CBSE Class 10 Maths Chapter 1 - Real Numbers Model Questions - 1 Marks - Part 1
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CBSE Class 10 Maths Chapter 1 - Real Numbers Model Questions - 1 Marks - Part 1
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SECTION A — Multiple Choice Questions
[1 Mark Each]
1 MarkQ1. If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = xy^3$, where $x, y$ are prime numbers, then $\text{HCF}(a, b)$ is:
(a) $xy$
(b) $xy^2$
(c) $x^3y^3$
(d) $x^2y^2$
Solution:
$\text{HCF}$ is the product of the smallest power of each common prime factor involved in the numbers.
$a = x^3 y^2$, $b = xy^3$
Common prime factors are $x$ and $y$. Smallest power of $x$ is $x^1$ and of $y$ is $y^2$.
$\text{HCF}(a, b) = xy^2$. Answer: (b) $xy^2$
1 MarkQ2. The LCM of smallest two-digit composite number and smallest composite number is:
1 MarkQ3. Given that $\text{HCF}(306, 657) = 9$, then $\text{LCM}(306, 657)$ is:
(a) $22338$
(b) $22328$
(c) $22348$
(d) $22358$
Solution:
We know that $\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b$
$\text{LCM}(306, 657) = \frac{306 \times 657}{\text{HCF}(306, 657)} = \frac{306 \times 657}{9} = 34 \times 657 = 22338$. Answer: (a) $22338$
1 MarkQ4. Which of the following is a rational number?
(a) $\sqrt{3}$
(b) $\pi$
(c) $3 + \sqrt{5}$
(d) $\frac{3\sqrt{7}}{\sqrt{7}}$
Solution:
$\frac{3\sqrt{7}}{\sqrt{7}} = 3$, which can be expressed in the form $p/q$ where $q \neq 0$. Thus, it is a rational number, whereas the others are irrational. Answer: (d) $\frac{3\sqrt{7}}{\sqrt{7}}$
1 MarkQ5. The decimal expansion of the rational number $\frac{143}{1100}$ will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Solution:
Simplify the fraction: $\frac{143}{1100} = \frac{13 \times 11}{100 \times 11} = \frac{13}{100} = \frac{13}{10^2}$.
Since the denominator is of the form $2^n 5^m$ with max exponent 2, it terminates after two decimal places. Answer: (b) two decimal places
1 MarkQ6. If $n$ is any natural number, then $6^n$ always ends with the digit:
(a) $3$
(b) $6$
(c) $5$
(d) $0$
Solution:
For any natural number $n$, $6^n = (2 \times 3)^n$. Since it contains both 2 and 3 as prime factors but no 5, $6^n$ will always end with the digit 6 for any $n \in \mathbb{N}$. Answer: (b) $6$
1 MarkQ7. The product of a non-zero rational and an irrational number is always:
(a) always rational
(b) always irrational
(c) rational or irrational
(d) one
Solution:
The product of a non-zero rational number and an irrational number is always an irrational number. Answer: (b) always irrational
1 MarkQ8. The HCF of two numbers is 18 and their product is 1296. Their LCM is:
1 MarkQ9. Which of the following rational numbers has a terminating decimal expansion?
(a) $\frac{17}{8}$
(b) $\frac{11}{15}$
(c) $\frac{35}{50}$
(d) Both (a) and (c)
Solution:
A rational number has a terminating decimal if the denominator's prime factorization is of the form $2^n 5^m$.
(a) $\frac{17}{8} = \frac{17}{2^3}$ (terminating)
(b) $\frac{11}{15} = \frac{11}{3 \times 5}$ (non-terminating)
(c) $\frac{35}{50} = \frac{7}{10} = \frac{7}{2 \times 5}$ (terminating) Answer: (d) Both (a) and (c)
1 MarkQ10. If two positive integers $p$ and $q$ can be expressed as $p = ab^2$ and $q = a^3b$, where $a, b$ are prime numbers, then $\text{LCM}(p, q)$ is:
(a) $ab$
(b) $a^2b^2$
(c) $a^3b^2$
(d) $a^3b^3$
Solution:
$\text{LCM}$ is the product of the highest power of each prime factor involved.
For $p = ab^2$ and $q = a^3b$, the highest power of $a$ is $a^3$ and of $b$ is $b^2$.
$\text{LCM}(p, q) = a^3b^2$. Answer: (c) $a^3b^2$
1 MarkQ11. The exponent of 2 in the prime factorization of 144 is:
(a) $2$
(b) $4$
(c) $6$
(d) $8$
Solution:
$144 = 12^2 = (2^2 \times 3)^2 = 2^4 \times 3^2$.
