Mathematics and Science PYQs

Given Expression:

\( 31096 \div (2704 \div 2) + 0.6738 + 0.63 + 0.059 - 0.79 \)

Step 1: Simplify the Division Inside Parentheses

2704 ÷  2 = 1352

Step 2: Perform the Main Division

\( 31096 \div 1352 \)

Simplify by dividing numerator and denominator by 8:

\( \frac{31096 \div 8}{1352 \div 8} = \frac{3887}{169} = 23 \)

Step 3: Add the Decimal Terms

0.6738 + 0.63 = 1.3038

1.3038 + 0.059 = 1.3628

Step 4: Subtract the Final Term

1.3628 - 0.79 = 0.5728

Step 5: Combine Results

23 + 0.5728 = 23.5728

Final Answer:

23.5728

Option 1: AB = 3 cm, BC = 4 cm, AC = 7 cm

Check Triangle Inequality:

\(3 + 4 > 7 \quad \Rightarrow \quad 7 > 7\) (False)

Fails the triangle inequality. Cannot be constructed.

Option 2: AB = 8 cm, BC = 4 cm, \(\angle BCA = 90^\circ\), \(\angle ABC = 60^\circ\)

Angle Sum: \(\angle BAC = 30^\circ\) (Valid)

Law of Sines Check: \(\frac{8}{1} = \frac{4}{0.5} \quad \Rightarrow \quad 8 = 8\) (Consistent)

Valid angles and sides. Can be constructed.

Option 3: \(\angle ACB = 120^\circ\), \(\angle ABC = 25^\circ\), AC = BC = 4 cm

Angle Sum: \(\angle BAC = 35^\circ\) (Valid)

Law of Sines Check: AB ≈ 8.19 cm (Valid), but AC cannot be equal to BC, as opposite angles are different.

Valid angles and sides. Cannot be constructed.

Option 4: AB = 5 cm, BC = 6 cm, \(\angle ABC = 130^\circ\), \(\angle BAC = 25^\circ\)

Angle Sum: \(\angle ACB = 25^\circ\) (Valid)

Law of Sines Check: \(\frac{5}{0.423} \approx \frac{6}{0.423}\) → Inconsistent

Fails the Law of Sines. Cannot be constructed.

Final Answer: \(\boxed{\text{Options 2}}\)

Given:

Polynomial: 20pq² + 4pq + 45pr² - 60pqr - 6rp

Calculation:

⇒ 20pq² + 4pq + 45pr² - 60pqr - 6rp

Take p as a common factor from the entire expression.

⇒ p (20q² - 60qr + 45r² + 4q - 6r) ---- (1)

Now, let's factor the quadratic expression in the parentheses.

⇒ 20q² - 60qr + 45r²

Take 5 as a common factor from these three terms.

⇒ 5 (4q² - 12qr + 9r²)

The expression inside the parenthesis is a perfect square trinomial.

⇒ (2q)² - 2 × (2q) × (3r) + (3r)²

⇒ (2q - 3r)²

So, the expression (1) becomes:

⇒ p [5 (2q - 3r)² + 4q - 6r]

⇒ p [5 (2q - 3r)² + 2 (2q - 3r)]

Now, we can take (2q - 3r) as a common factor from the terms inside the square brackets.

⇒ p (2q - 3r) [5 (2q - 3r) + 2]

⇒ p (2q - 3r) (10q - 15r + 2)

The factors of the polynomial are p, (2q - 3r), and (10q - 15r + 2).

∴ One of the factors is (2q - 3r) or (10q - 15r + 2).

Hence, the correct answer is "Option 2".

Assessment in education serves multiple purposes; it can be diagnostic, formative, summative, or even used as a tool for self and peer learning.

Key Points

• Assessment as learning is a process where students actively evaluate their own and their peers' learning.
• It promotes metacognition, where learners reflect on their thinking and understanding.
• This approach encourages self-assessment and peer-assessment to foster independence, responsibility, and deeper engagement with the material.

