LCM and HCF
- If a number is exactly divisible by 11 and 13, which of the following types the number must be?
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LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 x 13, then the same number must be exactly divisible by their LCM or by (11 x 13).
Correct Option: C
LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 x 13, then the same number must be exactly divisible by their LCM or by (11 x 13).
- The LCM of two numbers is 48. The numbers are in the ratio of 2 : 3. Find the sum of the numbers.
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Let numbers are 2N and 3N.
According to the question, 6N = 48
⇒ N = 8 ( ∵ LCM = 6N )
∴ Required sum = ( 2N + 3N ) = 5NCorrect Option: C
Let numbers are 2N and 3N.
According to the question, 6N = 48
⇒ N = 8 ( ∵ LCM = 6N )
∴ Required sum = ( 2N + 3N ) = 5N
= 5 x 8 = 40
- The Product of two whole numbers is 1500 and their HCF is 10, Find the LCM.
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According to the formula,
Product of two numbers = HCF x LCMCorrect Option: B
Given that, product of two numbers =1500
HCF = 10
According to the formula,
Product of two numbers = HCF x LCM
⇒ 1500 = 10 x LCM
∴ LCM = 1500/10 =150
- The least number which should be added to 2497, so that the sum is exactly divisible by 5, 6, 4 and 3, is
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LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added = 60 - 37Correct Option: C
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added = 60 - 37 = 23
- The least number which when divided by 12,16 and 18 leaves 5 as remainder in each case. Find the number.
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Given, a = 12, b =16 , c =18, k = 5
According to the formula,
Required number =(LCM of a, b and c) + k
= (LCM Of 12, 16, 18) + 5
LCM of 12, 16, 18 is
∴ LCM = 2 x 2 x 3 x 4 x 3 = 144
∴ Required number =144 + 5 = 149Correct Option: C
Given, a = 12, b =16 , c =18, k = 5
According to the formula,
Required number =(LCM of a, b and c) + k
= (LCM Of 12, 16, 18) + 5
LCM of 12, 16, 18 is
∴ LCM = 2 x 2 x 3 x 4 x 3 = 144
∴ Required number =144 + 5 = 149