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Math Olympiad Test: Real Numbers- 2 - Class 10 MCQ


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15 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Real Numbers- 2

Math Olympiad Test: Real Numbers- 2 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Real Numbers- 2 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Real Numbers- 2 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- 2 below.
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Math Olympiad Test: Real Numbers- 2 - Question 1

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

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

6n − 5n
For n = 1
61 − 51 = 6 − 5 = 01
⇒ Unit digit is 1, i.e., ends with 1.
For n = 2
62 − 52 = 36 − 25 = 11
⇒ Unit digit is 1, i.e., ends with 1.
For n=3
63 − 53 = 216 − 125 = 91
⇒ Unit digit is 1, i.e., ends with 1.
By continuity, we can say that (6n − 5n) always ends with 1.

Math Olympiad Test: Real Numbers- 2 - Question 2

If n is a natural number, then 92n - 42n is always divisible by ____.

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

Solve for divisor of given expression: 
Given, 92n - 42n is in the form of a2n -b2n where n is a natural number.
Since a2n - b2n is always divisible by (a - b) and (a + b),
92n - 42n is always divisible by:
9 - 4 = 5
9 + 4 = 13
∴ 92n - 42n is always divisible by both 5 and 13.

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Math Olympiad Test: Real Numbers- 2 - Question 3

What is product of H.C.F. and L.C.M. of the numbers 81 and 50?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 3
  • The H.C.F. of 81 and 50 is 1, as they have no common factors other than 1.
  • The L.C.M. is found by multiplying the two numbers since H.C.F. is 1: 81 × 50 = 4050.
  • Therefore, the product of H.C.F. and L.C.M. is 1 × 4050 = 4050.

The correct answer is 4050.
Math Olympiad Test: Real Numbers- 2 - Question 4

If H.C.F. of 306 and 657 is 9, then what is the L.C.M. of 306 and 657?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 4

L.C.M. × H.C.F. = 306 × 657
⇒ L.C.M. × 9 = 306 × 657
⇒ 

Math Olympiad Test: Real Numbers- 2 - Question 5

The product of H.C.F. and L.C.M. of the smallest prime number and smallest composite number is

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 5

Smallest prime number = 2
Smallest composite number = 4
∴ Required product = 2 × 4 = 8

Math Olympiad Test: Real Numbers- 2 - Question 6

If the H.C.F. of 210 and 55 is expressible in the form 210 × 5 + 55y then what is the value of y?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 6

H.C.F. of 210 and 55 = 5
210 × 5 + 55y = 5
⇒ 55y = 5 - 1050
⇒ 55y = -1045
⇒ y = 1045/55
= -19

Math Olympiad Test: Real Numbers- 2 - Question 7

Which of the following number has terminating decimal expansion?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 7

Math Olympiad Test: Real Numbers- 2 - Question 8

The decimal expansion of the rational numberwill terminate after

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 8

Math Olympiad Test: Real Numbers- 2 - Question 9

In the prime factorization of 13915 what is difference between largest factor and smallest factor?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 9

13915 = 5 × 11 × 11 × 23
Required difference = 23 - 5 = 18

Math Olympiad Test: Real Numbers- 2 - Question 10

Find the least number that is divisible by all the numbers between 1 and 10 both inclusive?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 10

The LCM of numbers from 1 to 10 need to be found out to find the least number We know that 2, 3, 5 and 7 are prime numbers
Prime factorization of
4 = 2 × 2
6 = 2 × 3
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
So we get required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520
Therefore, the least number that is divisible by all the numbers between 1 and 10 is 2520.

Math Olympiad Test: Real Numbers- 2 - Question 11

Find the smallest number by which √27 should be multiplied so as to get a rational number.

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 11

√27 x √3 = √81 = 9

Math Olympiad Test: Real Numbers- 2 - Question 12

What is the difference of exponents of prime factors in prime factorization of 1225?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 12

Here 1225 = 52 × 72
∴ Required difference = 2 - 2 = 0

Math Olympiad Test: Real Numbers- 2 - Question 13

What is the H.C.F of 95 and 152?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 13

95 = 5 × 19
and 152 = 23 × 19
H.C.F = 19

Math Olympiad Test: Real Numbers- 2 - Question 14

What is the sum of exponents of the prime factors in the prime factorization of 576?

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 14

576 = 26 × 34
6 + 2 = 8

Math Olympiad Test: Real Numbers- 2 - Question 15

Find L.C.M. of 42 and 63.

Detailed Solution for Math Olympiad Test: Real Numbers- 2 - Question 15

324 = 22 × 32

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