f(x)

. Step 1: Isolate the term containing

f(x)

.

81^{f(x)} = 3 - 81^x

Step 2: Analyze the domain and range constraints. For any real number

x

, an exponential function with a positive base is strictly positive:

81^x > 0

. Since

81^{f(x)}

is also strictly positive, its equal counterpart must be positive:

3 - 81^x > 0 \implies 81^x < 3

Because

81^x

is strictly bounded between

0

and

3

, the expression

(3 - 81^x)

maps continuously across the open interval

(0, 3)

. Step 3: Translate the range to

f(x)

. Because

81^{f(x)} = 3 - 81^x

, the function

81^{f(x)}

takes all values in the interval

(0, 3)

.

0 < 81^{f(x)} < 3

The lower bound

0 < 81^{f(x)}

holds true for all real values of

f(x)

(an exponential function has no lower limit restriction other than

0

). Solve the upper bound:

81^{f(x)} < 3 \implies (3^4)^{f(x)} < 3^1 \implies 3^{4f(x)} < 3^1

Because base

3 > 1

, we can equate the exponents directly:

4f(x) < 1 \implies f(x) < 0.25

Thus,

f(x)

covers all real numbers strictly less than

0.25

. Final Answer:

(-\infty, 0.25)

Q4:ipmat indore 2022QA › TrigonometryMediumMCQ · MCQ

Ayesha is standing atop a vertical tower

200 m

high and observes a car moving away from the tower on a straight, horizontal road from the foot of the tower. At 11:00 AM, she observes the angle of depression of the car to be

45^{\circ}

. At 11:02 AM, she observes the angle of depression of the car to be

30^{\circ}

. The speed at which the car is moving is approximately

A6.3 km per hour
B8.45 km per hour
C10.6 km per hour
D4.39 km per hour

Pick an option to attempt

The Setup: A car is tracked from a

200\text{m}

tower. Its angle of depression changes from

45^\circ

to

30^\circ

over exactly

2

minutes. We must use right-triangle trigonometry to find its speed in

\text{km/h}

. Step 1: Calculate the initial horizontal distance. Let the tower base be the origin. At 11:00 AM, the angle of depression is

45^\circ

. Using trigonometry,

\tan(45^\circ) = \frac{\text{Height}}{\text{Distance}_1}

.

1 = \frac{200}{d_1} \implies d_1 = 200\text{ meters}

Step 2: Calculate the subsequent horizontal distance. At 11:02 AM, the angle of depression is

30^\circ

.

\tan(30^\circ) = \frac{\text{Height}}{\text{Distance}_2}

\frac{1}{\sqrt{3}} = \frac{200}{d_2} \implies d_2 = 200\sqrt{3}\text{ meters}

Step 3: Calculate the distance traveled. The distance the car traveled in the 2-minute interval is the difference:

\Delta d = d_2 - d_1 = 200\sqrt{3} - 200 = 200(\sqrt{3} - 1)

Using the approximation

\sqrt{3} \approx 1.732

:

\Delta d \approx 200(1.732 - 1) = 200(0.732) = 146.4\text{ meters}

Step 4: Calculate the speed in km/h. The car traveled

146.4

meters in

2

minutes.

\text{Speed} = \frac{146.4\text{ meters}}{2\text{ minutes}} = 73.2\text{ meters/minute}

Convert to kilometers per hour (

\times 60 \text{ minutes/hour}, \div 1000 \text{ meters/km}

):

\text{Speed} = \frac{73.2 \times 60}{1000} = \frac{4392}{1000} = 4.392\text{ km/h}

Final Answer: 4.39 km per hour

Q5:ipmat indore 2022QA › Straight LinesMediumMCQ · MCQ

The value of

k

for which the following lines are concurrent is

x-y-1=0 \newline 2x+3y-12=0 \newline 2x-3y+k=0

A1
B-1
C0
D12

Pick an option to attempt

The Setup: Three straight lines are concurrent (they intersect at a single unified point). We must find the intersection of the first two lines and use it to solve for the missing constant

k

in the third line. Step 1: Find the intersection point of the first two lines. Line 1:

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

Line 2:

2x + 3y - 12 = 0

Substitute

x

into the second equation:

2(y + 1) + 3y - 12 = 0

2y + 2 + 3y - 12 = 0

5y - 10 = 0 \implies y = 2

Substitute

y

back to find

x

:

x = 2 + 1 = 3

The shared intersection point is

(3, 2)

. Step 2: Solve for

k

. For the three lines to be perfectly concurrent, Line 3 (

2x - 3y + k = 0

) must also pass precisely through

(3, 2)

. Substitute

x = 3

and

y = 2

into the third equation:

2(3) - 3(2) + k = 0

6 - 6 + k = 0 \implies k = 0

Final Answer: 0

Q6:ipmat indore 2022QA › ModulusMediumMCQ · MCQ

The sum of the squares of all the roots of the equation

x^{2}+|x+4|+|x-4|-35=0

is

A74
B175
C50
D148

Pick an option to attempt

The Setup: The equation

x^{2}+|x+4|+|x-4|-35=0

relies on absolute value boundaries. We must partition the domain to find all valid real roots, then calculate the sum of their squares. Step 1: Notice the inherent symmetry. Since

|x+4| + |x-4|

is an even function, if

x

is a valid root,

-x

is also a root. Step 2: Evaluate Case 1 (

x \ge 4

). Both absolute values evaluate to their positive counterparts:

x^2 + (x + 4) + (x - 4) - 35 = 0

x^2 + 2x - 35 = 0

(x + 7)(x - 5) = 0

The potential roots are

-7

and

5

. Because our domain is strictly

x \ge 4

,

x = 5

is a valid root. Step 3: Evaluate Case 2 (

-4 < x < 4

). The first term remains positive, but the second evaluates to its negative counterpart:

x^2 + (x + 4) + (4 - x) - 35 = 0

x^2 + 8 - 35 = 0

x^2 - 27 = 0 \implies x = \pm\sqrt{27} \approx \pm 5.196

Because

\pm 5.196

falls strictly outside the bounded domain of

(-4, 4)

, no valid roots exist in this region. Step 4: Evaluate Case 3 (

x \le -4

). Both absolute values flip sign:

x^2 - (x + 4) - (x - 4) - 35 = 0

x^2 - 2x - 35 = 0

(x - 7)(x + 5) = 0

The potential roots are

7

and

-5

. Because our domain is strictly

x \le -4

,

x = -5

is a valid root. This agrees with the symmetry noted in Step 1. The two valid real roots are

5

and

-5

. Step 5: Calculate the sum of squares.

