RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
This chapter takes the numbers you have used since primary school and puts them on a proper logical footing. The Fundamental Theorem of Arithmetic says every composite number breaks into primes in exactly one way, and that single fact powers HCF, LCM and the proofs that numbers like √5 can never be written as a fraction. Expect short, high-scoring questions that reward clean prime factorisation.
Two numbers have HCF 9 and LCM 540. If one number is 45, the other number is:
Answer
The other number is 108. For two positive integers, HCF × LCM = product of the numbers. So 9 × 540 = 45 × (other number). 4860 = 45 × (other number), giving other number = 4860/45 = 108. Check: 45 = 3² × 5 and 108 = 2² × 3³, so HCF = 3² = 9 and LCM = 2² × 3³ × 5 = 540. Both conditions hold.
Which of the following rational numbers has a terminating decimal expansion?
Answer
17/200 has a terminating decimal expansion. A fraction in lowest terms terminates only if its denominator is of the form 2ᵐ × 5ⁿ. 200 = 2³ × 5², which contains only 2s and 5s, so 17/200 terminates (it equals 0.085). 75 = 3 × 5², 42 = 2 × 3 × 7 and 60 = 2² × 3 × 5 each carry a prime factor other than 2 and 5, so those three expansions are non-terminating and recurring.
The LCM of 24 and 90 is:
Answer
The LCM is 360. Prime factorise: 24 = 2³ × 3 and 90 = 2 × 3² × 5. LCM takes the highest power of every prime that appears: 2³ × 3² × 5 = 8 × 9 × 5 = 360. As a check, HCF = 2 × 3 = 6 and HCF × LCM = 6 × 360 = 2160 = 24 × 90.
Which one of the following numbers is rational?
Answer
√18/√2 is rational. √18/√2 = √(18/2) = √9 = 3, an integer and therefore rational. 2√7 is a non-zero rational times an irrational, so it is irrational. 5 - √3 is a rational minus an irrational, so it is irrational. π is irrational, so π + 1 is irrational as well.
Assertion (A): The HCF of 120 and 45 is 15. Reason (R): The HCF of two numbers is the product of the smallest power of each prime factor common to both numbers.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the prime-factorisation rule for HCF correctly. Applying it: 120 = 2³ × 3 × 5 and 45 = 3² × 5. The primes common to both are 3 and 5, and the smallest powers are 3¹ and 5¹. So HCF = 3 × 5 = 15, which is exactly what Assertion A claims. Hence R explains A.
Find the HCF and the LCM of 156 and 234 using prime factorisation, and verify that their product equals the product of the two numbers.
Answer
Prime factorise both numbers. 156 = 2² × 3 × 13 and 234 = 2 × 3² × 13. HCF = smallest power of each common prime = 2 × 3 × 13 = 78. LCM = highest power of every prime = 2² × 3² × 13 = 4 × 9 × 13 = 468. Verification: HCF × LCM = 78 × 468 = 36504, and 156 × 234 = 36504. The two agree.
Given that √3 is irrational, prove that 5 + 2√3 is irrational.
Answer
Assume, to the contrary, that 5 + 2√3 is rational. Then we may write 5 + 2√3 = a/b, where a and b are integers and b ≠ 0. Rearranging, 2√3 = a/b - 5 = (a - 5b)/b, so √3 = (a - 5b)/(2b). The right-hand side is a ratio of two integers with non-zero denominator, so it is rational. That would make √3 rational. This contradicts the given fact that √3 is irrational. Hence our assumption is wrong, and 5 + 2√3 is irrational.
Three athletes jog around a circular track and complete one lap in 48 seconds, 60 seconds and 72 seconds respectively. If they start together from the same point, after how many minutes will they next be together at the starting point?
Answer
They meet again after a time that is a common multiple of all three lap times, and the first such time is the LCM. 48 = 2⁴ × 3, 60 = 2² × 3 × 5, 72 = 2³ × 3². LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720 seconds. 720 seconds = 720/60 = 12 minutes. They are next together at the starting point after 12 minutes.
(a) Prove that √11 is irrational. (b) Using part (a), show that 4 + √11 is also irrational.
Answer
(a) Suppose √11 is rational. Then √11 = a/b where a and b are coprime integers and b ≠ 0. Squaring, 11 = a²/b², so a² = 11b². Thus 11 divides a². Since 11 is prime, 11 must divide a, so write a = 11c for some integer c. Substituting, (11c)² = 11b² gives 121c² = 11b², so b² = 11c². Thus 11 divides b², and again since 11 is prime, 11 divides b. So 11 is a common factor of a and b, contradicting the assumption that they are coprime. Hence √11 is irrational. (b) Suppose 4 + √11 were rational, say 4 + √11 = p/q with integers p, q and q ≠ 0. Then √11 = p/q - 4 = (p - 4q)/q, which is a ratio of integers and therefore rational. That contradicts part (a). Hence 4 + √11 is irrational.
A florist has 264 roses and 429 lilies. She wants to make identical bouquets using all the flowers, with every bouquet containing the same number of roses and the same number of lilies. Find the greatest number of bouquets she can make and the composition of each bouquet. Also find the LCM of 264 and 429 and verify the relation between the HCF, the LCM and the two numbers.
Answer
Step 1 — The number of bouquets must divide both 264 and 429, and we want the greatest such number, so we need HCF(264, 429). Step 2 — Prime factorise: 264 = 2³ × 3 × 11 and 429 = 3 × 11 × 13. Step 3 — Common primes are 3 and 11, each to the power 1, so HCF = 3 × 11 = 33. The greatest number of bouquets is 33. Step 4 — Roses per bouquet = 264/33 = 8, lilies per bouquet = 429/33 = 13. So each bouquet has 8 roses and 13 lilies. Step 5 — LCM = 2³ × 3 × 11 × 13 = 8 × 3 × 11 × 13 = 3432. Step 6 — Verification: HCF × LCM = 33 × 3432 = 113256, and 264 × 429 = 113256. The relation HCF × LCM = product of the numbers holds.
Read the following and answer the questions that follow: A shopfront has three decorative LED signs. The first flashes every 18 seconds, the second every 24 seconds and the third every 30 seconds. On a particular morning all three flash together exactly at 9:00 am. (a) After how many seconds will all three flash together again? (b) At what time will they next flash together, and how many times will all three flash together between 9:00 am and 10:00 am, not counting the 9:00 am flash? (c) Find the HCF of 18, 24 and 30 and state what it represents here. (d) Verify that HCF(18, 24) × LCM(18, 24) = 18 × 24.
Answer
(a) They flash together after a common multiple of 18, 24 and 30, and the first one is the LCM. 18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360 seconds. (b) 360 seconds = 6 minutes, so they next flash together at 9:06 am. In the hour from 9:00 am to 10:00 am there are 60/6 = 10 such moments (9:06, 9:12, …, 10:00), so all three flash together 10 times after the 9:00 am flash. (c) Common primes of 18, 24 and 30 are 2 and 3, each to the power 1, so HCF = 2 × 3 = 6. It is the largest time interval (6 seconds) that divides all three flashing periods exactly. (d) 18 = 2 × 3², 24 = 2³ × 3, so HCF(18, 24) = 2 × 3 = 6 and LCM(18, 24) = 2³ × 3² = 72. HCF × LCM = 6 × 72 = 432, and 18 × 24 = 432. Verified.
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