# Enhanced Curriculum Support

**Sample question paper**

Complete above steps to enable this content

This is a comprehensive educational resource designed to provide students with the tools and guidance necessary to excel. This support system is structured to cater to various aspects of learning, ensuring that students are well-prepared for academic challenges and practical applications of mathematical concepts. Some are the key benefits are mentioned below:

**Comprehensive Learning:** This holistic approach helps students gain a thorough understanding of the subject. Practical Application: The resources encourage students to apply mathematical concepts to real-life scenarios, enhancing their practical understanding and problem-solving skills.

**Critical Thinking and Reasoning:** Value-Based and HOTS questions promote critical thinking and reasoning abilities. These skills are crucial for students to tackle complex problems and make informed decisions.

**Exam Preparedness:** Sample Question Papers and NCERT Exemplar Solutions provide ample practice for exams. They help students familiarize themselves with the exam format and types of questions, reducing exam anxiety.

**Ethical and Moral Development:** Value-Based Questions integrate ethical and moral lessons into the learning process, helping in the overall development of students' character and social responsibility. By incorporating these diverse elements, Enhanced Curriculum Support aims to provide a robust and well-rounded knowledge, preparing students for both academic success and real-world challenges.

## Sample Questions/ Previous year Questions

**Sample Question Paper:** These are designed to mimic actual exam papers, providing students with a practice platform to gauge their understanding and readiness. They cover a wide range of topics and question types that students might encounter. Regular practice with these papers helps in boosting confidence and improving exam performance.

**Quick Points:**

Practice for real exam scenarios.

Includes various types of questions.

Helps in time management.

Identifies areas of improvement.

**SecA**

**1.What is the ratio of HCF to LCM of the smallest prime number and the smallest composite number?**

**2.If two positive integers a and b are written as a = x**

**3.Let a and b be two positive integers such that a =**

**4. If LCM(x, 18) =36 and HCF(x, 18) =2, then x is (a) 2 (b) 3 (c) 4 (d) 5**

**5. If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are (a) 2 (b) 3 (c) 4 (d) 5**

**6. The LCM of two prime numbers p and q (p > q) is 221. Find the value of 3p – q. (a) 4 (b) 28 (c) 38 (d) 48**

**7. The LCM of ‘2^3 X 3^2’ and 22 X 33**

**8. The HCF of two numbers is 18 and their product is 12960. Their LCM will be (a) 420 (b) 600 (c) 720 (d) 800**

**9. The prime factorisation of 3825 is **

**10.The least number that is divisible by all the numbers from 1 to 10 (both inclusive) (a) 100 (b) 1000 (c) 2520 (d) 5040**

**12.If xy=180 and HCF(x,y)=3, then find the LCM(x,y).**

**13. The LCM of smallest two digit composite number and smallest composite number is a) 12 b) 4 c) 20 d) 44**

**Sol**

**Ratio of HCF to LCM of the smallest prime number (2) and the smallest composite number (4):**HCF = 2, LCM = 4

**Ratio = HCF / LCM =**2 4 = 1 2 **If**a = x 3 andy 2 b = x , then dividing the product ofy 3 a andb by LCM(a ,b ) results in:LCM(

a ,b ) =x 3 y 3

Product ofa andb =x 4 y 5

Result =x 4 y 5 x 3 y 3 = xy 2

**Answer: (b) xy²****If**a = p 3 andq 4 b = p 2 , and HCF(a, b) =q 3 p m and LCM(a, b) =q n p r , then ((m+n)(r+s) =):q s HCF =

p 2 (so,q 3 m = 2 ,n = 3 )

LCM =p 3 (so,q 4 r = 3 ,s = 4 )

((m+n)(r+s) = (2+3)(3+4) = 5 \cdot 7 = 35`

**Answer: (c) 35****If LCM(x, 18) = 36 and HCF(x, 18) = 2, then**x is:x · 18 = LCM x , 18 · HCF x , 18 x · 18 = 36 · 2 x = 4 **Answer: (d) 4****If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are:**Let the numbers be

81 a and81 b 81 a + b = 1215 a + b = 15

Possible pairs ((a, b)) that add up to 15 are (1, 14), (2, 13), (3, 12), (4, 11), (5, 10), (6, 9), (7, 8)