Hint

• Summative Assessment indeed evaluates if instructional objectives are met, but it is usually done at the end of a unit or term.
• The statement that “Assessment for learning is a process with no scope of giving feedback” is incorrect. Assessment for learning emphasizes continuous feedback to guide and improve learning during the teaching process.
• Formative Assessment is not used after course completion; it occurs during the learning process to inform ongoing teaching and learning. Using it only to assign grades after completion would make it summative, not formative.

Hence, the correct answer is Assessment as learning generates opportunities for self-assessment and peer assessment.

Step 1: Total Number of Cards

Numbers range from 12 to 71 (inclusive):

\(71 - 12 + 1 = 60\) cards

Step 2: Prime Numbers Between 12 and 71

Primes: 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71

\(\text{Total primes} = 15\)

Step 3: Non-Prime Numbers

\(60 - 15 = 45\) non-prime numbers

Step 4: Probability Calculation

\(P(\text{Non-prime}) = \frac{45}{60} = \frac{3}{4}\)

Mathematics is an essential discipline with applications across numerous fields and professions. The value of mathematics extends beyond its academic pursuit and is seen in its practical uses in everyday life, career opportunities, and societal functions.

Key Points

• The statement focuses on how mathematics is used directly or indirectly in various professions such as accountancy, banking, shop-keeping, and tailoring.
• This reflects the utilitarian value of mathematics, meaning its usefulness in real-world applications and practical problem-solving.
• The primary purpose of mathematics in these fields is not theoretical but serves to facilitate day-to-day tasks and decision-making processes.

Hint

• Disciplinary value refers to mathematics as a formal discipline with its own principles, theories, and methods. It focuses on the internal structure and beauty of the subject rather than its practical application.
• Cultural value would refer to how mathematics influences or is influenced by cultural practices, traditions, or intellectual history, not its use in professions.
• Professional value implies how mathematics directly impacts specific professions but doesn’t capture the broader, general practical application across a variety of jobs as the utilitarian value does.

Hence, the correct answer is Utilitarian value of Mathematics.

In the classroom, various types of records and assessment tools are used to track student progress, skills, and abilities. These tools help teachers understand how well students are mastering particular concepts, and provide insights into areas where students might need additional support or instruction.

Key Points

• To assess problem-solving ability, measurement ability, and experimentation ability, the Cumulative Record is the most appropriate tool.
• This record provides a comprehensive overview of a student's progress over time, including their performance in various skills such as problem-solving, measurements, and experiments.
• It keeps a detailed account of the student's achievements and difficulties in specific academic areas, giving a complete picture of their development in these key areas.

Hint

• Report Card is generally used to communicate students' grades and academic performance in a specific period (e.g., a term or semester), but it doesn’t provide a detailed breakdown of skills like problem-solving or experimentation.
• Anecdotal Record refers to brief, qualitative descriptions of individual students' behaviors or incidents observed by the teacher. While helpful for tracking certain behaviors or interactions, it doesn't provide a structured method to assess specific abilities like measurement or problem-solving.
• Teacher Diary is a personal reflection tool for teachers to record daily observations, thoughts, and teaching strategies. It’s not designed to specifically assess student abilities in areas like experimentation or problem-solving.

Hence, the correct answer is Cumulative Record.

Option 1: \(45 \times 10^3\)

Coefficient: 45 (must be ≥1 and <10).

Not in standard form.

Option 2: \(8.5 \times 10^{-12}\)

Coefficient: 8.5 (valid). Exponent: −12 (integer).

Valid standard form.

Option 3: \(0.6 \times 10^{15}\)

Coefficient: 0.6 (must be ≥1).

Not in standard form.

Option 4: \(83.7 \times 10^{-9}\)

Coefficient: 83.7 (must be <10).

Not in standard form.

Final Answer: \(\boxed{8.5 \times 10^{-12}}\)

Given:

• Parallel sides: AB (shorter) and CD = \(24\) cm
• Distance between AB and CD: \(h = 12\) cm
• Non-parallel sides: AD = BC = \(13\) cm

Diagram of trapezium ABCD with AB || CD, height = 12 cm

Step 1: Find Length of AB

Drop perpendiculars from A and B to CD, meeting at P and Q respectively.

