PGDBA 2024 Solutions

These solutions are in interactive format. If you wish to attempt them as timed mock, then please visit this link.

Important Links

Important Links before we start

Interactive Quiz - Question 1

Reading the passage and answer the following questions:
The human brain's billions of neurons present a menagerie of cells that are among both the most highly specialized and variable ones in our bodies. Neurons convert electrical signals to chemical signals, and in humans, their lengths can be so tiny as to span just the tip of a sharpened pencil or, in some cases, even stretch the width of a doorway...Most animals depend on their allotment of neurons for survival. It might stand to reason, then, that the common ancestor of all of these animals also moved about the Earth millions of years ago under the guidance of electrochemical signals transmitted and received by networks of neurons. The idea that these pivotal cells evolved multiple times seems implausible because neurons are highly complex cells and they are also quite similar among animal lineages. But a series of recent evolutionary biology studies are straining the assumption that all animal neurons have a single origin.

The first such finding came from studying relationships among early animals, with a focus on two particular types of organisms: sponges (including sea sponges and freshwater varieties) and ctenophores, invertebrates often known as comb jellies, though they are unrelated to jellyfish. For roughly 15 years, evolutionary biologists have been divided over whether ctenophores or sponges were the first animals to branch from all other animals in the evolutionary tree. Hundreds of millions of years ago the common ancestor to all living animals branched into two species. On one side was the common ancestor of all groups of animals except for one. On the other side was that "one" - the "sister group" that was the first to diverge from all other animals. A persistent question has been whether the sister group was the sponges or ctenophores.

A compelling paper published last year lends strong support to the hypothesis that ctenophores are, in fact, the long-sought sister group. Ctenophores, the researchers found, branched off before sponges and are therefore the group most distantly related to all other animals. Yet despite the new evidence, what exactly happened in evolutionary history is still unsettled because of the puzzle it poses in explaining the evolution of neurons.

Neurons are absent in sponges and present in ctenophores and nearly every other animal on the planet. If ctenophores branched off before sponges in the tree of life, that suggests one of two scenarios for neuron evolution. In one scenario, the precursor to all animals, which lived nearly a billion years ago, had neurons, and every single animal species inherited them. That would mean that sponges must have lost their neurons at some point, because they no longer have the neurons that their ancestors inherited.

An alternative posits that the ancestor to all animals lacked neurons, which explains why early-diverging animals such as sponges have no neurons. Neurons in most animals, then, must have arisen later, after sponges diverged - except for neurons in ctenophores. If the common ancestor lacked neurons, and neurons in most animals arose after ctenophores and sponges had already branched off, then the neurons in ctenophores must have evolved independently. Neurons evolve twice in this scenario-once in ctenophores and then later in other animals - which calls a single origin of neurons into question.

Q1. It was assumed that the common ancestors of all animals were guided by neurons because:

Interactive Quiz - Question 2

Q2. The "common ancestor of all groups of animals except for one" in the passage refers to:

Interactive Quiz - Question 3

Q3. The hypothesis that ctenophores were the long-sought sister group was based on evidence that:

Interactive Quiz - Question 4

Q4. A hypothesis that the precursor to all animals, which lived nearly a billion years ago, had neurons would be strengthened if:

Interactive Quiz - Question 5

Q5. The alternative hypothesis that neurons evolved twice was proposed because:

Interactive Quiz - Question 6

Reading the passage and answer the following questions:

The dream of a planet of almost 8 billion people all living in material comfort will be unachievable if it is based on an economy powered by coal, oil and natural gas. The harms from the cumulative emissions of carbon dioxide would eventually pile up so rapidly that fossil-fuel-fired development would stall.
Nowhere is this logic more pressing than in Asia. About 1.5 billion Asians live in the tropics. Hundreds of millions of them live near the coasts. For their economies to continue to grow, they will need ever more energy. If this comes in the fossil-fuelled manner of past decades they will have to bear the mounting costs of adapting to and living with floods, storms, heatwaves and droughts long before they get rich. As the world heats up, they will have to run faster just to stay in the same place. Zero-emissions technology could free them from this dismal bind: in principle, they can tap into a supply of development-promoting energy that is, in effect, unlimited.

In the long run, therefore, the only way to keep growing is by leaving fossil fuels behind. That requires Asian countries, in most of which emissions are still surging, to forgo much more by way of future emissions than the countries of the developed world, where emissions are already declining. India is vocal in pointing to the unfairness of this, so far refusing to embrace carbon neutrality. Let others with more responsibility for historical emissions do more, it says.

However just that may be, the problem for India-and for everyone else-is that the daunting cost of limiting emissions is falling on a few generations, most of whose members live in developing countries. All of them live in a fractious world where there is a dearth of leadership. America 's government is not suddenly a reliable partner just because it has now rejoined the Paris agreement. Nor is China, the world's largest emitter. Though its capacity for action is great, its pledges thus far are more about posturing than substance. The multilateral institutions created to spread the cost between countries equitably are weak and hostage to procedures based on consensus and unanimity....

As so often in climate change, the task is not choosing between options so much as finding how to press ahead with all of them at once. A commitment to large, fast reductions in methane emissions is vital. More money for developing-country decarbonisation, in which government investment can lower risks for the private sector, must flow alongside increased aid for adaptation. Innovation should be encouraged in various ways. America's 45 -billion-dollar tax incentives for carbon capture could be expanded at home and copied by Europe.

Investment in fossil fuels has fallen faster than replacements have come on line, aggravating the dramatic recent price rises. In the long term it is necessary that fossil fuels become increasingly expensive, but peaks and volatility are destructive. Governments need to build more buffers into the current system as well as hasten the alternatives. When prices fall those still subsidizing fossil fuels will have an excellent opportunity to stop.

