Mathematics and Science PYQs

Given:

Two sides of a triangle are 19 cm and 15 cm.

Concept:

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Step-by-Step Calculation:

1. Let the third side be x cm.
2. Apply the Triangle Inequality Theorem:

• 19 + 15 > x

x < 34

• 19 + x > 15

x > -4 (Always true since lengths are positive.)

• 15 + x > 19

x > 4
3. Combine the inequalities:

4 < x < 34

Final Answer:

The third side must be greater than 4 cm and less than 34 cm.

4 < x < 34. So correct answer is 33 cm

Given:

The angles of quadrilateral ABCD are:

• Angle A = 3x-10°
• Angle B = x+30°
• Angle C = 2x+30°
• Angle D = 2x-10°

Concept:

The sum of the interior angles of any quadrilateral is \( 360^\circ \).

Calculation:

1. Set up the equation for the sum of angles:

3x-10 + x+30 + 2x+30 + 2x-10 = 360°
2. Simplify the equation:  8x + 40 = 360° 3. Solve for x :

8x = 320° 

x = 40°
4. Find each angle:

• Angle A = 110°
• Angle B = 70°
• Angle C = 110°
• Angle D = 70°

Analysis of the Quadrilateral:

• Adjacent Angles:
• Angle A (110°)  + Angle B (70°)  = 180°
• Angle B (70°) + Angle C (110°) = 180°
• Angle C (110°) + Angle D (70°) = 180°
• Angle D (70°) + Angle A (110°) = 180°

Conclusion:

The quadrilateral has:

1. Two distinct pairs of adjacent sides where one pair is equal
2. One pair of equal opposite angles (110° and 110°)
3. The other pair of equal opposite angles (70° and 70°)
4. Adjacent angles are supplementary (add up to 180°)

These properties are characteristic of a Parallelogram.

NOTE:

As per the official answer key, Option 3 and 4 both are correct.

Calculation:

Side (24, 45, 51) is a triplet. So this is right angle triangle.

According to the question,

Area of triangle = Area of rectangle

⇒ 1/2 × base × perpendicular = Length × Breadth

⇒ 1/2 × 24 × 45 = 45 × Breadth

⇒ Breadth = 24/2 = 12 cm

So, the perimeter of rectangle = 2(l + b) = 2(45 + 12) = 2 × 57 = 114 cm

Hence, the correct answer is 114 cm.

Given:

A circular sheet of diameter 16 cm has a circle of radius 6 cm removed.

Concept:

The area of a circle is given by the formula: \( A = \pi r^2 \)

Calculation:

1. Area of the large circle (initial circular sheet):

The radius of the large circle is half of the diameter, i.e., \( \text{Radius of the large circle} = \frac{16}{2} = 8 , \text{cm} \)

Therefore, the area of the large circle is:

\( A_{\text{large}} = \pi \times 8^2 = \frac{22}{7} \times 64 = \frac{1408}{7} = 201.14 , \text{cm}^2 \) 2. Area of the small circle (removed circle):

The radius of the small circle is 6 cm, so its area is:

\( A_{\text{small}} = \pi \times 6^2 = \frac{22}{7} \times 36 = \frac{792}{7} = 113.14 , \text{cm}^2 \) 3. Remaining area:

The remaining area is the area of the large circle minus the area of the small circle:

\( A_{\text{remaining}} = A_{\text{large}} - A_{\text{small}} = 201.14 , \text{cm}^2 - 113.14 , \text{cm}^2 = 88 , \text{cm}^2 \)

Conclusion:

The area of the remaining sheet is 88 cm².

Given:

Dimensions of the cuboidal box are 200 cm × 50 cm × 100 cm.

The cuboidal box is melted and recast into a solid cube.