The exponent of 2 in this prime factorization is 4. Answer: (b) $4$
1 MarkQ12. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:
(a) $10$
(b) $100$
(c) $2520$
(d) $5040$
Solution:
The least number divisible by numbers from 1 to 10 is their LCM: $\text{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2520$. Answer: (c) $2520$
1 MarkQ13. If $\text{HCF}(165, 255) = 15$, then $\text{LCM}(165, 255)$ is:
1 MarkQ14. Which of the following rational numbers has a non-terminating repeating decimal expansion?
(a) $\frac{13}{3125}$
(b) $\frac{7}{80}$
(c) $\frac{64}{455}$
(d) $\frac{3}{160}$
Solution:
$455 = 5 \times 7 \times 13$. Since the denominator has prime factors other than 2 and 5 (namely 7 and 13), $\frac{64}{455}$ has a non-terminating repeating decimal expansion. Answer: (c) $\frac{64}{455}$
1 MarkQ15. The total number of distinct prime factors of the number 1729 is:
(a) $3$
(b) $4$
(c) $5$
(d) $6$
Solution:
$1729 = 7 \times 13 \times 19$.
The distinct prime factors are 7, 13, and 19 (total 3 distinct prime factors). Answer: (a) $3$
1 MarkQ16. Three bells toll together at intervals of 9, 12, and 15 minutes respectively. If they toll together now, after what time will they next toll together?
1 MarkQ17. If $p$ and $q$ are two positive co-prime integers, then their $\text{HCF}(p, q)$ is:
(a) $p$
(b) $q$
(c) $pq$
(d) $1$
Solution:
By definition, two integers are co-prime if their only common positive factor is 1. Thus, their HCF is always 1. Answer: (d) $1$
1 MarkQ18. The decimal expansion of $\frac{93}{1500}$ will terminate after how many decimal places?
(a) $1$
(b) $2$
(c) $3$
(d) $4$
Solution:
Simplify: $\frac{93}{1500} = \frac{31}{500} = \frac{31}{2^2 \times 5^3}$.
The maximum exponent between 2 and 5 in the denominator is 3, so it terminates after 3 decimal places. Answer: (c) $3$
1 MarkQ19. Which of the following is always true for any positive integer $n$ if $n^2 - n$ is considered?
(a) It is always divisible by $2$
(b) It is always divisible by $3$
(c) It is always odd
(d) None of these
Solution:
$n^2 - n = n(n-1)$, which is the product of two consecutive integers. The product of two consecutive integers is always even, meaning it is always divisible by 2. Answer: (a) It is always divisible by $2$
1 MarkQ20. If $a = 2^3 \times 3$, $b = 2 \times 3 \times 5$, $c = 3^n \times 5$ and $\text{LCM}(a, b, c) = 2^3 \times 3^2 \times 5$, then the value of $n$ is:
(a) $1$
(b) $2$
(c) $3$
(d) $4$
Solution:
The LCM considers the highest power of each prime factor. For factor 3, the powers are $3^1$ (from $a$), $3^1$ (from $b$), and $3^n$ (from $c$). Since the LCM has $3^2$, the maximum power $n$ must be 2. Answer: (b) $2$
1 MarkQ21. In a factor tree, if the top number $x$ has branches $3$ and $y$, and under $y$ are the branches $5$ and $7$, then the value of $x$ is:
1 MarkQ22. Which of the following is an irrational number?
(a) $\sqrt{4}$
(b) $3\sqrt{5}$
(c) $\frac{2\sqrt{3}}{\sqrt{3}}$
(d) $\sqrt{9}$
Solution:
$\sqrt{4} = 2$, $\frac{2\sqrt{3}}{\sqrt{3}} = 2$, and $\sqrt{9} = 3$ are all rational numbers. $3\sqrt{5}$ is the product of a non-zero rational and irrational, making it irrational. Answer: (b) $3\sqrt{5}$
1 MarkQ23. Sonia and Ravi take 18 minutes and 12 minutes respectively to drive one round of a circular field. If they start at the same point and time and go in the same direction, after how many minutes will they meet again at the starting point?
