CBSE Class 10 · Maths
CBSE Class 10 Maths Real Numbers Competency Based Questions
Help your child build a rock-solid foundation with these CBSE Class 10 Maths Real Numbers competency based questions, crafted to match the latest CBSE exam pattern. Every question comes with a detailed, step-by-step solution so your child understands exactly where marks come from — not just the final answer.
CBSE Class 10 Maths Real Numbers — Questions with Solutions
The exponent of 5 in the prime factorisation of 3750 is
Solution
Option (B) is correct.
Explanation: According to the prime factorisation, 3750 can be written as
$3750=5 \times 5 \times 5 \times 5 \times 3 \times 2=5^{4} \times 3^{1} \times 2^{1}$
It is clear from above, that exponent of 5 in the prime factorisation of 3750 is 4.
Explanation: According to the prime factorisation, 3750 can be written as
$3750=5 \times 5 \times 5 \times 5 \times 3 \times 2=5^{4} \times 3^{1} \times 2^{1}$
It is clear from above, that exponent of 5 in the prime factorisation of 3750 is 4.
What is the greatest possible speed at which a girl can walk 95 m and 171 m in an exact number of minutes?
Solution
Option (B) is correct.
Explanation: As the girl needs to walk 95 m and 171 m at the exact number of minutes, we have to find HCF of 95 and 171. According to prime factorisation of 95 and 171:
$$95 = 5 \times 19$$
$$171 = 3 \times 3 \times 19$$
$$\operatorname{HCF}(95,171) = 19$$
Hence, greatest possible speed is $19 \mathrm{~m} / \mathrm{min}$.
Explanation: As the girl needs to walk 95 m and 171 m at the exact number of minutes, we have to find HCF of 95 and 171. According to prime factorisation of 95 and 171:
$$95 = 5 \times 19$$
$$171 = 3 \times 3 \times 19$$
$$\operatorname{HCF}(95,171) = 19$$
Hence, greatest possible speed is $19 \mathrm{~m} / \mathrm{min}$.
Three alarm clocks ring their alarms at regular intervals of 20 min., 25 min. and 30 min. respectively. If they first beep together at 12 noon, at what time will they beep again for the first time?
Solution
Option (C) is correct.
Explanation: Time when they ring together = $\operatorname{LCM}(20, 25, 30)$
According to prime factorisation:
$$20 = 2 \times 2 \times 5$$
$$25 = 5 \times 5$$
$$30 = 2 \times 3 \times 5$$
$$\operatorname{LCM} = 2 \times 2 \times 3 \times 5 \times 5 = 300$$
Thus, 3 bells ring together after 300 minutes or 5 hours. Since they rang together first at 12 noon, they ring together again at 5:00 pm.
Explanation: Time when they ring together = $\operatorname{LCM}(20, 25, 30)$
According to prime factorisation:
$$20 = 2 \times 2 \times 5$$
$$25 = 5 \times 5$$
$$30 = 2 \times 3 \times 5$$
$$\operatorname{LCM} = 2 \times 2 \times 3 \times 5 \times 5 = 300$$
Thus, 3 bells ring together after 300 minutes or 5 hours. Since they rang together first at 12 noon, they ring together again at 5:00 pm.
The greatest number which when divides 1251, 9377 and 15628 leaves remainder 1, 2 and 3 respectively is
Solution
Option (D) is correct.
First subtract the remainders from their respective numbers:
$$1251-1 = 1250$$
$$9377-2 = 9375$$
$$15628-3 = 15625$$
According to the prime factorisation:
$$1250 = 2 \times 5 \times 5 \times 5 \times 5$$
$$9375 = 3 \times 5 \times 5 \times 5 \times 5 \times 5$$
$$15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5$$
$$\operatorname{HCF}(1250, 9375, 15625) = 5 \times 5 \times 5 \times 5 = 625$$
First subtract the remainders from their respective numbers:
$$1251-1 = 1250$$
$$9377-2 = 9375$$
$$15628-3 = 15625$$
According to the prime factorisation:
$$1250 = 2 \times 5 \times 5 \times 5 \times 5$$
$$9375 = 3 \times 5 \times 5 \times 5 \times 5 \times 5$$
$$15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5$$
$$\operatorname{HCF}(1250, 9375, 15625) = 5 \times 5 \times 5 \times 5 = 625$$
If $a$ and $b$ are two coprime numbers, then $a^{3}$ and $b^{3}$ are
Solution
Option (A) is correct.
Explanation: As $a$ and $b$ are co-prime then $a^{3}$ and $b^{3}$ are also co-prime. We can understand this with an example.
Let $a=3$ and $b=4$.
$a^{3}=3^{3}=27$ and $b^{3}=4^{3}=64$
$\operatorname{HCF}(a, b)=\operatorname{HCF}(3,4)=1$
$\operatorname{HCF}(a^{3}, b^{3})=\operatorname{HCF}(27,64)=1$
Explanation: As $a$ and $b$ are co-prime then $a^{3}$ and $b^{3}$ are also co-prime. We can understand this with an example.
Let $a=3$ and $b=4$.
$a^{3}=3^{3}=27$ and $b^{3}=4^{3}=64$
$\operatorname{HCF}(a, b)=\operatorname{HCF}(3,4)=1$
$\operatorname{HCF}(a^{3}, b^{3})=\operatorname{HCF}(27,64)=1$
If $n$ is a natural number, then $2(5^{n}+6^{n})$ always ends with
Solution
Option (D) is correct.
Explanation: Let us take examples of different powers of 5:
$$5^{1}=5, \quad 5^{2}=25, \quad 5^{3}=125, \quad 5^{4}=625 \ldots$$
It is clear that $5^{n}$ will always end with 5. Similarly, $6^{n}$ will always end with 6.
So, $5^{n}+6^{n}$ will always end with $5+6=11$.
Also, $2(5^{n}+6^{n})$ will always end with $2 \times 11=22$, i.e., it will always end with 2.
Explanation: Let us take examples of different powers of 5:
$$5^{1}=5, \quad 5^{2}=25, \quad 5^{3}=125, \quad 5^{4}=625 \ldots$$
It is clear that $5^{n}$ will always end with 5. Similarly, $6^{n}$ will always end with 6.
So, $5^{n}+6^{n}$ will always end with $5+6=11$.
Also, $2(5^{n}+6^{n})$ will always end with $2 \times 11=22$, i.e., it will always end with 2.
The LCM of two numbers is 2400. Which of the following cannot be their HCF?
Solution
Option (C) is correct.
Explanation: According to the property, HCF of two numbers is also a factor of LCM of the same two numbers. Out of all the options, only (C) 500 is not a factor of 2400. Therefore, 500 cannot be the HCF.
Explanation: According to the property, HCF of two numbers is also a factor of LCM of the same two numbers. Out of all the options, only (C) 500 is not a factor of 2400. Therefore, 500 cannot be the HCF.
Which of these numbers can be expressed as a product of two or more prime numbers?
(i) 15
(ii) 34568
(iii) $(15 \times 13)$
(i) 15
(ii) 34568
(iii) $(15 \times 13)$
Solution
Option (D) is correct.
Explanation:
Prime factors of $15 = 3 \times 5$
Prime factors of $34568 = 2^{3} \times 29 \times 149$
Prime factors of $15 \times 13 = 3 \times 5 \times 13$
It is clear that all three numbers in parts (i), (ii) and (iii) are products of two or more prime numbers.
Explanation:
Prime factors of $15 = 3 \times 5$
Prime factors of $34568 = 2^{3} \times 29 \times 149$
Prime factors of $15 \times 13 = 3 \times 5 \times 13$
It is clear that all three numbers in parts (i), (ii) and (iii) are products of two or more prime numbers.
1245 is a factor of the numbers $p$ and $q$. Which of the following will always have 1245 as a factor?
