CBSE Class 10 Mathematics Real Numbers Worksheet Set 02

Fill in the blanks/tables with suitable information:

Question. (2+√5/3) is .................... number.
Answer : irrational

Question. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, the other number is .................... .
Answer : 81

Question. If a = (2² × 3³ × 5⁴) and b = (2³ × 3² × 5) then HCF (a, b) = .................... .
Answer : 180

Question. A decimal number 0.8 can be expressed in its simplest form as .................... .
Answer : x = 8/9

Question. Product of two numbers is 18144 and their HCF is 6, then their LCM is .................... .
Answer : 324

Question. The decimal expression of the rational number 23/2² × 5 will terminate after ............. decimal place(s).
Answer : 2

Question. The HCF of smallest composite number and the smallest prime number is .................... .
Answer : 2

Question. If a and b are positive integers, then
HCF (a,b) x LCM (a,b) /ab = ....................
Answer : 1

Question. ................... is the H.C.F. of two consecutive even numbers.
Answer : 2

Question. If two positive integers p and q can be expressed as p = a²b³ and q = a⁴b; a, b being prime numbers, then LCM (p, q) is.....................
Answer : a⁴b³

Short Answer Type Questions

Question. There are 44 boys and 32 girls in a class. These students arranged in rows for a prayer in such a way that each row consists of only either boys or girls, and every row contains an equal number of students. Find the minimum number of rows in which all students can be arranged.
Answer : 44 = 2² x 11
32 = 2⁵
HCF = 2² = 4
Therefore, minimum number of rows in which all srudents can be arranged = 44/4 + 32/4 = 11 + 8 = 19 rows

Question. Find the largest number which divides 70 and 125 leaving reminder 5 and 8 respectively.
Answer : 70 – 5 = 65
125 – 8 = 117
65 = 5 x 13
117 = 3² x 13
HCF = 13
i.e., 13 is the largest number that will divide 65 and 117.

Question. The LCM of two number is 14 times their HCF. The sum of LCM and HCF is 600.If one number is 280, then find the other number.
Answer : HCF = x
LCM = 14 x HCF = 14x
LCM + HCF = 600
14x + x = 500
15x = 600
x = 40
HCF = 40 and LCM = 14 x 40 = 560
Since, LCM x HCF = product of the numbers
560 x 40 = 280 x second number
Second number = 80

Question. 144 Cartons of coke can and 90 cartons of Pepsi can are to be stacked in a canteen. If each stack is of the same height and is to contain cartons of the same drink. What would be the greater number of cartons each stack would have?
Answer : 144 = 2⁴ x 3²
90 = 2 x 3² x 5
HCF = 2 x 3² = 18 cartons

Question. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons
Answer : No, two numbers cannot have 15 as their HCF and 175 as LCM because,
HCF of the numbers must be a factor of the LCM. Therefore, LCM = k x HCF (k ∈ N)
175 = k x 15
k = 175/15 = 35/3 ∉ N

Question. Prove that √𝑛 is not a rational number if n is not a perfect square.
Answer : Let on the contrary say it is rational .
Then
√𝑛 = 𝑝/𝑞, 𝑞 ≠ 0 where p and q are coprime integers.
so 𝑛 = 𝑝2𝑞2
𝑝2 = 𝑛𝑞2
This shows p divides q
which is a contradiction.
Hence √𝑛 is irrational if n is not a perfect square.

Question. Two bells toll at intervals of 24 minutes and 36 minutes respectively. If they toll together at 9 am, after how many minutes do they toll together again, at the earliest?
Answer : 24 = 2³ x 3
36 = 2² x 3²
LCM = 23 x 3² = 8 x 9 = 72
After 72 minutes = 1 hr 12 minutes they toll together.

Question. Prove that the difference and quotient of ( 3 + 2√3) and ( 3 - 2√3) are irrational.
Answer : The difference of ( 3 + 2√3) and ( 3 - 2√3) is 4√3 which is irrational.
Dividing ( 3 + 2√3) by ( 3 - 2√3) we get -7 - 4√3 which is irrational

Question. Explain why 17 x 5 x 11 x 3 x 2 + 2 x11 is a composite number.
Answer : 17 x 5 x 11 x 3 x 2 + 2 x 11 = 2 x 11 (17 x 5 x 3 + 1)
= 2 x 11 (255 + 1)
= 2 x 11 x 256
= 2 x 11 x 2⁸
This number has more than 2 prime factors.
Therefore, 17 x 5 x 11 x 3 x 2 + 2 x 11 is a composite number.

