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Math Olympiad Test: Real Numbers- 1 - Class 10 MCQ


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10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Real Numbers- 1

Math Olympiad Test: Real Numbers- 1 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Real Numbers- 1 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Real Numbers- 1 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Real Numbers- 1 below.
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Math Olympiad Test: Real Numbers- 1 - Question 1

What is the L.C.M. of 144, 180 and 192 by prime factorization method?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 1

L.C.M. of 144, 180, 192
144 = 24 × 32
180 = 22 × 32 × 5
192 = 26 × 3
∴ L.C.M. = 26 × 32 × 5 = 2880

Math Olympiad Test: Real Numbers- 1 - Question 2

What is the largest number that divides 445, 572 and 699 having remainders 4, 5, 6 respectively?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 2

Here 445 - 4 = 441; 572 - 5 = 567, 699 - 6 = 693
Now we have to find H.C.F. of 441, 567, 693 is
441 = 3 × 3 × 7 × 7
567 = 3 × 3 × 3 × 3 × 7
693 = 3 × 3 × 7 × 11
The common factors are 3 × 3 × 7 = 63.

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Math Olympiad Test: Real Numbers- 1 - Question 3

What is the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 3

Solve for the HCF of the given numbers by deducting the given remainders
Subtracting the remainders from numbers we get,
⇒ 398 − 7 = 391 [As the remainder is 7]
⇒ 436 − 11 = 425 [As the remainder is 11]
⇒ 542 − 15 = 527 [As the remainder is 15]
HCF of these new numbers is the largest number that divides 398, 436, 542 leaving respective remainders.
Factors of 391 = 17 × 23
Factors of 425 = 17 × 52
Factors of 527 = 17 × 31
The HCF of 391, 425, 527 is17.
Hence, the largest positive integer that will divide 398, 436 and 542, leaving remainders7, 11 and 15 respectively is 17.

Math Olympiad Test: Real Numbers- 1 - Question 4

What is the largest number which divides 615 and 963 leaving remainder 6 in each case?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 4

In order to find the largest number which divides 615 and 963 leaving remainder 6 in each case, we need to deduct the remainder 6 from both the numbers and find their HCF.
615 − 6 = 609
963 − 6 = 957
Prime Factorising both 609 and 657 we get,
609 = 3 x 3 x 29  and
957 = 3 x 11 x 29
HCF of 609, 957 = 3 × 29
= 3 x 29 = 87
∴ The largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

Math Olympiad Test: Real Numbers- 1 - Question 5

What is the smallest number that when divided by 35, 56, and 91 leaves remainder 7 in each case?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 5

The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91
35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13
LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.

Math Olympiad Test: Real Numbers- 1 - Question 6

If the H.C.F. of 408 and 1032 is expressed in the form of 1032m - 408 × 5. What is the value of m?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 6

Given H.C.F. of 408 and 1032 = 24 and
1032m - 408 × 5 = 24
⇒ 1032m = 24 + 2040
⇒ 1032m = 2064
⇒ m = 2064/1032
= 2

Math Olympiad Test: Real Numbers- 1 - Question 7

Find the sum of the exponents of the prime factors in the prime factorization of 196.

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 7

Using the factor tree for prime factorization, we have:

We have 196 = 22 × 72
2 + 2  = 4

Math Olympiad Test: Real Numbers- 1 - Question 8

If 5005 is expressed in the term of product of price factors, then which prime factor is the largest?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 8

5005 can be expressed as product of prime factor as:
5005 = 5 × 7 × 11 × 13
Thus, 5005 can be expressed as product of prime factors as 5005 = 5 × 7 × 11 × 13.

Math Olympiad Test: Real Numbers- 1 - Question 9

The H.C.F of two numbers is 145, their L.C.M. is 2175. If one number is 725, then what is the other number?

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 9

product of number = LCM x HCF
a x b = LCM x HCF
Other number

Math Olympiad Test: Real Numbers- 1 - Question 10

If n  is any natural number then 6n - 5n always ends with

Detailed Solution for Math Olympiad Test: Real Numbers- 1 - Question 10

6n - 5n = 61 - 51 = 1
62 - 52 = 36 - 25 = 11 and so on

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