(i) $p+q$ (ii) $p \times q$ (iii) $p \div q$
(i) $p+q$ (ii) $p \times q$ (iii) $p \div q$
Solution
Option (B) is correct.
Explanation: Since 1245 is a factor of both $p$ and $q$, we can write $p = 1245a$ and $q = 1245b$.
Now, $p+q = 1245(a+b)$. Hence $(p+q)$ always has factor 1245.
Now, $p \times q = (1245)^{2}(ab)$. Hence $(p \times q)$ always has factor 1245.
$p \div q = \frac{a}{b}$, which need not have 1245 as a factor.
Explanation: Since 1245 is a factor of both $p$ and $q$, we can write $p = 1245a$ and $q = 1245b$.
Now, $p+q = 1245(a+b)$. Hence $(p+q)$ always has factor 1245.
Now, $p \times q = (1245)^{2}(ab)$. Hence $(p \times q)$ always has factor 1245.
$p \div q = \frac{a}{b}$, which need not have 1245 as a factor.
To enhance the reading skills of Grade X students, the school nominates you and two of your friends to set up a class library. There are two sections — Section A and Section B of Grade X. There are 32 students in Section A and 36 students in Section B.
(I) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?
(A) 144 (B) 128 (C) 288 (D) 272
(II) If the product of two positive integers is equal to the product of their HCF and LCM, then the HCF(32, 36) is:
(A) 2 (B) 4 (C) 6 (D) 8
(III) 36 can be expressed as a product of its primes as:
(A) $2^{2} \times 3^{2}$ (B) $2^{1} \times 3^{3}$ (C) $2^{3} \times 3^{1}$ (D) $2^{0} \times 3^{0}$
(IV) $7 \times 11 \times 13 \times 15 + 15$ is a:
(A) Prime number (B) Composite number (C) Neither prime nor composite (D) None of the above
(V) If $p$ and $q$ are positive integers such that $p = ab^{2}$ and $q = a^{2}b$, where $a$, $b$ are prime numbers, then the LCM$(p, q)$ is:
(A) $ab$ (B) $a^{2}b^{2}$ (C) $a^{3}b^{2}$ (D) $a^{3}b^{3}$
(I) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?
(A) 144 (B) 128 (C) 288 (D) 272
(II) If the product of two positive integers is equal to the product of their HCF and LCM, then the HCF(32, 36) is:
(A) 2 (B) 4 (C) 6 (D) 8
(III) 36 can be expressed as a product of its primes as:
(A) $2^{2} \times 3^{2}$ (B) $2^{1} \times 3^{3}$ (C) $2^{3} \times 3^{1}$ (D) $2^{0} \times 3^{0}$
(IV) $7 \times 11 \times 13 \times 15 + 15$ is a:
(A) Prime number (B) Composite number (C) Neither prime nor composite (D) None of the above
(V) If $p$ and $q$ are positive integers such that $p = ab^{2}$ and $q = a^{2}b$, where $a$, $b$ are prime numbers, then the LCM$(p, q)$ is:
(A) $ab$ (B) $a^{2}b^{2}$ (C) $a^{3}b^{2}$ (D) $a^{3}b^{3}$
Answer
(I) Option (C) is correct.
We have to find the LCM of 32 and 36.
$$32 = 2^{5}$$
$$36 = 2^{2} \times 3^{2}$$
$$\operatorname{LCM}(32,36) = 2^{5} \times 3^{2} = 32 \times 9 = 288$$
Hence, the minimum number of books required is 288.
(II) Option (B) is correct.
$$\operatorname{HCF}(32,36) = \frac{32 \times 36}{288} = 4$$
(III) Option (A) is correct.
Prime factorisation of $36 = 2 \times 2 \times 3 \times 3 = 2^{2} \times 3^{2}$
(IV) Option (B) is correct.
$$7 \times 11 \times 13 \times 15 + 15 = 15(7 \times 11 \times 13 + 1) = 15 \times 1002 = 3 \times 5 \times 1002$$
It is a composite number because it has more than 2 factors.
(V) Option (B) is correct.
$$p = ab^{2} = a \times b \times b, \quad q = a^{2}b = a \times a \times b$$
$$\operatorname{LCM}(p,q) = a^{2}b^{2}$$
We have to find the LCM of 32 and 36.
$$32 = 2^{5}$$
$$36 = 2^{2} \times 3^{2}$$
$$\operatorname{LCM}(32,36) = 2^{5} \times 3^{2} = 32 \times 9 = 288$$
Hence, the minimum number of books required is 288.
(II) Option (B) is correct.
$$\operatorname{HCF}(32,36) = \frac{32 \times 36}{288} = 4$$
(III) Option (A) is correct.
Prime factorisation of $36 = 2 \times 2 \times 3 \times 3 = 2^{2} \times 3^{2}$
(IV) Option (B) is correct.
$$7 \times 11 \times 13 \times 15 + 15 = 15(7 \times 11 \times 13 + 1) = 15 \times 1002 = 3 \times 5 \times 1002$$
It is a composite number because it has more than 2 factors.
(V) Option (B) is correct.
$$p = ab^{2} = a \times b \times b, \quad q = a^{2}b = a \times a \times b$$
$$\operatorname{LCM}(p,q) = a^{2}b^{2}$$
A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
(I) In each room, the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can be accommodated in each room is:
(A) 14 (B) 12 (C) 16 (D) 18
(II) What is the minimum number of rooms required during the event?
(A) 11 (B) 31 (C) 41 (D) 21
(III) The LCM of 60, 84 and 108 is:
(A) 3780 (B) 3680 (C) 4780 (D) 4680
(IV) The product of HCF and LCM of 60, 84 and 108 is:
(A) 55360 (B) 35360 (C) 45500 (D) 45360
(V) 108 can be expressed as a product of its primes as:
(A) $2^{3} \times 3^{2}$ (B) $2^{3} \times 3^{3}$ (C) $2^{2} \times 3^{2}$ (D) $2^{2} \times 3^{3}$
(I) In each room, the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can be accommodated in each room is:
(A) 14 (B) 12 (C) 16 (D) 18
(II) What is the minimum number of rooms required during the event?
(A) 11 (B) 31 (C) 41 (D) 21
(III) The LCM of 60, 84 and 108 is:
(A) 3780 (B) 3680 (C) 4780 (D) 4680
(IV) The product of HCF and LCM of 60, 84 and 108 is:
(A) 55360 (B) 35360 (C) 45500 (D) 45360
(V) 108 can be expressed as a product of its primes as:
(A) $2^{3} \times 3^{2}$ (B) $2^{3} \times 3^{3}$ (C) $2^{2} \times 3^{2}$ (D) $2^{2} \times 3^{3}$
Answer
(I) Option (B) is correct.
$$60 = 2 \times 2 \times 3 \times 5, \quad 84 = 2 \times 2 \times 3 \times 7, \quad 108 = 2 \times 2 \times 3 \times 3 \times 3$$
$$\operatorname{HCF} = 12$$
(II) Option (D) is correct.
$$\text{No. of rooms} = \frac{60+84+108}{12} = \frac{252}{12} = 21$$
(III) Option (A) is correct.
$$60 = 2^{2} \times 3 \times 5, \quad 84 = 2^{2} \times 3 \times 7, \quad 108 = 2^{2} \times 3^{3}$$
$$\operatorname{LCM}(60,84,108) = 2^{2} \times 3^{3} \times 5 \times 7 = 4 \times 27 \times 35 = 3780$$
(IV) Option (D) is correct.
$$\operatorname{HCF} \times \operatorname{LCM} = 12 \times 3780 = 45360$$
(V) Option (D) is correct.
$$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^{2} \times 3^{3}$$
$$60 = 2 \times 2 \times 3 \times 5, \quad 84 = 2 \times 2 \times 3 \times 7, \quad 108 = 2 \times 2 \times 3 \times 3 \times 3$$
$$\operatorname{HCF} = 12$$
(II) Option (D) is correct.
$$\text{No. of rooms} = \frac{60+84+108}{12} = \frac{252}{12} = 21$$
(III) Option (A) is correct.
$$60 = 2^{2} \times 3 \times 5, \quad 84 = 2^{2} \times 3 \times 7, \quad 108 = 2^{2} \times 3^{3}$$
$$\operatorname{LCM}(60,84,108) = 2^{2} \times 3^{3} \times 5 \times 7 = 4 \times 27 \times 35 = 3780$$
(IV) Option (D) is correct.
$$\operatorname{HCF} \times \operatorname{LCM} = 12 \times 3780 = 45360$$
(V) Option (D) is correct.
$$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^{2} \times 3^{3}$$
Read the following text and answer the following questions:
A Mathematics Exhibition is being conducted in your school and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
(I) What will be the value of $X$?