\text{Sum} = (5)^2 + (-5)^2 = 25 + 25 = 50

Final Answer: 50

Q7:ipmat indore 2022QA › TrianglesEasyMCQ · MCQ

The lengths of the sides of a triangle are

x, 21

and

40

, where

x

is the shortest side. A possible value of

x

is:

A18
B20
C19
D16

Pick an option to attempt

The Setup: A triangle is composed of side lengths

x, 21

and

40

, where

x

is stated to be the shortest side. We must determine a viable integer value for

x

bounded by the triangle inequality. Step 1: Apply the triangle inequality theorem. For three lengths to successfully form a closed triangle, the sum of the two shorter sides must strictly exceed the longest side.

x + 21 > 40 \implies x > 19

Step 2: Apply the specific length constraint. The problem states that

x

is the shortest side. Therefore,

x

must be less than or equal to

21

. Combining both strict inequalities:

19 < x \le 21

Step 3: Evaluate the given options. The mathematically possible integer values for

x

are

20

and

21

. Reviewing the multiple-choice options provided (

18, 20, 19, 16

), the only valid candidate listed is

20

. Final Answer: 20

Q8:ipmat indore 2022QA › Divisibility RulesEasyMCQ · MCQ

If the five-digit number abcde is divisible by 6, then which of the following numbers is not necessarily divisible by 6?

Aedcba
Beee
Cbbadcacede
Dcdbae

Pick an option to attempt

The Setup: We are given a

5

-digit number

abcde

that is fully divisible by

6

. We need to evaluate four altered configurations to see which is *not necessarily* divisible by

6

. Step 1: Establish the divisibility rule for 6. A number is divisible by

6

if and only if it is mathematically divisible by both

2

(its unit digit is even) and

3

(the sum of its digits is a multiple of

3

). For

abcde

: 1.

a+b+c+d+e \equiv 0 \pmod 3

2.

e

is an even digit. Step 2: Test the proposed configurations. * **Option 2 (

eee

):** Sum is

3e

(divisible by

3

). Ends in

e

(even). Always divisible by

6

. * **Option 3 (

bbadcacede

):** Sum is

2a + 2b + 2c + 2d + 2e = 2(\text{sum})

. Since the original sum is divisible by

3

,

2 \times \text{sum}

is also divisible by

3

. Ends in

e

(even). Always divisible by

6

. * **Option 4 (

cdbae

):** Sum is the same as the original (divisible by

3

). Ends in

e

(even). Always divisible by

6

. * **Option 1 (

edcba

):** The sum of the digits remains the same (divisible by

3

). However, the unit digit is now

a

. We only know

e

is even;

a

can be either odd or even. Therefore,

edcba

is not guaranteed to be even, meaning it is not necessarily divisible by

6

. Final Answer: edcba

Q9:ipmat indore 2022QA › Permutation & CombinationMediumMCQ · MCQ

In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together?

A75600
B25200
C37800
D21600

Pick an option to attempt

The Setup: We must arrange the

10

letters of MANAGEMENT so that no two vowels are adjacent. This relies on the Gap Method of combinatorics. Step 1: Audit the frequency of the letters. The word MANAGEMENT consists of

10

letters: Vowels (

4

total): A (

2

), E (

2

). Consonants (

6

total): M (

2

), N (

2

), G (

1

), T (

1

). Step 2: Arrange the consonants. First, place the

6

consonants in a row. The number of unique permutations accounting for identical letters is:

\frac{6!}{2! \times 2!} = \frac{720}{4} = 180 \text{ ways}

Step 3: Place the vowels in the gaps. Arranging

6

consonants in a line creates exactly

7

gaps (including the outside ends) where vowels can be slotted to ensure they never touch. We must choose

4

gaps for our

4

vowels:

\binom{7}{4} = 35 \text{ ways}

Next, arrange the

4

vowels (A, A, E, E) within those

4

chosen gaps:

\frac{4!}{2! \times 2!} = \frac{24}{4} = 6 \text{ ways}

Step 4: Calculate the total configurations. Multiply the independent combinatorics together:

\text{Total Ways} = 180 \times 35 \times 6 = 180 \times 210 = 37800

Final Answer: 37800

Q10:ipmat indore 2022QA › Mixture & AlligationEasyMCQ · MCQ

In a room, there are

n

persons whose average height is

160 \ \text{cm}

. If

m

more persons, whose average height is

172 \ \text{cm}

, enter the room, then the average height of all persons in the room becomes

164 \ \text{cm}

. Then

m: n

is

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

Pick an option to attempt

The Setup: This is a weighted average problem mapping two groups of distinct heights into a combined new average height. We can isolate the ratio of group sizes using the Alligation method. Step 1: Define the parameters for Alligation. Group 1 (

n

persons): Average height =

160 \text{ cm}

. Group 2 (

m

persons): Average height =

172 \text{ cm}

. Mixture (Combined group): Average height =

164 \text{ cm}

. Step 2: Apply the Alligation cross-difference formula. The ratio of the quantities (

n : m

) is inversely proportional to the distances their specific averages sit from the combined mixture mean.