Question. Find the largest number that will divide 398 , 436 and 542 leaving reminders 7,11 and 15 respectively.
Answer : 398 – 7 = 391
436 – 11 = 425
542 – 15 = 527
391 = 17 x 23
425 = 5² x 17
527 = 17 x 31
HCF = 17
i.e., 17 is the largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.

1. If 7x5x3x2 + 3 is composite number? Justify your answer

2. Show that any positive odd integer is of the form 4q + 1 or 4q +3 where q is a positive integer

3. Prove that √2 + √5 is irrational

4. Prove that 5 - 2√3 is an irrational number

5. Prove that √2 is irrational

6. Use Euclid’s Division Algorithms to find the H.C.F of a) 135 and 225 (45)

b) 4052 and 12576 (4)

c) 270, 405 and 315 (45)

7. Find the HCF and LCM of 26 and 91 and verify that LCM X HCF = Product of two numbers (13,182)

8. Explain why 29 is a terminating decimal expansion 23 x 53

9. 163 will have a terminating decimal expansion. State true or false .Justify your answer.150

10. Find HCF of 96 and 404 by prime factorization method. Hence, find their LCM. (4, 9696)

11. Using prime factorization method find the HCF and LCM of 72, 126 and 168 (6, 504)

12. If HCF (6, a) = 2 and LCM (6, a) = 60 then find a (20)

13. given that LCM (77, 99) = 693, find the HCF (77, 99) (11)

14. Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 3 (138)

15. The LCM of two numbers is 64699, their HCF is 97 and one of the numbers is 2231. Find the other (2813)

16. Two numbers are in the ratio 15: 11. If their HCF is 13 and LCM is 2145 then find the numbers (195,143)

17. Express 0.363636………… in the form a/b (4/11)

18. Write the HCF of smallest composite number and smallest prime number

19. Write whether 2√45 + 3√20 on simplification give a rational or an irrational number 2√5

20. State whether 10.064 is rational or not. If rational, express in p/q form

21. Write a rational number between √2 and √3

22. State the fundamental theorem of arithmetic

Question 1. If 7x5x3x2 + 3 is composite number? Justify your answer
Answer:
Let us simplify the given expression by taking out the common factor:
\( 7 \times 5 \times 3 \times 2 + 3 = 3(7 \times 5 \times 2 + 1) \)
\( = 3(70 + 1) \)
\( = 3 \times 71 \)
Since the number can be expressed as a product of prime factors other than 1 and itself, it has more than two factors. Therefore, by definition, it is a composite number.

In simple words: When we pull out the common number 3, we can see that this expression is equal to 3 multiplied by 71. Since it can be written as a product of two smaller numbers, it must be composite.

Exam Tip: Do not waste time multiplying out the whole expression first. Taking out the common factor is much faster and directly proves the number is composite.

Question 2. Show that any positive odd integer is of the form 4q + 1 or 4q +3 where q is a positive integer
Answer:
Let \( a \) be any positive integer and let \( b = 4 \).
By Euclid's division lemma, we can write:
\( a = 4q + r \), where \( q \) is a positive integer and \( 0 \le r < 4 \).
This means the possible remainders are \( r = 0, 1, 2, 3 \).

Let us analyze each possible case:
- If \( r = 0 \), then \( a = 4q = 2(2q) \), which is divisible by 2 and therefore an even integer.
- If \( r = 1 \), then \( a = 4q + 1 \). Since \( 4q \) is even, \( 4q + 1 \) is an odd integer.
- If \( r = 2 \), then \( a = 4q + 2 = 2(2q + 1) \), which is divisible by 2 and therefore an even integer.
- If \( r = 3 \), then \( a = 4q + 3 \). Since \( 4q + 2 \) is even, \( 4q + 3 \) is an odd integer.

Since \( a \) is an odd integer, it cannot take the even forms \( 4q \) or \( 4q + 2 \). Therefore, any positive odd integer must be of the form \( 4q + 1 \) or \( 4q + 3 \).

