MCQ Questions for Class 10 Maths Chapter 1 Real Numbers

Real Numbers Class 10 MCQ Questions with Answers

Question 1.

n²– 1 is divisible by 8, if n is:

(A) an integer.
(B) a natural number.
(C) an odd integer.
(D) an even integer.
Answer:
(C) an odd integer.

Explanation:
Any odd integer can be written as 2m + 1.
put n = 2m + 1 in n²– 1
n²– 1 =(n + 1)(n-1) (2m + 2)(2m) = 4m(m + 1 ) The product of two consecutive numbers is divisible by 2. Thus, m(m + 1) is divisible bv 2. Let m(m + 1) = 2k
n²– 1 =(n + 1)(n-1)= (2m + 2)(2m)
4m(m + 1) = 4 × 2k =8k
1Thus, if n is an odd integer then n²-1 is divisible by 8.

Question 2.

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is:

(A) 13.
(B) 65.
(C) 875.
(D) 1,750.
Answer:
(A) 13.

Explanation:
Required largest number = HCF 1 of (70 – 5) and (125 – 8) = HCF of 65 and I 117 = 13

Question 3.

If two positive integers a and b are written as a = x³y² and b = xy²; x, y are prime numbers, then HCF (a, b) is:

(A) xy.
(B) xy²
(C) x³y³
(D) x²y²
Answer:
(B) xy²

Explanation:
Since n-x³y³x × x × x × y × y and b – xy³=x × y × y × y, Thus, HCF: of a and b = x × y × y =x y²

Question 4.

If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is

(A) 4.
(B) 2.
(C) 1.
(D) 3.
Answer:
(B) 2.

Explanation:
By the Ludid’s division algorithm,
HCF of (65, 117) = 13
Since 65m – 117 = 13 › m = 2

Question 5.

If two positive integers p and q can be expressed as p = ab² and q = a³b; a, b being prime numbers, then LCM (p, q) is

(A) ab
(B) a²b²
(C) a³b³
(D) a²b²
Answer:
(C) a³b³

Explanation:
Since p = ab²=a x b x b
q=a³b = a x a x a x b x b =
Thus, LCM of p and q = a x a x a x b x b = a³b²

Question 6.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 10.
(B) 100.
(C) 504.
(D) 2,520.
Answer:
(D) 2,520.

Explanation:
Required number = LCM( 1,2, 3,4, 5, 6,8, 9,10)
=1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520

Question 7.

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

(A) always irrational.
(B) always rational.
(C) rational or irrational.
(D) one.
Answer:
(A) always irrational.

Explanation:
The product of a non-zo rational with and an irrational number alwavs irrational.

Question 8.

The decimal expansion of the rational number 14587 wyj terminal after : 1250

(A) one decimal place.
(B) two decimal places.
(C) three decimal places.
(D) four decimal places.
Answer:
(D) four decimal places.

Explanation:
\(\frac{14587}{1250}=\frac{14587}{2 \times 5^{4}}=\frac{14587}{2 \times 5^{4}} \times \frac{2^{4}}{2^{4}}\)
=\(\frac{14587 \times(2)^{3}}{10^{4}}=\frac{116,696}{10000}=11.6696\)

Question 9.

\(\frac{1}{\sqrt{2}}\)= is a/an

(A) fraction
(B) rational number
(C) irrational number
(D) none of these
Answer:
(C) irrational number

Explanation:
\(\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2} \cdot \sqrt{2} \)
We have \(\frac{1}{2}\) is a rational but ?2 is irrational. So, the product of a rational and an irrational is irrational.
Hence, \(\frac{1}{\sqrt{2}}\) = = \(\frac{1}{2} \cdot \sqrt{2}\) is irrational.

Question 10.

(3 + √5) is:

(A) an integer
(B) a rational number
(C) an irrational number
(D) none of these
Answer:
(C) an irrational number

Explanation:
Here 3 is a rational and √5 is an irrational.
?The sum (3 + √5) i.e., rational and irrational is irrational.Hence (3 + √5) is an irrational number.

Question 11.

The number 1.732 is:

(A) an irrational number
(B) a rational number
(C) an integer
(D) a whole number
Answer:
(B) a rational number

Explanation:
We have \(1.732=\frac{1732}{1000}=\frac{433}{250}\) Which is a rational number.

Question 12.