Using Pythagorean theorem in right triangles APD and BQC:

\(PD = \sqrt{AD^2 - AP^2} = \sqrt{13^2 - 12^2} = \sqrt{25} = 5\) cm

\(CQ = \sqrt{BC^2 - BQ^2} = \sqrt{13^2 - 12^2} = 5\) cm

Since CD = AB + PD + CQ:

\(24 = AB + 5 + 5 \implies AB = 14\) cm

Step 2: Calculate Area

Area of trapezium formula:

\(A = \frac{1}{2} \times (AB + CD) \times h\)

\(A = \frac{1}{2} \times (14 + 24) \times 12 = \frac{1}{2} \times 38 \times 12 = 228\) cm²

Final Answer: \(\boxed{228}\) cm²

At the elementary level, mathematics textbooks play a crucial role in shaping how young learners grasp new ideas. Effective textbooks balance clarity with engagement and guide students through a logical progression of concepts.

 Key Points

• A key principle in early mathematics education is to build understanding from tangible experiences before moving to abstract representations.
• Explanation of Concepts Presented from Concrete to Abstract: Presenting ideas in a concrete form allows students to manipulate objects and visualize mathematical relationships.
• By using hands‑on activities such as counting blocks or drawing shapes, learners can internalize the underlying concept.
• Once the concrete foundation is secure, the teacher can introduce pictorial representations that bridge to symbols.
• Gradually, students gain familiarity with numerals and operations as abstractions of those concrete models. This progression mirrors cognitive development and reduces confusion when symbols replace objects.
• As a result, learners develop a deeper understanding and retain concepts more effectively.

Hint

• Using highly precise and technical language can overwhelm young readers and hinder comprehension.
• Filling books with many similar practice questions may lead to rote repetition rather than genuine understanding.
• Introducing concepts from complex to simple reverses the natural learning sequence and can confuse students who lack foundational experiences.

Hence, the correct answer is ‘Concepts should be presented from concrete to abstract’.

Given:

• Original cube side length: \(12 \text{ cm}\)
• Small cube side length: \(4 \text{ cm}\)

Concept:

• Total Surface Area (TSA) of a cube: \(6 \times (\text{side length})^2\)
• When a cube is divided into smaller cubes, the total surface area changes based on the number of small cubes formed.

Calculation:

1. TSA of the Original Cube:

\(\text{TSA}_{\text{original}} = 6 \times (12)^2 = 6 \times 144 = 864 \text{ cm}^2\)
2. Number of Small Cubes Formed:

\(\text{Along one edge: } \frac{12}{4} = 3\)

\(\text{Total small cubes: } 3 \times 3 \times 3 = 27\)
3. TSA of One Small Cube:

\(\text{TSA}_{\text{small}} = 6 \times (4)^2 = 6 \times 16 = 96 \text{ cm}^2\)
4. Total TSA of All Small Cubes:

\(\text{TSA}_{\text{new}} = 27 \times 96 = 2592 \text{ cm}^2\)
5. Ratio of New TSA to Original TSA:

\(\text{Ratio} = \frac{\text{TSA}_{\text{new}}}{\text{TSA}_{\text{original}}} = \frac{2592}{864} = 3\)

Result:

The new total surface area is thrice the original total surface area.

When a set of numbers is arranged in ascending order, each number is greater than the one before. Multiplying every term in such an inequality by a negative number reverses the direction of each comparison. This happens because negative multiplication flips the sign of each product, altering their relative magnitudes.

Key PointsMultiplying the ordered list by –1/2 transforms each term:

• 2 becomes – -1
• 7 becomes – -3.5
• 9 becomes – -4.5
• 12 becomes – -6
• This new ascending order is the exact reverse of the original sequence (2, 7, 9, 12).
• Therefore, the original ascending order is reversed, resulting in a descending order of the original numbers.

Hint

• Mean remains affected by scaling and changes magnitude and sign, so it does not stay the same.
• The total count of terms is unaffected by multiplication, so the number of terms remains four.
• Retaining the same ascending order cannot occur because the direction of each inequality flips under multiplication by a negative.

Hence, the correct answer is ‘Ascending order will now be changed to descending order’.