In simple words: When we divide any positive integer by 4, the remainder can only be 0, 1, 2, or 3. The cases with remainders of 0 and 2 are even, so the odd numbers must have remainders of 1 or 3.

Exam Tip: Be sure to write down the conditions for the remainder \( 0 \le r < b \) clearly, as examiners award specific step marks for this mathematical statement.

Question 3. Prove that √2 + √5 is irrational
Answer:
Let us assume on the contrary that \( \sqrt{2} + \sqrt{5} \) is a rational number, say \( x \).
We can write:
\( x = \sqrt{2} + \sqrt{5} \)
\( x - \sqrt{2} = \sqrt{5} \)

Squaring both sides of the equation:
\( (x - \sqrt{2})^2 = (\sqrt{5})^2 \)
\( x^2 - 2x\sqrt{2} + 2 = 5 \)
\( x^2 - 3 = 2x\sqrt{2} \)
\( \sqrt{2} = \frac{x^2 - 3}{2x} \)

Since \( x \) is a rational number, the expression \( \frac{x^2 - 3}{2x} \) must also be rational. However, this implies that \( \sqrt{2} \) is rational, which contradicts the established fact that \( \sqrt{2} \) is an irrational number.
Therefore, our assumption is false, and \( \sqrt{2} + \sqrt{5} \) is irrational.

In simple words: If we assume the sum is rational and rearrange the algebra by squaring both sides, we end up showing that the square root of 2 is rational. This is impossible, proving our initial assumption was wrong.

Exam Tip: When squaring both sides, be careful to correctly expand the quadratic identity \( (a-b)^2 = a^2 - 2ab + b^2 \).

Question 4. Prove that 5 - 2√3 is an irrational number
Answer:
Let us assume on the contrary that \( 5 - 2\sqrt{3} \) is a rational number, say \( \frac{a}{b} \), where \( a \) and \( b \) are coprime integers and \( b \ne 0 \).
We can rearrange the terms as follows:
\( 5 - 2\sqrt{3} = \frac{a}{b} \)
\( 2\sqrt{3} = 5 - \frac{a}{b} \)
\( 2\sqrt{3} = \frac{5b - a}{b} \)
\( \sqrt{3} = \frac{5b - a}{2b} \)

Since \( a \) and \( b \) are integers, \( \frac{5b - a}{2b} \) is a rational number. This implies that \( \sqrt{3} \) is rational, which contradicts the known fact that \( \sqrt{3} \) is irrational.
Thus, our assumption is incorrect, and \( 5 - 2\sqrt{3} \) is indeed an irrational number.

In simple words: If we assume the expression is rational and isolate the square root of 3 on one side, we get a fraction made of normal integers. This would mean the square root of 3 is rational, which we know is false.

Exam Tip: Clearly state that \( a \) and \( b \) are coprime integers so that the step-by-step logic of your contradiction is completely sound.

Question 5. Prove that √2 is irrational
Answer:
Let us assume on the contrary that \( \sqrt{2} \) is rational. Therefore, we can write:
\( \sqrt{2} = \frac{a}{b} \), where \( a \) and \( b \) are coprime integers and \( b \ne 0 \).

Squaring both sides:
\( 2 = \frac{a^2}{b^2} \implies a^2 = 2b^2 \) - (i)
Since 2 divides \( a^2 \), it must also divide \( a \) (by theorem).
Let \( a = 2c \) for some integer \( c \).

Substituting \( a = 2c \) into equation (i):
\( (2c)^2 = 2b^2 \)
\( 4c^2 = 2b^2 \implies b^2 = 2c^2 \)
Since 2 divides \( b^2 \), it must also divide \( b \).

This means that both \( a \) and \( b \) have 2 as a common factor, which contradicts our assumption that \( a \) and \( b \) are coprime integers. Thus, our assumption is incorrect, and \( \sqrt{2} \) is an irrational number.

In simple words: If we assume the square root of 2 is a fraction in its simplest form, we can prove that both the top and bottom numbers must be even. This contradicts our assumption that the fraction cannot be simplified further.

Exam Tip: This is a standard proof. Learn the sequence of steps perfectly, especially the theorem stating that if a prime divides \( a^2 \), it must divide \( a \).