2.1311 3111 311113…………is

(A) an integer
(B) a rational number
(C) an irrational number
(D) none of these.
Answer:
(C) an irrational number

Explanation:
Given number is non-terminating, non-repeating decimal, So, it is an irrational number.

Question 13.

Which of the following rational numbers is expressible as a terminating decimal?

\(\text { (A) } \frac{2027}{625}\)
\(\text { (B) } \frac{1625}{462}\)
\(\text { (C) } \frac{131}{35}\)
\(\text { (D) } \frac{124}{165}\)
Answer:
\(\text { (A) } \frac{2027}{625}\)

Explanation:
Given,\(\frac{2027}{625}=\frac{2027}{5 \times 5 \times 5 \times 5} \times \frac{2^{4}}{2^{4}}\)
\(=\frac{2027 \times 2^{4}}{5^{4} \times 2^{4}}=\frac{32,432}{10^{4}}\)
=3.2432
Thus, it is expressible as a terminating decimal.

Question 14.

The decimal expansion of the sum of rational numbers\(\frac{15}{4}\) and\(\frac{5}{40}\) will terminate after.

(A) one decimal place
(B) two decimal places
(C) three decimal places
(D) four decimal places
Answer:
(C) three decimal places

Explanation:
The sum of rational numbers
\(\begin{aligned} &=\frac{15}{4}+\frac{5}{40} \\
&=\frac{15 \times 25}{4 \times 25}+\frac{5 \times 25}{40 \times 25} \\
&=\frac{375}{100}+\frac{125}{1000}
\end{aligned}\)
= 3.75 + 0.125 = 3.875.
So, it will terminate after 3 decimal places.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(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 and R is True

Question 1.

Assertion (A): (7 × 13 × 11) + 11 and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 have exactly composite numbers.
Reason (R): (3 × 12 × 101) + 4 is not a composite number.

Answer:
(C) A is true but R is false

Explanation:
firstly consider the assertion, Since (7 × 13 × 11) + 11 = 11 × (7 × 13 + 1)
= 11× (91 + 1)
=11×92 ⇒11 × 2 × 2 × 23 and (7 × 6 × 5 × 4 × 3 × 2 × l) + 3
= 3(7 x 6 x 5 x 4 x 2 x l + l)
= 3 x (1681) ⇒ 3 × 41 × 41 Given numbers have more than two prime factors. So, both the numbers are composite. Hence, assertion is correct.
Now let us consider the reason:
3× 12× 101 + 4 = 4(3 × 3 × 101 + 1)
= 4(909 + 1)
= 4(910)
= 2 × 2 × 2 × 5 × 7 × 13 = a composite number
[∴ Product of more than two prime factors] Thus, reason is not correct. Thus, assertion is correct but reason is incorrect.

Question 2.

Assertion (A): If HCF (336,54) = 6, then LCM (336,54) = 3000.
Reason (R): The sum of exponents of prime factors in the prime factorisation of 196 is 4.

Answer:
(D) A is false and R is True

Explanation:
Let us consider the assertion,
∴ HCF × LCM = Product of numbers
∴ 6 x LCM = 336 x 54

⇒ LCM = \(\frac{336 \times 54}{6}\)

= 3024
Thus, the assertion is incorrect:
Now, let us consider the reason:
Prime factors of 196 = 2² x 7²
∴ The sum of exponents of prime factors = 2 + 2 = 4.
So, the reason is correct: Thus, assertion is incorrect but reason is correct.

Question 3.

Assertion (A): HCF of two or more numbers = Product of the smallest power of each common prime factor, involved in the numbers.
Reason (R): The HCF of 12,21 and 15 is 3.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Let a, a² and a2 be three numbers, then we have the smallest power of a¹, a² and a³ is 1. So, HCF is a. Now, let us consider the reason:
Prime factors of 12 = 22 × 3
Prime factors of 21 =3×7
Prime factors of 15 =3×5
?HCF of 12,21 and 15 = 3, which is a common prime factor. Thus both assertion and reason are correct and reason is the correct explanation for assertion.

Question 4.

Assertion (A): The product of two consecutive positive integers is divisible by 2.
Reason (R): 13233343563715 is a composite number.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion. Since, in the product of two consecutive positive integers, p = n(n +1), one of n or (n + 1) is an even number. Hence, the product of two consecutive positive integers is divisible by 2. So, it is correct. Now, let us consider the reason: Since, the given number ends in 5. It is a multiple of 5. Therefore, it is a composite number. Thus, both assertion and reason are correct and reason is not the correct explanation for assertion.