**Answer: (d) 5**

**The LCM of two prime numbers**p andq (wherep > q ) is 221. Find the value of3 p − q :Prime factors of 221 are 13 and 17

Ifp = 17 andq = 13 ,3 p − q = 3 · 17 − 13 = 51 − 13 = 38 **Answer: (c) 38****The LCM of**2 3 x and3 2 2 2 x is:3 3 LCM =

2 3 x 3 3 **Answer: (c) 2^3 x 3^3****The HCF of two numbers is 18 and their product is 12960. Their LCM will be:**Product of numbers = HCF x LCM

12960 = 18 · LCM LCM = 12960 18 = 720 **Answer: (c) 720****The prime factorization of 3825 is:**3825 =

3 2 x 5 2 x 17 **Answer: (c) 3^2 x 5^2 x 17****The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:**

To find this number, calculate the LCM of the numbers from 1 to 10:

Prime factorizations:

- 1 =
1 - 2 =
2 - 3 =
3 - 4 =
2 2 - 5 =
5 - 6 =
2 x 3 - 7 =
7 - 8 =
2 3 - 9 =
3 2 - 10 =
2 x 5

- 1 =
The LCM will be:

2 3 x 3 2 x 5 x 7 = 2520

**Answer: (c) 2520**

**If**xy = 180 andHCF x , y = 3 , then find the LCM(x, y):

Using the formula:

Rearranging to find LCM:

Substitute the values:

**Answer: 60**

**The LCM of the smallest two-digit composite number and the smallest composite number is:**

- Smallest two-digit composite number is
10 (factorization:2 x 5 ) - Smallest composite number is
4 (factorization: )2 2

Calculate the LCM of

- LCM =
2 2 x 5 = 20

**Answer: (c) 20**

**SecB**

**1.Three bells ring at intervals of 4, 7 and 14 minutes. All three rang at 6 AM. When will they ring together again?(a) 6:07 AM (b) 6:14 AM (c) 6:28 AM (d) 6:25 AM**

**2.Prove that 2-**

**3.The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other number.**

**4.Show that 7 − **

**5.If two positive integers p and q are written as p = a**

**6.Given that **

**7.**

**Step 1: Let (3 + **

**Step 2 : Hence, **

**Step 3 : Since p and q are integers, **

**Step 4 : Since **

**Step 5 : But this contradicts the fact that **

**Hence, (3 + **

**She made an error in one step due to which her subsequent steps were incorrect too. In which step did she make that error? Justify your answer.**

**8. Ajay has a box of length 3.2 m, breadth 2.4 m, and height 1.6 m. What is the length of the longest ruler that can exactly measure the three dimensions of the box? Show your steps and give valid reasons.**

**9. m is a positive integer. HCF of m and 450 is 25. HCF of m and 490 is 35. Find the HCF of m, 450 and 490. Show your steps.**

**10.Check whether the statement below is true or false.“The square root of every composite number is rational.”Justify your answer by proving rationality or irrationality as applicable.**

**Sol**

**Three bells ring at intervals of 4, 7, and 14 minutes. All three rang at 6 AM. When will they ring together again?**To find when all three bells will ring together again, calculate the LCM of the intervals:

- Intervals:
4 ,7 , and14 - Prime factorizations:
4 = 2 2 7 = 7 14 = 2 x 7

- LCM =
2 2 x 7 = 28

Therefore, they will ring together again after

28 minutes.If they rang together at

6 : 0 AM , they will ring together again at**Answer: (c) 6:28 AM**.- Intervals:
**Prove that 2 -** is irrational, given that3 is irrational.3 Suppose 2 -

is rational. Let 2 -3 = r where r is a rational number.3 Rearranging,

= 2 - r.3 Since 2 and r are rational, 2 - r is also rational, which implies that

is rational.3 This contradicts the fact that

is irrational. Hence, 2 -3 must be irrational.3 **The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other number.**Use the relationship:

HCF a , b × LCM a , b = a × b .Let the other number be

x .9 × 360 = 45 × x 3240 = 45 × x x = 3240 45 x = 72 **Answer: 72****Show that 7 -** is irrational, given that5 is irrational.5 Suppose 7 -

is rational. Let 7 -5 = r where r is a rational number.5 Rearranging,

= 7 - r.5 Since 7 and r are rational, 7 - r is also rational, which implies that

is rational.5 This contradicts the fact that

is irrational. Hence, 7 -5 must be irrational.5 **If two positive integers p and q are written as**p = a 2 andb 3 q = a 3 b , where a and b are prime numbers, then verify:LCM p , q × HCF p , q = p × q .p = a 2 b 3 q = a 3 b