Question 6. Use Euclid’s Division Algorithms to find the H.C.F of a) 135 and 225
Answer:
(a) To find the HCF of 135 and 225:
Since \( 225 > 135 \), we apply Euclid's division lemma:
\( 225 = 135 \times 1 + 90 \)
Since the remainder \( 90 \ne 0 \), we apply the lemma to 135 and 90:
\( 135 = 90 \times 1 + 45 \)
Since the remainder \( 45 \ne 0 \), we apply the lemma to 90 and 45:
\( 90 = 45 \times 2 + 0 \)
The remainder is now 0, so our algorithm stops. The divisor at this stage is 45.
Therefore, \( \text{HCF}(135, 225) = 45 \).

(b) To find the HCF of 4052 and 12576:
Since \( 12576 > 4052 \), we apply the division lemma:
\( 12576 = 4052 \times 3 + 420 \)
Since the remainder \( 420 \ne 0 \), we apply the lemma to 4052 and 420:
\( 4052 = 420 \times 9 + 272 \)
Since the remainder \( 272 \ne 0 \), we apply the lemma to 420 and 272:
\( 420 = 272 \times 1 + 148 \)
Since the remainder \( 148 \ne 0 \), we apply the lemma to 272 and 148:
\( 272 = 148 \times 1 + 124 \)
Since the remainder \( 124 \ne 0 \), we apply the lemma to 148 and 124:
\( 148 = 124 \times 1 + 24 \)
Since the remainder \( 24 \ne 0 \), we apply the lemma to 124 and 24:
\( 124 = 24 \times 5 + 4 \)
Since the remainder \( 4 \ne 0 \), we apply the lemma to 24 and 4:
\( 24 = 4 \times 6 + 0 \)
The remainder is now 0, so the divisor at this final step is 4.
Therefore, \( \text{HCF}(4052, 12576) = 4 \).

(c) To find the HCF of 270, 405, and 315:
First, let us find the HCF of 270 and 405:
\( 405 = 270 \times 1 + 135 \)
\( 270 = 135 \times 2 + 0 \)
So, \( \text{HCF}(270, 405) = 135 \).

Now, we find the HCF of 135 and the third number, 315:
\( 315 = 135 \times 2 + 45 \)
\( 135 = 45 \times 3 + 0 \)
The divisor when the remainder becomes 0 is 45.
Therefore, the HCF of 270, 405, and 315 is 45.

In simple words: We repeatedly divide the larger number by the smaller number and replace the numbers with the divisor and the remainder until the remainder becomes zero. The final divisor we used is the HCF.

Exam Tip: For three numbers, always find the HCF of any two numbers first, then find the HCF of that result and the third number.

Question 7. Find the HCF and LCM of 26 and 91 and verify that LCM X HCF = Product of two numbers
Answer:
Let us find the prime factorizations of both numbers:
\( 26 = 2 \times 13 \)
\( 91 = 7 \times 13 \)

The HCF is the product of the lowest powers of common prime factors:
\( \text{HCF} = 13 \)

The LCM is the product of the highest powers of all involved prime factors:
\( \text{LCM} = 2 \times 7 \times 13 = 182 \)

Now, let us verify the relation:
\( \text{LCM} \times \text{HCF} = 182 \times 13 = 2366 \)
\( \text{Product of the two numbers} = 26 \times 91 = 2366 \)
Since both values are equal, the relation is verified.

In simple words: The factors show that the highest common divisor is 13 and the lowest common multiple is 182. Multiplying these two gives 2366, which is exactly the same as multiplying 26 by 91.

Exam Tip: Always show both multiplication calculations clearly on separate lines to make the verification step easy for the examiner to read.

Question 8. Explain why 29 is a terminating decimal expansion
23 x 53
Answer:
Let the given fraction be \( \frac{29}{2^3 \times 5^3} \).
According to the decimal expansion theorem, a rational number has a terminating decimal expansion if the prime factorization of its denominator is of the form \( 2^m \times 5^n \), where \( m \) and \( n \) are non-negative integers.

In this case, the denominator is \( 2^3 \times 5^3 \), which is explicitly in the form \( 2^m \times 5^n \) with \( m = 3 \) and \( n = 3 \).
Since there are no prime factors other than 2 and 5 in the denominator, the fraction will have a terminating decimal expansion.