(II) What will be the value of $Y$? OR What will be the value of $Z$?
(III) Find LCM($X$, $Y$ and $Z$).
A Mathematics Exhibition is being conducted in your school and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
(I) What will be the value of $X$?
(II) What will be the value of $Y$? OR What will be the value of $Z$?
(III) Find LCM($X$, $Y$ and $Z$).
Answer
(I) $$X = 2783 \times 5 = 13915$$
(II) $$2783 = Y \times 253 \implies Y = \frac{2783}{253} = 11$$
OR
$$253 = 11 \times Z \implies Z = \frac{253}{11} = 23$$
(III) Here, $X = 13915$, $Y = 11$, $Z = 23$.
$$\operatorname{LCM}(13915, 11, 23) = 5 \times 11^{2} \times 23 = 13915$$
(II) $$2783 = Y \times 253 \implies Y = \frac{2783}{253} = 11$$
OR
$$253 = 11 \times Z \implies Z = \frac{253}{11} = 23$$
(III) Here, $X = 13915$, $Y = 11$, $Z = 23$.
$$\operatorname{LCM}(13915, 11, 23) = 5 \times 11^{2} \times 23 = 13915$$
In a morning walk three students step off together, their steps measure 80 cm, 85 cm and 90 cm respectively.
(I) What is the minimum distance each should walk so that they can cover the distance in complete steps? OR What is the product of HCF and LCM of 80, 85 and 90?
(II) What is the HCF of 80 and 90?
(III) Find the sum of exponents of the prime factors of total distance.
(I) What is the minimum distance each should walk so that they can cover the distance in complete steps? OR What is the product of HCF and LCM of 80, 85 and 90?
(II) What is the HCF of 80 and 90?
(III) Find the sum of exponents of the prime factors of total distance.
Answer
(I) We find the LCM of 80, 85 and 90:
$$80 = 2^{4} \times 5, \quad 85 = 5 \times 17, \quad 90 = 2 \times 3^{2} \times 5$$
$$\operatorname{LCM}(80,85,90) = 2^{4} \times 3^{2} \times 5 \times 17 = 12240 \text{ cm}$$
Hence, the minimum distance each should walk is 12240 cm or 122 m 40 cm.
OR
$\operatorname{LCM}(80,85,90) = 12240$ and $\operatorname{HCF}(80,85,90) = 5$.
Product of HCF and LCM $= 12240 \times 5 = 61200$.
(II) Prime factors: $80 = 2^{4} \times 5$ and $90 = 2 \times 3^{2} \times 5$.
$$\operatorname{HCF}(80,90) = 2 \times 5 = 10$$
(III) Total distance $= (80+85+90) \text{ cm} = 255 \text{ cm}$.
Prime factors of $255 = 3 \times 5 \times 17$.
Sum of exponents $= 1+1+1 = 3$.
$$80 = 2^{4} \times 5, \quad 85 = 5 \times 17, \quad 90 = 2 \times 3^{2} \times 5$$
$$\operatorname{LCM}(80,85,90) = 2^{4} \times 3^{2} \times 5 \times 17 = 12240 \text{ cm}$$
Hence, the minimum distance each should walk is 12240 cm or 122 m 40 cm.
OR
$\operatorname{LCM}(80,85,90) = 12240$ and $\operatorname{HCF}(80,85,90) = 5$.
Product of HCF and LCM $= 12240 \times 5 = 61200$.
(II) Prime factors: $80 = 2^{4} \times 5$ and $90 = 2 \times 3^{2} \times 5$.
$$\operatorname{HCF}(80,90) = 2 \times 5 = 10$$
(III) Total distance $= (80+85+90) \text{ cm} = 255 \text{ cm}$.
Prime factors of $255 = 3 \times 5 \times 17$.
Sum of exponents $= 1+1+1 = 3$.
Ashish supplies bread and jams to a hospital and a school. Bread and jam are supplied in equal number of pieces. Bread comes in a bunch of $8$ pieces and jam comes in a pack of $6$ pieces. On a particular day, Ashish has supplied $x$ packets of bread and $y$ packets of jam to the school. On the same day, Ashish has supplied $3x$ packets of bread along with sufficient packets of jam to the hospital. The number of students in the school are between $500$ and $550$.
(I) How many students are there in the school?
(A) $508$ (B) $504$ (C) $512$ (D) $548$
(II) How many packets of bread are supplied in the school?
(A) $63$ packets (B) $86$ packets (C) $65$ packets (D) $84$ packets
(III) How many packets of jam are supplied in the school?
(A) $63$ packets (B) $86$ packets (C) $65$ packets (D) $84$ packets
(IV) How many packets of bread are supplied in the hospital?
(A) $189$ packets (B) $252$ packets (C) $165$ packets (D) $288$ packets
(V) How many packets of jam are supplied in the hospital?
(A) $248$ packets (B) $252$ packets (C) $165$ packets (D) $288$ packets
(I) How many students are there in the school?
(A) $508$ (B) $504$ (C) $512$ (D) $548$
(II) How many packets of bread are supplied in the school?
(A) $63$ packets (B) $86$ packets (C) $65$ packets (D) $84$ packets
(III) How many packets of jam are supplied in the school?
(A) $63$ packets (B) $86$ packets (C) $65$ packets (D) $84$ packets
(IV) How many packets of bread are supplied in the hospital?
(A) $189$ packets (B) $252$ packets (C) $165$ packets (D) $288$ packets
(V) How many packets of jam are supplied in the hospital?
(A) $248$ packets (B) $252$ packets (C) $165$ packets (D) $288$ packets
Answer
(I) Correct option: (B) $504$
$\operatorname{LCM}(8,6) = 24$. A multiple of $24$ between $500$ and $550$ is $504$. Thus there are $504$ students.
(II) Correct option: (A) $63$ packets
$\frac{504}{8} = 63$ packets of bread.
(III) Correct option: (D) $84$ packets
$\frac{504}{6} = 84$ packets of jam.
(IV) Correct option: (A) $189$ packets
$3x = 3 \times 63 = 189$ packets of bread to the hospital.
(V) Correct option: (B) $252$ packets
$189 \times 8 = 1512$ bread pieces. Jam packets $= \frac{1512}{6} = 252$.
$\operatorname{LCM}(8,6) = 24$. A multiple of $24$ between $500$ and $550$ is $504$. Thus there are $504$ students.
(II) Correct option: (A) $63$ packets
$\frac{504}{8} = 63$ packets of bread.
(III) Correct option: (D) $84$ packets
$\frac{504}{6} = 84$ packets of jam.
(IV) Correct option: (A) $189$ packets
$3x = 3 \times 63 = 189$ packets of bread to the hospital.