Concept:

• The volume of a cuboid is given by: \( V_{\text{cuboid}} = \text{length} \times \text{breadth} \times \text{height} \)
• The volume of a cube is given by: \( V_{\text{cube}} = \text{side}^3 \)
• The surface area of a cuboid is given by: \( A_{\text{cuboid}} = 2(lw + lh + wh) \)
• The surface area of a cube is given by: \( A_{\text{cube}} = 6s^2 \)

Calculation:

1. Volume of the cuboidal box:

\( V_{\text{cuboid}} = 200 \times 50 \times 100 = 1,000,000 , \text{cm}^3 \)

2. Volume of the cube:

Since the volume of the cuboid is melted and recast into the cube, the volume of the cube will be the same.

\( V_{\text{cube}} = 1,000,000 , \text{cm}^3 \)

Let the side of the cube be \( s \).

So, \( s^3 = 1,000,000 \)

\( s = \sqrt[3]{1,000,000} = 100 , \text{cm} \)

3. Surface area of the cuboid:

The dimensions of the cuboid are 200 cm, 50 cm, and 100 cm. So,  A_cuboid = 2(200 × 50 + 200 × 100 + 50 × 100) = 2(10000 + 20000 + 5000) = 2 × 35000 = 70,000 cm²

4. Surface area of the cube:

The side of the cube is 100 cm. \( A_{\text{cube}} = 6 \times 100^2 = 6 \times 10,000 = 60,000 , \text{cm}^2 \)

5. Difference in surface areas:

The difference in surface areas is: \( \text{Difference} = A_{\text{cuboid}} - A_{\text{cube}} = 70,000 - 60,000 = 10,000 , \text{cm}^2 \)

Conclusion:

The difference between the surface areas of the two solids is 10,000 cm².

Given:

The observations (arranged in ascending order) are 42, 48, x+15, x+17, x+20, 58, and 60. The mean of these observations is 53.

Concept:

• The formula for the mean is: \( \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
• The formula for the median for an odd number of observations is the middle value when the data is arranged in ascending order.

Calculation:

1. Mean Calculation:

There are 7 observations, so the mean is given by:

\( \frac{42 + 48 + (x + 15) + (x + 17) + (x + 20) + 58 + 60}{7} = 53 \)

Simplifying the sum of the observations:

→  42 + 48 + (x + 15) + (x + 17) + (x + 20) + 58 + 60 = 53×  7

→ 42 + 48 + x + 15 + x + 17 + x + 20 + 58 + 60 = 371

→ 260 + 3x = 371

→ 3x = 371 - 260 = 111

→ x = 111/3 = 37

2. Substituting x = 37 into the observations:

• x + 15 = 37 + 15 = 52

•  x + 17 = 37 + 17 = 54

• x + 20 = 37 + 20 = 57

The observations become: 42, 48, 52, 54, 57, 58, 60. 3. Median Calculation:

The data is now: 42, 48, 52, 54, 57, 58, 60. Since there are 7 observations, the median is the 4^th value in the ascending order.

The 4^th value is 54.

Conclusion:

The median of the data is 54.

he data is 54.

Given:

The time spent by a student during a day in different activities is represented by sectors of central angles in a pie chart. The angles for each activity are as follows:

• School: 90°
• Homework: 60°
• Play: 60°
• Sleep: 120°
• Others: 30°

Concept:

• The total angle of a circle is 360°. 
• The time spent in each activity is proportional to the central angle in the pie chart.
• The difference between the time spent on two activities can be found by comparing their central angles.

Calculation:

1. Time spent on sleep and school:

The central angle for sleep is 120°,  and the central angle for school is 90°. 2. Difference between time spent on sleep and school:

The difference between the central angles is: Difference = 90° - 60° = 30°  

Conclusion:

The difference between the time spent by the student on sleep and school is represented by a central angle of 30°.

30/360 = 1/12.

1/12 × 24 = 2 hours.

Given:

A student was asked to find the value of x in the equation x-4 =3. He completed the task by subtracting 4 from 3.

Concept:

To solve the equation x-4=3, the correct method involves isolating x by adding 4 to both sides of the equation:

x -4+4 = 3+ 4 → x= 7.

However, the student incorrectly subtracted 4 from 3 instead of adding 4 to both sides.

Calculation:

If the student had subtracted 4 from 3, the student would have done the following:

x-4 = 3 → x = 3-4 = -1.

