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Let the number of columns be x.

x is the largest number, which should divide both 104 and 96 

104 = 96 x 1 + 8 

96 = 8 x 12 + 0 

.'.     HCF of 104 and 96 is 8

Hence, 8 columns are required.

Total number of pens = 1001 

Total number pencils = 910 

∴ Maximum number of students who get the same number of pens and pencils = HCF (1001, 910) 

Prime factorization: 

1001 = 11 × 91 

910 = 10 × 91 

∴ HCF = 91 

Hence, 91 students receive same number of pens and pencils.

Given:

Total number of pens = 1001

Total number pencils = 910

Maximum number of students who get the same number of pens and the same number of pencils = HCF (1001 and 910)

Using prime factorization:

910 = 2 x 5 x 7 x 13

1001 = 7x 11 x 13

Now, HCF(1001 and 910) = 91

Answer: 91 students get same number of pens and pencils.

3 1/3 = 10/3

0.3 = 3/10

a * 1/a =1

then 1/a is multiplicative inverse of a

0.3* 3 1/3 = 3/10 * 10/3 =1

therefore 0.3 is multiplicative inverse of 3 1/3

Let’s first choose 184 and 230 to find the HCF by using Euclid’s division lemma. 

Thus, we obtain 

230 = 184 x 1 + 46 

Since the remainder 46 ≠ 0. So we apply the division lemma to the divisor 184 and remainder 46. We get, 

184 = 46 x 4 + 0 

The remainder at this stage is 0, the divisor will be the HCF i.e., 46 for 184 and 230. 

Now, we again use Euclid’s division lemma to find the HCF of 46 and 276. And we get, 

276 = 46 x 6 + 0 

So, this stage has remainder 0. Thus, the HCF of the third number 276 and 46 is 46. 

Hence, the HCF of 184, 230 and 276 is 46.

Irrational numbers can’t be expressed in p/q form where p and q are integers and q ≠ 0.

E.g. √2, √3; √5, √7 etc.

Where as a rational can be expressed in p/q form

E.g. :- -3 = -3/1 and 5/4 etc.

3/4, 5/9, 2/7

Let ‘a’ and ‘b’ are positive integers, 

a = b × q + r 

Let b = 3. 

∴ a = 3q + r 

(i) If r = 0 then a = 3q 

(ii) If r = l then a = 3q + 1 

(iii) If r = 2 then a = 3q + 2 

If we consider cubes of these, 

(i) If a = 3q, then 

a³ = (3)³ = 27q³ = 9(3q)³ = 9m (∵ m = 3q²) 

(ii) If a = 3q + 1, then (a)³ = (3q + 1)³ 

= 27q³ + 1 + 27q³ + 9q 

= 9(3q³ + 3q² + q) + 1 

= 9m + 1 

(∵ m = 3q³+ 3q² + q) 

iii) If a = 3q + 2, then 

(a)³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 

= 9(3q³ + 6q² + 4q) + 8 

= 9m + 8 

∴ Cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.

Correct option is (D) 1/3

Let x = \(0.\bar{3}\)

i.e., x = 0.333 ..... _________(1)

Multiply equation (1) by 10, we get

10x = 3.333 .....    _________(2)

Subtract equation (1) from (2), we get

9x = 3

\(\Rightarrow\) x = \(\frac39=\frac{1}{3}.\)

Correct option is  D 1/3

Correct option is (C) 5/2

(A) \(\frac{2}{3}\) = \(0.\bar6\) < 2,

(B) \(\frac{3}{2}\) = 1.5 < 2,

(C) \(\frac{5}{2}\) = 2.5 lies between 2 and 3,

(D) \(\frac{5}{3}\) = \(1.\bar6\) < 2

Correct option is  C) 5/3 

Correct option is  A) 1,3, 4, 2

Correct option is (A) 2√2

\(x^2=2\)

\(\Rightarrow\) x = \(\sqrt2\)

\(\therefore\) \(x^3\) = \((\sqrt2)^3\) = \((\sqrt2)^2\sqrt2\) = \(2\sqrt{2}\)

Correct option is  A) 2√2

Correct option is (A) \(0.0\bar{5}\)

\(\frac{1}{18}=\frac1{2\times9}=\frac{0.5}9\)

\(=0.5\times0.1111....\)

= 0.05555.....