(V) Correct option: (B) $252$ packets
$189 \times 8 = 1512$ bread pieces. Jam packets $= \frac{1512}{6} = 252$.
Directions: A statement of Assertion (A) is followed by a statement of Reason (R).
A number $q$ is prime factorised as $3^{2} \times 7^{2} \times b$, where $b$ is a prime number other than 3 and 7.
Assertion (A): $q$ is definitely an odd number.
Reason (R): $3^{2} \times 7^{2}$ is an odd number.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
A number $q$ is prime factorised as $3^{2} \times 7^{2} \times b$, where $b$ is a prime number other than 3 and 7.
Assertion (A): $q$ is definitely an odd number.
Reason (R): $3^{2} \times 7^{2}$ is an odd number.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (D) is correct.
Consider the Assertion: Let $b=2$ (a prime number). Then $q = 3^{2} \times 7^{2} \times 2 = 9 \times 49 \times 2 = 882$, which is even. Hence, Assertion is false.
Consider the Reason: $3^{2} \times 7^{2} = 9 \times 49 = 441$, which is odd. Hence, Reason is true.
Thus, Assertion is false but Reason is true.
Consider the Assertion: Let $b=2$ (a prime number). Then $q = 3^{2} \times 7^{2} \times 2 = 9 \times 49 \times 2 = 882$, which is even. Hence, Assertion is false.
Consider the Reason: $3^{2} \times 7^{2} = 9 \times 49 = 441$, which is odd. Hence, Reason is true.
Thus, Assertion is false but Reason is true.
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.
Statement R (Reason): HCF is always a factor of LCM.
Answer
Option (B) is correct.
Assertion: $\operatorname{HCF} \times \operatorname{LCM} = 17 \times 340 = 5780$, which equals the product of the two numbers. So Assertion is true.
Reason: HCF is always a factor of LCM — this is true.
But the Reason is not the correct explanation of the Assertion.
Assertion: $\operatorname{HCF} \times \operatorname{LCM} = 17 \times 340 = 5780$, which equals the product of the two numbers. So Assertion is true.
Reason: HCF is always a factor of LCM — this is true.
But the Reason is not the correct explanation of the Assertion.
Statement A (Assertion): The HCF of two numbers is $18$ and their product is $3072$. Then their LCM is $169$.
Statement R (Reason): If $a$ and $b$ are two positive integers, then $\text{HCF} \times \text{LCM} = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Statement R (Reason): If $a$ and $b$ are two positive integers, then $\text{HCF} \times \text{LCM} = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (D) is correct.
Using the relation: $\text{HCF} \times \text{LCM} = \text{Product of the numbers}$
$$18 \times \text{LCM} = 3072 \implies \text{LCM} = \frac{3072}{18} = 170.67$$
Since LCM must be a whole number, $\text{LCM} = 169$ is incorrect. Assertion is false.
The Reason ($\text{HCF} \times \text{LCM} = a \times b$) is a correct mathematical identity — Reason is true.
Using the relation: $\text{HCF} \times \text{LCM} = \text{Product of the numbers}$
$$18 \times \text{LCM} = 3072 \implies \text{LCM} = \frac{3072}{18} = 170.67$$
Since LCM must be a whole number, $\text{LCM} = 169$ is incorrect. Assertion is false.
The Reason ($\text{HCF} \times \text{LCM} = a \times b$) is a correct mathematical identity — Reason is true.
Statement A (Assertion): $12^n$ ends with the digit zero, where $n$ is any natural number.
Statement R (Reason): Any number ends with the digit zero if its prime factorisation is of the form $2^m \times 5^n$, where $m$ and $n$ are natural numbers.
(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 but 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.
Statement R (Reason): Any number ends with the digit zero if its prime factorisation is of the form $2^m \times 5^n$, where $m$ and $n$ are natural numbers.
(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 but 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.
Answer
Option (D) is correct.
Prime factorisation of $12 = 2^2 \times 3$, so $12^n = 2^{2n} \times 3^n$.
The factor $5$ is absent. Since a number must contain both $2$ and $5$ as factors to end with zero, $12^n$ does not always end with zero. Assertion is false.
The Reason is true: any number of the form $2^m \times 5^n$ will end with the digit zero.
Prime factorisation of $12 = 2^2 \times 3$, so $12^n = 2^{2n} \times 3^n$.
The factor $5$ is absent. Since a number must contain both $2$ and $5$ as factors to end with zero, $12^n$ does not always end with zero. Assertion is false.
The Reason is true: any number of the form $2^m \times 5^n$ will end with the digit zero.
Statement A (Assertion): Denominator of $12.145$ when expressed in the form $\dfrac{p}{q}$, $q \neq 0$, is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.
Statement R (Reason): $12.145$ is a terminating decimal fraction.
(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 but 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.
Statement R (Reason): $12.145$ is a terminating decimal fraction.
(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 but 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.
Answer
Option (A) is correct.
$$12.145 = \frac{12145}{1000}, \quad 1000 = 2^3 \times 5^3$$
The denominator is of the form $2^m \times 5^n$. Assertion is true.
A decimal fraction is terminating if and only if its denominator (in lowest form) has prime factors only $2$ and $5$. Therefore Reason is true and it correctly explains the Assertion.
$$12.145 = \frac{12145}{1000}, \quad 1000 = 2^3 \times 5^3$$
The denominator is of the form $2^m \times 5^n$. Assertion is true.
A decimal fraction is terminating if and only if its denominator (in lowest form) has prime factors only $2$ and $5$. Therefore Reason is true and it correctly explains the Assertion.
Let $a$ and $b$ be two positive integers such that $a=p^{3}q^{4}$ and $b=p^{2}q^{3}$, where $p$ and $q$ are prime numbers. If HCF$(a, b)=p^{m}q^{n}$ and LCM$(a, b)=p^{r}q^{s}$, then $(m+n)(r+s)=$
Solution
Option (C) is correct.
$\operatorname{HCF}(a,b) = p^{2}q^{3}$, so $m=2$ and $n=3$.
$\operatorname{LCM}(a,b) = p^{3}q^{4}$, so $r=3$ and $s=4$.
$$(m+n)(r+s) = (2+3)(3+4) = 5 \times 7 = 35$$
$\operatorname{HCF}(a,b) = p^{2}q^{3}$, so $m=2$ and $n=3$.
$\operatorname{LCM}(a,b) = p^{3}q^{4}$, so $r=3$ and $s=4$.
$$(m+n)(r+s) = (2+3)(3+4) = 5 \times 7 = 35$$
The product of a non-zero rational and an irrational number is:
Solution
Option (A) is correct.
Let $r$ be a non-zero rational number and $x$ be an irrational number. If $r \times x$ were rational, then $x = \frac{r \times x}{r}$ would be a ratio of two rational numbers, making $x$ rational — a contradiction. Hence, $r \times x$ must be irrational.
Let $r$ be a non-zero rational number and $x$ be an irrational number. If $r \times x$ were rational, then $x = \frac{r \times x}{r}$ would be a ratio of two rational numbers, making $x$ rational — a contradiction. Hence, $r \times x$ must be irrational.
In a formula racing competition, the time taken by two racing cars A and B to complete 1 round of the track is 30 minutes and $p$ minutes respectively. If the cars meet again at the starting point for the first time after 90 minutes and the HCF$(30, p)=15$, then the value of $p$ is
Solution
Option (A) is correct.
The time after which both cars meet again is the LCM of their times.
$$\operatorname{LCM}(30, p) = \frac{30 \times p}{\operatorname{HCF}(30, p)} = \frac{30p}{15} = 2p$$
Given that they meet after 90 minutes: $2p = 90 \implies p = 45$ minutes.
The time after which both cars meet again is the LCM of their times.
$$\operatorname{LCM}(30, p) = \frac{30 \times p}{\operatorname{HCF}(30, p)} = \frac{30p}{15} = 2p$$
Given that they meet after 90 minutes: $2p = 90 \implies p = 45$ minutes.