(a) $18$ minutes
(b) $24$ minutes
(c) $36$ minutes
(d) $72$ minutes
Solution:
The time they meet again is given by $\text{LCM}(18, 12)$ = $\text{LCM}(2 \times 3^2, 2^2 \times 3) = 2^2 \times 3^2 = 36$ minutes. Answer: (c) $36$ minutes
1 MarkQ24. Can two numbers have $18$ as their HCF and $380$ as their LCM?
(a) Yes
(b) No
(c) Only if both numbers are even
(d) Cannot be determined
Solution:
The HCF of two numbers must always exactly divide their LCM. Here, $380$ is not divisible by $18$ ($380 \div 18 = 21.11$), so such numbers cannot exist. Answer: (b) No
1 MarkQ25. The sum of the exponents of the prime factors in the prime factorization of $196$ is:
(a) $2$
(b) $3$
(c) $4$
(d) $5$
Solution:
$196 = 14^2 = (2 \times 7)^2 = 2^2 \times 7^2$.
Sum of the exponents = $2 + 2 = 4$. Answer: (c) $4$
1 MarkQ26. The decimal expansion of the rational number $\frac{33}{2^2 \times 5}$ will terminate after:
(a) $1$ decimal place
(b) $2$ decimal places
(c) $3$ decimal places
(d) Will not terminate
Solution:
$\frac{33}{2^2 \times 5} = \frac{33}{20} = \frac{33 \times 5}{20 \times 5} = \frac{165}{100} = 1.65$.
The maximum exponent between 2 and 5 in the denominator is 2, so it terminates after two decimal places. Answer: (b) $2$ decimal places
1 MarkQ27. For any positive integer $n$, $n^3 - n$ is always divisible by:
(a) Only $3$
(b) Only $6$
(c) Both $2$ and $3$
(d) $6$ always
Solution:
$n^3 - n = n(n^2 - 1) = n(n-1)(n+1)$, which is the product of three consecutive integers. The product of three consecutive integers is always divisible by both 2 and 3, meaning it is always divisible by 6. Answer: (d) $6$ always
1 MarkQ28. If $a$ and $b$ are two coprime numbers, then $a^2$ and $b^2$ are always:
(a) Coprime
(b) Not coprime
(c) Even numbers
(d) Odd numbers
Solution:
If $a$ and $b$ share no common prime factors (coprime), their squares $a^2$ and $b^2$ will also share no common prime factors and are therefore coprime. Answer: (a) Coprime
1 MarkQ29. If $p_1$ and $p_2$ are two odd prime numbers such that $p_1 > p_2$, then $p_1^2 - p_2^2$ is always:
(a) an even number
(b) an odd number
(c) a prime number
(d) a composite odd number
Solution:
The square of any odd prime number is odd. The difference of two odd numbers is always an even number. Answer: (a) an even number
1 MarkQ30. If $\text{HCF}(a, b) = 12$ and the product of the numbers $a \times b = 1800$, then $\text{LCM}(a, b)$ is:
(a) $150$
(b) $300$
(c) $120$
(d) $15$
Solution:
$\text{LCM}(a, b) = \frac{\text{Product of } a \text{ and } b}{\text{HCF}(a, b)} = \frac{1800}{12} = 150$. Answer: (a) $150$
1 MarkQ31. If the HCF of 65 and 117 is expressible in the form $65m - 117$, then the value of $m$ is:
1 MarkQ32. The decimal expansion of the rational number $\frac{27}{2^3 \times 5^4}$ will terminate after how many decimal places?
(a) $3$
(b) $4$
(c) $5$
(d) $7$
Solution:
The maximum exponent between 2 and 5 in the denominator is 4, so the decimal expansion terminates after 4 decimal places. Answer: (b) $4$
1 MarkQ33. If two positive integers $a$ and $b$ are written as $a = x^4 y$ and $b = x^2 y^3$ where $x, y$ are prime numbers, then the value of $\frac{\text{LCM}(a, b)}{\text{HCF}(a, b)}$ is:
1 MarkQ34. What is the largest number that divides 245 and 1029, leaving a remainder of 5 in each case?
(a) $12$
(b) $15$
(c) $16$
(d) $24$
Solution:
Subtract the remainder from each number: $245 - 5 = 240$ and $1029 - 5 = 1024$.
$\text{HCF}(240, 1024) = 16$. Answer: (c) $16$
1 MarkQ35. For any positive integer $n$, $3^{2n} - 1$ is always divisible by:
(a) Only $3$
(b) Only $6$
(c) $8$
(d) $9$
Solution:
$3^{2n} - 1 = (9)^n - 1^2$, which is always divisible by $9 - 1 = 8$ for any positive integer $n$. Answer: (c) $8$
1 MarkQ36. The ratio of the HCF to the LCM of the least prime number and the least composite number is:
(a) $1 : 2$
(b) $2 : 1$
(c) $1 : 4$
(d) $4 : 1$
Solution:
Least prime number = 2, Least composite number = 4.
$\text{HCF}(2, 4) = 2$, $\text{LCM}(2, 4) = 4$.