n : m = (172 - 164) : (164 - 160)

n : m = 8 : 4

n : m = 2 : 1

Step 3: Evaluate the requested ratio. We must explicitly provide the ratio

m : n

. Since

n : m = 2 : 1

, the ratio

m : n

is exactly

1 : 2

. Final Answer:

1 : 2

Q11:ipmat indore 2022QA › Progression & SeriesMediumMCQ · MCQ

The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350. Then the common difference is

A $\frac{7}{9}$
B $\frac{2}{3}$
C $\frac{4}{9}$
D $\frac{1}{3}$

Pick an option to attempt

The Setup: We are given two continuous summation blocks of an Arithmetic Progression (

S_{1-15}

and

S_{16-30}

). We must build a system of linear equations mapping the progression formula to isolate the common difference

d

. Step 1: Translate the sums into standard AP formulation. Let the first term be

a

and the common difference be

d

. The sum of an AP is

S_n = \frac{n}{2}(2a + (n-1)d)

. Sum of the first 15 terms (

S_{15}

) =

200

.

200 = \frac{15}{2}(2a + 14d) = 15(a + 7d) \implies a + 7d = \frac{200}{15} = \frac{40}{3}

Step 2: Evaluate the total cumulative sum constraint. The sum of the *next* 15 terms is

350

. This means the total sum of the first 30 terms (

S_{30}

) is

200 + 350 = 550

.

550 = \frac{30}{2}(2a + 29d) = 15(2a + 29d) \implies 2a + 29d = \frac{550}{15} = \frac{110}{3}

Step 3: Solve the linear system for

d

. Multiply the first simplified equation by

2

to align coefficients:

2a + 14d = \frac{80}{3}

Subtract this from the second equation:

(2a + 29d) - (2a + 14d) = \frac{110}{3} - \frac{80}{3}

15d = \frac{30}{3} = 10

d = \frac{10}{15} = \frac{2}{3}

Final Answer:

\frac{2}{3}

Q12:ipmat indore 2022QA › Matrices & DeterminantsMediumMCQ · MCQ

Suppose

a, b

and

c

are integers such that

a>b>c>0

, and

A=\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]

. Then the value of the determinant of

A

is

Acan be positive or negative
Bis positive
Cis negative
Dis zero

Pick an option to attempt

The Setup: A specific circulant

3 \times 3

matrix

A

is populated with integers

a > b > c > 0

. We need to evaluate the algebraic sign of its determinant. Step 1: Expand the determinant of Matrix

A

.

A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}

Calculate the determinant using the standard top-row expansion:

\det(A) = a(bc - a^2) - b(b^2 - ac) + c(ab - c^2)

\det(A) = abc - a^3 - b^3 + abc + abc - c^3

\det(A) = 3abc - a^3 - b^3 - c^3

Step 2: Factor the resulting polynomial. The expression

a^3 + b^3 + c^3 - 3abc

possesses a highly standardized factorization identity:

a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]

Notice that our determinant is the exact algebraic negative of this expression:

\det(A) = - \left( \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] \right)

Step 3: Evaluate the bounds of the factored terms. We are strictly given

a > b > c > 0

. Because all terms are strictly positive, their sum

(a+b+c)

is strictly positive. Because the terms are strictly unequal, the squared differences

(a-b)^2, (b-c)^2, (c-a)^2

are strictly positive non-zeros. Thus, the entire bracketed term evaluates to a strictly positive number. Because a negative sign sits in front of it, the total determinant is strictly negative. Final Answer: is negative

Q13:ipmat indore 2022QA › TrianglesMediumMCQ · MCQ

In a right-angled triangle ABC, the hypotenuse AC is of length 13 cm. A line drawn connecting the midpoints D and E of sides AB and AC is found to be 6 cm in length. The length of BC is

A $12$ cm
B $5$ cm
C $2 \sqrt{3}$ cm
D $8$ cm

Pick an option to attempt

The Setup: In a right-angled triangle

ABC

, the hypotenuse

AC

is

13\text{ cm}

. The segment

DE

joining the midpoint of the leg

AB

to the midpoint of the hypotenuse

AC

measures

6\text{ cm}

. We must find the length of

BC

. Step 1: Map the geometry using the Midpoint Theorem. The problem defines a line segment drawn connecting the midpoints (

D

and

E

) of two sides of a triangle (

AB

and

AC

). According to the Triangle Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is perfectly parallel to the third side and exactly half of its length.

DE = \frac{1}{2} BC

Step 2: Calculate the base. We are explicitly given that

DE = 6\text{ cm}

.

6 = \frac{1}{2} BC \implies BC = 12\text{ cm}

*(The hypotenuse measurement of 13 is consistent data: with

BC = 12

and

AC = 13

, the remaining leg

AB = 5

, forming the standard 5-12-13 Pythagorean triple.)* Final Answer:

12

cm

Q14:ipmat indore 2022QA › Matrices & DeterminantsMediumMCQ · MCQ

If

A=\left[\begin{array}{ll}1 & 0 \\ \frac{1}{2} & 0\end{array}\right]

then

A^{2022}

is

A $\left[\begin{array}{cc}1 & 0 \\ \frac{1}{2^{2022}} & 0\end{array}\right]$
B $\left[\begin{array}{cc}1 & 0 \\ 1011 & 0\end{array}\right]$
C $\left[\begin{array}{cc}1 & 0 \\ 2022 & 0\end{array}\right]$
DNone of these

Pick an option to attempt

The Setup: We are given

A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}

and must identify

A^{2022}

among the listed options. The decisive detail is the bottom-right entry, which is

0

. Step 1: Compute

A^2

.

