LCM and HCF
 The sum of a pair of positive integer is 336 and their H.C.F. is 21. The number of such possible pairs is

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Let the numbers be 21p and 21q where p and q are prime to each other.
∴ 21p + 21q = 336
⇒ 21 (p + q) = 336⇒ p + q = 336 = 16 21
Correct Option: C
Let the numbers be 21p and 21q where p and q are prime to each other.
∴ 21p + 21q = 336
⇒ 21 (p + q) = 336⇒ p + q = 336 = 16 21
∴ Possible pairs = (1 , 15), (5 , 11), (7 , 9), (3 , 13)
Hence , The number of such possible pairs is 4 .
 The sum of two numbers is 84 and their HCF is 12. Total number of such pairs of number is

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Given , HCF = 12
∴ Numbers = 12p and 12q , where p and q are prime to each other.
∴ 12p + 12q = 84
⇒ 12 (p + q) = 84⇒ p + q = 84 = 7 12
Correct Option: B
Given , HCF = 12
∴ Numbers = 12p and 12q , where p and q are prime to each other.
∴ 12p + 12q = 84
⇒ 12 (p + q) = 84⇒ p + q = 84 = 7 12
∴ Possible pairs of numbers satisfying this condition = (1 , 6), (2 , 5) and (3 , 4).
Hence three pairs are of required numbers.
 The sum of the H.C.F. and L.C.M of two numbers is 680 and the L.C.M. is 84 times the H.C.F. If one of the number is 56, the other is :

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Let HCF be H and LCM be L.
According to question ,
Then, L = 84H and L + H = 680
⇒ 84H + H = 680⇒ H = 680 = 8 85
∴ L = 680 – 8 = 672
Correct Option: D
Let HCF be H and LCM be L.
According to question ,
Then, L = 84H and L + H = 680
⇒ 84H + H = 680⇒ H = 680 = 8 85
∴ L = 680 – 8 = 672∴ Other number = 672 × 8 = 96 56
 Two numbers, both greater than 29, have HCF 29 and LCM 4147. The sum of the numbers is :

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Let the number be 29p and 29q respectively , where p and q are prime to each other.
∴ LCM of 29p and 29q = 29pq
Now, 29pq = 4147∴ pq = 4147 = 143 29
Thus pq = 11 × 13
Here , p = 11 , q = 13
Correct Option: B
Let the number be 29p and 29q respectively , where p and q are prime to each other.
∴ LCM of 29p and 29q = 29pq
Now, 29pq = 4147∴ pq = 4147 = 143 29
Thus pq = 11 × 13
Here , p = 11 , q = 13
∴ Numbers are 29 × 11 = 319 and 29 × 13 = 377
∴ Required sum = 377 + 319 = 696
 The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is :

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Suppose 1st number is p then, 2nd number = 100 – p
As we know that ,
∴ LCM × HCF = 1st number × 2nd number
⇒ 495 × 5 = p × (100 – p)
⇒ 495 × 5 = 100p – p^{2}
⇒ p^{2} – 55p – 45p – 2475 = 0
⇒ (p – 45) (p – 55) = 0
⇒ p = 45 or p = 55Correct Option: A
Suppose 1st number is p then, 2nd number = 100 – p
As we know that ,
∴ LCM × HCF = 1st number × 2nd number
⇒ 495 × 5 = p × (100 – p)
⇒ 495 × 5 = 100p – p^{2}
⇒ p^{2} – 55p – 45p – 2475 = 0
⇒ (p – 45) (p – 55) = 0
⇒ p = 45 or p = 55
Then, difference = 55 – 45 = 10