Given Data:

Number of wickets taken in last 11 matches:

1, 0, 2, 4, 0, 3, 1, 1, 2, 4, 1

Step 1: Calculate the Mean

Mean = (Sum of all values) / (Number of values)

Sum = 1 + 0 + 2 + 4 + 0 + 3 + 1 + 1 + 2 + 4 + 1 = 19

Number of matches = 11

Mean = \(\frac{19}{11} \approx 1.727\)

Step 2: Determine the Mode

Mode = Most frequently occurring value

Frequency count:

• 0 appears 2 times
• 1 appears 4 times
• 2 appears 2 times
• 3 appears 1 time
• 4 appears 2 times

Mode = \(1\) (highest frequency)

Step 3: Calculate the Difference

Difference = Mean - Mode

= 1.727 - 1 = 0.727

Given:

Two corresponding angles formed by the intersection of two parallel lines by a transversal:

\((3x - 20^\circ)\) and \((5x - 60^\circ)\)

Concept:

Corresponding angles formed when a transversal intersects two parallel lines are equal.

Step 1: Set the angles equal to each other

\(3x - 20^\circ = 5x - 60^\circ\)

Step 2: Solve for x

Subtract 3x from both sides:

\(-20^\circ = 2x - 60^\circ\)

Add 60° to both sides:

\(40^\circ = 2x\)

Divide both sides by 2:

\(x = 20^\circ\)

Step 3: Calculate (4x + 30°)

Substitute x = 20°:

\(4(20^\circ) + 30^\circ = 80^\circ + 30^\circ = 110^\circ\)

Given:

A nine-digit number \(565x9274y\) divisible by 88.

Concept:

For a number to be divisible by 88, it must be divisible by both 8 and 11 (since 88 = 8 × 11).

Step 1: Apply divisibility rule for 8

A number is divisible by 8 if its last three digits form a number divisible by 8.

Last three digits: \(74y\)

Possible values for y (0-9) that make 74y divisible by 8:

\(744 ÷ 8 = 93\) → y = 4

\(752 ÷ 8 = 94\) → y = 2

Possible y values: \(2, 4\)

Step 2: Apply divisibility rule for 11

A number is divisible by 11 if the difference between the sum of digits in odd positions and even positions is a multiple of 11 (including 0).

Number: 5 6 5 x 9 2 7 4 y

Sum of odd positions (1^st,3^rd,5^th,7^th,9^th): \(5 + 5 + 9 + 7 + y = 26 + y\)

Sum of even positions (2^nd,4^th,6^th,8^th): \(6 + x + 2 + 4 = 12 + x\)

Difference: \((26 + y) - (12 + x) = 14 + y - x\)

This must be divisible by 11.

Step 3: Find valid (x,y) pairs

Case 1: y = 4

\(14 + 4 - x = 18 - x\) must be divisible by 11

Possible x: 7 (since 18-7=11)

Case 2: y = 2

\(14 + 2 - x = 16 - x\) must be divisible by 11

Possible x: 5 (since 16-5=11)

Valid pairs: \((x,y) = (7,4)\) or \((5,2)\)

Step 4: Calculate (2x + 3y)

For (7,4): \(2(7) + 3(4) = 14 + 12 = 26\)

For (5,2): \(2(5) + 3(2) = 10 + 6 = 16\)

Verification:

Both numbers \(565792744\) and \(565592742\) are divisible by 88.

Given:

Parallelogram PQRS with sides:

• \(PQ = (4x + 1)\) cm
• \(QR = 26\) cm
• \(RS = 25\) cm
• \(PS = (3y - 1)\) cm

Concept:

In a parallelogram:

1. Opposite sides are equal in length
2. Opposite angles are equal
3. Consecutive angles are supplementary

Step 1: Apply property of opposite sides

Since PQ = RS and QR = PS:

\(4x + 1 = 25\) (Equation 1)

\(26 = 3y - 1\) (Equation 2)

Step 2: Solve for x and y

From Equation 1:

\(4x + 1 = 25\)

\(4x = 24\)

\(x = 6\)

From Equation 2:

\(26 = 3y - 1\)

\(3y = 27\)

\(y = 9\)