A^2 = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} = \begin{bmatrix} 1(1) + 0\left(\frac{1}{2}\right) & 1(0) + 0(0) \\ \frac{1}{2}(1) + 0\left(\frac{1}{2}\right) & \frac{1}{2}(0) + 0(0) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} = A

Because the second column of

A

is entirely zero, multiplying by

A

reproduces

A

exactly. Step 2: Extend the result to every power. The matrix is therefore idempotent (

A^2 = A

). By induction, every higher power collapses back to

A

:

A^3 = A^2 A = A A = A, \qquad A^4 = A, \qquad \dots, \qquad A^n = A \text{ for all } n \ge 1

The bottom-left entry never grows.

A^{2022} = A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}

Step 3: Compare against the given options. The three explicit choices offer bottom-left entries of

\frac{1}{2^{2022}}

,

1011

and

2022

. Our result has bottom-left entry

\frac{1}{2}

, which matches none of them. *A note on the trap:* had the bottom-right entry been

1

rather than

0

, the matrix would satisfy

A^n = \begin{bmatrix} 1 & 0 \\ \frac{n}{2} & 1 \end{bmatrix}

, giving a bottom-left entry of

\frac{2022}{2} = 1011

. That is precisely the decoy offered in the second option. With the printed

0

in the bottom-right, that reasoning does not apply and the power never accumulates. Final Answer: None of these

Q15:ipmat indore 2022QA › FunctionsMediumMCQ · MCQ

Let

A=\{1,2,3\}

and

B=\{a, b\}

. Assuming all relations from set

A

to set

B

are equally likely, what is the probability that a relation from

A

to

B

is also a function?

A $\frac{1}{8}$
B $\frac{1}{2}$
C1
D $\frac{3^{2}}{2^{6}}$

Pick an option to attempt

The Setup: Set

A = \{1, 2, 3\}

and Set

B = \{a, b\}

. If we pull randomly from the pool of all possible mathematical relations mapping A to B, we need the probability that the relation is specifically a valid function. Step 1: Calculate the size of the total sample space (all relations). A mathematical relation from

A

to

B

is defined as any possible subset of the Cartesian product

A \times B

. The cardinality of the Cartesian product is

|A| \times |B| = 3 \times 2 = 6

. The total number of unique relations (subsets) is

2^{|A \times B|} = 2^6 = 64

. Step 2: Calculate the size of the restricted target group (all functions). A function is a highly specific relation where every element in set

A

maps to exactly one valid element in set

B

. Each of the

3

elements in

A

has exactly

2

choices to map to in

B

. Total number of functions =

|B|^{|A|} = 2^3 = 8

. Step 3: Calculate the resulting probability.

\text{Probability} = \frac{\text{Target Outcomes}}{\text{Total Outcomes}} = \frac{8}{64} = \frac{1}{8}

Final Answer:

\frac{1}{8}

Q16:ipmat indore 2022QA › Time, Speed & DistanceMediumMCQ · MCQ

In a 400-metre race, Ashok beats Bipin and Chandan respectively by 15 seconds and 25 seconds. If Ashok beats Bipin by 150 metres, by how many metres does Bipin beat Chandan in the race?

A80
B100
C150
D50

Pick an option to attempt

The Setup: Three runners (Ashok, Bipin, Chandan) race over a

400\text{m}

distance. Using their offset finishing times and a distance offset, we must evaluate the specific physical distance between Bipin and Chandan when Bipin finishes the race. Step 1: Establish the time relationships. Let Ashok's finishing time for

400\text{m}

be

t_A

. Ashok beats Bipin by

15

seconds

\implies t_B = t_A + 15

. Ashok beats Chandan by

25

seconds

\implies t_C = t_A + 25

. Step 2: Calculate Ashok's finishing time using the distance discrepancy. Ashok beats Bipin by exactly

150\text{m}

. This means when Ashok finished at time

t_A

, Bipin had only run

400 - 150 = 250\text{m}

. We know Bipin completes the full

400\text{m}

at a constant speed in time

(t_A + 15)

. Bipin's Speed =

\frac{250}{t_A} = \frac{400}{t_A + 15}

Cross-multiply to solve for

t_A

:

250(t_A + 15) = 400 t_A

250 t_A + 3750 = 400 t_A \implies 150 t_A = 3750 \implies t_A = 25\text{ seconds}

Step 3: Calculate constant speeds for Bipin and Chandan. Bipin's finishing time =

25 + 15 = 40\text{ s}

. Speed

V_B = \frac{400}{40} = 10\text{ m/s}

. Chandan's finishing time =

25 + 25 = 50\text{ s}

. Speed

V_C = \frac{400}{50} = 8\text{ m/s}

. Step 4: Calculate the distance gap at Bipin's finish. Bipin finishes the race precisely at the

40

-second mark. At

40

seconds, Chandan has traveled:

40\text{ s} \times 8\text{ m/s} = 320\text{m}

. The gap distance is

400 - 320 = 80\text{m}

. Thus, Bipin beats Chandan by

80\text{m}

. Final Answer: 80

Q17:ipmat indore 2022QA › Mixture & AlligationMediumMCQ · MCQ

In a bowl containing 60 ml orange juice, 40 ml of water is poured. Thereafter, 100 ml of apple juice is poured to make a fruit punch. Madhu drinks 50 ml of this fruit punch and comments that the proportion of orange juice needs to be higher for better taste. How much orange juice should be poured into the fruit punch that remained, in order to bring up the level of orange juice to 50 percentage?