Ratio of HCF to LCM = $2 : 4 = 1 : 2$. Answer: (a) $1 : 2$
1 MarkQ37. If $\text{HCF}(a, 8) = 4$ and $\text{LCM}(a, 8) = 24$, then the value of $a$ is:
1 MarkQ40. If $x$ and $y$ are two coprime numbers, then $x^3$ and $y^3$ are always:
(a) Coprime
(b) Composite numbers
(c) Even numbers
(d) Divisible by 3
Solution:
Since $x$ and $y$ share no common prime factors, their powers $x^3$ and $y^3$ will also share no common prime factors and are therefore coprime. Answer: (a) Coprime
1 MarkQ41. The ratio between the LCM and HCF of 5, 15, and 20 is:
(a) $9 : 1$
(b) $4 : 3$
(c) $11 : 1$
(d) $12 : 1$
Solution:
$\text{HCF}(5, 15, 20) = 5$, $\text{LCM}(5, 15, 20) = 60$.
Ratio of LCM to HCF = $60 : 5 = 12 : 1$. Answer: (d) $12 : 1$
1 MarkQ42. If two positive integers $a$ and $b$ are written as $a = xy^3$ and $b = x^4yz$, where $x, y, z$ are prime numbers, then $\text{LCM}(a, b)$ is:
(a) $xy^2$
(b) $x^4y^2z$
(c) $x^4y^3$
(d) $x^4y^3z$
Solution:
$\text{LCM}$ is the product of the highest power of each prime factor ($x, y, z$): $x^4 y^3 z$. Answer: (d) $x^4y^3z$
1 MarkQ43. The HCF of 8, 9, and 25 is:
(a) $8$
(b) $9$
(c) $25$
(d) $1$
Solution:
Since 8, 9, and 25 are pairwise coprime and share no common factor other than 1, their HCF is 1. Answer: (d) $1$
1 MarkQ44. The product of three consecutive positive integers is always divisible by:
(a) $4$
(b) $6$
(c) No common factor
(d) Only $1$
Solution:
The product of three consecutive integers always contains a multiple of 2 and a multiple of 3, making it always divisible by 6. Answer: (b) $6$
1 MarkQ45. For positive integers $a$ and $3$, there exist unique integers $q$ and $r$ such that $a = 3q + r$, where $r$ must satisfy:
(a) $0 \le r < 3$
(b) $1 < r < 3$
(c) $0 < r < 3$
(d) $0 < r \le 3$
Solution:
By Euclid's Division Lemma, for any integer $a$ and divisor $b$, the remainder $r$ satisfies $0 \le r < b$. Here $b = 3$, so $0 \le r < 3$. Answer: (a) $0 \le r < 3$
1 MarkQ46. If $a$ and $b$ are two positive integers such that the least prime factor of $a$ is 3 and the least prime factor of $b$ is 5, then the least prime factor of $(a + b)$ is:
(a) $1$
(b) $2$
(c) $3$
(d) $5$
Solution:
Since the least prime factor of $a$ is 3 and of $b$ is 5, both $a$ and $b$ must be odd numbers. The sum of two odd numbers ($a + b$) is an even number, and the least prime factor of any even number is 2. Answer: (b) $2$
1 MarkQ47. If the LCM of 12 and 42 is given by $10m + 4$, then the value of $m$ is:
1 MarkQ48. The HCF of the numbers $k, 2k, 3k, 4k$, and $5k$, where $k$ is a positive integer, is:
(a) $k$
(b) $2k$
(c) $3k$
(d) $5k$
Solution:
All terms share a common factor $k$, and the remaining numbers $(1, 2, 3, 4, 5)$ have an HCF of 1. Thus, the HCF of the sequence is $k$. Answer: (a) $k$
1 MarkQ49. Two natural numbers whose difference is 66 and whose least common multiple is 360 are:
(a) $120 \text{ and } 54$
(b) $90 \text{ and } 24$
(c) $180 \text{ and } 114$
(d) $130 \text{ and } 64$
Solution:
Checking the options for a difference of 66 and an LCM of 360:
For 90 and 24: $90 - 24 = 66$ and $\text{LCM}(90, 24) = 360$. Answer: (b) $90 \text{ and } 24$
1 MarkQ50. Which of the following numbers is divisible by 11?
(a) $1516$
(b) $1452$
(c) $1011$
(d) $1121$
Solution:
A number is divisible by 11 if the difference between the sum of digits at odd places and even places is 0 or a multiple of 11.
For 1452: $(2 + 4) - (5 + 1) = 6 - 6 = 0$. Answer: (b) $1452$
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English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
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