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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=x3y2 and b=xy3, where x, y are prime numbers, then the result obtained by dividing the product of the positive integers by the LCM (a, b) is (a) xy (b) xy2 (c) x3y3 (d) x2y2

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

4. 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

5. 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

6. The LCM of 23 X 32’and2^2X3^3isa2^3b3^3c2^3X3^3d2^2X3^2`

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

8. The prime factorisation of 3825 is (a) 3 x 52 x 21 (b) 32 x 52 x 35 (c) 32 x 52 x 17 (d) 32 x 25 x 17

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

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

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

Sol

1. Ratio of HCF to LCM of the smallest prime number (2) and the smallest composite number (4):

HCF = 2, LCM = 4
Ratio = HCF / LCM = 24=12

2. If a=x3y2 and b=xy3, then dividing the product of a and b by LCM(a, b) results in:

LCM(a, b) = x3y3
Product of a and b = x4y5
Result = x4y5x3y3 = xy2
Answer: (b) xy²

3. If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is:

x·18=LCMx,18·HCFx,18
x·18=36·2
x=4

Answer: (d) 4

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 81a and 81b
81a+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

5. The LCM of two prime numbers p and q (where p>q) is 221. Find the value of 3p−q:

Prime factors of 221 are 13 and 17
If p=17 and q=13,
3p−q=3·17−13=51−13=38
Answer: (c) 38

6. The LCM of 23x32 and 22x33 is:

LCM = 23x33
Answer: (c) 23 x 33

7. 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=1296018=720
Answer: (c) 720

8. The prime factorization of 3825 is:

3825 = 32 x 52 x 17
Answer: (c) 32 x 52 x 17

9. 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 = 22
  • 5 = 5
  • 6 = 2x3
  • 7 = 7
  • 8 = 23
  • 9 = 32
  • 10 = 2x5

• The LCM will be: 23x32x5x7=2520

Answer: (c) 2520

10. If xy=180 and HCFx,y=3, then find the LCM(x, y):

Using the formula: xy=HCFx,yxLCMx,y

Rearranging to find LCM: LCMx,y=xyHCFx,y

Substitute the values: LCMx,y=1803=60

Answer: 60

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

• Smallest two-digit composite number is 10 (factorization: 2x5)
• Smallest composite number is 4 (factorization: 22)

Calculate the LCM of 10 and 4:

• LCM = 22x5=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 is irrational, given that 3 is irrational.

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 is irrational, give that 5 is irrational.

5. If two positive integers p and q are written as p=a2 b3 and q=a3 b; a, b are prime numbers, then verify: LCM (p, q) × HCF (p, q) = pq

6. Given that 3 is irrational, show by contradiction that the sum of 3 and 2 is irrational. Show your steps.

7. 5 is an irrational number. Meera was asked to prove that (3 + 5) is an irrational number. Shown below are the steps of Meera's proof:

Step 1: Let (3 + 5) be a rational number. Then (3 + 5) can be written as p q , where p and q ( q ≠ 0) are co-primes.

Step 2 : Hence, 5 = ( pq - 3).

Step 3 : Since p and q are integers, pq−3is also an integer.

Step 4 : Since pq−3is an integer and every integer is a rational number, pq−3is a rational number. It implies that 5 is a rational number.

Step 5 : But this contradicts the fact that 5 is an irrational number.

Hence, (3 + 5) is an irrational number.

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

1. 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, and 14
• Prime factorizations:
• 4=22
• 7=7
• 14=2x7
• LCM = 22x7=28

Therefore, they will ring together again after 28 minutes.

If they rang together at 6:00 AM, they will ring together again at

Answer: (c) 6:28 AM.

2. Prove that 2 - 3 is irrational, given that 3 is irrational.

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

Rearranging, 3 = 2 - r.

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

This contradicts the fact that 3 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: HCFa,b×LCMa,b=a×b.

Let the other number be x. 9×360=45×x 3240=45×x x=324045 x=72

Answer: 72

4. Show that 7 - 5 is irrational, given that 5 is irrational.

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

Rearranging, 5 = 7 - r.

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

This contradicts the fact that 5 is irrational. Hence, 7 - 5 must be irrational.

5. If two positive integers p and q are written as p = a2 b3 and q = a3 b, where a and b are prime numbers, then verify: LCMp,q×HCFp,q=p×q.

• p=a2b3

• q=a3b

  HCF(p, q) = a2b (taking the lowest powers) LCM(p, q) = a3b3 (taking the highest powers)

  HCFp,q×LCMp,q=a2b×a3b3=a5b4 p×q=a2b3×a3b=a5b4

  Therefore, LCM(p, q) × HCF(p, q) = p × q.