If $p^{2}=\dfrac{32}{50}$, then $p$ is a/an
Solution
Option (C) is correct.
$$p^{2}=\frac{32}{50} = \frac{16}{25}$$
$$p = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Since $p$ is in the form $\frac{P}{Q}$ where $Q \neq 0$, $p$ is a rational number.
$$p^{2}=\frac{32}{50} = \frac{16}{25}$$
$$p = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Since $p$ is in the form $\frac{P}{Q}$ where $Q \neq 0$, $p$ is a rational number.
Let $p$ and $q$ be two natural numbers such that $p > q$. When $p$ is divided by $q$, then remainder is $r$.
(i) $r$ CANNOT be $(p-q)$.
(ii) $r$ CAN either be $q$ or $(p-q)$.
(iii) $r$ is DEFINITELY less than $q$.
Which of the above statements is/are true?
(A) only (ii) (B) only (iii) (C) only (i) and (iii) (D) cannot be determined without knowing the values of $p$, $q$ and $r$
(i) $r$ CANNOT be $(p-q)$.
(ii) $r$ CAN either be $q$ or $(p-q)$.
(iii) $r$ is DEFINITELY less than $q$.
Which of the above statements is/are true?
(A) only (ii) (B) only (iii) (C) only (i) and (iii) (D) cannot be determined without knowing the values of $p$, $q$ and $r$
Solution
Option (B) is correct.
Explanation: $q$ is divisor and $r$ is remainder. By the division algorithm, the remainder $r$ is always less than the divisor $q$.
Explanation: $q$ is divisor and $r$ is remainder. By the division algorithm, the remainder $r$ is always less than the divisor $q$.
If two positive integers $a$ and $b$ are written as $a=x^{3}y^{2}$ and $b=xy^{3}$; $x$, $y$ are prime numbers, then HCF$(a, b)$ is:
Solution
Option (C) is correct.
$$a = x^{3}y^{2} = x \times x \times x \times y \times y$$
$$b = xy^{3} = x \times y \times y \times y$$
HCF takes the lowest power of each common prime factor:
$$\operatorname{HCF}(a,b) = x \times y^{2} = xy^{2}$$
(Note: The solution labels this as $p^{3}q^{2}$ in the context of the LCM calculation.)
$$a = x^{3}y^{2} = x \times x \times x \times y \times y$$
$$b = xy^{3} = x \times y \times y \times y$$
HCF takes the lowest power of each common prime factor:
$$\operatorname{HCF}(a,b) = x \times y^{2} = xy^{2}$$
(Note: The solution labels this as $p^{3}q^{2}$ in the context of the LCM calculation.)
The decimal form of $\dfrac{129}{2^2 \cdot 5^7 \cdot 7^5}$ is
Solution
Option (D) is correct.
The denominator is $2^2 \times 5^7 \times 7^5$. Since the prime factor $7$ is present (not just $2$ or $5$), the decimal is non-terminating. Since the number is rational, its decimal must be repeating. Therefore it is non-terminating but repeating — none of the other options states this correctly.
The denominator is $2^2 \times 5^7 \times 7^5$. Since the prime factor $7$ is present (not just $2$ or $5$), the decimal is non-terminating. Since the number is rational, its decimal must be repeating. Therefore it is non-terminating but repeating — none of the other options states this correctly.
HCF of $8$, $9$, $25$ is
Solution
Option (D) is correct.
$$8 = 2^3, \quad 9 = 3^2, \quad 25 = 5^2$$
There is no common prime factor among $8$, $9$, and $25$. Hence, $\operatorname{HCF}(8,9,25) = 1$.
$$8 = 2^3, \quad 9 = 3^2, \quad 25 = 5^2$$
There is no common prime factor among $8$, $9$, and $25$. Hence, $\operatorname{HCF}(8,9,25) = 1$.
Which of the following is not irrational?
Solution
Option (C) is correct.
(A) $(2-\sqrt{3})^2 = 4-4\sqrt{3}+3 = 7-4\sqrt{3}$ — irrational
(B) $(\sqrt{2}+\sqrt{3})^2 = 2+3+2\sqrt{6} = 5+2\sqrt{6}$ — irrational
(C) $(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3}) = 2-3 = -1$ — rational, so not irrational ✓
(D) $\dfrac{2\sqrt{7}}{7}$ — irrational
(A) $(2-\sqrt{3})^2 = 4-4\sqrt{3}+3 = 7-4\sqrt{3}$ — irrational
(B) $(\sqrt{2}+\sqrt{3})^2 = 2+3+2\sqrt{6} = 5+2\sqrt{6}$ — irrational
(C) $(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3}) = 2-3 = -1$ — rational, so not irrational ✓
(D) $\dfrac{2\sqrt{7}}{7}$ — irrational
The product of a rational and an irrational number is
Solution
Option (B) is correct.
Let $r$ be a non-zero rational number and $x$ be an irrational number. Assume $r \times x$ is rational. Then $x = \frac{r \times x}{r}$ would be rational — a contradiction. Hence, the product is always irrational.
Let $r$ be a non-zero rational number and $x$ be an irrational number. Assume $r \times x$ is rational. Then $x = \frac{r \times x}{r}$ would be rational — a contradiction. Hence, the product is always irrational.
Amar, Akbar, and Anthony are playing a game. Amar climbs $5$ stairs and gets down $2$ stairs in one turn. Akbar goes up $7$ stairs and comes down $2$ stairs every time. Anthony goes up $10$ stairs and $3$ stairs down each time. They have to reach the nearest point of the $100^{\text{th}}$ stair and stop once it’s impossible to go forward.
(I) Who reaches the nearest point?
(A) Amar (B) Akbar (C) Anthony (D) All together reach the nearest point
(II) How many times can they meet on the same step?
(A) $3$ (B) $4$ (C) $5$ (D) No, they cannot meet in between on the same step
(III) Who takes the least number of steps to reach near $100$?
(A) Amar (B) Akbar (C) Anthony (D) All of them take equal number of steps
(IV) What is the first stair where any two out of three will meet together?
(A) Amar and Akbar will meet for the first time after $15$ steps
(B) Akbar and Anthony will meet for the first time after $35$ steps
(C) Amar and Anthony will meet for the first time after $21$ steps
(D) Amar and Akbar will meet for the first time after $21$ steps
(V) What is the second stair where any two out of three will meet together?
(A) Amar and Akbar will meet after $21$ steps
(B) Akbar and Anthony will meet after $35$ steps
(C) Amar and Anthony will meet after $21$ steps
(D) Amar and Anthony will meet after $35$ steps
(I) Who reaches the nearest point?
(A) Amar (B) Akbar (C) Anthony (D) All together reach the nearest point
(II) How many times can they meet on the same step?
(A) $3$ (B) $4$ (C) $5$ (D) No, they cannot meet in between on the same step
(III) Who takes the least number of steps to reach near $100$?
(A) Amar (B) Akbar (C) Anthony (D) All of them take equal number of steps
(IV) What is the first stair where any two out of three will meet together?
(A) Amar and Akbar will meet for the first time after $15$ steps
(B) Akbar and Anthony will meet for the first time after $35$ steps
(C) Amar and Anthony will meet for the first time after $21$ steps
(D) Amar and Akbar will meet for the first time after $21$ steps
(V) What is the second stair where any two out of three will meet together?
(A) Amar and Akbar will meet after $21$ steps
(B) Akbar and Anthony will meet after $35$ steps
(C) Amar and Anthony will meet after $21$ steps
(D) Amar and Anthony will meet after $35$ steps
Answer
(I) Option (A) is correct.