In simple words: Since the bottom part of the fraction only contains factors of 2 and 5, it can easily be converted into a power of 10. This ensures the decimal ends after a finite number of digits.

Exam Tip: Mention the standard theorem about the denominator's factorization \( 2^m \times 5^n \) to secure full marks for the explanation.

Question 9. 163 will have a terminating decimal expansion. State true or false .Justify your answer.
150
Answer:
The given rational number is \( \frac{163}{150} \).
Let us analyze the prime factorization of the denominator, 150:
\( 150 = 2 \times 3 \times 5^2 \)

For a rational number to have a terminating decimal expansion, the prime factorization of its denominator must be strictly of the form \( 2^m \times 5^n \).
Since the denominator of \( \frac{163}{150} \) contains the prime factor 3 (which cannot be canceled with the numerator 163 as 163 is prime), it is not in the form \( 2^m \times 5^n \).
Therefore, it will have a non-terminating repeating decimal expansion. This means the given statement is False.

In simple words: The bottom of the fraction contains the prime factor 3. Because of this extra factor of 3, the decimal will go on forever without ending, making the statement false.

Exam Tip: Always check if the numerator and denominator have any common factors to cancel out before checking the prime factors of the denominator.

Question 10. Find HCF of 96 and 404 by prime factorization method. Hence, find their LCM.
Answer:
Let us write the prime factorizations of both numbers:
\( 96 = 2^5 \times 3 \)
\( 404 = 2^2 \times 101 \)

The HCF is the product of the lowest powers of common prime factors:
\( \text{HCF} = 2^2 = 4 \)

Now, using the relationship between HCF and LCM:
\( \text{LCM} = \frac{\text{Product of the two numbers}}{\text{HCF}} \)
\( \text{LCM} = \frac{96 \times 404}{4} \)
\( \text{LCM} = 96 \times 101 = 9696 \).
Therefore, the HCF is 4 and the LCM is 9696.

In simple words: By looking at their prime factors, we find they share a common factor of 4. We then use this HCF to quickly divide and calculate their lowest common multiple of 9696.

Exam Tip: When a question says "Hence, find their LCM", you must use the HCF formula to find the LCM rather than calculating the LCM independently from prime factors.

Question 11. Using prime factorization method find the HCF and LCM of 72, 126 and 168
Answer:
First, let us write the prime factorizations of the three numbers:
\( 72 = 2^3 \times 3^2 \)
\( 126 = 2 \times 3^2 \times 7 \)
\( 168 = 2^3 \times 3 \times 7 \)

To find the HCF, we take the product of the lowest powers of the common prime factors:
Common prime factors are 2 and 3.
\( \text{HCF} = 2^1 \times 3^1 = 6 \)

To find the LCM, we take the product of the highest powers of all the prime factors involved:
The prime factors involved are 2, 3, and 7.
\( \text{LCM} = 2^3 \times 3^2 \times 7^1 = 8 \times 9 \times 7 = 504 \).
Therefore, the HCF is 6 and the LCM is 504.

In simple words: The only prime numbers that divide all three values are 2 and 3, making the HCF equal to 6. Combining the maximum counts of all factors gives us the LCM of 504.

Exam Tip: Be careful when identifying common factors. The prime factor 7 is only present in 126 and 168, so it must not be included in the HCF calculation.

Question 12. If HCF (6, a) = 2 and LCM (6, a) = 60 then find a
Answer:
We know the fundamental relationship between two numbers and their HCF and LCM:
\( \text{Product of two numbers} = \text{HCF} \times \text{LCM} \)

Given:
\( \text{HCF}(6, a) = 2 \)
\( \text{LCM}(6, a) = 60 \)
One of the numbers is 6, and the other is \( a \).

Substituting these values:
\( 6 \times a = 2 \times 60 \)
\( 6a = 120 \)
\( a = \frac{120}{6} = 20 \).
Therefore, the value of \( a \) is 20.

In simple words: Multiplying two numbers gives the same result as multiplying their HCF and LCM. Since \( 2 \times 60 = 120 \), the other number must be \( 120 / 6 = 20 \).

Exam Tip: This simple formula is a favorite for one-mark and two-mark questions. Learn to rearrange it quickly for any of the four variables.

