Chapter 1: Real Numbers (Competency-Based Questions) [1]
Q1. Show that the numbers 143 and 187 are not co-prime. [1]
Solution: Two numbers are co-prime if their Highest Common Factor (HCF) is equal to 1 [1].
Finding prime factorisations:
\\(143 = 11 \times 13\\)
\\(187 = 11 \times 17\\)
\\(\text{HCF}(143, 187) = 11 \neq 1\\) [1].
Since the HCF is 11 (which is not equal to 1), 143 and 187 share a common factor other than 1 and are not co-prime [1].
Finding prime factorisations:
\\(143 = 11 \times 13\\)
\\(187 = 11 \times 17\\)
\\(\text{HCF}(143, 187) = 11 \neq 1\\) [1].
Since the HCF is 11 (which is not equal to 1), 143 and 187 share a common factor other than 1 and are not co-prime [1].
Q2. Pens are sold in packs of 8 and notepads are sold in packs of 12. Find the least number of packs of each type that one should buy so that there are an equal number of pens and notepads. [2]
Solution: To find the equal number of pens and notepads, we compute the Least Common Multiple (LCM) of 8 and 12 [2].
\\(\text{LCM}(8, 12) = 24\\) [2].
Least number of packs of pens = \\(\frac{24}{8} = 3\\) packs [2].
Least number of packs of notepads = \\(\frac{24}{12} = 2\\) packs [2].
\\(\text{LCM}(8, 12) = 24\\) [2].
Least number of packs of pens = \\(\frac{24}{8} = 3\\) packs [2].
Least number of packs of notepads = \\(\frac{24}{12} = 2\\) packs [2].
Q3. A boy with a collection of marbles realizes that if he makes a group of 5 or 6 marbles at a time, there are always 2 marbles left. Can you explain why the boy cannot have a prime number of marbles? [2]
Solution:
\\(\text{LCM}(5, 6) = 5 \times 6 = 30\\) [2].
The number of marbles the boy has can be expressed in the form \\(30n + 2\\), where \\(n \in \mathbb{N}\\) [3].
Factoring out 2: \\(30n + 2 = 2(15n + 1)\\) [3].
Since \\(2(15n + 1)\\) has 2 as a factor other than 1 and itself, it is a composite number and cannot be a prime number [3].
\\(\text{LCM}(5, 6) = 5 \times 6 = 30\\) [2].
The number of marbles the boy has can be expressed in the form \\(30n + 2\\), where \\(n \in \mathbb{N}\\) [3].
Factoring out 2: \\(30n + 2 = 2(15n + 1)\\) [3].
Since \\(2(15n + 1)\\) has 2 as a factor other than 1 and itself, it is a composite number and cannot be a prime number [3].
Q4. A rectangular courtyard is \\(18\text{ m } 72\text{ cm}\\) long and \\(13\text{ m } 20\text{ cm}\\) broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles. [3]
Solution:
Length = \\(1872\text{ cm}\\), Breadth = \\(1320\text{ cm}\\) [3].
Size of largest square tile = \\(\text{HCF}(1872, 1320) = 24\text{ cm}\\) [3].
Least number of tiles required = \\(\frac{\text{Area of courtyard}}{\text{Area of one tile}} = \frac{1872 \times 1320}{24 \times 24} = 78 \times 55 = 4290\\) tiles [3].
Length = \\(1872\text{ cm}\\), Breadth = \\(1320\text{ cm}\\) [3].
Size of largest square tile = \\(\text{HCF}(1872, 1320) = 24\text{ cm}\\) [3].
Least number of tiles required = \\(\frac{\text{Area of courtyard}}{\text{Area of one tile}} = \frac{1872 \times 1320}{24 \times 24} = 78 \times 55 = 4290\\) tiles [3].
Figure 1.1: Courtyard paving geometry with square tiles of side length equal to HCF [3].