Net movement: Amar = 3 steps/turn → $3 \times 32 = 96$ steps. Akbar = 5 steps/turn → $5 \times 19 = 95$ steps. Anthony = 7 steps/turn → $7 \times 13 = 91$ steps. Amar reaches nearest to 100.
(II) Option (D) is correct.
$\operatorname{LCM}(3,5,7) = 105 > 100$. Since the staircase only has 100 stairs, they cannot meet on the same step.
(III) Option (C) is correct.
Amar takes 32 steps, Akbar takes 19 steps, Anthony takes 13 steps. Anthony takes the least number of steps.
(IV) Option (A) is correct.
$\operatorname{LCM}(3,5) = 15$, $\operatorname{LCM}(5,7) = 35$, $\operatorname{LCM}(3,7) = 21$. The smallest is 15, so Amar and Akbar meet first.
(V) Option (C) is correct.
$\operatorname{LCM}(3,7) = 21$. Amar and Anthony meet at the second meeting point (21 steps).
Net movement: Amar = 3 steps/turn → $3 \times 32 = 96$ steps. Akbar = 5 steps/turn → $5 \times 19 = 95$ steps. Anthony = 7 steps/turn → $7 \times 13 = 91$ steps. Amar reaches nearest to 100.
(II) Option (D) is correct.
$\operatorname{LCM}(3,5,7) = 105 > 100$. Since the staircase only has 100 stairs, they cannot meet on the same step.
(III) Option (C) is correct.
Amar takes 32 steps, Akbar takes 19 steps, Anthony takes 13 steps. Anthony takes the least number of steps.
(IV) Option (A) is correct.
$\operatorname{LCM}(3,5) = 15$, $\operatorname{LCM}(5,7) = 35$, $\operatorname{LCM}(3,7) = 21$. The smallest is 15, so Amar and Akbar meet first.
(V) Option (C) is correct.
$\operatorname{LCM}(3,7) = 21$. Amar and Anthony meet at the second meeting point (21 steps).
In a classroom activity on real numbers, students pick a number card and frame a question if it is not a rational number.
(i) Suraj picked up $\sqrt{8}$. Which of the following is true about $\sqrt{8}$?
(A) It is a natural number (B) It is a rational number (C) It is an irrational number (D) None of these
(ii) Shreya picked up ‘BONUS’. Which of the following is not irrational?
(A) $3 – 4\sqrt{5}$ (B) $\sqrt{7} – 6$ (C) $2 + 2\sqrt{9}$ (D) $4\sqrt{11} – 6$
(iii) Ananya picked up $\sqrt{15} – \sqrt{10}$. It is a ______ number.
(A) natural number (B) rational number (C) irrational number (D) whole number
(iv) Suman picked up $\dfrac{1}{\sqrt{5}}$. It is a ______ number.
(A) natural number (B) rational number (C) irrational number (D) whole number
(i) Suraj picked up $\sqrt{8}$. Which of the following is true about $\sqrt{8}$?
(A) It is a natural number (B) It is a rational number (C) It is an irrational number (D) None of these
(ii) Shreya picked up ‘BONUS’. Which of the following is not irrational?
(A) $3 – 4\sqrt{5}$ (B) $\sqrt{7} – 6$ (C) $2 + 2\sqrt{9}$ (D) $4\sqrt{11} – 6$
(iii) Ananya picked up $\sqrt{15} – \sqrt{10}$. It is a ______ number.
(A) natural number (B) rational number (C) irrational number (D) whole number
(iv) Suman picked up $\dfrac{1}{\sqrt{5}}$. It is a ______ number.
(A) natural number (B) rational number (C) irrational number (D) whole number
Answer
(i) Option (C) is correct.
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. Since $\sqrt{2}$ is irrational, $2\sqrt{2}$ is also irrational.
(ii) Option (C) is correct.
$\sqrt{9} = 3$, so $2 + 2\sqrt{9} = 2 + 6 = 8$, which is rational. Hence not irrational.
(iii) Option (C) is correct.
Both $\sqrt{15}$ and $\sqrt{10}$ are irrational numbers, and their difference is also irrational.
(iv) Option (C) is correct.
$\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$. Since $\sqrt{5}$ is irrational, $\dfrac{\sqrt{5}}{5}$ is also irrational.
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. Since $\sqrt{2}$ is irrational, $2\sqrt{2}$ is also irrational.
(ii) Option (C) is correct.
$\sqrt{9} = 3$, so $2 + 2\sqrt{9} = 2 + 6 = 8$, which is rational. Hence not irrational.
(iii) Option (C) is correct.
Both $\sqrt{15}$ and $\sqrt{10}$ are irrational numbers, and their difference is also irrational.
(iv) Option (C) is correct.
$\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$. Since $\sqrt{5}$ is irrational, $\dfrac{\sqrt{5}}{5}$ is also irrational.
HCF and LCM are widely used in the number system. The product of two positive integers is equal to the product of their HCF and LCM.
(i) If two positive integers $x$ and $y$ are expressible in terms of primes as $x = p^2q^3$ and $y = p^3q$, then which of the following is true?
(A) $\mathrm{HCF} = pq^2 \times \mathrm{LCM}$ (B) $\mathrm{LCM} = pq^2 \times \mathrm{HCF}$ (C) $\mathrm{LCM} = p^2q \times \mathrm{HCF}$ (D) $\mathrm{HCF} = p^2q \times \mathrm{LCM}$
(ii) A boy with a collection of marbles realizes that if he makes a group of $5$ or $6$ marbles, there are always $2$ marbles left. Which of the following is correct if the number of marbles is $p$?
(A) $p$ is odd (B) $p$ is even (C) $p$ is not prime (D) both (b) and (c)
(iii) Find the largest possible positive integer that will divide $398$, $436$, and $542$ leaving remainders $7$, $11$, and $15$ respectively.
(A) $3$ (B) $1$ (C) $34$ (D) $17$
(i) If two positive integers $x$ and $y$ are expressible in terms of primes as $x = p^2q^3$ and $y = p^3q$, then which of the following is true?
(A) $\mathrm{HCF} = pq^2 \times \mathrm{LCM}$ (B) $\mathrm{LCM} = pq^2 \times \mathrm{HCF}$ (C) $\mathrm{LCM} = p^2q \times \mathrm{HCF}$ (D) $\mathrm{HCF} = p^2q \times \mathrm{LCM}$
(ii) A boy with a collection of marbles realizes that if he makes a group of $5$ or $6$ marbles, there are always $2$ marbles left. Which of the following is correct if the number of marbles is $p$?
(A) $p$ is odd (B) $p$ is even (C) $p$ is not prime (D) both (b) and (c)
(iii) Find the largest possible positive integer that will divide $398$, $436$, and $542$ leaving remainders $7$, $11$, and $15$ respectively.
(A) $3$ (B) $1$ (C) $34$ (D) $17$
Answer
(i) Option (B) is correct.
$\mathrm{HCF} = p^2q$ (lowest powers of common primes).
$\mathrm{LCM} = p^3q^3$ (highest powers of all primes).
Therefore $\mathrm{LCM} = pq^2 \times \mathrm{HCF}$.
(ii) Option (D) is correct.
$p – 2$ is divisible by $\mathrm{LCM}(5,6) = 30$, so $p = 32$. The number $32$ is even and not prime.
(iii) Option (D) is correct.
Subtract remainders: $398-7=391$, $436-11=425$, $542-15=527$.
$391 = 17 \times 23$, $425 = 17 \times 25$, $527 = 17 \times 31$.
$\mathrm{HCF}(391,425,527) = 17$.
$\mathrm{HCF} = p^2q$ (lowest powers of common primes).
$\mathrm{LCM} = p^3q^3$ (highest powers of all primes).
Therefore $\mathrm{LCM} = pq^2 \times \mathrm{HCF}$.