Q6. Which of the following statements is NOT true according to the passage?

Interactive Quiz - Question 7

Q7. What would free Asian economies of the "dismal bind" of fossil fuel energy?

Interactive Quiz - Question 8

Q8. Which among the following has to forgo much more by way of future emissions?

Interactive Quiz - Question 9

Q9. Which among the following is vocal in opposing the responsibility of carbon neutrality on the developing world?

Interactive Quiz - Question 10

Q10. Which of the following options pertaining to addressing climate change is NOT mentioned in the passage?

Interactive Quiz - Question 11

Q11. Para Jumbles

Since another effect of these drugs is to reduce inflammation, the suspicion has been that this where the protection is coming from.
Trials show that even people who do not have high cholesterol benefit from being prescribed cheap, cholesterol lowering drugs.
About half of heart attacks happen to those who have what are considered normal levels of cholesterol.
But the cholesterol lowering properties of statins have made it impossible to isolate the effect.

Interactive Quiz - Question 12

Q12. Para Jumbles

Filtration through beds of sand needs no membranes, but does need chemicals called flocculants to persuade pollutants to coagulate so that they can be caught by the filter.
Filtering water may be done through porous membranes, but that requires pressure, and thus needs costly pumps.
An alternative 'slow-sand' filtration employs the layers of algae and bacteria that develop on wet sand grains to remove pollutants and require fewer chemicals.
Also the membranes foul quickly, so require frequent replacement.

Interactive Quiz - Question 13

Q13. Para Jumbles

So, a recent surge of attacks on the vessels in the Red Sea poses a grave threat to global trade.
For the world to prosper, ships must reach their ports.
They are most vulnerable when passing through narrow passages.
The Houthi militants in Yemen have fired over 100 drones and missiles at ships linked to more than 35 countries.

Interactive Quiz - Question 14

Q14. Fill in the blanks

Cancer is a disease that _________ many who are chronic smokers.

Interactive Quiz - Question 15

Q15. Fill in the blanks

___________ I don't usually like spicy food, I enjoyed the food served at the restaurant.

Quiz - Question 157

Logical Reasoning Set

There are six specializations SP1, SP2, SP3, SP4, SP5 and SP6 in a postgraduate degree programme of a university. Each specialization has five specified subjects out of ten subjects S1, S2, S3, S4, S5, S6, S7, S8, S9 and S10. Specified subjects for specializations are given in the following table.

For each subject, there must be 4 hours of class per week (Monday to Friday). The class hours are $9-10,10-11,11-12,12-13,14-15,15-16,16-17$. There is lunch break during 13-14 hours. There can be two consecutive classes for a subject. The class hours 12-13 and 14-15 are not considered consecutive. No subject can be taught on consecutive days in a week or for more than two hours in a day. The class schedule for the 10 subjects has the following features. Classes of S1 are held from 9-11 on Mondays and Wednesdays. For Subject S2, there are consecutive classes during 9-11 on Fridays and consecutive classes during 11-13 on another day. For Subject S3, there are consecutive classes on Mondays and Thursdays, and the Thursday classes are not during 11-13. The classes of Subject S4 are held during 1416 on two days of the week other than Tuesday. There are consecutive classes of Subject S5 on Mondays and Thursdays. The classes of Subject S6 are held during 911 on two days of the week. There are consecutive classes of Subject S7 on two days of the week. For Subject S8, two classes are from 11-13 on Wednesdays and the other classes need not be consecutive. Free class hours of SP3 coincide with those of SP5 and SP6.

Q16. Which subjects are taught on Tuesdays?

Quiz - Question 158

Q17. Which subjects of SP2 are taught on Mondays?

Quiz - Question 159

Q18. The subject S4 is taught on:

Solution:
- The description for S4 states that its classes are "held during

14 - 16

on two days of the week other than Tuesday."
- By further elimination (and by cross-checking with the available answer choices) the only valid pair that fits the overall schedule is Wednesdays and Fridays. (This also ensures that S4 is not on a day immediately adjacent to a day on which many other subjects are fixed, preserving the "no consecutive days" rule.)

Answer: C. Wednesdays and Fridays

Quiz - Question 160

Q19. Which day is an off day (no class) for some specialization?

Quiz - Question 161

Q20. The number of different class schedules that fits the above description is

Interactive Math Quiz

Q26. If the alphabets of the word 'RANDOM' are arranged in a random order, then the probability that there will be exactly two letters between $R$ and $M$ is:

Solution: To find the probability that there are exactly two letters between

R

and

M

in the word "RANDOM," we can solve it as follows:
1. Total Arrangements: The word RANDOM has 6 distinct letters, so there are

6! = 720

possible arrangements.
2. Desired Arrangements: We want exactly two letters between

R

and

M

. Consider the following cases:
-

R

is in position 1, so

M

can only be in position 4.
-

R

is in position 2, so

M

can be in position 5.
-

R

is in position 3, so

M

can be in position 6.
- Likewise, the same holds if we switch

R

and

M

in these positions.
Hence, there are

6

desired pairs of positions for

R

and

M

.
3. Arrangements for Remaining Letters: For each pair of positions for

R

and

M

, the remaining 4 letters can be arranged in

4! = 24

ways.
4. Calculating Probability: Total favorable outcomes

= 6 \times 24 = 144

.
Probability =

\frac{Favorable outcomes}{Total outcomes} = \frac{144}{720} = \frac{1}{5}

.
So, the probability is

1 / 5

Interactive Math Quiz - Question 27

Q27. Two persons C and D are sitting at a round table having a capacity of six persons. The number of ways in which C and D will not sit together is:

Solution: To solve this problem, we first calculate the total number of seating arrangements for six people around a round table and then subtract the cases where C and D sit together.