A100 ml
B40 ml
C80 ml
D60 ml

Pick an option to attempt

The Setup: A mixed volume fruit punch starts out with unequal ratios. After some is consumed, pure orange juice is added back to strictly raise its proportion to

50\%

of the newly defined mixture. Step 1: Establish the baseline volume ratios of the mixture. Initial ingredients:

60\text{ml}

orange juice (OJ),

40\text{ml}

water (W). Subtotal =

100\text{ml}

. Add

100\text{ml}

apple juice (AJ). Grand total =

200\text{ml}

. The proportional composition of the punch is:

\text{OJ} = \frac{60}{200} = 30\%

\text{W} = \frac{40}{200} = 20\%

\text{AJ} = \frac{100}{200} = 50\%

Step 2: Calculate volumes remaining after consumption. Madhu drinks

50\text{ml}

of the punch. Remaining total volume =

200\text{ml} - 50\text{ml} = 150\text{ml}

. Because mixtures are perfectly homogenous, the

30\%

ratio is maintained in the leftover fluid. Volume of OJ remaining =

30\%

of

150 = 45\text{ml}

. Step 3: Calculate the required OJ addition. Let

x

be the volume in

\text{ml}

of pure OJ poured back into the punch. The new total volume of OJ is

45 + x

. The new total volume of the entire punch is

150 + x

. We require the new OJ ratio to be exactly

50\%

:

\frac{45 + x}{150 + x} = 0.5

45 + x = 0.5(150 + x)

45 + x = 75 + 0.5x

0.5x = 30 \implies x = 60\text{ml}

Final Answer: 60 ml

Q18:ipmat indore 2022QA › Conic SectionsMediumMCQ · MCQ

The curve represented by the equation

\dfrac{x^{2}}{\sin \sqrt{2}-\sin \sqrt{3}}+\dfrac{y^{2}}{\cos \sqrt{2}-\cos \sqrt{3}}=1

is

Aan ellipse with the foci on the y-axis
Ban ellipse with the foci on the x-axis
Ca hyperbola with the foci on the x-axis
Da hyperbola with the foci on the y-axis

Pick an option to attempt

The Setup: We are asked to classify a conic section represented by an equation with complex trigonometric constants in its denominators. Step 1: Analyze the standard conic equation format. The equation takes the structure

\frac{x^2}{A} + \frac{y^2}{B} = 1

. The specific conic shape depends rigidly on the algebraic signs of

A

and

B

. Step 2: Evaluate the sign of denominator

A

.

A = \sin(\sqrt{2}) - \sin(\sqrt{3})

. Note that

\sqrt{2} \approx 1.414

rad and

\sqrt{3} \approx 1.732

rad.

\pi/2 \approx 1.571

rad. Thus,

\sqrt{2}

resides in the first quadrant and

\sqrt{3}

resides in the second quadrant. The sine function acts symmetrically around its peak at

\pi/2

. Distance of

\sqrt{2}

from

\pi/2

is

|1.414 - 1.571| = 0.157

. Distance of

\sqrt{3}

from

\pi/2

is

|1.732 - 1.571| = 0.161

. Because

\sqrt{2}

sits slightly closer to the absolute peak than

\sqrt{3}

,

\sin(\sqrt{2}) > \sin(\sqrt{3})

. Thus,

A

is strictly positive. Step 3: Evaluate the sign of denominator

B

.

B = \cos(\sqrt{2}) - \cos(\sqrt{3})

. The cosine function is strictly and continuously decreasing on the interval

(0, \pi)

. Because

\sqrt{2} < \sqrt{3}

, it follows rigidly that

\cos(\sqrt{2}) > \cos(\sqrt{3})

. Thus,

B

is strictly positive. Because both

A

and

B

are positive, the equation represents an ellipse. Let

A = a^2

and

B = b^2

. Step 4: Find the major axis. To locate the foci, we must compare the magnitudes of

a^2

and

b^2

.

a^2 - b^2 = (\sin\sqrt{2} - \sin\sqrt{3}) - (\cos\sqrt{2} - \cos\sqrt{3}) = (\sin\sqrt{2} - \cos\sqrt{2}) - (\sin\sqrt{3} - \cos\sqrt{3})

Let

f(x) = \sin x - \cos x

. Its derivative is

f'(x) = \cos x + \sin x

. On the interval containing

\sqrt{2}

and

\sqrt{3}

,

\sin x

stays near

1

while

\cos x

is small in magnitude, so

f'(x) > 0

and

f(x)

is increasing there. Because

f(x)

is increasing,

f(\sqrt{2}) < f(\sqrt{3})

. Thus,

a^2 - b^2 < 0 \implies a^2 < b^2

. Because the

y

-denominator is larger, the major axis is vertical, placing the foci strictly on the

y

-axis. Final Answer: an ellipse with the foci on the y-axis

Q19:ipmat indore 2022QA › FunctionsEasyMCQ · MCQ

A set of all possible values the function

f(x)=\dfrac{x}{|x|}

, where

x \neq 0

, takes is

A $\{1\}$
B $\{1,-1\}$
C $\{1,0\}$
D $\{1,0,-1\}$

Pick an option to attempt

The Setup: We need to specify the range (the set of all possible outputs) for the absolute value ratio function

f(x)=\frac{x}{|x|}

. Step 1: Analyze the piecewise domains of the function. The function possesses a domain restricting

x \neq 0

. If

x > 0

(positive numbers), then

|x| = x

.

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

If

x < 0

(negative numbers), then

|x| = -x

.

f(x) = \frac{x}{-x} = -1

Step 2: Formulate the set of possible values. Since the function mathematically collapses into exactly two static outputs dependent on algebraic sign, its absolute range is limited exclusively to those two integers. The set of outputs is

\{1, -1\}

. Final Answer:

\{1,-1\}

Q20:ipmat indore 2022QA › HCF & LCMEasyMCQ · MCQ

When the square of the difference of two natural numbers is subtracted from the square of the sum of the same two numbers and the result is divided by four, we get

Athe product of the LCM and HCF of the two numbers
Bthe HCF of the two numbers
Cthe LCM of the two numbers
Dthe square of the product of the two numbers

Pick an option to attempt

The Setup: Two natural numbers are subjected to a specific string of algebraic operations. We must simplify the operations to find the resulting identity. Step 1: Map the text to algebraic formulation. Let the two natural numbers be

a

and

b

. 'Square of the difference' =

(a-b)^2

. 'Square of the sum' =

(a+b)^2

. The text requires:

\frac{(a+b)^2 - (a-b)^2}{4}

. Step 2: Expand and simplify the expression. Expand the squares:

\frac{(a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)}{4}

Distribute the negative sign:

\frac{a^2 + 2ab + b^2 - a^2 + 2ab - b^2}{4}

The squares cancel out perfectly:

\frac{4ab}{4} = ab

The result is simply the direct mathematical product of the two original natural numbers. Step 3: Relate to the Number System rules. A fundamental property of arithmetic states that the product of any two natural numbers equals the product of their Least Common Multiple (LCM) and their Highest Common Factor (HCF).

a \times b = \text{LCM}(a,b) \times \text{HCF}(a,b)

Final Answer: the product of the LCM and HCF of the two numbers

Q21:ipmat indore 2022QA › Ratio, Proportion & VariationEasyMCQ · MCQ

The cost of a piece of jewellery is proportional to the square of its weight. A piece of jewellery weighing 10 grams is INR 3600. The cost of a piece of jewellery of the same kind weighing 4 grams is

AINR 1220
BINR 600
CINR 576
DINR 1440

Pick an option to attempt

The Setup: A piece of jewellery's cost is strictly proportional to the square of its weight. Using a known data point, we must calculate the cost for a lighter piece. Step 1: Establish the mathematical proportionality rule. Let

C

be the cost and

W

be the weight.

C \propto W^2 \implies C = k \cdot W^2

where

k

is the constant of proportionality. Step 2: Isolate the constant

k

. We are given a data point: a

10

-gram piece costs INR

3600

.

3600 = k \times (10)^2

3600 = 100k \implies k = 36

Step 3: Evaluate the secondary cost. Calculate the cost for a piece weighing

4

grams:

C = 36 \times (4)^2

C = 36 \times 16 = 576

Final Answer: INR 576

Q22:ipmat indore 2022QA › Set TheoryEasyMCQ · MCQ

Let

A

and

B

be two sets such that the Cartesian product

A \times B

consists of four elements. If two elements of

A \times B

are

(1,4)

and

(4,1)

, then

ANone of these
B $A \times B \neq B \times A$
C $\emptyset \in A \times B$
D $A \times B=B \times A$

Pick an option to attempt

The Setup: For sets

A

and

B

, the Cartesian product

A \times B

yields exactly four elements, two of which are specifically

(1, 4)

and

(4, 1)

. We evaluate the mathematical relationship between the parent sets. Step 1: Define the cardinality constraints. The total size of the Cartesian product is

|A \times B| = |A| \times |B| = 4

. By definition, for a Cartesian product

(x,y) \in A \times B

, the first element belongs to

A

and the second belongs to

B

. The presence of

(1, 4)

dictates that

1 \in A

and

4 \in B

. The presence of

(4, 1)

dictates that

4 \in A

and

1 \in B

. Step 2: Formulate the parent sets. We have confirmed that set

A

inherently contains at least two distinct elements:

\{1, 4\}

. Thus

|A| \ge 2

. We have confirmed that set

B

inherently contains at least two distinct elements:

\{1, 4\}

. Thus

|B| \ge 2

. Because

|A| \times |B|

must rigidly equal

4

, the only mathematically viable integer size for both subsets is precisely

2

. Therefore, there are no hidden elements.

A = \{1, 4\}

and

B = \{1, 4\}

. Step 3: Evaluate the target relation. Because

A

exactly equals

B

, their Cartesian products yield completely identical mappings.

A \times B = B \times A

Final Answer:

A \times B=B \times A

Q23:ipmat indore 2022QA › FunctionsHardMCQ · MCQ

If

f\left(x^{2}+f(y)\right)=x f(x)+y

for all non-negative integers

x

and

y

, then the value of

[f(0)]^{2}+f(0)

equals _________.

A2
B0
C6
D1

Pick an option to attempt

The Setup: A functional mapping rule dictates

f(x^2 + f(y)) = x f(x) + y

for all non-negative integers

x, y

. We must solve for

f(0)

to evaluate

[f(0)]^2 + f(0)

. Step 1: Exploit substitution to define the function. Let

x = 0

. The equation becomes:

f(0^2 + f(y)) = 0 \cdot f(0) + y \implies f(f(y)) = y

This proves that the function is an involution (it acts identically as its own inverse). Consequently, substituting

y = 0

yields

f(f(0)) = 0

. Step 2: Set up a root evaluation using the involution property. Let

c = f(0)

. Substituting this into our involution proof implies

f(c) = 0

. Now, set

y = 0

in the original primary equation:

f(x^2 + f(0)) = x f(x)

Substitute

x = c

:

f(c^2 + c) = c \cdot f(c)

Because we proved

f(c) = 0

:

f(c^2 + c) = c(0) = 0

Step 3: Solve for

c

. Because the function maps

c^2 + c

to

0

, and we previously established that

f(c)

uniquely maps to

0

(since

f

acts as a strict bijection due to

f(f(x))=x

), the internal arguments must be rigidly equal:

c^2 + c = c \implies c^2 = 0 \implies c = 0

Thus,

f(0) = 0

. Step 4: Evaluate the requested expression. We must calculate

[f(0)]^2 + f(0)

.

0^2 + 0 = 0

Final Answer: 0

Q24:ipmat indore 2022QA › FactorisationMediumMCQ · MCQ

If one of the factors of the number

3^{7} 2^{8} 17^{3}

is randomly chosen, then the probability that the chosen factor will be a perfect square is

A $\frac{5}{36}$
B $\frac{1}{12}$
C $\frac{3}{40}$
D $\frac{5}{32}$

Pick an option to attempt

The Setup: A factor is pulled randomly from the large integer

N = 3^7 \cdot 2^8 \cdot 17^3

. We must identify the probability that this chosen factor acts as a perfect square. Step 1: Calculate the size of the sample space (total factors). Using the prime factorization exponent rule, the total number of divisors of

p^a q^b r^c

is

(a+1)(b+1)(c+1)

.