Question 5.

Assertion (A): The decimal expansion of assertion: \(\frac{15}{16000}\) is 0.09375
Reason (R): The decimal expansion of \(\frac{23}{2^{3} 5^{2}}\)

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
\(\frac{15}{1600}\) = \(\begin{aligned}
&\frac{15}{2^{4} \times 100} \\ &\frac{15 \times 5^{4}}{2^{4} \times 5^{4} \times 100} \\ &\frac{9375}{(2 \times 5)^{4} \times 100} \\ &\frac{9375}{1000000}=0.009375
\end{aligned}\)

So, assertion is incorrect: Now, in case of Reason
\(\frac{23}{2^{3} 5^{2}}\) = \(\begin{aligned} &\frac{23 \times 5}{2^{3} \times 5^{2} \times 5} \\ &\frac{115}{2^{3} \times5^{3}}=\frac{115}{(2 \times 5)^{3}} \\
&\frac{115}{1000}=0.115
\end{aligned}\)
So, reason is correct. Thus, asset tion incorrect but reason is correct.

Question 6.

Assertion (A): The decimal expansion of the rational number \(\frac{57}{2^{2} \times 5}\) will terminate after 2 decimal places. 2 x 5
Reason (R): The decimal expansion of the rational number \(\frac{13}{3125}\)will terminate after 3 decimal places.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
\(\frac{57}{2^{2} \times 5}\)
\(=\frac{57 \times 5}{2^{2} \times 5^{2}}\)
\(=\frac{285}{(2 \times 5)^{2}}\)
\(=\frac{285}{100}\)
So, U will terminate after 2 decimal places.
∴ Assertion is correct.
In case of Reason:
\(\frac{13}{3125}=\frac{13}{5^{5}}\)
[∴ prim factor of 3125 = 5 × 5 × 5 × 5]
\(=\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}\)
\(=\frac{416}{(5 \times 2)^{5}}=\frac{416}{100000}\)
= 0.00416.
So. it will terminate after 5 decimal places.
∴Reason is incorrect.
Thus, assertion is correct but reason is incorrect.

Case -Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
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.

Question 1.

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
Answer:
(C) 288

Explanation:
We have to find the LCM of 32 and 36.
LCM(32, 36) = 2⁵ × 9 = 288
Hence, the minimum number of books required to distribute equally among students of section A and section B are 288.

Question 2.

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32,36) is

(A) 2
(B) 4
(C) 6
(D) 8
Answer:
(B) 4

Question 3.
36 can be expressed as a product of its primes as

(A) 2² × 3²
(B) 2¹ × 3⁵
(C) 2³ ×3¹
(D) 2° x 3°
Answer:
(A) 22 x 32

Question 4.

7 ×11× 13× 15 + 15 is a

(A) Prime number
(B) Composite number
(C) Neither prime nor composite
(D) None of the above
Answer:
(B) Composite number

Question 5.

If p and q are positive integers such that p = ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is

(A) ab
(B) a²b²
(C) a³b²
(D)a³b³
Answer:
(D)a³b³

Explanation: Given,p = ab² = a × b×b
q = a²b= a×a×b
LCM of (p, q) =a²b²

II. Read the following text and answer the following questions on the basis of the same: 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.

Question 1.

In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are

(A) 14
(B) 12
(C) 16
(D) 18
Answer:
(B) 12

Explanation:
No. of participants seated in each room would be HCF of all the three values above.
60 = 2 × 2 × 3 × 5
84 = 2 × 2 × 3 × 7
108 = 2 × 2 × 3 × 3 × 3
Hence, HCF = 12.

Question 2.

What is the minimum number of rooms required during the event?

(A) 11
(B) 31
(C) 41
(D) 21
Answer:
(D) 21

Explanation:
Minimum no. of rooms required are total number of students divided by number of students in each robin.
No. of rooms \(\begin{aligned} &=\frac{60+84+108}{12} \\ &=21 \end{aligned}\)

Question 3.

The LCM of 60, 84 and 108 is

(A) 3780
(B) 3680
(C) 4780
(D) 4680
Answer:
(A) 3780

Question 4.