The student would end up with x= -1, which is incorrect.

Conclusion:

The situation most appropriately describes an incorrect method of solving the equation. The student misinterpreted the equation and made a mistake by subtracting 4 from 3 instead of correctly adding 4 to both sides of the equation.

In mathematics, factors and multiples are basic concepts that help us understand the relationship between numbers. A factor of a number divides it exactly without leaving a remainder, while a multiple of a number is obtained by multiplying it by whole numbers.  

Key Points

• The statement that claims every factor of a given number is always greater than the given number is not consistent with the definition of a factor. A factor of a number is always less than or equal to that number.
• For example, 1, 2, and 3 are factors of 6, and all of them are less than or equal to 6.
• No factor of 6 is greater than 6. Hence, this statement contradicts the basic definition and is not true.

Hint

• Every multiple of a given number is equal to or greater than the number itself because we get multiples by multiplying the number with whole numbers starting from 1.
• The number of factors of any given number is always finite since only a limited number of whole numbers can divide a number exactly.
• It is also true that each factor of a number is either equal to or smaller than that number, because factors divide the number exactly without leaving any remainder.

Hence, the correct answer is 'every factor of a given number is always greater than the given number' is not true.

Experimental probability is a key concept in probability that is based on actual experiments and real outcomes. 

Key Points

• Tossing a fair coin several times and noting down the outcomes is the most suitable way to introduce experimental probability.
• This method involves direct participation, allowing students to experience how probability emerges from repeated trials.
• They can record heads and tails, calculate the ratio of outcomes, and understand how probability approaches a predictable value as the number of trials increases.

Hint

• Using events like tossing a coin or drawing cards can support the concept later, but without actually doing the experiment, students may not fully grasp the idea.
• Defining experimental probability on the board and then giving examples is less engaging and might feel abstract at the start.
• Solving statistics-based problems is more appropriate after students have understood the concept through firsthand experiences.

Hence, the correct answer is tossing a fair coin several times and noting down the outcomes.

Assessment at the primary stage is a crucial part of the teaching-learning process. Its primary purpose is to support learning, identify learners' progress, and adapt instruction to meet their individual needs. At this stage, the goal is not competition but to foster understanding, curiosity, and confidence in mathematics.

Key Points

• Identifying individual and specific needs of the learner helps teachers modify their teaching strategies to suit different learning levels and styles.
• This supports inclusive and personalized learning. Similarly, assessing whether the aims and objectives of the mathematics curriculum are being met ensures that learning is aligned with intended educational goals, not just rote procedures.

 Hint

• While understanding whether students have learned standard procedures may be part of checking conceptual clarity, it is not the most holistic or primary goal of assessment at this level.
• Selecting students for competitive exams is not appropriate at the primary stage, as the focus here should be on developing foundational skills, not on comparison or selection.

Hence, the correct answer is A and B.

A closed-ended problem is one that has a single, correct answer or a limited set of predetermined correct answers. There isn't much room for interpretation or creativity in the solution.

 Key Points

• Write four integers less than 5: This is an open-ended problem. There are infinitely many integers less than 5 (e.g., 4, 3, 2, 1, 0, -1, -2, ...). Any set of four integers chosen from this infinite set would be correct.
• Write four equivalent fractions for 5/7: This is an open-ended problem. You can multiply the numerator and denominator by any non-zero integer to get an equivalent fraction (e.g., 10/14, 15/21, 20/28, 50/70, etc.). There are infinitely many correct answers.
• List four rational numbers between 5/11 and 10/11: This is an open-ended problem. While the range is specific, there are infinitely many rational numbers between any two distinct rational numbers. You could pick 6/11, 7/11, 8/11, 9/11, or you could pick fractions with larger denominators like 51/110, 52/110, etc.
• List four natural numbers between 104 and 109: This is a closed-ended problem. Natural numbers are positive integers (1, 2, 3, ...). The natural numbers between 104 and 109 are 105, 106, 107, and 108. There are only four specific natural numbers that fit this criterion, making it a problem with a limited and definitive set of correct answers.