Step 3: Calculate (3x + 2y)

\(3x + 2y = 3(6) + 2(9) = 18 + 18 = 36\)

Given Equation:

\[\frac{x-4}{7} - 1 = \frac{5-x}{3} + x\]

Step 1: Solve for x = k

1. Multiply both sides by 21 (LCM of 7 and 3):

\[21 \left( \frac{x-4}{7} - 1 \right) = 21 \left( \frac{5-x}{3} + x \right)\]
2. Expand and simplify:

\[3(x - 4) - 21 = 7(5 - x) + 21x\]

\[3x - 12 - 21 = 35 - 7x + 21x\]

\[3x - 33 = 35 + 14x\]
3. Solve for x:

\[3x - 14x = 35 + 33\]

\[-11x = 68\]

\[x = -\frac{68}{11}\]

Therefore, \(k = -\frac{68}{11}\)

Step 2: Compute \(\frac{11k + 18}{11k - 12}\)

1. Calculate 11k:

\[11k = 11 \times \left(-\frac{68}{11}\right) = -68\]
2. Evaluate the expression:

\[\frac{11k + 18}{11k - 12} = \frac{-68 + 18}{-68 - 12} = \frac{-50}{-80} = \frac{5}{8}\]

Teaching methods can be broadly categorized by who holds the primary responsibility for directing learning. In teacher-centered approaches, the instructor is the focal point, delivering information directly to students who play a more passive role.

Key Points

• The lecture method positions the teacher as the primary source of knowledge, structuring the lesson around a prepared presentation of concepts.
• Students listen, take notes, and absorb information, with limited opportunities for interaction or inquiry.
• Classroom activities are organized around the teacher’s delivery, and assessments often focus on recall of presented material.
• This approach allows efficient coverage of content, ensures consistency across learners, and relies on the teacher’s expertise to guide understanding. In mathematics, lectures can introduce new formulas or theorems before practice problems are assigned.

Hint

• Problem solving method engages students in tackling real-world problems, fostering active reasoning.
• Heuristic method guides learners through discovery using hints and questions rather than direct instruction.
• Group discussion method relies on peer interaction and shared exploration of ideas.

Hence, the correct answer is ‘Lecture Method’.

Conceptual errors in mensuration arise when students misunderstand the underlying principles of measuring areas, volumes, and perimeters. Effective remediation requires strategies that reinforce the concrete foundations of these concepts, helping learners bridge the gap between abstract formulas and real‑world shapes.

Key Points Course of Action (I)

• Giving Anamika plenty of hands‑on and visual practice directly targets her misconceptions.
• By manipulating physical models such as cutting out shapes, using grid paper, or building three‑dimensional solids, she can see and feel how dimensions relate to measurements.
• Visual aids like diagrams and dynamic geometry software further solidify her understanding of how lengths, heights, and bases combine in area and volume formulas.
• This multi‑sensory approach helps replace faulty mental images with accurate ones, enabling her to apply formulas correctly.

Course of Action (II)

• While open‑ended questions and discussions can encourage deeper thinking, they do not specifically address the concrete misunderstandings in mensuration.
• Without a solid hands‑on foundation, discussion alone may leave her conceptual gaps unfilled.

Hence, the correct answer is Only course of action (I).

Understanding integer addition often relies on concrete models that make positive and negative movements intuitive. Physical representations help students grasp how numbers combine, reinforcing abstract symbols with real‑world experiences. 

Key Points

• When you think of moving one step at a time to add or subtract, a lift does not reflect that gradual process.
• An elevator skips multiple floors at once, making it hard to illustrate the uniform jumps between successive integers.
• It cannot easily show why adding a negative number reverses direction in single‐unit increments.
• The unpredictability of lift travel undermines students’ sense of steady progression along the number line. Because of these drawbacks, it fails to model the continuous one‑unit shifts that integer addition requires.

Hint

• Stairs mimic the number line by moving one step per integer change.
• A drawn number line directly shows each integer position and the effect of moving forward or backward.
• Chart paper can display integer operations visually through arrows or counters.

Hence, for teaching addition of integers at the upper primary stage use of lift is least useful.