HCF(p, q) =

a 2 b (taking the lowest powers) LCM(p, q) =a 3 (taking the highest powers)b 3 HCF p , q × LCM p , q = a 2 b × a 3 b 3 = a 5 b 4 p × q = a 2 b 3 × a 3 b = a 5 b 4 Therefore, LCM(p, q) × HCF(p, q) = p × q.

**Given that** is irrational, show by contradiction that the sum of3 and 2 is irrational.3 Suppose

+ 2 is rational. Let3 + 2 = r where r is a rational number.3 Rearranging,

= r - 2.3 Since r and 2 are rational, r - 2 is also rational, which implies that

is rational.3 This contradicts the fact that

is irrational. Hence,3 + 2 must be irrational.3 **Meera's proof steps contain an error. Identify the error:****Step 1**: Correctly states that 3 + is rational if5 is rational.5 **Step 2**: Correctly expresses = (5 - 3).p q **Step 3**: - 3 is indeed an integer if p and q are integers.p q **Step 4**: The error is here. The step incorrectly concludes that is rational from5 p q − 3 being rational. The mistake is in Step 4.

Hence,

**Step 4**contains the error.**Ajay has a box of length 3.2 m, breadth 2.4 m, and height 1.6 m. What is the length of the longest ruler that can exactly measure the three dimensions of the box?**Find the HCF of the dimensions:

- Convert to integers: 32, 24, and 16 (by multiplying by 10)
- HCF of 32, 24, and 16:
- Prime factorizations:
32 = 2 5 24 = 2 3 x 3 16 = 2 4

- HCF =
2 3 = 8

- Prime factorizations:

The length of the longest ruler that can measure all dimensions is

0.8 meters (since HCF = ).8 10 **Answer: 0.8 meters****Let m be a positive integer. HCF of m and 450 is 25, and HCF of m and 490 is 35. Find the HCF of m, 450, and 490.**- Prime factorizations:
450 = 2 x 3 2 x 5 2 490 = 2 x 5 x 7 2 - HCF(m, 450) =
25 = 5 2 - HCF(m, 490) =
35 = 5 x 7

Combining:

- Common factors: 5 (since 25 and 35 share a factor of 5)
- Hence, HCF of m, 450, and 490 is 5.

**Answer: 5**- Prime factorizations:
**Check whether the statement below is true or false: "The square root of every composite number is rational."**A composite number is a number with factors other than 1 and itself. For a composite number, its square root is not necessarily rational. For instance:

- 4 is a composite number, and √4 = 2, which is rational.
- 9 is a composite number, and √9 = 3, which is rational.
- However, 18 is a composite number, and √18 = 3
, which is irrational.2

Therefore, the statement is

**false**. Not every composite number has a rational square root.

**SecC**

**1.Given that **

**2.Given that **

**3.If HCF of 144 and 180 is expressed in the form 13m-16. Find the value of m.**

**4. Prove that **

**5. Prove that **

**6. Write two rational numbers each between the following pair: i) **

**7. The number 3837425721 is divided by a number between 5621 and 5912. State true or false for the below statements about the remainder and justify your answer. i) The remainder can be more than 5912. ii) The remainder cannot be less than 5621. iii) The remainder is always between 5621 and 5912.**

**8.Given 2 positive integers a and b expressed as powers of x and y, finding HCF(a,b) or LCM(a,b). x=**

**Sol**

**Given that** is irrational, prove that 5 + 23 is irrational.3 Suppose 5 + 2

is rational. Let 5 + 23 = r where r is a rational number.3 Rearranging, 2

= r - 5.3 Dividing by 2,

= (r - 5) / 2.3 Since r and 5 are rational, (r - 5) / 2 is also rational, which implies that

is rational.3 This contradicts the fact that

is irrational. Hence, 5 + 23 must be irrational.3 **Answer: 5 + 2** is irrational3 **Given that** is irrational, prove that 25 − 3 is an irrational number.5 Suppose 2