= \(0.0\bar{5}\)

Correct option is   A) \(0.0\overline5\)

Correct option is (C) \(\sqrt[3]{225}\)

\(15^\frac23=(15^2)^\frac13\)

\(=(225)^\frac13=\sqrt[3]{225}\)

Correct option is  C) \(\sqrt[3]{225}\)

Correct option is (B) 5/4

1.25 = \(\frac{125}{100}=\frac{125\div25}{100\div25}\)

= \(\frac{5}{4}\)

Correct option is  B) 5/4

Correct option is  A)  -7/5

Correct option is (A) 2

\((128)^{\frac17}=(2^7)^{\frac17}\)

= 2

Correct option is  A) 2

Correct option is (B) 9^2/4

\(\sqrt[4]{81}=(81)^\frac14=(3^4)^\frac14\)

\(=(3^2)^{\frac24}=9^{\frac24}\)

Correct option is  B) 9^2/4

Correct option is (A) \(\sqrt[3]{36}\)

\(6^\frac23=(6^2)^\frac13=(36)^\frac13\)

= \(\sqrt[3]{36}\)

Correct option is   A) \(\sqrt[3]{36}\)

Correct option is (D) Both B and C

1.121231234…… is non-terminating and non-recurring decimal.

Thus, 1.121231234…… is an rational number.

D) Both B and C

Correct option is (B) 2

\(\sqrt[5]{32}=(32)^\frac15=(2^5)^\frac15\)

= 2

Correct option is  B) 2

Correct option is  C) 1.96

Correct option is (D) \(\sqrt[5]{27}\)

\(27^{\frac15}\) = \(\sqrt[5]{27}\)

Correct option is  D) \(\sqrt[5]{27}\)

Correct option is (D) \(\frac{a+b}{2}\)

\(\frac{a+b}{2}\) is a rational number lying between two rational numbers a and b.

Correct option is   D) \(\frac{a+b}{2}\)

Correct option is (C) \(\frac{a+b}{2}\)

Rational number between a and b is \(\frac{a+b}{2}.\)

 Correct option is  C) \(\frac{a+b}{2}\)

Correct option is (A) \(\frac{13}{4}\)

3.25 = \(\frac{325}{100}=\frac{13}{4}\)

Correct option is   A) \(\frac{13}{4}\)

Correct option is (B} \((\frac{3}{4})^6\)

\((\frac{3}{4})^{-3}\times(\frac{3}{4})^3\times(\frac{3}{4})^6\)

\(=(\frac{3}{4})^{-3+3+6}=(\frac{3}{4})^6\)

Correct option is  B) (3/4)⁶

Correct option is (C) \(\sqrt{3}\)

\(\because\) \(\frac{1}{\sqrt{27}}\times\sqrt3\) \(=\frac{\sqrt3}{3\sqrt{3}}\) \(=\frac13\) which is an irrational number.

\(\therefore\) \(\sqrt{3}\) is the rationalising factor of \(\frac{1}{\sqrt{27}}.\)

Correct option is  C) \(\sqrt{3}\) 

Correct option is (B) a

\(\because\) \(a^n=b\)

\(\Rightarrow\) \(a=b^\frac1n\)

Thus, \(\sqrt[n]{b}\) \(=b^\frac1n\) = a

Correct option is  B) a

Correct option is (D) a rational number

\(p^3=216=6^3\)

\(\therefore\) p = 6 which is a rational number but not an odd number or an irrational number or a perfect square.

D) a rational number

Correct option is (B) an irrational number

\(x^3=10\)

\(\Rightarrow\) x = \(10^\frac13\) which is an irrational number.

B) an irrational number

Correct option is (C) \(\frac{1}{5+\sqrt{3}}\)

\(\because\) \(\frac{1}{5-\sqrt{3}}\times\frac{1}{5+\sqrt{3}}\) \(=\frac{1}{(5-\sqrt{3})(5+\sqrt{3})}\)

\(=\frac{1}{5^2-(\sqrt3)^2}\) \(=\frac1{25-3}\)

\(=\frac1{22}\) which is a rational number.

\(\therefore\) Rationalising factor of \(\frac{1}{5-\sqrt{3}}\) is \(\frac{1}{5+\sqrt{3}}.\)

Correct option is  C) \(\frac{1}{5+\sqrt3}\)

 Correct option is (A) \(\sqrt[n]{a}\)

\(a^\frac1n\) = \(\sqrt[n]{a}\)

Correct option is   A) \(\sqrt[n]{a}\)

To show:

the square of an odd positive integer is of the form 8q + 1, for some integer q.