Q5. If two positive integers \\(x\\) and \\(y\\) are expressible in terms of primes as \\(x = p^2 q^3\\) and \\(y = p^3 q\\), what can you say about their LCM and HCF? Is LCM a multiple of HCF? Explain. [3]
Solution:
\\(\text{HCF}(x, y) = p^2 q\\) [4].
\\(\text{LCM}(x, y) = p^3 q^3\\) [4].
Yes, LCM is a multiple of HCF because \\(\text{LCM} = pq^2 \times \text{HCF}\\) [4]. In general, the LCM of any two numbers is always a multiple of their HCF [4].
\\(\text{HCF}(x, y) = p^2 q\\) [4].
\\(\text{LCM}(x, y) = p^3 q^3\\) [4].
Yes, LCM is a multiple of HCF because \\(\text{LCM} = pq^2 \times \text{HCF}\\) [4]. In general, the LCM of any two numbers is always a multiple of their HCF [4].
Q6. Dhudnath has two vessels containing \\(720\text{ mL}\\) and \\(405\text{ mL}\\) of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled. [4]
Solution:
Maximum capacity of each glass = \\(\text{HCF}(720, 405) = 45\text{ mL}\\) [4].
Minimum number of glasses = \\(\frac{720}{45} + \frac{405}{45} = 16 + 9 = 25\\) glasses [4].
Maximum capacity of each glass = \\(\text{HCF}(720, 405) = 45\text{ mL}\\) [4].
Minimum number of glasses = \\(\frac{720}{45} + \frac{405}{45} = 16 + 9 = 25\\) glasses [4].
Q7. Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy? [5]
Solution:
Equal number of chocolates = \\(\text{LCM}(24, 15) = 120\\) chocolates.
Number of boxes for pack of 24 = \\(\frac{120}{24} = 5\\) boxes [5].
Number of boxes for pack of 15 = \\(\frac{120}{15} = 8\\) boxes [5].
Equal number of chocolates = \\(\text{LCM}(24, 15) = 120\\) chocolates.
Number of boxes for pack of 24 = \\(\frac{120}{24} = 5\\) boxes [5].
Number of boxes for pack of 15 = \\(\frac{120}{15} = 8\\) boxes [5].
Q8. Find the largest number which divides 70 and 125 leaving remainders 5 and 8 respectively. [5]
Solution:
The required number is \\(\text{HCF}(70 - 5, 125 - 8) = \text{HCF}(65, 117)\\) [5].
\\(65 = 5 \times 13\\)
\\(117 = 3^2 \times 13\\)
\\(\text{HCF}(65, 117) = 13\\) [5].
The required number is \\(\text{HCF}(70 - 5, 125 - 8) = \text{HCF}(65, 117)\\) [5].
\\(65 = 5 \times 13\\)
\\(117 = 3^2 \times 13\\)
\\(\text{HCF}(65, 117) = 13\\) [5].
Q9. Can the number \\(6^n\\), \\(n\\) being a natural number, end with the digit 5? Give reasons. [5]
Solution: No [5].
For a number to end with the digit 5, its prime factorisation must contain 5 as a prime factor.
The prime factorisation of \\(6^n = (2 \times 3)^n = 2^n \times 3^n\\).
By the Fundamental Theorem of Arithmetic, the prime factors of \\(6^n\\) are only 2 and 3. Since 5 is not in the prime factorisation, \\(6^n\\) can never end with the digit 5 [5].
For a number to end with the digit 5, its prime factorisation must contain 5 as a prime factor.
The prime factorisation of \\(6^n = (2 \times 3)^n = 2^n \times 3^n\\).
By the Fundamental Theorem of Arithmetic, the prime factors of \\(6^n\\) are only 2 and 3. Since 5 is not in the prime factorisation, \\(6^n\\) can never end with the digit 5 [5].