− 3 is rational. Let 25 − 3 = r where r is a rational number.5 Rearranging, 2

= r + 3.5 Dividing by 2,

= (r + 3) / 2.5 Since r and 3 are rational, (r + 3) / 2 is also rational, which implies that

is rational.5 This contradicts the fact that

is irrational. Hence, 25 − 3 must be irrational.5 **Answer: 2** − 3 is irrational5 **If HCF of 144 and 180 is expressed in the form 13m - 16, find the value of m.**First, find the HCF of 144 and 180:

- Prime factorizations:
- 144 = 2^4 x 3^2
- 180 = 2^2 x 3^2 x 5

- HCF = 2^2 x 3^2 = 36

Given HCF = 13m - 16:

- 36 = 13m - 16
- 13m = 36 + 16
- 13m = 52
- m = 52 / 13
- m = 4

**Answer: m = 4**- Prime factorizations:
**Prove that** is irrational.7 Suppose

is rational. Then it can be expressed as7 = p/q where p and q are coprime integers and q ≠ 0.7 Squaring both sides, 7 = p^2 / q^2.

Rearranging, p^2 = 7q^2.

This implies that p^2 is divisible by 7, so p must be divisible by 7. Let p = 7k.

Substituting, (7k)^2 = 7q^2 49k^2 = 7q^2 7k^2 = q^2

This implies that q^2 is divisible by 7, so q must be divisible by 7.

Hence, both p and q are divisible by 7, which contradicts the fact that p and q are coprime.

Therefore,

is irrational.7 **Answer:** is irrational7 **Prove that 1/** is irrational.2 Suppose 1/

is rational. Then it can be expressed as 1/2 = p/q where p and q are coprime integers and q ≠ 0.2 Rearranging,

= q/p.2 Squaring both sides, 2 = q^2 / p^2.

Rearranging, p^2 = 2q^2.

This implies that p^2 is divisible by 2, so p must be divisible by 2. Let p = 2k.

Substituting, (2k)^2 = 2q^2 4k^2 = 2q^2 2k^2 = q^2

This implies that q^2 is divisible by 2, so q must be divisible by 2.

Hence, both p and q are divisible by 2, which contradicts the fact that p and q are coprime.

Therefore, 1/

is irrational.2 **Answer: 1/** is irrational2 **Write two rational numbers each between the following pair:**i)

and3 10 - Approximate values:
≈ 1.732,3 ≈ 3.16210

**Two rational numbers between these values: 2 and 3**ii)

**7 and**64 = 864 **Two rational numbers between these values: 7.5 and 7.75**

iii)

and 615 - Approximate value:
≈ 3.87315

**Two rational numbers between these values: 4 and 5**- Approximate values:
**The number 3837425721 is divided by a number between 5621 and 5912. State true or false for the following statements about the remainder and justify your answer.**i)

**The remainder can be more than 5912.****False**: The remainder when dividing by a number between 5621 and 5912 must be less than the divisor. So, the remainder cannot be more than 5912.

ii)

**The remainder cannot be less than 5621.****False**: The remainder can be less than 5621. For instance, if 3837425721 divided by a number close to 5912, the remainder could be any value from 0 up to just below 5912.

iii)

**The remainder is always between 5621 and 5912.****False**: As stated above, the remainder must be less than the divisor, so it can be less than 5621.

**Answer: All statements are false.**

8.Given two positive integers

x = a 3 b 2 y = a 4 b 5

-Finding the HCF

The HCF (Highest Common Factor) is found by taking the minimum power of each prime factor common to both numbers.

For

- The HCF of
a terms:a min 3 , 4 = a 3 - The HCF of
b terms:b min 2 , 5 = b 2

So, the HCF is:

-Finding the LCM

The LCM (Lowest Common Multiple) is found by taking the maximum power of each prime factor present in either number.

For

- The LCM of
a terms:a max 3 , 4 = a 4 - The LCM of
b terms:b max 2 , 5 = b 5

So, the LCM is:

**Answer: - HCF: **

**- LCM:**a 4 b 5

**SecD**

**1. On the two real numbers a = 2 + **

**i) Calculate the sum ( a + b ).**

**ii) Calculate the product ( ab ).**

**iii) Find the additive inverse of a.**

**iv) Rationalise 1/b .**

**v) Verify whether the numbers a and b are rational or irrational. Provide a valid reason for your answer.**

**2. i) Find the LCM and HCF of 78, 91, and 195.**

**ii) Check whether LCM( a, b , c ) × HCF( a, b , c ) = a × b × c where a, b and c are natural numbers. Show your work**

**3.Find the HCF & LCM of 60 and 48 and verify the relationship- LCM x HCF = a x b**

**Sol**

Given two real numbers a = 2 +

i) Calculate the sum (a+b).

ii) Calculate the product (ab).

iii) Find the additive inverse of a.