Total Arrangements Around a Round Table:
When arranging

n

people around a round table, the total number of arrangements is

(n - 1)!

. For six people, the total number of arrangements is:

5! = 120

Arrangements Where C and D Sit Together:
If

C

and

D

are to sit together, we can consider them as a single "block," reducing the problem to arranging five "blocks" (the C-D block and the other four people).
- The number of ways to arrange five blocks around a round table is

4! = 24

.
- Within the

C - D

block, C and D can switch places, giving us 2 arrangements per block arrangement.

Therefore, the number of ways in which C and D sit together is:

24 \times 2 = 48

Arrangements Where C and D Do Not Sit Together:
To find the number of ways in which C and D do not sit together, we subtract the cases where they sit together from the total arrangements:

120 - 48 = 72

Interactive Math Quiz - Question 28

Q28. A, B, and C roll a dice independently in succession with the understanding that the first one to throw a '6' wins. Then the probability that $B$ wins is:

Solution: If they are throwing the dice in the order of

A, B,

and

C

, then:
- Probability of getting 6 (winning) =

\frac{1}{6}

- Probability of not getting 6 (losing) =

\frac{5}{6}

B

can win in the following arrangements only:

\begin{aligned} B_{1} = A_{loses} B_{Wins} \\ B_{2} = A_{loses} B_{loses} C_{loses} A_{loses} B_{Wins} \\ B_{3} = A_{loses} B_{loses} C_{loses} A_{loses} B_{loses} C_{loses} A_{loses} B_{Wins}, \dots \end{aligned}

If we reproduce this arrangement in terms of probability:

\begin{aligned} P (B_{1}) = P (A_{loses}) \cdot P (B_{Wins}) \\ P (B_{2}) = P (A_{loses}) \cdot P (B_{loses}) \cdot P (C_{loses}) \cdot P (A_{loses}) \cdot P (B_{Wins}) \\ P (B_{3}) = P (A_{loses}) \cdot P (B_{loses}) \cdot P (C_{loses}) \cdot P (A_{loses}) \cdot P (B_{loses}) \cdot P (C_{loses}) \cdot P (A_{loses}) \cdot P (B_{Wins}) \end{aligned}

And so on.
Since all these events are mutually exclusive, we sum them up to find the total probability:

\begin{aligned} P (B) & = P (B_{1}) + P (B_{2}) + P (B_{3}) + \dots \\ P (B) & = \frac{5}{6} \cdot \frac{1}{6} + {(\frac{5}{6})}^{4} \cdot \frac{1}{6} + {(\frac{5}{6})}^{7} \cdot \frac{1}{6} + \dots (This forms a GP) \end{aligned}

Solving this GP:

\frac{\frac{5}{6} \cdot \frac{1}{6}}{1 - {(\frac{5}{6})}^{3}} = \frac{30}{91}

Therefore, the probability that

B

wins is

\frac{30}{91}

Interactive Math Quiz - Question 29

Q29. Consider an isosceles trapezium $A B C D$ where $A B$ is parallel to $D C$ . The three vertices $A, B, C$ taken in order are given by $A (2, 0), B (0, 2)$ , and $C (0, 7)$ . Then the coordinates of $D$ will be:

Solution: In this problem, we are given an isosceles trapezium

A B C D

with

A B ∥ D C

, and points

A (2, 0)

B (0, 2)

, and

C (0, 7)

. We need to find the coordinates of point

D

such that

A B ∥ D C

and

A D = B C

(since it's isosceles).

**Step 1:** Identifying the Parallel Sides:
Since

A B ∥ D C

, and we know the coordinates of

A, B,

and

C

, we can infer that

D

must lie on a line parallel to

A B

passing through

C

.

**Step 2:** Slope of

A B

:
To ensure

A B ∥ D C

, we calculate the slope of

A B

and use it to determine the line through

C

that is parallel to

A B

.
- Slope of

A B = \frac{2 - 0}{0 - 2} = - 1

.
Therefore, the line through

C

that is parallel to

A B

also has a slope of

- 1

.

**Step 3:** Equation of Line

D C

:
Since

D

lies on a line parallel to

A B

through

C (0, 7)

, the equation of line

D C

is:

y - 7 = - 1 (x - 0) ⟹ y = - x + 7

**Step 4:** Ensuring

A D = B C

:
- Distance

B C = 5

.
- Therefore,

A D

must also be 5.

**Step 5:** Solving for

D

:
Since

D

lies on the line

y = - x + 7

and must be 5 units away from

A (2, 0)

, we can set up the distance formula for

A D = 5

A D^{2} = 5^{2}

Substituting

D (x, - x + 7)

(x - 2)^{2} + (- x + 7)^{2} = 25

Expanding and solving gives two possible values for

x

x = 7

and

x = 2

.
Finding corresponding

y

-values:
- If

x = 7

y = - 7 + 7 = 0

, so

D = (7, 0)

.
- If

x = 2

y = - 2 + 7 = 5

, so

D = (2, 5)

.

**Conclusion:**
Thus, the possible coordinates for

D

are

(7, 0)

(2, 5)

Interactive Math Quiz - Question 30

Q30. If the perpendicular from the origin $O$ to a line $A B$ meets the line at the point $P$ whose coordinates are $(- 2, 9)$ , then the equation of the line $A B$ is:

Solution:
Slope of

O P \times

slope of

A B = - 1

(Since both are perpendicular).
**Step 1:** Slope of

O P

:
Using the formula for the slope between two points

(y_{2} - y_{1}) / (x_{2} - x_{1})

, the slope of

O P

is:

slope of O P = \frac{9 - 0}{- 2 - 0} = - \frac{9}{2}

Hence, equation (1) becomes:

(- \frac{9}{2}) \times slope of A B = - 1

Solving, we get the slope of

A B = \frac{2}{9}

.

