Problem Solving and Data Analysis MCQ Quiz in తెలుగు - Objective Question with Answer for Problem Solving and Data Analysis - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Given that the ratio \(\frac{a}{b} = 3\), we can express this as \(a = 3b\). We also have \(\frac{9a}{mb} = 3\). Substituting \(a = 3b\) into this equation gives \(\frac{9(3b)}{mb} = 3\). Simplifying, we have \(\frac{27b}{mb} = 3\). Dividing both sides by \(b\) results in \(\frac{27}{m} = 3\). Solving for \(m\), we multiply both sides by \(m\) to get \(27 = 3m\). Dividing both sides by 3, we find \(m = 9\). Therefore, the value of \(m\) is 9.

The given ratio \(\frac{p}{q} = 5\) implies that \(p = 5q\). Substituting this into the second equation \(\frac{35p}{tq} = 5\) yields \(\frac{35(5q)}{tq} = 5\). Simplifying, we get \(\frac{175q}{tq} = 5\). Canceling \(q\) gives \(\frac{175}{t} = 5\). Solving for \(t\), we multiply both sides by \(t\) to obtain \(175 = 5t\). Dividing both sides by 5 results in \(t = 35\). Thus, the value of \(t\) is 35.

To find out how many widgets the factory produces in a day, we first calculate how many widgets it produces in one hour. At 30 widgets per minute, in one hour (60 minutes), the factory produces 30 × 60 = 1,800 widgets. Since the factory operates for 8 hours a day, the total production in a day is 1,800 × 8 = 14,400 widgets. Therefore, the correct answer is option 1, 14,400 widgets. The other options are incorrect as they do not correctly calculate the widgets produced in a day based on the given rate and operating hours.

The even numbers between 1 and 14 are 2, 4, 6, 8, 10, 12, and 14, making a total of 7 even numbers. Therefore, the probability is \(\frac{7}{14}\). Option 1 is the correct choice. Options 2, 3, and 4 incorrectly represent the count of even numbers.

The problem-solving method is an instructional approach that engages students in identifying, analyzing, and solving problems through critical thinking and inquiry.

Key Points

• Being a structured and inquiry-based approach, the problem-solving method is generally not a time-saving method.
• In fact, it often requires more time than traditional methods because students explore multiple possibilities, gather information, test solutions, and reflect on their findings.
• The focus is on deep understanding rather than quick coverage of content.
• While it is highly beneficial in promoting meaningful learning, it demands considerable classroom time for planning, execution, and discussion.

Hint

•  Helping in the development of scientific attitude is a key strength, as students learn to think logically, question assumptions, and make evidence-based conclusions.
• Training in the scientific method occurs naturally in this approach since learners go through steps such as problem identification, hypothesis formation, experimentation, and evaluation.
• Being a learner-centered method, it places the student at the core of the learning process, encouraging autonomy, curiosity, and active participation.

Hence, the correct answer is time saving method.

Explanation:

Given:

\(P(A) = P(\frac{A}{B}) = 0.25\)

and \(P(\frac{B}{A}) = 0.5\)

I. \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(A)}\)

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(B)}\)

⇒ \(P(B)= \frac{P(A∩ B)}{P(\frac{A}{B})}\)

⇒ \(P(B) = \frac{0.125}{0.25} = 0.5\)

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. \(P(\overline A\cup \overline B ) = 1 – P(A ∩ B)\)

= 1 – 0.125 = 0.875

III. \(P(\overline A∩ \overline B ) = 1 – P(A \cup B)\)

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

m = x₂−x₁/ y_2​ − y_1​​

Substituting the points (x1,y1)=(2,18)Unknown node type: span" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">(x1,y1)=(2,18)Unknown node type: span(x1,y1)=(2,18)<merror><mtext>Unknown node type: span</mtext></merror> (x1,y1)=(2,18)<span style="font-family: var(--bs-body-font-family); font-size: var(--bs-body-font-size); font-weight: var(--bs-body-font-weight); text-align: var(--bs-body-text-align);"> </span> (x2,y2)=(5,22) (x_2, y_2) = (5, 22)" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">(x2,y2)=(5,22)$(x_2, y_2) = (5, 22)$ $(x_2, y_2) = (5, 22)$

 

$$m = \frac{22 - 18}{5 - 2}$$ m=43=1.33

 

The slope m m" id="MathJax-Element-13-Frame" role="presentation" style="position: relative;" tabindex="0">m$m$ $m$  of the best-fit line is 1.33, meaning for every 1 unit increase in X, Y increases by 1.33 units. $$m = \frac{4}{3} = 1.33$$
 

Since the population decreases by a fixed percentage each month, this follows an exponential decay model, making option A (Decreasing exponential) the correct choice.

Solution:

• (A) Incorrect: The data points show that individuals weighing 65 pounds have more offspring than those weighing 50 pounds. So, this statement is false.
• (B) Correct: The number of offspring increases as weight increases, which aligns with what we observe in the graph.
• (C) Incorrect: The number of offspring is not constant—it changes with weight, so this statement is false.
• (D) Incorrect: The lowest weight (40 pounds) corresponds to the lowest number of offspring (5), not the highest.

Solution:

We need to count the days where the probability of rain is 70% or more.
From the given data:

Tuesday → 60% ( Does not qualify)
Wednesday → 90% (Qualifies)
Thursday → 30% (Does not qualify)
Friday → 70% ( Qualifies)
There are 2 favorable days (Wednesday & Friday) out of 4 total days.
Thus, the probability is: 2/4 = 1/2
​
Final Answer:
Option B) 1/2 (or 50%)