The additive inverse of a =

__{.m-green}So, the additive inverse of a is

iv) Rationalise

To rationalize

Multiply by the conjugate

v) Verify whether the numbers a and b are rational or irrational. Provide a valid reason for your answer.

Both

**- The square root of a non-perfect square (such as **

**- Adding or subtracting a rational number (such as 2 or 3) to an irrational number remains irrational.**

2.i) To find the LCM and HCF of 78, 91, and 195.

Prime factorization of 78: 78 = 2 x 3 x 13

Prime factorization of 91: 91 = 7 x 13

Prime factorization of 195: 195 = 3 x 5 x 13

HCF is the product of the smallest powers of common prime factors:

HCF(78, 91, 195) = 13

LCM is the product of the highest powers of all prime factors:

LCM(78, 91, 195) = 2 x 3 x 5 x 7 x 13 = 2730

ii) Check whether LCM(a, b, c) x HCF(a, b, c) = a x b x c where a, b and c are natural numbers. Show your work.

Given:

a = 78

b = 91

c = 195

Calculate a x b x c:

a x b x c = 78 x 91 x 195 = 1381950`

Calculate LCM(78, 91, 195) x HCF(78, 91, 195):

LCM(78, 91, 195) = 2730

HCF(78, 91, 195) = 13

LCM(78, 91, 195) x HCF(78, 91, 195) = 2730 x 13 = 35490

Since 138195` is not equal to 35490, the statement is false.

**Answer:**

**- HCF78,91,195 = 13**

**- LCM78,91,195 = 2730**

**- The statement LCM a**

3.Step 1: Prime Factorization

60:

48:

**Step 2: Calculate HCF (Greatest Common Divisor)**

HCF is the product of the smallest power of each common prime factor.

- For 2: the smallest power is
2 2 - For 3: the smallest power is
3 1

**Step 3: Calculate LCM (Least Common Multiple)**

LCM is the product of the highest power of each prime factor present in any of the numbers.

- For 2: the highest power is
2 4 - For 3: the highest power is
3 1 - For 5: the highest power is
5 1

**Step 4: Verify the Relationship LCM ·HCF=axb**

Given a = 60 and b = 48:

- LCM = 240
- HCF = 12

LCM x HCF = 240 x 12

Calculate the product:

240 x 12 = 2880

Calculate a x b:

a x b = 60 x 48

60 x 48 = 2880

Both results are equal:

LCM x HCF = a x b

**Answer: The relationship is verified -- LCM * HCF = a * b.**

**Value based questions**

Complete above steps to enable this content

**Value-Based Questions:** They integrate moral and ethical values into the learning process, encouraging students to think beyond just academic knowledge. These questions aim to develop a student's character and social responsibility through mathematics. They connect mathematical concepts with everyday life and moral lessons.

**Quick Points:**

Promotes critical thinking.

Encourages ethical reasoning.

Relates mathematics to real-life scenarios.

Enhances decision-making skills.

**Problem1**

**Situation:** A forester, Rajesh, wants to plant 66 apple trees, 88 banana trees and 110 mango trees in equal rows(in terms of number of trees). Also he wants to make distinct rows of trees (i.e. only the type of tree in the row).

**Find the number of minimum rows required ? What value you learn from Rajesh?**

**Sol**

**Solution**

Calculation: To find the minimum number of rows, we need to determine the greatest common divisor (GCD) of 66, 88, and 110.

The prime factors of 66 are 2 x 3 x 11 The prime factors of 88 are

Therefore, the minimum number of rows required is: ** 22**

**Value Learned:** Rajesh’s approach to organizing the planting of trees teaches us the value of planning and efficiency. By finding the optimal way to plant the trees, Rajesh ensures that resources are used effectively and the planting process is well-organized. This reflects the importance of careful planning and problem-solving in achieving efficient and successful outcomes.