Question 13. given that LCM (77, 99) = 693, find the HCF (77, 99)
Answer:
Using the standard relationship between HCF and LCM of two numbers:
\( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \)

Here, the two numbers are \( a = 77 \) and \( b = 99 \), and \( \text{LCM}(77, 99) = 693 \).

Rearranging the formula to solve for HCF:
\( \text{HCF}(77, 99) = \frac{77 \times 99}{693} \)
\( \text{HCF}(77, 99) = \frac{7623}{693} = 11 \).
Therefore, the HCF of 77 and 99 is 11.

In simple words: We can calculate the common divisor by multiplying the two numbers (7623) and dividing the product by their LCM (693), which gives us 11.

Exam Tip: You can simplify the fraction before multiplying the numerator: \( \frac{77 \times 99}{693} = \frac{77 \times 1}{7} = 11 \). This prevents large multiplication steps.

Question 14. Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 3
Answer:
The required greatest number must divide \( 280 - 4 \) and \( 1245 - 3 \) exactly.
Therefore, we need to find the HCF of 276 and 1242.

Let us use the prime factorization method:
\( 276 = 2^2 \times 3 \times 23 \)
\( 1242 = 2 \times 3^3 \times 23 \)

The HCF is the product of the lowest powers of the common prime factors:
\( \text{HCF} = 2^1 \times 3^1 \times 23^1 = 6 \times 23 = 138 \).
Therefore, the greatest number is 138.

In simple words: We first subtract the respective remainders from the two numbers to find the parts that must divide perfectly (276 and 1242). Finding the highest common factor of these two numbers gives us 138.

Exam Tip: Always subtract the remainders from the original numbers first before calculating the HCF. Doing it the other way around is a common mistake.

Question 15. The LCM of two numbers is 64699, their HCF is 97 and one of the numbers is 2231. Find the other
Answer:
Let the other number be \( x \).
Using the formula:
\( \text{First Number} \times \text{Second Number} = \text{HCF} \times \text{LCM} \)

Substitute the given values:
\( 2231 \times x = 97 \times 64699 \)
\( x = \frac{97 \times 64699}{2231} \)
\( x = \frac{6275803}{2231} = 2813 \).
Therefore, the other number is 2813.

In simple words: Since the product of the two numbers is equal to the product of their HCF and LCM, we multiply 97 by 64699 and divide the result by 2231 to find the second number is 2813.

Exam Tip: Since 2231 is a multiple of 97 (\( 97 \times 23 = 2231 \)), you can simplify the division first: \( x = \frac{64699}{23} = 2813 \).

Question 16. Two numbers are in the ratio 15: 11. If their HCF is 13 and LCM is 2145 then find the numbers
Answer:
Let the two numbers be \( 15k \) and \( 11k \), where \( k \) represents their common factor.
Since the HCF of the two numbers is given as 13, the common factor \( k \) must be equal to 13.

Therefore, the two numbers are:
- First number = \( 15 \times 13 = 195 \)
- Second number = \( 11 \times 13 = 143 \)

Let us verify with the given LCM:
LCM of 195 and 143:
\( 195 = 3 \times 5 \times 13 \)
\( 143 = 11 \times 13 \)
\( \text{LCM} = 3 \times 5 \times 11 \times 13 = 2145 \), which matches the given LCM.
Hence, the numbers are 195 and 143.

In simple words: The ratio tells us the relative size of the numbers. Since their highest common factor is 13, we simply multiply both parts of the ratio by 13 to get the actual numbers: 195 and 143.

Exam Tip: If the numbers are coprime in their ratio (like 15 and 11), the actual numbers are simply obtained by multiplying the ratio terms by their HCF.

Question 17. Express 0.363636………… in the form a/b
Answer:
Let \( x = 0.363636\dots \) - (i)
Since two digits are repeating, we multiply both sides of the equation by 100:
\( 100x = 36.363636\dots \) - (ii)

Subtracting equation (i) from equation (ii):
\( 100x - x = (36.363636\dots) - (0.363636\dots) \)
\( 99x = 36 \)
\( x = \frac{36}{99} \)

Simplifying the fraction by dividing the numerator and denominator by 9:
\( x = \frac{4}{11} \).
Therefore, the rational form of the decimal is \( \frac{4}{11} \).