(ii) Option (D) is correct.
$p – 2$ is divisible by $\mathrm{LCM}(5,6) = 30$, so $p = 32$. The number $32$ is even and not prime.
(iii) Option (D) is correct.
Subtract remainders: $398-7=391$, $436-11=425$, $542-15=527$.
$391 = 17 \times 23$, $425 = 17 \times 25$, $527 = 17 \times 31$.
$\mathrm{HCF}(391,425,527) = 17$.
In a mathematics competition, a group of students chose the topic of Fibonacci series. The sequence they got was $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \ldots$
(i) The next two numbers in the sequence are:
(A) $223$ and $367$ (B) $233$ and $367$ (C) $223$ and $377$ (D) $233$ and $377$
(ii) A rational number has terminating decimal expansion if the denominator is of the form (where $m, n$ are non-negative integers):
(A) $2^n \times 5^n$ (B) $3^n \times 5^n$ (C) $2^n \times 3^n$ (D) $3^n \times 4^n$
(iii) Which of the following rational numbers (ratio of a term to its previous term) has terminating decimal expansion?
(A) $\dfrac{5}{3}$ (B) $\dfrac{21}{13}$ (C) $\dfrac{13}{8}$ (D) $\dfrac{34}{21}$
(iv) HCF of two consecutive numbers in the given sequence (except the first three terms) is:
(A) $1$ (B) $0$ (C) $2$ (D) $3$
(i) The next two numbers in the sequence are:
(A) $223$ and $367$ (B) $233$ and $367$ (C) $223$ and $377$ (D) $233$ and $377$
(ii) A rational number has terminating decimal expansion if the denominator is of the form (where $m, n$ are non-negative integers):
(A) $2^n \times 5^n$ (B) $3^n \times 5^n$ (C) $2^n \times 3^n$ (D) $3^n \times 4^n$
(iii) Which of the following rational numbers (ratio of a term to its previous term) has terminating decimal expansion?
(A) $\dfrac{5}{3}$ (B) $\dfrac{21}{13}$ (C) $\dfrac{13}{8}$ (D) $\dfrac{34}{21}$
(iv) HCF of two consecutive numbers in the given sequence (except the first three terms) is:
(A) $1$ (B) $0$ (C) $2$ (D) $3$
Answer
(i) Option (D) is correct.
$89 + 144 = 233$ and $144 + 233 = 377$.
(ii) Option (A) is correct.
A rational number $\frac{p}{q}$ in lowest form has a terminating decimal if and only if $q$ has only $2$ and/or $5$ as prime factors, i.e., $q = 2^m \times 5^n$.
(iii) Option (C) is correct.
$\frac{13}{8}$: denominator $8 = 2^3$ — only factor of $2$ ✓. The others have denominators with prime factors other than $2$ and $5$.
(iv) Option (A) is correct.
Any two consecutive Fibonacci numbers are coprime. Hence $\mathrm{HCF} = 1$.
$89 + 144 = 233$ and $144 + 233 = 377$.
(ii) Option (A) is correct.
A rational number $\frac{p}{q}$ in lowest form has a terminating decimal if and only if $q$ has only $2$ and/or $5$ as prime factors, i.e., $q = 2^m \times 5^n$.
(iii) Option (C) is correct.
$\frac{13}{8}$: denominator $8 = 2^3$ — only factor of $2$ ✓. The others have denominators with prime factors other than $2$ and $5$.
(iv) Option (A) is correct.
Any two consecutive Fibonacci numbers are coprime. Hence $\mathrm{HCF} = 1$.
Real numbers are extremely useful in everyday life. After knowing the importance of real numbers, answer the following questions based on real-life situations.
(i) Three people go for a morning walk together. Their step lengths are $80$ cm, $85$ cm, and $90$ cm respectively. What is the minimum distance travelled when they meet for the first time?
(A) $6120$ cm (B) $12240$ cm (C) $4080$ cm (D) None of these
(ii) In a school Independence Day parade, a group of $594$ students needs to march behind a band of $189$ members. What is the maximum number of columns?
(A) $9$ (B) $6$ (C) $27$ (D) $29$
(iii) Two tankers contain $768$ litres and $420$ litres of fuel. Find the maximum capacity of a container which can measure either tanker exactly.
(A) $1$ litre (B) $7$ litres (C) $12$ litres (D) $18$ litres
(iv) The dimensions of a room are $825$ cm, $675$ cm, and $450$ cm. Find the length of the largest measuring rod which can measure all dimensions exactly.
(A) $1$ m $25$ cm (B) $75$ cm (C) $90$ cm (D) $1$ m $35$ cm
(v) Pens are sold in packs of $8$ and notepads are sold in packs of $12$. Find the least number of packs of each type to buy so that the total number of pens and notepads is equal.
(A) $3$ and $2$ (B) $2$ and $5$ (C) $3$ and $4$ (D) $4$ and $5$
(i) Three people go for a morning walk together. Their step lengths are $80$ cm, $85$ cm, and $90$ cm respectively. What is the minimum distance travelled when they meet for the first time?
(A) $6120$ cm (B) $12240$ cm (C) $4080$ cm (D) None of these
(ii) In a school Independence Day parade, a group of $594$ students needs to march behind a band of $189$ members. What is the maximum number of columns?
(A) $9$ (B) $6$ (C) $27$ (D) $29$
(iii) Two tankers contain $768$ litres and $420$ litres of fuel. Find the maximum capacity of a container which can measure either tanker exactly.
(A) $1$ litre (B) $7$ litres (C) $12$ litres (D) $18$ litres
(iv) The dimensions of a room are $825$ cm, $675$ cm, and $450$ cm. Find the length of the largest measuring rod which can measure all dimensions exactly.
(A) $1$ m $25$ cm (B) $75$ cm (C) $90$ cm (D) $1$ m $35$ cm
(v) Pens are sold in packs of $8$ and notepads are sold in packs of $12$. Find the least number of packs of each type to buy so that the total number of pens and notepads is equal.
(A) $3$ and $2$ (B) $2$ and $5$ (C) $3$ and $4$ (D) $4$ and $5$
Answer
(i) Option (B) is correct.
$80 = 2^4 \times 5$, $85 = 5 \times 17$, $90 = 2 \times 3^2 \times 5$.
$\mathrm{LCM} = 2^4 \times 3^2 \times 5 \times 17 = 12240$ cm.
(ii) Option (C) is correct.
$594 = 2 \times 3^3 \times 11$, $189 = 3^3 \times 7$. $\mathrm{HCF}(594,189) = 3^3 = 27$.
(iii) Option (C) is correct.
$768 = 2^8 \times 3$, $420 = 2^2 \times 3 \times 5 \times 7$. $\mathrm{HCF} = 2^2 \times 3 = 12$ litres.
(iv) Option (B) is correct.
$825 = 3 \times 5^2 \times 11$, $675 = 3^3 \times 5^2$, $450 = 2 \times 3^2 \times 5^2$. $\mathrm{HCF} = 3 \times 5^2 = 75$ cm.
(v) Option (A) is correct.
$\mathrm{LCM}(8,12) = 24$. Pen packs $= \frac{24}{8} = 3$, Notepad packs $= \frac{24}{12} = 2$.
$80 = 2^4 \times 5$, $85 = 5 \times 17$, $90 = 2 \times 3^2 \times 5$.
$\mathrm{LCM} = 2^4 \times 3^2 \times 5 \times 17 = 12240$ cm.
(ii) Option (C) is correct.
$594 = 2 \times 3^3 \times 11$, $189 = 3^3 \times 7$. $\mathrm{HCF}(594,189) = 3^3 = 27$.
(iii) Option (C) is correct.
$768 = 2^8 \times 3$, $420 = 2^2 \times 3 \times 5 \times 7$. $\mathrm{HCF} = 2^2 \times 3 = 12$ litres.