**Problem 2**

**Situation:** An army contingent of 1000 members is to march behind an army band of 56 members in a parade. The two groups are to march in the same number of columns.

**What is the maximum number of columns in which they can march?What value you learn from Amy soliders?**

**Sol**

**Solution**

Calculation: To find the maximum number of columns, we need to determine the Highest Common Factor (HCF) of 1000 and 56.

The prime factors of 1000 are

Therefore, the maximum number of columns in which they can march is: ** 8**

**Value Learned:** The army soldiers' ability to coordinate and march in the same number of columns teaches us the value of unity and discipline. By aligning their movements and working together, they demonstrate the importance of teamwork and coordination in achieving a common goal. This reflects the significance of cooperation and discipline in both military and everyday life.

**Hots**

Complete above steps to enable this content

**HOTS (Higher Order Thinking Skills):** They require students to apply, analyze, synthesize, and evaluate information rather than just recall facts. These questions are designed to challenge students and stimulate intellectual growth. Engaging with HOTS questions helps students to develop a deeper understanding and prepares them for complex problem-solving.

**Quick Points:**

Develops advanced problem-solving skills.

Encourages deep understanding.

Fosters creativity and innovation.

Enhances analytical abilities.

**Q1**

**1.Prove that **

**Sol**

**Solution:**

- Let's consider x as an arbitrary positive integer. There are two cases to consider: x is even or x is odd.

**Case 1: x is even.**

Let x = 2k for some integer k.

Then,

=

Since

**Case 2: x is odd.**

Let x = 2k + 1 for some integer k.

Then,

=

Since

**Therefore, x**

**Q2**

**If m and n are odd positive integers, then m2 + n 2 is even, but not divisible by 4. Justify.**

**Sol**

**Solution:**

**If m and n are odd positive integers, then m2 + n 2 is even, but not divisible by 4. Justify.**

Let m and n be odd positive integers. We can represent any odd integer as

2 k + 1 , wherek is an integer.Let

m = 2 a + 1 andn = 2 b + 1 for some integers a and b.Then,

m 2 = 2 a + 1 2 = 4 a 2 + 4 a + 1 = 4 a a + 1 + 1 .Similarly,

n 2 = 2 b + 1 2 = 4 b 2 + 4 b + 1 = 4 b b + 1 + 1 .

Now, consider the sum

Since 4a(a + 1) and 4b(b + 1) are both multiples of 4, their sum 4a(a + 1) + 4b(b + 1) is also a multiple of 4.

Therefore,

This means

**Hence, if m and n are odd positive integers, then m2 + n 2 is even but not divisible by 4.**

**Q3**

**Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively.**

**Sol**

**Solution:**

28−8=20 and 32−12=20

Prime factors of 28=

Prime factors of 32=

LCM(28,32)=

**Therefore the required smallest number=224−20=204**

**Verification** ⇒28×7=196⇒204−196=8

⇒36×6=192⇒204−192=12

**Ncert exemplar solutions**

Complete above steps to enable this content

**NCERT Exemplar Solutions:** They provide detailed answers and explanations to problems in NCERT textbooks, aiding students in understanding complex concepts. These solutions serve as a valuable resource for clarifying doubts and reinforcing learning. They are essential for thorough exam preparation and achieving academic excellence.

**Quick Points:**

Comprehensive solutions for NCERT problems.

Clarifies difficult concepts.

Useful for exam preparation.

Provides step-by-step explanations

**Answer Ncert exemplar questions**

Complete above steps to enable this content

**Questions**

**Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.****The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.****A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.**

**Solutions**

**Solutions**

**No, not every positive integer can be of the form 4q +2.**

**justification:**

A number of the form 4q+2 will always leave a remainder of 2 when divided by 4. Therefore, only those integers which leave a remainder of 2 when divided by 4 can be written in this form, such as 2, 6, 10, etc. Other integers, like those leaving remainders of 0, 1, or 3 when divided by 4, cannot be expressed in this form.

- True.

**Reason:**

Among any two consecutive positive integers, one of them is always even. Since an even number is divisible by 2, the product of the two integers will always be divisible by 2.

- No, the square of a positive integer of the form 3q + 1 cannot be written in any form other than 3m + 1.