Q10. A sweet shopkeeper prepares 396 gulab jamuns and 342 ras-gullas. He packs them in containers. Each container consists of either gulab jamuns or ras-gullas but has an equal number of pieces. Find the number of pieces he should put in each box so that the number of boxes is least. [6]
Solution:
Number of pieces per box = \\(\text{HCF}(396, 342)\\) [6].
\\(396 = 2^2 \times 3^2 \times 11\\)
\\(342 = 2 \times 3^2 \times 19\\)
\\(\text{HCF}(396, 342) = 2 \times 3^2 = 18\\) pieces per box [6].
Number of pieces per box = \\(\text{HCF}(396, 342)\\) [6].
\\(396 = 2^2 \times 3^2 \times 11\\)
\\(342 = 2 \times 3^2 \times 19\\)
\\(\text{HCF}(396, 342) = 2 \times 3^2 = 18\\) pieces per box [6].
Q11. Find two numbers which on multiplication with \\(\sqrt{180}\\) give a rational number. Are these numbers rational or irrational? [6]
Solution:
Simplifying \\(\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 6\sqrt{5}\\) [6].
Two numbers that yield a rational number when multiplied with \\(6\sqrt{5}\\) are \\(\sqrt{5}\\) and \\(2\sqrt{5}\\) [7].
Check: \\(6\sqrt{5} \times \sqrt{5} = 30\\) (Rational), and \\(6\sqrt{5} \times 2\sqrt{5} = 60\\) (Rational).
These multiplying numbers (\\(\sqrt{5}\\) and \\(2\sqrt{5}\\)) are irrational numbers [7].
Simplifying \\(\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 6\sqrt{5}\\) [6].
Two numbers that yield a rational number when multiplied with \\(6\sqrt{5}\\) are \\(\sqrt{5}\\) and \\(2\sqrt{5}\\) [7].
Check: \\(6\sqrt{5} \times \sqrt{5} = 30\\) (Rational), and \\(6\sqrt{5} \times 2\sqrt{5} = 60\\) (Rational).
These multiplying numbers (\\(\sqrt{5}\\) and \\(2\sqrt{5}\\)) are irrational numbers [7].
Q12. Prove that the sum of a rational number and an irrational number is always irrational. [7]
Proof:
Let \\(x\\) be a rational number and \\(y\\) be an irrational number [7].
Assume on the contrary that \\(x + y\\) is rational, say \\(x + y = \frac{p}{q}\\) where \\(p, q \in \mathbb{Z}, q \neq 0\\) [7].
Then \\(y = \frac{p}{q} - x\\) [7].
Since \\(\frac{p}{q}\\) and \\(x\\) are both rational, their difference \\(\frac{p}{q} - x\\) must be a rational number [7].
This implies \\(y\\) is rational, which contradicts the fact that \\(y\\) is irrational [8].
Hence, our assumption is false. The sum \\(x + y\\) is always irrational [8].
Let \\(x\\) be a rational number and \\(y\\) be an irrational number [7].
Assume on the contrary that \\(x + y\\) is rational, say \\(x + y = \frac{p}{q}\\) where \\(p, q \in \mathbb{Z}, q \neq 0\\) [7].
Then \\(y = \frac{p}{q} - x\\) [7].
Since \\(\frac{p}{q}\\) and \\(x\\) are both rational, their difference \\(\frac{p}{q} - x\\) must be a rational number [7].
This implies \\(y\\) is rational, which contradicts the fact that \\(y\\) is irrational [8].
Hence, our assumption is false. The sum \\(x + y\\) is always irrational [8].
Q13. Is the square root of every non-square number always irrational? Find the smallest natural number which divides 2205 to make its square root a rational number. [8]
Solution:
Yes, the square root of every non-square natural number is always irrational [8].
Prime factorisation of 2205:
\\(2205 = 5 \times 21 \times 21 = 5 \times 3^2 \times 7^2\\) [8].
Dividing 2205 by 5 yields \\(21 \times 21 = 441\\), which is a perfect square whose square root is 21 (a rational number) [9].