Hence, the correct answer is list four natural numbers between 104 and 109.

An effective mathematics classroom is one where learners actively construct knowledge, explore diverse strategies, and develop a deeper understanding of concepts. It encourages reasoning, flexibility in thinking, and connects mathematics to real-life experiences and student perspectives.

Key Points

• Providing opportunities to students to solve problems in multiple ways is a key characteristic of such a classroom.
• It promotes creativity, deeper conceptual understanding, and helps students recognize that there can be more than one correct approach to a problem.
• This also fosters critical thinking and confidence in applying mathematics flexibly.

Hint

• Encouraging only individual problem-solving may limit collaboration and peer learning.
• Solving similar problems as demonstrated by the teacher promotes imitation but not necessarily understanding.
• Emphasizing the abstract nature of mathematics, even with examples, might alienate young learners if not balanced with concrete experiences.

Hence, the correct answer is provide opportunities to the students to solve problems in multiple ways.

Questioning is a powerful tool of assessment in the classroom that helps gauge student understanding, promote thinking, and identify gaps in learning. 

Key Points

• Asking questions for assessing errors and misconceptions is the most appropriate approach.
• This strategy helps in uncovering students' thought processes, clarifying misunderstandings, and guiding them towards correct reasoning.
• It creates an opportunity for formative assessment and responsive teaching.

Hint

• Asking too many questions without pause overwhelms students and doesn’t give them time to think.
• Answering your own questions discourages student engagement and makes the process one-sided.
• Targeting only low achievers may lead to bias and limit class-wide participation.

Hence, the correct answer is asking questions for assessing errors and misconceptions.

In algebra, properties of addition help us understand how expressions can be rearranged without changing their values.

 Key Points

• The commutative property states that the order of addition doesn't affect the sum, while the associative property allows grouping of terms in different ways.
• Recognizing these properties helps in simplifying and reorganizing algebraic expressions effectively.
• When the expression (x + 5) + 7y is written as 5 + (x + 7y), two properties are at play.
• First, the order of addition is changed — 5 is moved in place of (x + 5), which is an example of the commutative property.
• Second, the grouping of terms is altered — (x + 5) becomes (x + 7y), which shows the associative property in action.

Hint

• The distributive property is not involved here, as there is no multiplication over addition.
• Using only commutative or only associative wouldn't fully explain the transformation of the expression.

Hence, the correct answer is commutative and associative both.

The spiral approach in curriculum design, as promoted by the National Curriculum Framework (NCF) 2005, is based on the idea that students learn best when topics are revisited and expanded upon over time. 

Key Points

• Including concepts in the increasing order of difficulty by briefly revisiting concepts taught previously aligns best with the spiral approach.
• It ensures that students build on their earlier understanding while gradually tackling more challenging content.
• For example, a basic understanding of fractions in one grade can develop into operations with fractions in the next, and applications in real contexts later.

Hint

• Introducing both simple and difficult topics in the same grade may overwhelm students and disrupt progression.
• Teaching entirely different concepts at each grade can lead to gaps in understanding and weak foundations.
• Simply repeating concepts each year without increasing depth does not promote true conceptual growth.

Hence, the correct answer is including concepts in the increasing order of difficulty by briefly revisiting concepts taught previously.

In mathematics, numbers can be used in two main ways: cardinal and ordinal. A cardinal number tells us how many objects there are, while an ordinal number indicates the position or order of something in a sequence.

Key Points

• The number 6 is used in a cardinal sense in the statement "There are 6 elements in a set."
• This usage expresses quantity — it tells us how many elements are present, which is the defining feature of a cardinal number.

Hint

• Saying 6 is the successor of 5 or the predecessor of 7 refers to its position in the number sequence, which aligns with ordinal understanding.
• Mentioning the 6^th row of an auditorium clearly reflects an ordinal sense, indicating position rather than quantity.

Hence, the correct answer is there are 6 elements in a set.