(iv) Option (B) is correct.
$825 = 3 \times 5^2 \times 11$, $675 = 3^3 \times 5^2$, $450 = 2 \times 3^2 \times 5^2$. $\mathrm{HCF} = 3 \times 5^2 = 75$ cm.
(v) Option (A) is correct.
$\mathrm{LCM}(8,12) = 24$. Pen packs $= \frac{24}{8} = 3$, Notepad packs $= \frac{24}{12} = 2$.
Assertion (A): $\sqrt{x}$ is an irrational number, where $x$ is a prime number.
Reason (R): Square root of any prime number is an irrational number.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Reason (R): Square root of any prime number is an irrational number.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (A) is correct.
A prime number has exactly two factors: $1$ and itself. It cannot be a perfect square of any rational number. Therefore, $\sqrt{x}$ for any prime $x$ is irrational.
Both the Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
A prime number has exactly two factors: $1$ and itself. It cannot be a perfect square of any rational number. Therefore, $\sqrt{x}$ for any prime $x$ is irrational.
Both the Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Assertion (A): The HCF of two numbers is $18$ and their product is $3072$. Then their LCM is $169$.
Reason (R): If $a$ and $b$ are two positive integers, then $\text{HCF} \times \text{LCM} = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Reason (R): If $a$ and $b$ are two positive integers, then $\text{HCF} \times \text{LCM} = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (D) is correct.
$$18 \times \text{LCM} = 3072 \implies \text{LCM} = \frac{3072}{18} = 170.67$$
Since LCM must be a whole number, $\text{LCM} = 169$ is incorrect. Assertion is false.
The formula $\text{HCF} \times \text{LCM} = a \times b$ is always valid for positive integers. Reason is true.
$$18 \times \text{LCM} = 3072 \implies \text{LCM} = \frac{3072}{18} = 170.67$$
Since LCM must be a whole number, $\text{LCM} = 169$ is incorrect. Assertion is false.
The formula $\text{HCF} \times \text{LCM} = a \times b$ is always valid for positive integers. Reason is true.
Assertion (A): $12^n$ ends with the digit zero, where $n$ is a natural number.
Reason (R): Any number ends with the digit zero if its prime factorisation is of the form $2^m \times 5^n$, where $m$ and $n$ are natural numbers.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Reason (R): Any number ends with the digit zero if its prime factorisation is of the form $2^m \times 5^n$, where $m$ and $n$ are natural numbers.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (D) is correct.
$$12 = 2^2 \times 3 \implies 12^n = 2^{2n} \times 3^n$$
Factor $5$ is absent. Since a number must contain both $2$ and $5$ to end in zero, $12^n$ does not end with zero. Assertion is false.
Reason is true: numbers of the form $2^m \times 5^n$ end with zero.
$$12 = 2^2 \times 3 \implies 12^n = 2^{2n} \times 3^n$$
Factor $5$ is absent. Since a number must contain both $2$ and $5$ to end in zero, $12^n$ does not end with zero. Assertion is false.
Reason is true: numbers of the form $2^m \times 5^n$ end with zero.
Assertion (A): The HCF of two numbers is $5$ and their product is $150$. Then their LCM is $30$.
Reason (R): For any two positive integers $a$ and $b$, $\mathrm{HCF}(a,b) + \mathrm{LCM}(a,b) = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Reason (R): For any two positive integers $a$ and $b$, $\mathrm{HCF}(a,b) + \mathrm{LCM}(a,b) = a \times b$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (C) is correct.
$$5 \times \mathrm{LCM} = 150 \implies \mathrm{LCM} = 30$$
Assertion is true.
The Reason states $\mathrm{HCF} + \mathrm{LCM} = a \times b$, which is incorrect. The correct formula is $\mathrm{HCF} \times \mathrm{LCM} = a \times b$. Reason is false.
$$5 \times \mathrm{LCM} = 150 \implies \mathrm{LCM} = 30$$
Assertion is true.
The Reason states $\mathrm{HCF} + \mathrm{LCM} = a \times b$, which is incorrect. The correct formula is $\mathrm{HCF} \times \mathrm{LCM} = a \times b$. Reason is false.
Assertion (A): If $\mathrm{HCF}(336, 54) = 6$, then $\mathrm{LCM}(336, 54) = 3000$.
Reason (R): The sum of exponents of prime factors in the prime factorisation of $196$ is $4$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Reason (R): The sum of exponents of prime factors in the prime factorisation of $196$ is $4$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Answer
Option (D) is correct.
$$6 \times \mathrm{LCM} = 336 \times 54 = 18144 \implies \mathrm{LCM} = \frac{18144}{6} = 3024$$
The assertion states $\mathrm{LCM} = 3000$, which is incorrect. Assertion is false.
$196 = 2^2 \times 7^2$. Sum of exponents $= 2+2 = 4$. Reason is true.
$$6 \times \mathrm{LCM} = 336 \times 54 = 18144 \implies \mathrm{LCM} = \frac{18144}{6} = 3024$$
The assertion states $\mathrm{LCM} = 3000$, which is incorrect. Assertion is false.
$196 = 2^2 \times 7^2$. Sum of exponents $= 2+2 = 4$. Reason is true.
P and Q are two positive integers such that $P = p^{3}q$ and $Q = (pq)^{2}$, where $p$ and $q$ are prime numbers. What is LCM$(P, Q)$?
Solution
Option (C) is correct.
$$P = p^{3}q = p \times p \times p \times q$$
$$Q = (pq)^{2} = p \times p \times q \times q$$
$$\operatorname{LCM}(P,Q) = p \times p \times p \times q \times q = p^{3}q^{2}$$
$$P = p^{3}q = p \times p \times p \times q$$
$$Q = (pq)^{2} = p \times p \times q \times q$$
$$\operatorname{LCM}(P,Q) = p \times p \times p \times q \times q = p^{3}q^{2}$$
Explore More CBSE Class 10 Maths Chapters
Frequently Asked Questions
What is the Real Numbers chapter about in CBSE Class 10 Maths? ⌄
The Real Numbers chapter covers the fundamental properties of integers, rational and irrational numbers, and prime factorisation. Students learn key theorems like Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic, and explore concepts like HCF, LCM, and terminating vs non-terminating decimals — all essential for the Class 10 board exam.
How many marks does the Real Numbers chapter carry in CBSE Class 10 board exams? ⌄
Real Numbers is part of the Number Systems unit, which carries approximately 6 marks in the CBSE Class 10 Maths board exam. While the weightage may seem modest, the concepts form the backbone of many other chapters, so a thorough understanding is crucial for overall performance.
What are the most important topics in Real Numbers for Class 10? ⌄
The most exam-important topics are: prime factorisation (used to find HCF and LCM), the relationship HCF × LCM = Product of two numbers, identifying terminating and non-terminating decimals, and proving numbers irrational. Competency Based Questions increasingly test these through real-life scenarios, so understanding the logic — not just the steps — is key.
What are common mistakes students make in the Real Numbers chapter? ⌄
The most frequent mistakes include confusing HCF with LCM in word problems, incorrectly applying the formula HCF × LCM = product of two numbers (it does not extend directly to three numbers), and forgetting that a rational number has a terminating decimal only when the denominator’s prime factors are exclusively 2 and/or 5. Assertion-Reason questions also trip up students who don’t verify both the statement and the explanation separately.
How does Angle Belearn help students master Real Numbers? ⌄
Angle Belearn’s CBSE specialists curate competency based questions that mirror the latest board exam pattern, including case studies and assertion-reason formats. Every question comes with a detailed step-by-step solution so your child understands the reasoning behind each answer, not just the final result. This approach builds confidence and ensures they can handle any variation the board exam presents.