In simple words: We set the repeating decimal as \( x \). Multiplying by 100 shifts the decimal point by two places, allowing us to subtract the original equation and cancel out the infinite repeating decimal part.

Exam Tip: The multiplier depends on the number of repeating digits: multiply by 10 for 1 repeating digit, 100 for 2 digits, and 1000 for 3 digits.

Question 18. Write the HCF of smallest composite number and smallest prime number
Answer:
- The smallest prime number is 2.
- The smallest composite number is 4.

Now, we find the HCF of 2 and 4:
\( 2 = 2^1 \)
\( 4 = 2^2 \)
\( \text{HCF}(2, 4) = 2 \).
Therefore, the HCF is 2.

In simple words: The smallest prime number is 2, and the smallest composite number is 4. The greatest number that divides both 2 and 4 is 2.

Exam Tip: Make sure you know the basic definitions. 1 is neither prime nor composite, so the smallest prime is 2, and the smallest composite is 4.

Question 19. Write whether 2√45 + 3√20 on simplification give a rational or an irrational number
2√5
Answer:
Let us simplify the square roots in the numerator:
\( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \)
\( \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)

Substitute these values back into the expression:
\( \frac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} = \frac{2(3\sqrt{5}) + 3(2\sqrt{5})}{2\sqrt{5}} \)
\( = \frac{6\sqrt{5} + 6\sqrt{5}}{2\sqrt{5}} \)
\( = \frac{12\sqrt{5}}{2\sqrt{5}} \)
\( = \frac{12}{2} = 6 \).
Since 6 is an integer, it is a rational number. Therefore, on simplification, the expression gives a rational number.

In simple words: By simplifying the terms inside the square roots, we find that the numerator is equal to \( 12\sqrt{5} \). Dividing this by the denominator cancels the square root of 5, leaving us with the rational number 6.

Exam Tip: Do not assume a number is irrational just because it contains square roots. Always simplify the expression completely first before making a decision.

Question 20. State whether 10.064 is rational or not. If rational, express in p/q form
Answer:
The decimal number is \( 10.064 \).
Since the decimal expansion is terminating, the number is indeed a rational number.

Now, let us convert it into \( \frac{p}{q} \) form:
\( 10.064 = \frac{10064}{1000} \)

Simplifying the fraction by dividing the numerator and denominator by their HCF, which is 8:
\( 10064 \div 8 = 1258 \)
\( 1000 \div 8 = 125 \)
\( \frac{p}{q} = \frac{1258}{125} \).

In simple words: Since the decimal ends, it is a rational number. We write it as a fraction over 1000 and simplify the terms to get 1258/125.

Exam Tip: Terminating decimals are always rational. To write them in \( p/q \) form, place the digits over the appropriate power of 10 and simplify.

Question 21. Write a rational number between √2 and √3
Answer:
The approximate values of the given square roots are:
\( \sqrt{2} \approx 1.414 \)
\( \sqrt{3} \approx 1.732 \)

We need to find a rational number \( r \) such that:
\( 1.414 < r < 1.732 \)
Plausible terminating decimals in this range are \( 1.5 \) and \( 1.6 \).
Let us choose \( 1.5 \), which can be written in fractional form as:
\( 1.5 = \frac{15}{10} = \frac{3}{2} \).
Therefore, a rational number between \( \sqrt{2} \) and \( \sqrt{3} \) is \( 1.5 \) (or \( \frac{3}{2} \)).

In simple words: The value of the square root of 2 is about 1.41 and the square root of 3 is about 1.73. A simple rational number like 1.5 lies right between these two values.

Exam Tip: Converting irrational square roots to their approximate decimal values is the simplest way to find numbers in between them.

Question 22. State the fundamental theorem of arithmetic
Answer:
The Fundamental Theorem of Arithmetic states that:
Every composite number can be uniquely expressed (or factorized) as a product of prime numbers, except for the order in which the prime factors occur.
For example, the composite number 30 can be uniquely factorized as \( 2 \times 3 \times 5 \).

In simple words: This theorem says that any number that is not prime can be broken down into a unique set of prime numbers multiplied together, no matter what order we write them in.

Exam Tip: Be sure to write the word "uniquely" in your definition, as examiners look for this key term when grading this question.