Solution:

Let a be any positive integer and b = 4.

Applying the Euclid's division lemma with a and b = 4

we have a = 4p + r

where 0 ≤ r < 4 and p is some integer,

⇒ r can be 0,1,2,3

⇒ a = 4p + 0 , a = 4p + 1 , a = 4p+ 2 , a = 4p + 3,

Since a is odd integer,

So a = 4p + 1 or a = 4p + 3

So any odd integer is of the form a = 4p+ 1 or a = 4p+ 3.

Since any odd positive integer n is of the form 4p + 1 or 4p + 3.

When n = 4p + 1,

then n² = (4p + 1)²

Apply the formula (a + b)² = a² + b² + 2ab

⇒ (4p + 1)² = 16p² + 8p + 1 ..... (1)

Take 8p common out of 16p² + 8p ⇒ 8p

(2p+ 1) + 1 = 8q + 1 where q = p(2p + 1)

If n = 4p + 3, then n² = (4p + 3)²

Apply the formula (a + b)² = a² + b² + 2ab

⇒ (4p + 3)² = 16p² + 24p + 9

Now 16p² + 24p + 9 can be written as 16p² + 24p + 8 + 1

⇒ (4p + 3)² = 8(2p² + 3p + 1) + 1 = 8q + 1

where q = 2p² + 3p + 1

From above results we got that n² is of the form 8q + 1.

Note:

To show that the square of an odd positive integer is of the form 8q + 1

We have started the question from taking b = 4 initially because when we take square of any form of 4 such as 4p+1 we end up having the values which are the multiple of 8 as in (1).

While attempting these types of questions always remember to start with the value which would end up giving the value the question demands.and follow the above steps.

Correct option is (A) a^m/n

\(\sqrt[n]{a^m}\) = \((a^m)^\frac1n\)

= \(a^{m/n}\)

Correct option is  A) a^m/n 

Correct option is (C) 3^-2/15

= \(\cfrac{3^{\frac15}}{3^{\frac13}}\) \(=3^{\frac15-\frac13}=3^{\frac{3-5}{15}}\)

= \(3^\frac{-2}{15}\)

Correct option is  C) 3^-2/15 

we know,

That any positive integer N is of the form 3q, 3q + 1 or, 3q + 2.

When N = 3q, then

N² = 9q² = 3(3q)² = 3m where m = 3q²

And,

as q is an integer,

m = 3q² is also an integer.

When N = 3q + 1,

then N² = (3q + 1)² = 9q² + 6q + 1

⇒ 3q(3q + 2) + 1 = 3m + 1

where m = q(3q + 2)

And,

as q is an integer,

m = q(3q + 2) is also an integer.

When N = 3q + 2,

then N² = (3q + 2)² = 9q² + 12q + 4

⇒ 3(3q² + 4q + 1) + 1 3m + 1

where,

m = 3q² + 4q + 1

And,

as q is an integer,

m = 3q² + 4q + 1 is also an integer.

Therefore,

N² is of the form 3m, 3m + 1

but not of the form 3m + 2

Important Multiple Choice Questions (MCQ Questions) from Real Numbers are provided here for students who are preparing for the CBSE Class 10 Board Exam. Detailed solutions are also provided for all questions. These MCQs will help students clear all the fundamental concepts and prepare effectively for Class 10 Maths exam. 

Practice Class 10 Maths MCQ Question of Real Numbers

1. The decimal expansion of n is

(a) terminating
(b) non-terminating and non-recurring
(c) non-terminating and recurring
(d) does not exist.

2. For some integer m, every odd integer is of the form

(a) m
(b) m + 1
(c) 2m
(d) 2m + 1

3. The product of a non-zero number and an irrational number is:

(a) always irrational
(b) always rational
(c) rational or irrational
(d) one

4. The exponent of 2 in the prime factorisation of 144, is

(a) 4
(b) 5
(c) 6
(d) 3

5. The LCM of two numbers is 1200. Which of the following cannot be their HCF?

(a) 600
(b) 500
(c) 400
(d) 200

6. If two positive integers p and q can be expressed as
\(p = ab^2\) and \(q = a^3b\) a, b being prime numbers, then LCM (p, q) is

(a) \(ab\)
(b) \(a^2b^2\)
(c) \(a^3b^2\)
(d) \(a^3b^3\)

7. In a seminar, the number of participants in English, German and Sanskrit are 45,75 and 135. Find the number of rooms required to house them, if in each room, the same number of participants are to be accommodated and they should be of the same language.