6. Given that 3 is irrational, show by contradiction that the sum of 3 and 2 is irrational.

Suppose 3 + 2 is rational. Let 3 + 2 = r where r is a rational number.

Rearranging, 3 = r - 2.

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

This contradicts the fact that 3 is irrational. Hence, 3 + 2 must be irrational.

7. Meera's proof steps contain an error. Identify the error:

• Step 1: Correctly states that 3 + 5 is rational if 5 is rational.
• Step 2: Correctly expresses 5 = (pq - 3).
• Step 3: pq - 3 is indeed an integer if p and q are integers.
• Step 4: The error is here. The step incorrectly concludes that 5 is rational from pq−3 being rational. The mistake is in Step 4.

Hence, Step 4 contains the error.

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?

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=25
    • 24=23x3
    • 16=24
  • HCF = 23=8

The length of the longest ruler that can measure all dimensions is 0.8 meters (since HCF = 810).

Answer: 0.8 meters

9. 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=2x32x52
  • 490=2x5x72
  • HCF(m, 450) = 25=52
  • HCF(m, 490) = 35=5x7

Combining:

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

Answer: 5

10. 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 2, which is irrational.

Therefore, the statement is false. Not every composite number has a rational square root.

SecC

1. Given that 3 is irrational, prove that 5 + 23 is irrational.

2. Given that 5 is irrational, prove that 25 − 3is an irrational number.

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

4. Prove that 7 is irrational.

5. Prove that 12is irrational.

6. Write two rational numbers each between the following pair: i) 3 and 10 ii) 7 and 64 iii) 15 and 6

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=a3 b2, y=a4 b5

Sol

1. Given that 3 is irrational, prove that 5 + 23 is irrational.

Suppose 5 + 23 is rational. Let 5 + 23 = r where r is a rational number.

Rearranging, 23 = r - 5.

Dividing by 2, 3 = (r - 5) / 2.

Since r and 5 are rational, (r - 5) / 2 is also rational, which implies that 3 is rational.

This contradicts the fact that 3 is irrational. Hence, 5 + 23 must be irrational.

Answer: 5 + 23 is irrational

2. Given that 5 is irrational, prove that 25 − 3 is an irrational number.

Suppose 25 − 3 is rational. Let 25 − 3 = r where r is a rational number.

Rearranging, 25 = r + 3.

Dividing by 2, 5 = (r + 3) / 2.

Since r and 3 are rational, (r + 3) / 2 is also rational, which implies that 5 is rational.

This contradicts the fact that 5 is irrational. Hence, 25 − 3 must be irrational.

Answer: 25 − 3 is irrational

3. 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 = 24 x 32
  • 180 = 22 x 32 x 5
• HCF = 22 x 32 = 36

Given HCF = 13m - 16:

• 36 = 13m - 16
• 13m = 36 + 16
• 13m = 52
• m = 5213
• m = 4

Answer: m = 4

4. Prove that 7 is irrational.

Suppose 7 is rational. Then it can be expressed as 7 = pq where p and q are coprime integers and q ≠ 0.

Squaring both sides, 7 = p2q2.

Rearranging, p2 = 7q2.

This implies that p2 is divisible by 7, so p must be divisible by 7. Let p = 7k.

Substituting, 7k2 = 7q2 49k2 = 7q2 7k2 = q2

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, 7 is irrational.

Answer: 7 is irrational

5. Prove that 1/2 is irrational.

Suppose 1/2 is rational. Then it can be expressed as 1/2 = p/q where p and q are coprime integers and q ≠ 0.

Rearranging, 2 = qp.

Squaring both sides, 2 = q2p2.

Rearranging, p2 = 2q2.

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

Substituting, 2k2 = 2q2 4k2 = 2q2 2k2 = q2

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, 1/2 is irrational.

Answer: 1/2 is irrational

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

i) 3 and 10

• Approximate values: 3 ≈ 1.732, 10 ≈ 3.162

Two rational numbers between these values: 2 and 3

ii) 7 and 64

• 64 = 8

Two rational numbers between these values: 7.5 and 7.75

iii) 15 and 6

• Approximate value: 15 ≈ 3.873

Two rational numbers between these values: 4 and 5

7. 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 a and b expressed as powers of x and y, where:

• x=a3b2
• y=a4b5

Finding the HCF

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

For a=x and b=y:

• The HCF of a terms: amin3,4=a3
• The HCF of b terms: bmin2,5=b2

So, the HCF is:

HCFx,y=a3b2

Finding the LCM

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

For a=x and b=y:

• The LCM of a terms: amax3,4=a4
• The LCM of b terms: bmax2,5=b5

So, the LCM is:

LCMx,y=a4b5

Answer: - HCF: a3b2 - LCM: a4b5

SecD

1. On the two real numbers a = 2 + 5 and b = 3 - 7, perform the following operations:

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

1. Given two real numbers

a = 2 + 5 and b = 3 - 7:

i) Calculate the sum (a+b).

a+b=2+5+3−7

a+b=5+5−7

ii) Calculate the product (ab).

ab=2+53−7

ab=23+2−7+53+5−7

ab=6−27+35−35

ab=6+35−27−35

iii) Find the additive inverse of a.