Given Expression:

The value of the following expression is: \( \frac{0.125 × 57.8 × 0.729}{0.017 × 0.0081 × 0.25} \)

Solution:

0.125 = 125 × 10^-3

57.8 = 578 × 10^-1

0.729 = 729 × 10-3 

0.017 = 17 × 10-3

0.0081 = 81 × 10^-4

​0.25 = 25 × 10^-2

Numerator = 125 × 10-3 × 578 × 10-1 × 729 × 10-3 = 125 × 578 × 729 × 10^-7

Denominator = 17 × 81 × 25 × 10^-9

⇒ \(\frac{125}{25}×\frac{578}{17}×\frac{729}{81}×10^{-7}×10^{9}\)

⇒ 5 × 34 × 9 × 10²

So, \(\frac{0.125\times 57.8\times 0.729}{0.017\times 0.0081\times 0.25}\) = 153000

Given:

The LCM of two numbers is 392.

Concept:

The relationship between the LCM (Least Common Multiple) and HCF (Highest Common Factor) of two numbers \(a\) and \(b\) is given by the formula:

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

Since we know the LCM of the two numbers is 392, we need to determine which of the given values for the HCF is not possible.

Step 1: Prime factorization of LCM:

The prime factorization of 392 is:

\( 392 = 2^3 \times 7^2 \)

Thus, the HCF must be a divisor of 392. The possible divisors of 392 are:

 1, 2, 4, 7, 8, 14, 28, 49, 56, 98, 196, 392

Step 2: Check each given HCF:

Let's check each option:

• HCF = 56: 56 is a divisor of 392 (\(56 = 2^3 \times 7\)), so this can be the HCF.
• HCF = 196: 196 is a divisor of 392 (\(196 = 2^2 \times 7^2\)), so this can be the HCF.
• HCF = 28: 28 is a divisor of 392 (\(28 = 2^2 \times 7\)), so this can be the HCF.
• HCF = 42: 42 is not a divisor of 392, as 42 does not divide 392 evenly. The prime factorization of 42 is 2 × 3 × 7, and 3 is not a factor of 392.

Conclusion:

The HCF that cannot be the value is 42. Therefore, the correct answer is:

4) 42

Given:

We are asked to find the smallest positive integer that is not a factor of the product 264 ×  90 × 1680.

Concept:

To find the smallest integer that is not a factor of the product, we can start by determining the prime factorization of each number involved in the product and then analyze the product.

Step 1: Prime factorization of the numbers

• Prime factorization of 264:

\( 264 = 2^3 × 3 × 11 \)

• Prime factorization of 90:

\( 90 = 2 × 3^2 × 5 \)

• Prime factorization of 1680:

\( 1680 = 2^4 × 3 × 5 × 7 \)

Step 2: Prime factorization of the product

Now, we multiply the prime factorizations of the three numbers:

\( 264 × 90 × 1680 = (2^3 × 3 × 11) × (2 × 3^2 × 5) × (2^4 × 3 × 5 × 7) \)

Combining the powers of the common primes, we get:

\( = 2^{3+1+4} × 3^{1+2+1} × 5^{1+1} × 7^1 × 11^1 \)

\( = 2^8 × 3^4 × 5^2 × 7 × 11 \)

Step 3: Finding the smallest positive integer which is not a factor

The product 264 ×  90 × 1680264×90×1680264×90×1680264×90×1680264×90×1680264×90×1680264×90×1680

    includes all the prime factors with the powers 282828282828

  8  , 4343434343434

    , 2525252525252

    , 1717171717171

    , and 1111111111111111111

    . Therefore, any integer that is not a divisor of this product must have a prime factor that is either absent or raised to a higher power than what appears in the prime factorization of the product.

• The prime factors present in the product are 2,3,5,72,3,5,7,2,3,5,7,2,3,5,7,2,3,5,7,2,3,5,7,2,3,5,7,

    and 1111111111[Math Processing Error]11
  
    .  
  

• The smallest integer not dividing this product will be the smallest prime number greater than 11, which is 13.

Conclusion:

The smallest positive integer which is not a factor of the product 264 ×  90 × 1680 is 13.