**Step 2:** Equation of line

A B

:
Since

A B

passes through

(- 2, 9)

, the equation of the line with slope

\frac{2}{9}

is given by:

y - y_{1} = m (x - x_{1})

Substituting

(x_{1}, y_{1}) = (- 2, 9)

and

m = \frac{2}{9}

y - 9 = \frac{2}{9} (x + 2)

Simplifying:

9 (y - 9) = 2 (x + 2) ⟹ 9 y - 81 = 2 x + 4 ⟹ 2 x - 9 y + 85 = 0

**Conclusion:**
The equation of the line

A B

is:

2 x - 9 y + 85 = 0

Interactive Math Quiz - Question 31

Q31. The equations of the lines which cut off intercepts on the axes whose sum and product are 1 and -6, respectively, are:

Solution:
The general equation of a line with intercepts

a

and

b

on the

x

- and

y

-axes is:

\frac{x}{a} + \frac{y}{b} = 1

Given: - The sum of intercepts

a + b = 1

- The product of intercepts

a \cdot b = - 6

**Step 1:** Form the quadratic equation for

t

:
Using the relationships

a + b = 1

and

a \cdot b = - 6

, we form:

t^{2} - (a + b) t + a \cdot b = 0

Substituting values:

t^{2} - t - 6 = 0

Factoring gives:

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

So,

t = 3

t = - 2

.

**Step 2:** Possible intercept pairs

(a, b)

:
-

a = 3, b = - 2

a = - 2, b = 3

**Step 3:** Write the equations of the lines:
For

a = 3

and

b = - 2

\frac{x}{3} + \frac{y}{- 2} = 1 ⟹ \frac{x}{3} - \frac{y}{2} = 1

Multiplying through by 6:

2 x - 3 y = 6

For

a = - 2

and

b = 3

\frac{x}{- 2} + \frac{y}{3} = 1 ⟹ - \frac{x}{2} + \frac{y}{3} = 1

Multiplying through by 6:

- 3 x + 2 y = 6

**Conclusion:**
The equations of the lines are:

2 x - 3 y + 6 = 0 and - 3 x + 2 y + 6 = 0

Interactive Math Quiz - Question 32

Q32. The coefficient of $x^{10}$ in the binomial expansion of ${(2 x^{2} - \frac{5}{x})}^{11}$ , when $x \neq 0$ , is:

Solution:
To find the coefficient of

x^{10}

in the expansion of

{(2 x^{2} - \frac{5}{x})}^{11}

, we use the binomial theorem.

**Step 1:** The general term in the expansion is given by:

T_{k + 1} = (\binom{11}{k}) (2 x^{2})^{11 - k} {(- \frac{5}{x})}^{k}

Simplifying:

T_{k + 1} = (\binom{11}{k}) \cdot 2^{11 - k} \cdot (- 5)^{k} \cdot x^{2 (11 - k) - k}

**Step 2:** Find the power of

x

:
Set the exponent of

x

to 10:

2 (11 - k) - k = 10

Solving:

22 - 2 k - k = 10 ⟹ 22 - 3 k = 10 ⟹ 3 k = 12 ⟹ k = 4

**Step 3:** Substitute

k = 4

into the term:
Substituting

k = 4

T_{5} = (\binom{11}{4}) \cdot 2^{11 - 4} \cdot (- 5)^{4} \cdot x^{10}

Simplify:

T_{5} = (\binom{11}{4}) \cdot 2^{7} \cdot 5^{4}

**Conclusion:**
The coefficient of

x^{10}

is:

(\binom{11}{4}) \cdot 2^{7} \cdot 5^{4}

Interactive Math Quiz - Question 33

Q33. Let $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n} = 6 S_{n}$ , then $\frac{S_{3 n}}{S_{n}}$ is:

Solution:
Let

a

and

d

be the first term and common difference, respectively, of the A.P.

The sum of the first

n

terms of an A.P. is given by:

S_{n} = \frac{n}{2} \cdot [2 a + (n - 1) d]

**Step 1: Expressing

S_{2 n}

and

S_{3 n}

:**

S_{2 n} = \frac{2 n}{2} \cdot [2 a + (2 n - 1) d] = n \cdot [2 a + (2 n - 1) d]

S_{3 n} = \frac{3 n}{2} \cdot [2 a + (3 n - 1) d]

**Step 2: Using the given condition

S_{2 n} = 6 S_{n}

:**

n \cdot [2 a + (2 n - 1) d] = 6 \cdot \frac{n}{2} \cdot [2 a + (n - 1) d]

Simplifying:

2 a + (2 n - 1) d = 6 \cdot \frac{1}{2} \cdot [2 a + (n - 1) d]

2 a + (2 n - 1) d = 6 a + 3 (n - 1) d

Rearranging:

- 4 a = d (3 n - 3 - 2 n + 1) ⟹ 2 a = (n d - 2 d)

**Step 3: Finding

\frac{S_{3 n}}{S_{n}}

:**
Substituting

S_{n}

and

S_{3 n}

\frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} \cdot [2 a + (3 n - 1) d]}{\frac{n}{2} \cdot [2 a + (n - 1) d]}

Simplifying:

\frac{S_{3 n}}{S_{n}} = \frac{3 \cdot [2 a + (3 n - 1) d]}{[2 a + (n - 1) d]}

Substituting

2 a = n d - 2 d

\frac{S_{3 n}}{S_{n}} = 15

**Final Answer:**

15

Quiz - Question 111

Q34. If the system of linear equations
\[ \begin{aligned} & x+2y-3z=6 \\ & 2x - y + 4z = 10 \\ & -x + 3y + bz = c \end{aligned} \] has infinitely many solutions, then the value of $ b + c $ is equal to:

Interactive Math Quiz - Question 35

Q35. A circle of radius 6 units touches the coordinate axes in the first quadrant. The equation of its image with respect to the line $y = 0$ is given by:

Solution:
The given circle has a radius of 6, and its center

C

is at

(6, 6)

.

