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Class 5 Exam  >  Class 5 Tests  >  Olympiad Test: LCM And HCF - 2 - Class 5 MCQ

Olympiad Test: LCM And HCF - 2 - Class 5 MCQ


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20 Questions MCQ Test - Olympiad Test: LCM And HCF - 2

Olympiad Test: LCM And HCF - 2 for Class 5 2024 is part of Class 5 preparation. The Olympiad Test: LCM And HCF - 2 questions and answers have been prepared according to the Class 5 exam syllabus.The Olympiad Test: LCM And HCF - 2 MCQs are made for Class 5 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: LCM And HCF - 2 below.
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Olympiad Test: LCM And HCF - 2 - Question 1
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What is the least common multiple of 16, 24 and 30?

Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 1

The multiples of 16 are: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 
208, 224, 240, 256,...
The multiples of 24 are: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, ...
The multiples of 30 are: 30, 60, 90, 120, 150, 180, 210, 240, 270, ... 
So the least common multiple of 16, 24 and 30 is 240.

Olympiad Test: LCM And HCF - 2 - Question 2
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What is the least common multiple of 6, 9, 10 and 15?

Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 2

The multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 
90, 96, ...
The multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, ...
The multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ... 
The multiples of 15 are: 15, 30, 45, 60, 75, 90, 105, ... 
So the least common multiple of 6, 9, 10 and 15 is 90.

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Olympiad Test: LCM And HCF - 2 - Question 3
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What is the least common multiple of 8, 9, 12 and 16?

Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 3

The multiples of: 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 
144, 152, ...
9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, 
...
12 are: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, ... 
16 are: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, ... 
So the least common multiple of 8, 9, 12 and 16 is 144.

Olympiad Test: LCM And HCF - 2 - Question 4
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What is the greatest common factor of 6 and 15?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 4

Correct Answer is B.

Olympiad Test: LCM And HCF - 2 - Question 5
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The greatest number of four digits which is divisible by 15, 25, 40 and 75 is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 5

Greatest number of 4-digits is 9999.
and L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number=(9999 –399) = 9600.

Olympiad Test: LCM And HCF - 2 - Question 6
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The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 6

 

Olympiad Test: LCM And HCF - 2 - Question 7
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What is the greatest common factor of 45 and 105?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 7

Correct Answer is A

Olympiad Test: LCM And HCF - 2 - Question 8
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What is the greatest common factor of 27 and 126?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 8

Correct Answer is B

Olympiad Test: LCM And HCF - 2 - Question 9
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What is the greatest common factor of 81 and 180?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 9

Factors of 81 are 1, 3, 9, 27, 81
Factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 18, 20, 30, 36, 45, 60, 90, 180
Common Factors are : 1, 3, 9
Greatest Common Factor is 9.

Olympiad Test: LCM And HCF - 2 - Question 10
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Three numbers are in the ratio of 3: 4: 5 and their L.C.M. is 2400. Their H.C.F. is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 10

Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or × = 40.
The numbers are (3 × 40), (4 × 40) and (5 × 40).
Hence, required H.C.F. = 40.

Olympiad Test: LCM And HCF - 2 - Question 11
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What is the greatest common factor of 60 and 225?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 11

Correct Answer is B

Olympiad Test: LCM And HCF - 2 - Question 12
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What is the greatest common factor of 126 and 196?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 12

Correct Answer is D

Olympiad Test: LCM And HCF - 2 - Question 13
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What is the greatest common factor of 180, 225 and 270?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 13

Correct Answer is C

Olympiad Test: LCM And HCF - 2 - Question 14
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What is the greatest common factor of 72 and 252?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 14

Correct Answer is C
GCF of 72, 252 is 36.

Olympiad Test: LCM And HCF - 2 - Question 15
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What is the greatest common factor of 196 and 462?
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 15

Correct Answer is B

Olympiad Test: LCM And HCF - 2 - Question 16
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The G.C.D. of 1.08, 0.36 and 0.9 is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 16

Given numbers are 1.08, 0.36 and 0.90.
H.C.F. of 108, 36 and 90 is 18 so H.C.F.
of given numbers = 0.18.

Olympiad Test: LCM And HCF - 2 - Question 17
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The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 17

Let the numbers 13a and 13b. Then, 13a × 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4).
Clearly, there are 2 such pairs.

Olympiad Test: LCM And HCF - 2 - Question 18
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The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 18

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 × 4) + 4 = 364.

Olympiad Test: LCM And HCF - 2 - Question 19
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The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 19

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.

Olympiad Test: LCM And HCF - 2 - Question 20
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The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is :
Detailed Solution for Olympiad Test: LCM And HCF - 2 - Question 20

L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is
divisible by 9 is k = 2.
Required number = (840 × 2 + 3) = 1683.

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