\text{Total Factors} = (7+1)(8+1)(3+1) = 8 \times 9 \times 4 = 288

Step 2: Define the constraints for perfect square factors. For a factor to exist as a perfect square, the exponents of all of its individual prime bases must be even integers. Factor format:

3^{2x} \cdot 2^{2y} \cdot 17^{2z}

. We evaluate the valid even exponent boundaries: * Base 3:

0 \le 2x \le 7 \implies 2x \in \{0, 2, 4, 6\}

(4 valid choices). * Base 2:

0 \le 2y \le 8 \implies 2y \in \{0, 2, 4, 6, 8\}

(5 valid choices). * Base 17:

0 \le 2z \le 3 \implies 2z \in \{0, 2\}

(2 valid choices). Step 3: Calculate the target pool and the probability. Multiply the independent choices together to find the number of valid square factors:

\text{Perfect Square Factors} = 4 \times 5 \times 2 = 40

The resulting probability is:

\text{Probability} = \frac{40}{288} = \frac{5}{36}

Final Answer:

\frac{5}{36}

Q25:ipmat indore 2022QA › Divisibility RulesMediumMCQ · MCQ

The number of four-digit integers which are greater than 1000 and divisible by both 2 and 3, but not by 5, is

A1333
B1666
C1200
D1500

Pick an option to attempt

The Setup: We are asked to count the quantity of four-digit integers logically bounded by specific divisibility rules (divisible by

2

and

3

, but strictly not by

5

). Step 1: Establish the numerical boundaries. The domain represents four-digit integers greater than

1000

. The inclusive range is mathematically defined as

1001 \le N \le 9999

. Step 2: Apply the primary inclusion divisibility rule. A number mathematically divisible by both

2

and

3

is inherently divisible by their LCM, which is

6

. We calculate the count of numbers divisible by

6

within the domain. * Smallest multiple of

6

:

1002

(

6 \times 167

). * Largest multiple of

6

:

9996

(

6 \times 1666

). Count

= \frac{\text{Largest} - \text{Smallest}}{6} + 1 = \frac{9996 - 1002}{6} + 1 = 1499 + 1 = 1500

. Step 3: Apply the exclusion divisibility rule. We must exclude numbers within this pool that are also divisible by

5

. A number divisible by both

6

and

5

is divisible by their LCM, which is

30

. We calculate the count of numbers divisible by

30

within the domain to subtract them from our pool. * Smallest multiple of

30

:

1020

(

30 \times 34

). * Largest multiple of

30

:

9990

(

30 \times 333

). Count

= \frac{9990 - 1020}{30} + 1 = \frac{8970}{30} + 1 = 299 + 1 = 300

. Step 4: Calculate the final target count. Subtract the excluded intersecting pool from the primary included pool:

\text{Final Count} = 1500 - 300 = 1200

Final Answer: 1200

Q26:ipmat indore 2022LRDI › ArrangementsEasyMCQ · MCQ

A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage

Number of days Bhabha and Cathy work together in a week is

A0
B1
C3
D2

Pick an option to attempt

The Setup: This is a logical arrangement puzzle requiring us to map five employees onto a 7-day schedule. We deduce one Master Matrix satisfying every daily capacity and every shift rule; that single grid then answers all five questions in this set. Step 1: Check the shift arithmetic and fix the reading of 'consecutive'. Total shifts required

= 6 \text{ days} \times 2 + 1 \text{ Sunday} \times 3 = 15

. Five employees working exactly

3

days each also supply

5 \times 3 = 15

shifts, so the grid is exactly saturated: every slot is filled and nobody has spare capacity. A caselet like this is far easier to see than to hold in your head, so draw the grid first and fill cells in as they are forced. The margins carry the capacities, and they are what drive every deduction:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday						3
Monday						2
Tuesday						2
Wednesday						2
Thursday						2
Friday						2
Saturday						2
Works	3	3	3	3	3	15

Each employee's three days therefore consist of exactly one consecutive pair plus a third day adjacent to neither. The week is read as a cycle, so Saturday and Sunday count as consecutive. This is not a cosmetic detail: it means a run of Saturday, Sunday, Monday is a three-consecutive-day block and is barred. Read the week as a straight line instead and the puzzle admits three different grids, two of which contradict the stated answers to the later questions in this set, so the cyclic reading is the intended one. Step 2: Fix Cathy, then Alex. Alex works Tuesday and Wednesday. Neither Bhabha nor Cathy ever shares a day with Alex, so neither of them works Tuesday or Wednesday. Cathy is additionally barred from Saturday and Monday. That removes Monday, Tuesday, Wednesday and Saturday, leaving exactly three days: Cathy: Sunday, Thursday, Friday. (Thursday-Friday consecutive, Sunday isolated. Valid.) Alex cannot share a day with Cathy, so Alex is barred from Sunday, Thursday and Friday. His third day could not have been Monday or Thursday in any case, since either would extend Tuesday-Wednesday into a three-day run. Only one day survives: Alex: Tuesday, Wednesday, Saturday. Step 3: Force Monday, and with it Ethan. Tuesday and Wednesday are already full (Alex and Dilip), so no one else can work them. Now look at Monday. Alex works only Tuesday, Wednesday and Saturday; Cathy is barred from Monday; Dilip's third day cannot be Monday, as that would extend Tuesday-Wednesday to three in a row. That leaves only Bhabha and Ethan available, and Monday needs exactly two people. So Bhabha and Ethan both work Monday. Saturday needs one more person alongside Alex. Cathy is barred, Bhabha can never share a day with Alex, and Dilip's third day must differ from Alex's. So Ethan works Saturday. Ethan now holds Monday and Saturday, and his third day cannot be Tuesday or Wednesday. Test what remains: * Sunday would give

