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Test: Real Numbers (Medium) - Class 10 MCQ


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20 Questions MCQ Test - Test: Real Numbers (Medium)

Test: Real Numbers (Medium) for Class 10 2024 is part of Class 10 preparation. The Test: Real Numbers (Medium) questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Real Numbers (Medium) MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Real Numbers (Medium) below.
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Test: Real Numbers (Medium) - Question 1
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Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Detailed Solution for Test: Real Numbers (Medium) - Question 1
Denominator is 8, so the rational has a non-terminating decimal expansion
Test: Real Numbers (Medium) - Question 2
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Find the HCF of the following numbers: 18, 60.

Detailed Solution for Test: Real Numbers (Medium) - Question 2
18, 60

18 = 1, 2, 3, 6, 9, 18

60 = 1, 2, 3, 4, 6, 10, 12, 15, 30, 60

Common factors = 1, 2, 3, 6

∴ H.C.F.of 18 & 60 is 6

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Test: Real Numbers (Medium) - Question 3
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Find the HCF and LCM of the following integers by applying the prime factorization method 40, 36 and 126.

Detailed Solution for Test: Real Numbers (Medium) - Question 3
40 = 2 x 2 x 2 x 5 = 23 x 5

36 = 2 x 2 x 3 x 3 = 22 x 32

and 126 = 2 x 3 x 3 x 7 = 2 x 32 x 7

For H.C.F taking minimum power of common factor H.C.F = (2)1 = 2

For L.C.M, taking maximum power of prime factors LC.M = 23 x 32 x 5 x 7 = 8 x 9 x 5 x 7 = 2520

Hence, H.C.F = 2 L.C.M = 2520

Test: Real Numbers (Medium) - Question 4
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Find the HCF by prime factorization 45 and 75.
Detailed Solution for Test: Real Numbers (Medium) - Question 4
HCF of two numbers are

45 = 3 × 3 × 5

75 = 3 × 5 × 5

The common factors make the HCF: HCF = 3 × 5 = 15

Test: Real Numbers (Medium) - Question 5
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Find the H.C.F. of the following numbers using the prime factorization method: 84, 120, 138.
Detailed Solution for Test: Real Numbers (Medium) - Question 5

84 = 22 × 3 × 7

120 = 23 × 3 × 5

138 = 2 × 3 × 23

HCF = product of common terms with lowest power

HCF = 2 × 3 = 6

Test: Real Numbers (Medium) - Question 6
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Check whether 6 / 200 has terminating or non terminating repeating decimal expansion.
Detailed Solution for Test: Real Numbers (Medium) - Question 6
6 / 200 = 3 / 100 = 0.03 which is terminating decimal.
Test: Real Numbers (Medium) - Question 7
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Find the H.C.F of the following numbers by Prime Factorization: 120, 240, 300
Detailed Solution for Test: Real Numbers (Medium) - Question 7
240 = 24 x 3 x 5

300 = 22 x 3 x 52

120 = 23 x 3 x 5

H.C.F = 22 x 3 x 5

Highest Common Factor is : 60

Test: Real Numbers (Medium) - Question 8
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Find the H.C.F. of the following numbers using the prime factorization method: 225, 450.
Detailed Solution for Test: Real Numbers (Medium) - Question 8
We have, 225 = 32 × 52

450 = 2 × 32 × 52

HCF(225, 450) = Product of common terms with lowest power HCF(225, 450) = 32 × 52 HCF(225, 450) = 9 × 25

HCF(225, 450) = 225

∴ The required number or HCF is 225.

Test: Real Numbers (Medium) - Question 9
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If HCF of 336 and 54 is 6, then LCM is:

Explanations: According to Euclid division algorithm

LCM => HCF x LCM = product if two no. 6 x LCM = 54 x 336

LCM = 54 x 336 / 6

LCM = 3024
Test: Real Numbers (Medium) - Question 10
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Prime factors of 4050 is:
Detailed Solution for Test: Real Numbers (Medium) - Question 10

Thus, these are the numbers that make up the prime factorization of 4050,

So the prime factorization of 4050 is 2 × 3 × 3 × 3 × 3 × 5 × 5.