The product of HCF and LCM of 60,84 and 108 is

(A) 55360
(B) 35360
(C) 45500
(D) 45360
Answer:
(D) 45360

Question 5.

108 can be expressed as a product of its primes as

(A) 2³ × 3²
(B) 2³ × 3³
(C) 2² × 3²
(D) 2² × 3²
Answer:
(D) 2² × 3²

III. Read the following text and answer the following questions on the basis of the same: A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

Read carefully the above paragraph and answer the following questions:

Question 1.

What is the maximum number of columns in which they can be planted?

(A) 45
(B) 40
(C) 15
(D) 35
Answer:
(A) 45

Explanationi:
No. of rose plants = 135
No. of marigold plants =225
The maximum number of columns in which they can be planted = HCF of 135 and 225
∴ Prime factors of 135 = 3 x 3 x 3 x 5 and 225 = 3 × 3 × 5 × 5
∴ Prime factors of 135 = 3 x 3 x 5 = 45.

Question 2.

Find the total number of plants

(A) 135
(B) 225
(C) 360
(D) 45
Answer:
(C) 360

Explanation:
Total number of plants 135 + 225 = 360 plants

Question 3.

Find the sum of exponents of the prime factors of the maximum number of columns in which they can be planted.

(A) 5
(B) 3
(C) 4
(D) 6
Answer:
(B) 3

Explanation:
We have proved that the maximum number of columns = 45
So, prime factors of 45 = 3 × 3 × 5
= 3² × 5¹
?Sum of exponents =2 + 1 = 3

Question 4.

What is total numbers of row in which they can be planted

(A) 3
(B) 5
(C) 8
(D) 15
Answer:
(C) 8

Explanation:
Number of rows of Rose plants \(=\frac{135}{45}=3\)
Number of rows of marigold plants = \(=\frac{225}{45}=5\)
Total numbers of rows = 3 + 5 = 8

Question 5.

Find the sum of exponents of the prime factors of total number of plants

(A) 2
(B) 3
(C) 5
(D) 6
Answer:
(D) 6

Explanation:
Total number of plants = 135 + 225 =360
The prime factors of 360 = 2 × 2 × 2 × 3 × 3 × 5
= 2³ × 3² × 5¹
Sum of exponents = 3 + 2 + 1 = 6.

IV Read the following text and answer the following questions on the basis of the same: 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.
ig4
Question 1.

What will be the value of x?

(A) 15005
(B) 13915
(C) 56920
(D) 17429
Answer:
(B) 13915

Explanation:
x = 2783 × 5
x = 13915

Question 2.

What will be the value of y?

(A) 23
(B) 22
(C) 11
(D) 19
Answer:
(C) 11

Explanation:
2783 = y ×253
\(\begin{aligned} &y=\frac{2783}{253} \\ &y=11 \end{aligned}\)

Question 3.

What will be the value of z?

(A) 22
(B) 23
(C) 17
(D) 19
Answer:
(B) 23

Explanation: 253 = 11 x z
\(\begin{aligned} &z=\frac{253}{11} \\ &z=23 \end{aligned}\)

Sum of exponents = 3 + 2 + 1 = 6.

IV Read the following text and answer the following questions on the basis of the same:
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.

Question 1.

What will be the value of x?

(A) 15005
(B) 13915
(C) 56920
(D) 17429
Answer:
(B) 13915

Explanation:
x = 2783 × 5
x = 13915

Question 2.

What will be the value of y?

(A) 23
(B) 22
(C) 11
(D) 19
Answer:
(C) 11

Explanation:
2783 = y × 253
\(\begin{aligned} &y=\frac{2783}{253} \\ &y=11 \end{aligned}\)

Question 3.

What will be the value of z?

(A) 22
(B) 23
(C) 17
(D) 19
Answer:
(B) 23

Explanation:
253 = 11 × z
\(\begin{aligned} &z=\frac{253}{11} \\ &z=23 \end{aligned}\)

Question 4.

According to Fundamental Theorem of Arithmetic 13915 is a

(A) Composite number
(B) Prime number
(C) Neither prime nor composite
(D) Even number
Answer:
(A) Composite number

Question 5.

The prime factorisation of 13915 is

(A) 5 ×11³ × 13,²
(B) 5 × 11³ × 23²
(C) 5 × 11² × 23
(D) 5 × 11² × 23²
Answer:
(C) 5 × 11² × 23

MCQ Questions for Class 10 Maths with Answers