**Step 1: Finding the center of the image:**
Reflect the center

C (6, 6)

over the line

y = 0

(the x-axis). The reflected center

C_{1}

(6, - 6)

.

**Step 2: Equation of the circle with center

(6, - 6)

and radius 6:**
The general equation of a circle is:

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

Substituting

h = 6

k = - 6

, and

r = 6

(x - 6)^{2} + (y + 6)^{2} = 36

Expanding:

x^{2} - 12 x + 36 + y^{2} + 12 y + 36 = 36

Simplify:

x^{2} + y^{2} - 12 x + 12 y + 36 = 0

**Final Answer:**

x^{2} + y^{2} - 12 x + 12 y + 36 = 0

(Option B)

Interactive Math Quiz - Question 36

Q36. Consider a rectangle $A B C D$ whose sides satisfy the following conditions: - $A B$ lies on $y = - 1$ , - $D C$ lies on $y = 3$ , - $A D$ lies on $x = - 3$ , - $B C$ lies on $x = 6$ .

The equation of the circle with diameter $A C$ is:

Solution:
**Step 1: Determine the coordinates of points

A

and

C

:**
- Point

A

lies at the intersection of

y = - 1

and

x = - 3

, so

A (- 3, - 1)

.
- Point

C

lies at the intersection of

y = 3

and

x = 6

, so

C (6, 3)

.

**Step 2: Equation of the circle with diameter

A C

:**
The equation of a circle with diameter endpoints

A (x_{1}, y_{1})

and

C (x_{2}, y_{2})

is:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

Substituting

A (- 3, - 1)

and

C (6, 3)

(x + 3) (x - 6) + (y + 1) (y - 3) = 0

Expanding:

x^{2} - 3 x - 18 + y^{2} - 2 y - 3 = 0

Simplify:

x^{2} + y^{2} - 3 x - 2 y - 21 = 0

**Final Answer:**

x^{2} + y^{2} - 3 x - 2 y - 21 = 0

(Option D)

Interactive Math Quiz - Question 37

Q37. If the latus rectum of an ellipse is equal to half of the minor axis, the eccentricity of the ellipse is given by:

Solution:
Let the equation of the ellipse be:

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

- The length of the latus rectum is

\frac{2 b^{2}}{a}

.
- The length of the minor axis is

2 b

.

**Step 1: Relating the latus rectum to the minor axis**
Given:

\frac{2 b^{2}}{a} = b

Dividing through by

b

(

b \neq 0

\frac{b}{a} = \frac{1}{2}

**Step 2: Finding

\frac{b^{2}}{a^{2}}

:**
Squaring both sides:

\frac{b^{2}}{a^{2}} = \frac{1}{4}

**Step 3: Calculating the eccentricity

e

:**
The eccentricity

e

of the ellipse is:

e = \sqrt{1 - \frac{b^{2}}{a^{2}}}

Substituting

\frac{b^{2}}{a^{2}} = \frac{1}{4}

e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

**Final Answer:**

\frac{\sqrt{3}}{2}

(Option B)

Interactive Math Quiz - Question 38

Q38. A computer sales company reports that $70 %$ of their computers are sold with a Pen drive, $80 %$ with a CD drive, and $60 %$ with both. Let:

P_{1} = Probability that a computer without a Pen drive is sold,

P_{2} = Probability that a computer with a Pen drive or a CD drive is sold,

P_{3} = Probability that a computer with a Pen drive but not a CD drive is sold,

P_{4} = Probability that a computer with a Pen drive or a CD drive but not both is sold.

Which of the following is true?

Solution:
We are given the following probabilities: -

P (Pen drive) = 0.7

P (CD drive) = 0.8

P (Pen drive \cap CD drive) = 0.6

Step 1: Compute

P_{1}

P_{1} = P (No Pen drive) = 1 - P (Pen drive) = 1 - 0.7 = 0.3

Step 2: Compute

P_{2}

P_{2} = P (Pen drive \cup CD drive) = P (Pen drive) + P (CD drive) - P (Pen drive \cap CD drive)

Substituting values:

P_{2} = 0.7 + 0.8 - 0.6 = 0.9

Step 3: Compute

P_{3}

P_{3} = P (Pen drive \cap No CD drive) = P (Pen drive) - P (Pen drive \cap CD drive)

Substituting values:

P_{3} = 0.7 - 0.6 = 0.1

Step 4: Compute

P_{4}

P_{4} = P (Pen drive) + P (CD drive) - 2 \cdot P (Pen drive \cap CD drive)

Substituting values:

P_{4} = 0.7 + 0.8 - 2 \cdot 0.6 = 0.3

Final Values:

P_{1} = 0.3, P_{2} = 0.9, P_{3} = 0.1, P_{4} = 0.3

Step 5: Compare probabilities:

P_{3} < P_{1} = P_{4} < P_{2}

**Correct Answer:**

C

Interactive Math Quiz - Question 39

Q39. If $A = [\begin{matrix} 1 & 0 \\ - 1 & 6 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then the value of $k$ for which $A^{2} = 7 A + k I$ is:

Solution:
Step 1: Define the Matrix

A

and Calculate

A^{2}

A = [\begin{matrix} 1 & 0 \\ - 1 & 6 \end{matrix}]

Compute

A^{2}

A^{2} = A \cdot A = [\begin{matrix} 1 & 0 \\ - 1 & 6 \end{matrix}] \cdot [\begin{matrix} 1 & 0 \\ - 1 & 6 \end{matrix}] = [\begin{matrix} 1 & 0 \\ - 7 & 36 \end{matrix}]

Step 2: Set up the Matrix Equation:
According to the problem statement:

A^{2} = 7 A + k I

Substituting

A

and

I

[\begin{matrix} 1 & 0 \\ - 7 & 36 \end{matrix}] = 7 \cdot [\begin{matrix} 1 & 0 \\ - 1 & 6 \end{matrix}] + k \cdot [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Expanding:

[\begin{matrix} 1 & 0 \\ - 7 & 36 \end{matrix}] = [\begin{matrix} 7 & 0 \\ - 7 & 42 \end{matrix}] + [\begin{matrix} k & 0 \\ 0 & k \end{matrix}]

Step 3: Equate Corresponding Elements:
For the

(1, 1)

element:

1 = 7 + k ⟹ k = - 6

For the

(2, 2)

element:

36 = 42 + k ⟹ k = - 6

**Final Answer:**

- 6

(Option A)

Interactive Math Quiz - Question 40

Q40. If in any triangle, the ratio of the angles is $1 : 2 : 3$ , the corresponding sides will be in the ratio:

Solution:
Step 1: Let the angles of the triangle be

A = x

B = 2 x

, and

C = 3 x

.
From the angle sum property:

A + B + C = 180^{\circ} ⟹ x + 2 x + 3 x = 180^{\circ} ⟹ 6 x = 180^{\circ} ⟹ x = 30^{\circ}

Hence,

A = 30^{\circ}, B = 60^{\circ}, C = 90^{\circ}

.

Step 2: Use the sine rule to determine the ratio of sides:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Substituting the values of the angles:

\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}

Using

\sin 30^{\circ} = \frac{1}{2}

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

, and

\sin 90^{\circ} = 1

\frac{a}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{1}

Simplifying:

a = 1, b = \sqrt{3}, c = 2

Step 3: Conclude the ratio:

a : b : c = 1 : \sqrt{3} : 2

**Final Answer:**

1 : \sqrt{3} : 2

(Option C)

Interactive Math Quiz - Question 41

Q41. The value of $15 \sin \frac{5 π}{12} + 15 \cos \frac{5 π}{12} - 20 \sin^{3} \frac{5 π}{12} - 20 \cos^{3} \frac{5 π}{12}$ is:

Solution:
Step 1: Express the sum of sine and cosine terms:

15 \sin \frac{5 π}{12} + 15 \cos \frac{5 π}{12}

Using the identity:

\sin x + \cos x = \sqrt{2} \sin (x + \frac{π}{4})

This becomes:

15 \sqrt{2} \sin (\frac{5 π}{12} + \frac{π}{4})

Step 2: Express the cubic terms:

- 20 \sin^{3} \frac{5 π}{12} - 20 \cos^{3} \frac{5 π}{12}

Using the identity:

\sin^{3} x + \cos^{3} x = (\sin x + \cos x) (1 - \sin x \cos x)

Substitute

\sin x \cos x = \frac{1}{2} \sin (2 x)

:
For

x = \frac{5 π}{12}

, this becomes:

\sin x \cos x = \frac{1}{2} \sin (\frac{5 π}{6}) = \frac{1}{4}

Substituting back:

- 20 \cdot \frac{3 \sqrt{2}}{4} \sin (\frac{5 π}{12} + \frac{π}{4}) = - 15 \sqrt{2} \sin (\frac{5 π}{12} + \frac{π}{4})

Step 3: Combine terms:

15 \sqrt{2} \sin (\frac{5 π}{12} + \frac{π}{4}) - 15 \sqrt{2} \sin (\frac{5 π}{12} + \frac{π}{4}) = 0

**Final Answer:**

A

Interactive Math Quiz - Question 42

Q42. Which term of the G.P. $5, 10, 20, 40, \dots$ is 2560?

Solution:
The

n^{th}

term of a geometric progression (G.P.) is given by:

T_{n} = first term \times (common ratio)^{n - 1}

Here: - First term = 5 - Common ratio = 2

Substituting into the formula:

2560 = 5 \times 2^{n - 1}

Divide both sides by 5:

512 = 2^{n - 1}

Taking the base-2 logarithm:

n - 1 = \log_{2} (512) = 9

Therefore:

n = 10

**Final Answer:**

C

Interactive Math Quiz - Question 43

Q43. The solution set of the given inequation $| \frac{2}{x - 3} | > 1, x \neq 3$ in $R$ , is:

Solution:
Step 1: Rewrite the inequality:

| \frac{2}{x - 3} | > 1

Step 2: Multiply both sides by

| x - 3 |

:
Since

| x - 3 | > 0

(because

x \neq 3

), we can safely multiply:

2 > | x - 3 |

Step 3: Interpret the absolute value:

| x - 3 | < 2 implies - 2 < x - 3 < 2

Step 4: Solve the compound inequality:
Add 3 to all parts:

1 < x < 5

Step 5: Exclude

x = 3

:
Since

x \neq 3

, exclude this value from the interval:

(1, 3) \cup (3, 5)

**Final Answer:**

A

Interactive Math Quiz - Question 44

Q44. For any real number $x$ , let $[x]$ denote the greatest integer less than or equal to $x$ . For the function $f (x) = (x - [x])^{- \frac{1}{2}}$ , let $P$ and $Q$ denote the domain and range respectively. Then which of the following is correct?