Thus, the required smallest natural number is 5 [9].
Yes, the square root of every non-square natural number is always irrational [8].
Prime factorisation of 2205:
\\(2205 = 5 \times 21 \times 21 = 5 \times 3^2 \times 7^2\\) [8].
Dividing 2205 by 5 yields \\(21 \times 21 = 441\\), which is a perfect square whose square root is 21 (a rational number) [9].
Thus, the required smallest natural number is 5 [9].
Q14. The sum of LCM and HCF of two numbers is 7380. If the LCM of these numbers is 7340 more than their HCF, find the product of the two numbers. [9]
Solution:
Let \\(\text{LCM} = x\\) and \\(\text{HCF} = y\\) [9].
\\(x + y = 7380\\) ---(1) [9]
\\(x - y = 7340\\) ---(2) [9]
Adding (1) and (2): \\(2x = 14720 \implies x = 7360\\) [9].
From (1): \\(7360 + y = 7380 \implies y = 20\\) [10].
\\(\text{Product of numbers} = \text{HCF} \times \text{LCM} = 20 \times 7360 = 147200\\) [10].
Let \\(\text{LCM} = x\\) and \\(\text{HCF} = y\\) [9].
\\(x + y = 7380\\) ---(1) [9]
\\(x - y = 7340\\) ---(2) [9]
Adding (1) and (2): \\(2x = 14720 \implies x = 7360\\) [9].
From (1): \\(7360 + y = 7380 \implies y = 20\\) [10].
\\(\text{Product of numbers} = \text{HCF} \times \text{LCM} = 20 \times 7360 = 147200\\) [10].
Q15. Find the greatest number of six digits exactly divisible by 18, 24, and 36. [10]
Solution:
\\(\text{LCM}(18, 24, 36) = 72\\) [10].
The greatest 6-digit number is 999999 [10].
Dividing 999999 by 72 gives a remainder of 63 [10].
Required number = \\(999999 - 63 = 9,99,936\\) [10].
\\(\text{LCM}(18, 24, 36) = 72\\) [10].
The greatest 6-digit number is 999999 [10].
Dividing 999999 by 72 gives a remainder of 63 [10].
Required number = \\(999999 - 63 = 9,99,936\\) [10].
Q16. Jenny and Sally bought a special 360 days joint membership of a tennis club. Jenny will use the club every alternate day and Sally will use the club every third day. They both use the club on the first day. How many days will neither person use the club in the 360 days? [10]
Solution:
Jenny uses the club on \\(\frac{360}{2} = 180\\) days [11].
Sally uses the club on \\(\frac{360}{3} = 120\\) days [11].
Days both use the club = \\(\frac{360}{\text{LCM}(2, 3)} = \frac{360}{6} = 60\\) days [11].
Days either person uses the club = \\(180 + 120 - 60 = 240\\) days [11].
Days neither person uses the club = \\(360 - 240 = 120\\) days [11].
Jenny uses the club on \\(\frac{360}{2} = 180\\) days [11].
Sally uses the club on \\(\frac{360}{3} = 120\\) days [11].
Days both use the club = \\(\frac{360}{\text{LCM}(2, 3)} = \frac{360}{6} = 60\\) days [11].
Days either person uses the club = \\(180 + 120 - 60 = 240\\) days [11].
Days neither person uses the club = \\(360 - 240 = 120\\) days [11].
Q17. Write the HCF and LCM of the smallest odd composite number and the smallest odd prime number. If an odd number \\(p\\) divides \\(q^2\\), then will it divide \\(q^3\\) also? Explain. [11]
Solution:
Smallest odd composite number = 9
Smallest odd prime number = 3
\\(\text{HCF}(9, 3) = 3\\) and \\(\text{LCM}(9, 3) = 9\\) [11].