Test: Real Numbers (Medium) - Question 11
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The LCM of three numbers 28, 44, 132 is:
Detailed Solution for Test: Real Numbers (Medium) - Question 11
Prime factors of 28 = 22 x 7

Prime factors of 44 = 22 x 11

Prime factors of 132 = 22 x 3 x 11

LCM = 22 x 71 x 111 x 31 = 924

Test: Real Numbers (Medium) - Question 12
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If p2 is an even integer, them p is an:
Detailed Solution for Test: Real Numbers (Medium) - Question 12
"Even" and "odd" are properties of integers, not of arbitrary real numbers. The correct statement is "if p is an integer and p2 is even, then p is even."
Test: Real Numbers (Medium) - Question 13
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Every composite number can be expressed as a product of:
Detailed Solution for Test: Real Numbers (Medium) - Question 13
This is true as any composite number can be expressed in terms of its prime factors. The prime factors for 2 numbers can be the same but their combination is unique.

Example: 12 = 22 x 3 and 18 = 2 x 32

Both have prime factors as 2 and 3 but the combination that 12 has is different from that of 18.

Test: Real Numbers (Medium) - Question 14
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The decimal expansion of a rational number is always:
Detailed Solution for Test: Real Numbers (Medium) - Question 14
The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number.
Test: Real Numbers (Medium) - Question 15
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The largest number which divides 445, 572 and 699 leaving remainder 4, 5 and 6 respectively is :
Detailed Solution for Test: Real Numbers (Medium) - Question 15
445 - 4 = 441

572 - 5 = 567

699 - 6 = 693

Now find the greatest common factor of those 3 numbers:

441 = 3 x 3 x 7 x 7

572 = 3 x 3 x 3 x 3 x 7

693 = 3 x 3 x 7 x 11

The common factors are 3 x 3 x 7 = 63

HCF Of (441, 567, 693) = 63

445 / 63 = 7 remainder 4

572 / 63 = 9 remainder 5

699 / 63 = 11 remainder 6

∴ 63 is the largest divisor that will give the desired remainders.

Test: Real Numbers (Medium) - Question 16
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The product of a non-zero rational and irrational number is:
Detailed Solution for Test: Real Numbers (Medium) - Question 16
Product of a non-zero rational number and an irrational number is always irrational.
Test: Real Numbers (Medium) - Question 17
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The HCF of the smallest composite number and the smallest prime number is:
Detailed Solution for Test: Real Numbers (Medium) - Question 17
HCF of the smallest composite number and the smallest prime number = 2

Smallest composite number = 4 = 2 × 2

Smallest prime number = 2

Test: Real Numbers (Medium) - Question 18
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If HCF (306, 657) = 9, then LCM (306, 657) is;
Detailed Solution for Test: Real Numbers (Medium) - Question 18
Given that HCF = 9 and the numbers are 306 and 657.

LCM = ?

We know that LCM × HCF = product of two numbers

⇒ LCM × 9 = 306 × 657

⇒ LCM = 201042 / 9 = 22338

Test: Real Numbers (Medium) - Question 19
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Write the exponent of 2 in the prime factorisation of 288.
Detailed Solution for Test: Real Numbers (Medium) - Question 19
288 = 2 × 2 × 2 × 2 × 2 × 3 × 3

⇒ 25 × 3²

Hence, The exponent of 2 is 5 and exponent of 3 is 2.

Test: Real Numbers (Medium) - Question 20
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What is the sum of exponents of prime factors in the prime factorisation of 98.
Detailed Solution for Test: Real Numbers (Medium) - Question 20
98 = 2 × 7 × 7 = 21 × 72

Sum of exponents = 1 + 2 = 3

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