**justification:**

Let the integer be n = 3q + 1. Then its square is:

This shows

**Case-Based Questions**

Complete above steps to enable this content

**Case-Based Question:** They present real-life situations requiring students to apply their mathematical knowledge to solve problems, promoting practical understanding. These questions enhance the ability to connect theoretical knowledge with practical applications. They are instrumental in developing problem-solving skills relevant to real-world scenarios.

**Quick Points:**

Real-life application of concepts.

Encourages analytical thinking.

Enhances comprehension of practical problems.

Promotes interdisciplinary learning.

**Question**

**In a school, buses are hired to take the students of Std-X on a field trip. There are 156, 208, and 260 students in groups A, B, and C respectively. The goal is to determine the minimum number of buses needed, with each bus accommodating the same number of students, and each group traveling separately.**

**Based on your understanding of the above case study,answer all the five questions below:**

**(1) Maximum number of students to be accommodated in each bus is:**

**(2) How many buses will be required for group A students?**

**(3) If group C will not be taken along and 26 students of group A refuse to join, then the minimum number of buses required to accommodate the remaining students is:**

**(4) If group A students will not be taken along, then the minimum number of buses required to accommodate the students of group B and group C is in the form of **

**(5) What is the minimum number of buses required for all the students?**

**Solution**

**Solution**

**(1)Prime factorization:**

156: 156 =

x 3 x 132 2 208: 208 =

x 132 4 206: 260 =

x 5 x 132 2

**Common factors:**

**So, the maximum number of students per bus is 52.**

**(2)Number of buses required for group A students:**

Number of buses for group A =

**(3)If group C is not taken and 26 students of group A refuse to join:**

**Remaining students in group A:**156 - 26 = 130**Number of buses for group A and B together:****HCF of 130 and 208:**

130: 130 = 2 x 5 13

208: 208 =

Common factor: 2 x 13 = 26

Number of buses for group A =

Number of buses for group B =

**Total number of buses = 5 + 8 =13**

**(4)If group A students will not be taken along:**

Number of buses required for group B and group C together:

HCF of 208 and 260:

208:

260:

Common factor:

Number of buses for group B =

Number of buses for group C =

Total number of buses = 4+5=9

**The format of **

**(5)Minimum number of buses required for all students:**

Number of buses required for groups A, B, and C:

HCF of 156, 208, and 260 is 52.

Number of buses for group A =

Number of buses for group B =

Number of buses for group C =

**Total number of buses = 3 + 4 + 5 = 12**

**Assertion and Reasoning**

Complete above steps to enable this content

**Question**

ASSERTION AND REASONING:

**1.Choose the correct option Statement**

**A (Assertion): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340 Statement .**

**R( Reason): HCF is always a factor of LCM**

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

(c) Assertion (A) is true but reason (R) is false.

(d) Assertion (A) is false but reason (R) is true.

**2. Assertion (A): Product of HCF and LCM of THREE numbers is equal to the product of those numbers.**

**Reason (R): Product of HCF and LCM of TWO numbers is equal to the product of those numbers.**

1 Both (A) and (R) are true and (R) is the correct explanation for (A).

2 Both (A) and (R) are true and (R) is not the correct explanation for (A).

3 (A) is false but (R) is true.

4 Both (A) and (R) are false.

**Solution**

1.Assertion (A): If the product of two numbers is 5780 and their HCF is 17, then their LCM is 340.

Explanation: Use the formula:

Product of the two numbers=HCF×LCM

Solving for LCM:

LCM=

Thus, Assertion (A) is true.

Reason (R): HCF is always a factor of LCM.

Explanation: While it is true that HCF is always a factor of LCM, this fact does not explain why the LCM is 340 in this specific case. The correct explanation is based on the calculation of LCM using HCF and the product of the numbers, not just the fact that HCF is a factor of LCM.

**Answer: (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).**

- Statement and Options

Assertion (A): Product of HCF and LCM of THREE numbers is equal to the product of those numbers.

Explanation: This statement is false. The property that the product of HCF and LCM equals the product of the numbers applies only to TWO numbers, not three.

Reason (R): Product of HCF and LCM of TWO numbers is equal to the product of those numbers.

Explanation: This statement is true. For any two numbers, the product of their HCF and LCM equals the product of the two numbers.

**Answer: (3) (A) is false but (R) is true.**