(a) 45
(b) 17
(c) 75
(d) 135

8. If p = HCF (100,190) and q = LCM (100, 190); then \(p^2q^2\) is :

(a) 3.61 x \(10^5\)
(b) 361 x \(10^3\)
(c) 3.61 x \(10^6\)
(d) 3.61 x \(10^8\)

9. The largest positive integer which divides 434 and 539 leaving remainders 9 and 12 respectively is:

(a) 9
(b) 108
(c) 17
(d) 539

10. There is a circular path around a field. Reema takes 22 minutes to complete one round while her friend Saina takes 20 minutes to complete the same. If they both start at the same time and move in the same direction, after how many minutes will they meet again at the starting

(a) 220
(b) 3.4
(c) 440
(d) 4.4

11. If n = \(2^3 × 3^4 × 5^4 × 7\), then the number of consecutive zeros in n, where n is a natural number, is

(a) 2
(b) 3
(c) 4
(d) 7

12. The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1
(b) 2
(c) 4
(d) 6

13. 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

(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number

14. If two positive integers m and n are expressible in the form \(m = pq^3\) and \(n = p^3 q^2\) where p, q are prime numbers, then HCF (m, n) =

(a) \(pq\)
(b) \(pq^2\)
(c) \(p^3q^3\)
(d) \(p^2q^3\)

15. Euclid’s Lemma states that, for given positive integers a and b, there exist unique integers q and r, such that a = bq + r, where:

(a) 0 < r < b
(b) 0 ≤ r < b
(c) 0 < r ≤ b
(d) 0 ≤ r ≤ b

16. If a non-zero rational number is multiplied to an irrational number, we always get:

(a) an irrational number
(b) a rational number
(c) zero
(d) one

17. The values of the remainder r, when a positive integer a is divided by 3 are:

(a) 0, 1, 2, 3
(b) 0, 1
(c) 0, 1, 2
(d) 2, 3, 4

18. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

(a) 2520cm
(b) 2525cm
(c) 2555cm
(d) 2528cm

19. Write an irrational number between 2 and 3.

(a) 2.5
(b) 2.001
(c) 2.1333333456…
(d) 2.13

20. The greatest number which divides 87 and 97, leaving 7 as remainder is:

(a) 10
(b) 1
(c) 87 x 97
(d) 6300

Answers & Explanations

1. Answer: (b) non-terminating and non-recurring

2. Answer: (d) 2m + 1

Explanation: As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

3. Answer: (a) always irrational

Explanation: Product of a non-zero rational and an irrational number is always irrational i.e.,

\(\frac{3}{4}\times\sqrt2\) = (rational) \(\times\) (irrational) = irrational.

4. Answer: (a) 4

Explanation: The prime factorization of 144 is as follows:

144 = 2 × 2 × 2 × 2 × 3 × 3

⇒ 144 = 2⁴ × 3²

We know that the exponent of a number a^m is m.

∴ The exponent of 2 in the prime factorization of 144 is 4.

5. Answer: (b) 500

Explanation: We know that LCM of two or more numbers is always divisible by their HCF.

1200 is divisible by 600, 200 and 400 but not by 500.

6. Answer: (c) \(a^3b^2\)

Explanation: LCM is Product of the greatest power of each prime factor involved in the number

So LCM = \(a^3b^2\)

7. Answer: (b) 17

Explanation: Since, in each room, the same number of participants, of the same language, are to be accommodated, their number in each room

HCF of 45, 75 and 135.

HCF (45, 75,135) = 15

Each room accommodates 15 participants
\(\Rightarrow\) Total no. of rooms required for English = \(\frac{45}{15}\) = 3

Total no. of rooms required for German = \(\frac{75}{15}\) = 5

Total no. of rooms required for Sanskrit= \(\frac{135}{15}\) = 9

Total no. of rooms = 3 + 5 + 9 = 17

8. Answer: (d) 3.61 x \(10^8\)

Explanation: pq = (HCF) (LCM) = Product of given numbers.