\{

Saturday, Sunday, Monday

\}

, three consecutive days on the cyclic week. Barred. * Thursday would give

\{

Monday, Thursday, Saturday

\}

, which contains no consecutive pair at all. Barred. * Friday gives

\{

Monday, Friday, Saturday

\}

, exactly one pair (Friday-Saturday). Valid. Ethan: Monday, Friday, Saturday. Step 4: Close out Dilip and Bhabha. Friday now holds Cathy and Ethan, so it is full and Dilip's third day cannot be Friday. It also cannot be Monday or Thursday (three-in-a-row), nor Saturday (his third day must differ from Alex's). One option remains: Dilip: Sunday, Tuesday, Wednesday. Sunday needs three people and currently holds Cathy and Dilip, so Bhabha takes the last Sunday slot. Thursday needs two and holds only Cathy, so Bhabha works Thursday as well. Together with the Monday shift forced in Step 3: Bhabha: Sunday, Monday, Thursday. (Sunday-Monday consecutive, Thursday isolated. Valid.) Master Matrix:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday		✓	✓	✓		3
Monday		✓			✓	2
Tuesday	✓			✓		2
Wednesday	✓			✓		2
Thursday		✓	✓			2
Friday			✓		✓	2
Saturday	✓				✓	2
Works	3	3	3	3	3	15

Reading the grid down the columns: Alex takes Tuesday, Wednesday and Saturday; Bhabha Sunday, Monday and Thursday; Cathy Sunday, Thursday and Friday; Dilip Sunday, Tuesday and Wednesday; Ethan Monday, Friday and Saturday. Every day carries exactly two workers except Sunday, which carries three, and every employee has exactly one consecutive pair plus one isolated day. No other grid satisfies all the constraints. Step 5: Answer this question. Bhabha works Sunday, Monday and Thursday. Cathy works Sunday, Thursday and Friday. They overlap on exactly two days: Sunday and Thursday. Final Answer: 2

Q27:ipmat indore 2022LRDI › ArrangementsEasyMCQ · MCQ

A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage

Which among the following employees do not work together on any of the days?

ABhabha and Dilip
BBhabha and Ethan
CAlex and Dilip
DDilip and Ethan

Pick an option to attempt

The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday		✓	✓	✓		3
Monday		✓			✓	2
Tuesday	✓			✓		2
Wednesday	✓			✓		2
Thursday		✓	✓			2
Friday			✓		✓	2
Saturday	✓				✓	2
Works	3	3	3	3	3	15

Reading off the answer. The question asks which pair never shares a shift, so test each option against the grid: * Bhabha and Dilip both appear in the Sunday row. They work together. * Bhabha and Ethan both appear in the Monday row. They work together. * Alex and Dilip share Tuesday and Wednesday. The clues say so outright, so this pair could never have been the answer. * Dilip works Sunday, Tuesday, Wednesday. Ethan works Monday, Friday, Saturday. These two sets have nothing in common. Dilip and Ethan are the only pair with no day in common. Final Answer: Dilip and Ethan

Q28:ipmat indore 2022LRDI › ArrangementsEasyMCQ · MCQ

A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage

One of the days Alex works on is

AMonday
BFriday
CSunday
DSaturday

Pick an option to attempt

The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday		✓	✓	✓		3
Monday		✓			✓	2
Tuesday	✓			✓		2
Wednesday	✓			✓		2
Thursday		✓	✓			2
Friday			✓		✓	2
Saturday	✓				✓	2
Works	3	3	3	3	3	15

Reading off the answer. Take the Alex column: he works Tuesday, Wednesday and Saturday. Checking that against the options, Monday, Friday and Sunday are all ruled out, and each for a reason worth seeing: * Monday would sit next to Tuesday, extending Tuesday-Wednesday into a three-day run. * Sunday and Friday are both Cathy's days, and Alex never shares a shift with Cathy. That leaves Saturday, which is exactly the isolated third day the grid assigns him. Final Answer: Saturday

Q29:ipmat indore 2022LRDI › ArrangementsEasyMCQ · MCQ

A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage

The consecutive days on which Ethan works are

AThursday and Friday
BSaturday and Sunday
CFriday and Saturday
DSunday and Monday

Pick an option to attempt

The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday		✓	✓	✓		3
Monday		✓			✓	2
Tuesday	✓			✓		2
Wednesday	✓			✓		2
Thursday		✓	✓			2
Friday			✓		✓	2
Saturday	✓				✓	2
Works	3	3	3	3	3	15

Reading off the answer. Take the Ethan column: he works Monday, Friday and Saturday. Every employee's week is one consecutive pair plus one isolated day. Friday and Saturday are adjacent, so they are his pair, and Monday is the isolated day. Note that Ethan could not have taken Sunday instead: with Saturday and Monday already his, adding Sunday would give Saturday, Sunday, Monday, a run of three consecutive days on a week that wraps. Final Answer: Friday and Saturday

Q30:ipmat indore 2022LRDI › ArrangementsEasyMCQ · MCQ

A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage

Employees who work on Sunday are

ABhabha, Cathy and Dilip
BBhabha, Cathy and Ethan
CAlex, Dilip and Ethan
DCathy, Dilip and Ethan

Pick an option to attempt

The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:

Day	Alex	Bhabha	Cathy	Dilip	Ethan	Needed
Sunday		✓	✓	✓		3
Monday		✓			✓	2
Tuesday	✓			✓		2
Wednesday	✓			✓		2
Thursday		✓	✓			2
Friday			✓		✓	2
Saturday	✓				✓	2
Works	3	3	3	3	3	15

Reading off the answer. Sunday is the only day needing three people rather than two. Read across the Sunday row and the three ticks fall under Bhabha, Cathy and Dilip. Alex is excluded because Sunday is one of Cathy's days and he never works with her; Ethan is excluded because Saturday, Sunday, Monday would be three consecutive days. Final Answer: Bhabha, Cathy and Dilip

IPMAT Indore 2022 — MCQ

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