Solution:
The function

f (x) = (x - [x])^{- \frac{1}{2}} = \frac{1}{\sqrt{{x}}}

, where

{x}

is the fractional part of

x

.

- The fractional part

{x} \in [0, 1)

. Since it is in the denominator,

{x} \in (0, 1)

.
- Thus,

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

has a range

Q = (1, \infty)

. - The domain

P

excludes values where

{x} = 0

, which means it includes all real numbers except

x

that are integers.

Only Option D satisfies these conditions:

5 \in Q

.

**Final Answer:**

D

Interactive Math Quiz - Question 45

Q45. Let $f (x) = {\begin{cases} x^{12} - 1, & x \leq 1 \\ x^{4}, & x > 1 \end{cases}$ .
Then, which of the following is correct?

Solution:
Polynomial functions are continuous for all values in their domain.
For

x \leq 1

x^{12} - 1

(polynomial) is continuous.
For

x > 1

x^{4}

(polynomial) is continuous.

The only point where discontinuity may occur is at

x = 1

.
Step 1: Evaluate

f (x)

from the left at

x = 1

lim_{x \to 1^{-}} f (x) = 1^{12} - 1 = 0

Step 2: Evaluate

f (x)

from the right at

x = 1

lim_{x \to 1^{+}} f (x) = 1^{4} = 1

Since

LHL \neq RHL

f (x)

is not continuous at

x = 1

.

**Final Answer:**

C

Interactive Math Quiz - Question 46

Q46. Let $y (x) = 2^{2} \cdot (2^{2 \log_{2} x}), x > 0$ .
Then the value of $\frac{d y}{d x}$ at $x = 2$ is:

Solution:

y (x) = 2^{2} \cdot 2^{2 \log_{2} x}

Step 1: Simplify

y (x)

Using the properties of exponents

a^{m} \cdot a^{n} = a^{m + n}

, we get:

y (x) = 2^{2 + 2 \log_{2} x}

Factor the exponent:

y (x) = 2^{2 (1 + \log_{2} x)}

Using the identity

\log_{2} x + 1 = \log_{2} (2 x)

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

Simplify further:

y (x) = (2 x)^{2} = 4 x^{2}

Step 2: Differentiate

y (x) = 4 x^{2}

with respect to

x

\frac{d y}{d x} = 4 \cdot 2 x = 8 x

x = 2

\frac{d y}{d x} = 8 \cdot 2 = 16

**Final Answer:**

D

Interactive Math Quiz - Question 47

Q47. Let $f : R \to R$ be a differentiable function such that $f (1) = 1, f^{'} (1) = 2$ .
Then $lim_{x \to 1} \frac{\sqrt{4 f (x)} - 2}{\sqrt{x} - 1}$ is:

Solution:

lim_{x \to 1} \frac{\sqrt{4 f (x)} - 2}{\sqrt{x} - 1}

Step 1: Apply L'Hôpital's Rule since direct substitution gives an indeterminate form.
Differentiate the numerator and denominator with respect to

x

:
Numerator:

\frac{d}{d x} (\sqrt{4 f (x)} - 2) = \frac{1}{2 \sqrt{4 f (x)}} \cdot 4 f^{'} (x) = \frac{2 f^{'} (x)}{\sqrt{4 f (x)}}

Denominator:

\frac{d}{d x} (\sqrt{x} - 1) = \frac{1}{2 \sqrt{x}}

The limit becomes:

lim_{x \to 1} \frac{\frac{2 f^{'} (x)}{\sqrt{4 f (x)}}}{\frac{1}{2 \sqrt{x}}} = lim_{x \to 1} \frac{2 f^{'} (x) \cdot 2 \sqrt{x}}{\sqrt{4 f (x)}}

Step 2: Substitute the given values at

x = 1

f (1) = 1, f^{'} (1) = 2

Substitute into the expression:

\frac{2 \cdot 2 \cdot \sqrt{1}}{\sqrt{4 \cdot 1}} = \frac{4}{2} = 4

**Final Answer:**

C

Interactive Math Quiz - Question 48

Q48. The minimum distance from a point $P$ on the ellipse $\frac{x^{2}}{4} + y^{2} = 1$ to $(1, 0)$ is:

Solution:
To find the minimum distance from a point

P

on the ellipse given by the equation

\frac{x^{2}}{4} + y^{2} = 1

to the point

(1, 0)

, we can use calculus to minimize the distance function.

Step 1: Define the distance function:
The distance

d

between a point

(x, y)

on the ellipse and the point

(1, 0)

is given by:

d = \sqrt{(x - 1)^{2} + y^{2}}

Step 2: Use the ellipse equation to express

y^{2}

:
From the ellipse equation,

y^{2} = 1 - \frac{x^{2}}{4}

.
Substitute this into the distance function:

d = \sqrt{(x - 1)^{2} + 1 - \frac{x^{2}}{4}}

Step 3: Simplify the expression:
Simplify under the square root to form a function of

x

only:

d = \sqrt{1 - \frac{x^{2}}{4} + x^{2} - 2 x + 1}

Simplify further:

d = \sqrt{\frac{3 x^{2}}{4} - 2 x + 2}

Step 4: Differentiate

d^{2}

with respect to

x

:
Differentiating

d^{2}

(since minimizing

d

and

d^{2}

yields the same

x

values):

\frac{d}{d x} (d^{2}) = \frac{d}{d x} (\frac{3 x^{2}}{4} - 2 x + 2) = \frac{3}{2} x - 2

Step 5: Set the derivative equal to zero to find critical points:

\frac{3}{2} x - 2 = 0 ⟹ \frac{3}{2} x = 2 ⟹ x = \frac{4}{3}

x = \frac{4}{3}

, the corresponding

y

value is:

y^{2} = 1 - \frac{(\frac{4}{3})^{2}}{4} = \frac{5}{9} ⟹ y = \pm \sqrt{\frac{5}{9}}

The point

P

(\frac{4}{3}, \pm \sqrt{\frac{5}{9}})

.
Step 6: Calculate the minimum distance:
The minimum distance from

P

(1, 0)

is:

\sqrt{\frac{2}{3}}

**Final Answer:**

B

Interactive Math Quiz - Question 49

Q49. The integral $\int \frac{1}{\sqrt{\sin^{3} x \cos^{5} x}} d x$ is equal to:

Solution:
Let

\tan x = t

, so:

\sec^{2} x d x = d t, \sin x = \frac{t}{\sqrt{1 + t^{2}}}, \cos x = \frac{1}{\sqrt{1 + t^{2}}} .

Now substitute

d x

d x = \frac{d t}{1 + t^{2}} .

\sqrt{\sin^{3} x} = \sqrt{{(\frac{t}{\sqrt{1 + t^{2}}})}^{3}} = \frac{t^{3 / 2}}{(1 + t^{2})^{3 / 4}}, \sqrt{\cos^{5} x} = \sqrt{{(\frac{1}{\sqrt{1 + t^{2}}})}^{5}} = \frac{1}{(1 + t^{2})^{5 / 4}} .

Now the original integrand becomes:

\frac{1}{\sqrt{\sin^{3} x \cos^{5} x}} = \frac{1}{\frac{t^{3 / 2}}{(1 + t^{2})^{3 / 4}} \cdot \frac{1}{(1 + t^{2})^{5 / 4}}} .

Simplify:

\frac{1}{\sqrt{\sin^{3} x \cos^{5} x}} = \frac{(1 + t^{2})^{2}}{t^{3 / 2}} .

Substitute

d x = \frac{d t}{1 + t^{2}}

, so the integral becomes:

\int \frac{d x}{\sqrt{\sin^{3} x \cos^{5} x}} = \int \frac{(1 + t^{2})^{2}}{t^{3 / 2}} \cdot \frac{d t}{1 + t^{2}} .

\int \frac{(1 + t^{2})^{2}}{t^{3 / 2}} \cdot \frac{1}{1 + t^{2}} d t = \int \frac{1 + t^{2}}{t^{3 / 2}} d t .

Expand and simplify:

\int \frac{1 + t^{2}}{t^{3 / 2}} d t = \int \frac{1}{t^{3 / 2}} d t + \int \frac{t^{2}}{t^{3 / 2}} d t .

Simplify each term:
1.

\int \frac{1}{t^{3 / 2}} d t = \int t^{- 3 / 2} d t = - 2 t^{- 1 / 2} = - \frac{2}{\sqrt{t}} .

\int \frac{t^{2}}{t^{3 / 2}} d t = \int t^{1 / 2} d t = \frac{2}{3} t^{3 / 2} .

Add them together:

\int \frac{1 + t^{2}}{t^{3 / 2}} d t = - \frac{2}{\sqrt{t}} + \frac{2}{3} t^{3 / 2} + C .

Back-substitute

t = \tan x

\int \frac{1}{\sqrt{\sin^{3} x \cos^{5} x}} d x = - \frac{2}{\sqrt{\tan x}} + \frac{2}{3} (\tan x)^{3 / 2} + C .

Thus, the correct answer is

A

Interactive Math Quiz - Question 50

Q50. The area bounded between the lines $x = 1, x = 2$ and the curves $y_{1} (x) = - x^{2} + 4 x + 1$ , $y_{2} (x) = - x^{3} + 7 x^{2} - 10 x + 3$ is:

Solution:
The area is given by the integral of the difference between the upper curve and the lower curve over the interval

[1, 2]

.

The two curves are:
1.

y_{1} (x) = - x^{2} + 4 x + 1

(parabola)
2.

y_{2} (x) = - x^{3} + 7 x^{2} - 10 x + 3

(cubic)

First, determine which curve is above the other by comparing

y_{1}

and

y_{2}

in the interval

[1, 2]

.
At

x = 1

y_{1} = - (1)^{2} + 4 (1) + 1 = 4, y_{2} = - (1)^{3} + 7 (1)^{2} - 10 (1) + 3 = - 1.

So,

y_{1} > y_{2}

x = 1

.
At

x = 2

y_{1} = - (2)^{2} + 4 (2) + 1 = 1, y_{2} = - (2)^{3} + 7 (2)^{2} - 10 (2) + 3 = 1.

Thus,

y_{1} = y_{2}

x = 2

.
The parabola

y_{1}

is above the cubic

y_{2}

throughout the interval.

The area

A

is:

A = \int_{1}^{2} [(- x^{2} + 4 x + 1) - (- x^{3} + 7 x^{2} - 10 x + 3)] d x .

Simplify the integrand:

A = \int_{1}^{2} (x^{3} - 8 x^{2} + 14 x - 2) d x .

Integrate term by term:

\int x^{3} d x = \frac{x^{4}}{4}, \int x^{2} d x = \frac{x^{3}}{3}, \int x d x = \frac{x^{2}}{2}, \int 1 d x = x .

Applying this:

A = {[\frac{x^{4}}{4} - \frac{8 x^{3}}{3} + 7 x^{2} - 2 x]}_{1}^{2} .

Evaluating at the limits:

A = (\frac{16}{4} - \frac{64}{3} + 28 - 4) - (\frac{1}{4} - \frac{8}{3} + 7 - 2) .

Simplify to get:

A = \frac{49}{12} .

Thus, the area is

\frac{49}{12}

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