\(\Rightarrow\) pq = 190×100 =19000
\(\Rightarrow\) \(p^2q^2\) = 361 x \(10^6\) = 3.61 x \(10^8\)

9. Answer: (c) 17

Explanation: Required number is the HCF of (434 – 9) and (539 -12)

= HCF of 425 and 527.
= 17

10. Answer: (a) 220

Explanation: LCM of 20 and 22 = 220 (question state: after how many minutes will they meet)

11. Answer: (b) 3

Explanation: If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3⁴ × 5⁴ × 7.

It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5)

⇒ 10 × 10 × 10 = 1000

Thus, there are 3 zeros in n.

12. Answer: (c) 4

Explanation: The prime factorization of 196 is as follows:

196 = 2 × 2 × 7 × 7

⇒ 98 = 2² ×7²

We know that the exponent of a number a^m is m.

∴The sum of powers = 2 + 2 = 4

13. Answer: (a) an even number

Explanation: Let us take \(p_1\) = 5 and \(p_2\) = 3

Then \(p_1^2 – p_2^2\) = 25 – 9 = 16

16 is an even number

14. Answer: (b) \(pq^2\)

Explanation: We know that HCF = Product of the smallest power of each common prime factor in the numbers.

So, HCF (a, b) = \(pq^2\)

15. Answer: (b) 0 ≤ r < b

16. Answer: (a) an irrational number

Explanation: The product of a rational (non-zero) and m irrational number is always an irrational number.

17. Answer: (c) 0, 1, 2

Explanation: According to Euclid’s division lemma,

a = 3q + r, where 0 r < 3

As the number is divided by 3. So, the remainder cannot be greater than divisor 3 also r is an integer. Therefore, the values of r can be 0, 1 or 2.

18. Answer: (a) 2520cm

Explanation: We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.

40 = 2×2×2×5

42 = 2×3×7

45 = 3×3×5

L.C.M. = 2×3×5×2×2×3×7 = 2520

19. Answer: (c) 2.1333333456…

Explanation: Non terminating non repeating

20. Answer: (a) 10

Explanation: Greatest number which divides 87 and 97, leaving 7 as remainder = HCF of 80 and 90

Click here to practice more MCQ Questions from Chapter Real Numbers Class 10 Maths

The correct option is: (D)

The correct option is: (A) For natural numbers a and b, if a divides b² then a divides b.

Explanation:

If a (any prime number) divides b² and b is natural number, then a divides b.

The correct option is: (D) 9 a.m.

Explanation:

Convert all the hours into minutes to find L.C.M.
1 hour = 60 minutes
3/2 hour = 80 minutes
1 hour 45 minutes = 105 minutes
Hence we take LCM of (30, 60, 90, 105) = 1260 min = 21 hours. 
So if they ring together at 12 noon today, they will ring together at 9 am tomorrow.

Given rational number is \(\frac{129}{2^2 \times 5^7 \times 7^5}\)

Since the denominator is not of the form 2ⁿ5^m

\(\frac{129}{2^2 \times 5^7 \times 7^5}\) has a non-terminating repeating decimal expansion.

We know, 35/50

= 7/10

= 7/(2 x 5)

= 7/10

= 0.7

Given rational number is 29/343

29/243 = 29/3⁵

Since the denominator is not of the form 2ⁿ5^m.

29/243 has a non-terminating repeating decimal expansion.

We know, 29/343 = 29/7³

Given rational number is 29/343

p/q is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly we check co-prime

29 = 29 × 1

343 = 7 × 7 × 7

29 and 343 have no common factors

Therefore, 29 and 343 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

343 = 7³

So, the denominator is not of the form 2ⁿ5^m

Thus, 29/343 is a non-terminating repeating decimal.

Correct option is (A) \(\sqrt{20}\)

\(\because\) 16 < 20 < 25

\(\Rightarrow\) \(\sqrt{16}<\sqrt{20}<\sqrt{25}\)

\(\Rightarrow\) \(4<\sqrt{20}<5\) and \(\sqrt{20}\) is an irrational number.

Correct option is  A) \(\sqrt{20}\)

Correct option is  B) √2 Units

Correct option is  A) √3 – 1 Units

Correct option is  B) 6 Units

Correct option is  B) 2 and 3