The additive inverse of a = 2+5 is -a = −2+5

So, the additive inverse of a is −2−5.

iv) Rationalise 1b.

To rationalize 1b, multiply by the conjugate of b:

1b=13−7

Multiply by the conjugate 3+7:

13−7·3+73+7

=3+732−72

=3+79−7

=3+72

1b=3+72

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

Both a=2+5 and b=3−7 are irrational because:

- The square root of a non-perfect square (such as 5 and 7) is irrational.

- 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 LCMa,b,cxHCFa,b,c=axbxc is false.

3. Step 1: Prime Factorization

60:60=22x3x5

48:48=24x3

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 22
• For 3: the smallest power is 31

HCF=22x3=4x3=12

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 24
• For 3: the highest power is 31
• For 5: the highest power is 51

LCM=24x3x5=16x3x5=240

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.

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 23x11 The prime factors of 110 are 2 x 5 x 11 The GCD is 2 x 11 = 22.

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 23x53.

The prime factors of 56 are 23x7.

The GCD is 23 = 8.

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.

Q1

1. Prove that x2 – x is divisible by 2 for all positive integer x.

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, x2−x=2k2−2k

= 4k2−2k=22k2−k.

Since 22k2−k is clearly divisible by 2, x2−x is divisible by 2 when x is even.

Case 2: x is odd.

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

Then, x2−x=2k+12−2k+1=4k2+4k+1−2k−1

= 4k2+2k=22k2+k.

Since 22k2+k is clearly divisible by 2, x2−x is divisible by 2 when x is odd.

Therefore, x2−x is divisible by 2 for all positive integers x.

Q2

2. If m and n are odd positive integers, then m2+n2 is even, but not divisible by 4. Justify.

Sol

Solution:

If m and n are odd positive integers, then m2+n2 is even, but not divisible by 4. Justify.

• Let m and n be odd positive integers. We can represent any odd integer as 2k+1, where k is an integer.

• Let m=2a+1 and n=2b+1 for some integers a and b.

• Then, m2=2a+12=4a2+4a+1=4aa+1+1.

• Similarly, n2=2b+12=4b2+4b+1=4bb+1+1.

Now, consider the sum m2+n2:

m2+n2=4aa+1+1+4bb+1+1=4aa+1+4bb+1+2

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, m2+n2=4aa+1+4bb+1+2 is of the form 4k+2 for some integer k.

This means m2+n2 is even because it is 2 more than a multiple of 4, but it is not divisible by 4 since it is not of the form 4k.

Hence, if m and n are odd positive integers, then m2+n2 is even but not divisible by 4.

Q3

3. 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=22×7

Prime factors of 32=25

LCM(28,32)=25×7=224

Therefore the required smallest number=224−20=204

Verification ⇒28×7=196⇒204−196=8

⇒36×6=192⇒204−192=12

Questions

1. Write whether every positive integer can be of the form 4q + 2, where q is an integer? Justify your answer.

2. The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.

3. 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

1. 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.

2. 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.

3. 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:

n2 = 3q+12 = 9q2 + 6q + 1 = 3(3q2+2q) + 1

This shows n2 is of the form 3m + 1.

Question 1

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 xy then the value of x and y are respectively __ and

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

Solution

Solution

(1)Prime factorization:

• 156: 156 = 22 x 3 x 13

• 208: 208 = 24 x 13

• 206: 260 = 22 x 5 x 13

Common factors: 22 x 13 = 4 x 13 = 52

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 = 15252 = 3

(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 = 24 x 13

Common factor: 2 x 13 = 26

Number of buses for group A = 13026 = 5

Number of buses for group B = 20826= 8

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: 24 x 13

260: 22 x 5 x 13

Common factor: 22 x 13 = 4 x13 = 52

Number of buses for group B = 20852 = 4

Number of buses for group C = 26052 = 5

Total number of buses = 4+5=9

The format of xy seems unclear, but the total buses are 9.

(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 = 15652 = 3

Number of buses for group B = 20852 = 4

Number of buses for group C = 26052 = 5

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

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=578017=340

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).

2. 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.