Yes, if an odd prime number \\(p\\) divides \\(q^2\\), then by theorem, \\(p\\) divides \\(q\\) [11]. Since \\(p\\) divides \\(q\\), it will also divide \\(q^3\\) [11].
Smallest odd composite number = 9
Smallest odd prime number = 3
\\(\text{HCF}(9, 3) = 3\\) and \\(\text{LCM}(9, 3) = 9\\) [11].
Yes, if an odd prime number \\(p\\) divides \\(q^2\\), then by theorem, \\(p\\) divides \\(q\\) [11]. Since \\(p\\) divides \\(q\\), it will also divide \\(q^3\\) [11].
Q18. A dining hall has a length of \\(8.25\text{ m}\\), breadth of \\(6.75\text{ m}\\), and height of \\(4.50\text{ m}\\). What is the length of the longest unmarked ruler that can exactly measure the three dimensions of the hall? [11]
Solution:
Length = \\(825\text{ cm}\\), Breadth = \\(675\text{ cm}\\), Height = \\(450\text{ cm}\\) [11].
Required ruler length = \\(\text{HCF}(825, 675, 450)\\) [12].
\\(825 = 3 \times 5^2 \times 11\\)
\\(675 = 3^3 \times 5^2\\)
\\(450 = 2 \times 3^2 \times 5^2\\)
\\(\text{HCF}(825, 675, 450) = 3 \times 5^2 = 75\text{ cm}\\) (or \\(0.75\text{ m}\\)) [12].
Length = \\(825\text{ cm}\\), Breadth = \\(675\text{ cm}\\), Height = \\(450\text{ cm}\\) [11].
Required ruler length = \\(\text{HCF}(825, 675, 450)\\) [12].
\\(825 = 3 \times 5^2 \times 11\\)
\\(675 = 3^3 \times 5^2\\)
\\(450 = 2 \times 3^2 \times 5^2\\)
\\(\text{HCF}(825, 675, 450) = 3 \times 5^2 = 75\text{ cm}\\) (or \\(0.75\text{ m}\\)) [12].
Q19. The prime factorisation of a prime number is the number itself. How many factors and prime factors does the square of a prime number have? [12]
Solution:
Let \\(p\\) be a prime number. Its square is \\(p^2\\) [12].
Factors of \\(p^2\\) are \\(1, p,\\) and \\(p^2\\) \\(\implies\\) Total number of factors = 3 [12].
Prime factor of \\(p^2\\) is only \\(p\\) \\(\implies\\) Number of prime factors = 1 [12].
Let \\(p\\) be a prime number. Its square is \\(p^2\\) [12].
Factors of \\(p^2\\) are \\(1, p,\\) and \\(p^2\\) \\(\implies\\) Total number of factors = 3 [12].
Prime factor of \\(p^2\\) is only \\(p\\) \\(\implies\\) Number of prime factors = 1 [12].
Q20. If the prime factorisation of a natural number \\(n\\) is \\(2^3 \times 3^2 \times 5^2 \times 7\\), then write the number of consecutive zeroes in \\(n\\). [12]
Solution:
Consecutive zeroes are determined by pairs of \\(2 \times 5 = 10\\) [12].
\\(n = 2^3 \times 3^2 \times 5^2 \times 7 = 2^1 \times 3^2 \times 7 \times (2^2 \times 5^2)\\) [12]
\\(n = (2 \times 9 \times 7) \times 10^2 = 126 \times 100 = 12600\\) [12].
Therefore, there are 2 consecutive zeroes in \\(n\\) [12].
Consecutive zeroes are determined by pairs of \\(2 \times 5 = 10\\) [12].
\\(n = 2^3 \times 3^2 \times 5^2 \times 7 = 2^1 \times 3^2 \times 7 \times (2^2 \times 5^2)\\) [12]
\\(n = (2 \times 9 \times 7) \times 10^2 = 126 \times 100 = 12600\\) [12].
Therefore, there are 2 consecutive zeroes in \\(n\\) [12].
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