LCM and HCF
 The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are possible ?

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Given Here , HCF of two numbers = 4.
Hence, the numbers can be given by 4p and 4q , where p and q are coprime. Then,
4p + 4q = 36 ⇒ 4 (p + q) = 36
⇒ p + q = 9Correct Option: C
Given Here , HCF of two numbers = 4.
Hence, the numbers can be given by 4p and 4q , where p and q are coprime. Then,
4p + 4q = 36 ⇒ 4 (p + q) = 36
⇒ p + q = 9
Possible pairs satisfying this condition are : (1 , 8), (4 , 5), (2 , 7)
Hence , number of possible pairs is 3.
 Three numbers which are coprime to one another are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is :

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Let the numbers be p , q and r which are prime to one another.
Now, pq = 551
qr = 1073
∴ q = HCF of 551 and 1073
∴ q = 29∴ p = 551 = 19 29 and r = 1073 = 37 29
Correct Option: C
Let the numbers be p , q and r which are prime to one another.
Now, pq = 551
qr = 1073
∴ q = HCF of 551 and 1073
∴ q = 29∴ p = 551 = 19 29 and r = 1073 = 37 29
∴ Sum = 19 + 29 + 37 = 85
Hence , Sum of the three numbers is 85.
 L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers ?

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Let the numbers be 10p and 10q , where p and q are prime to each other.
∴ LCM = 10pq
⇒ 10pq = 120
⇒ pq = 12
Possible pairs = (3 , 4) or (1 , 12)
Numbers are : 10p = 10 × 3 = 30 and 10q = 10 × 4 = 40 or 10p = 10 × 1 = 10 and 10q = 10 × 12 = 120Correct Option: D
Let the numbers be 10p and 10q , where p and q are prime to each other.
∴ LCM = 10pq
⇒ 10pq = 120
⇒ pq = 12
Possible pairs = (3 , 4) or (1 , 12)
Numbers are : 10p = 10 × 3 = 30 and 10q = 10 × 4 = 40 or 10p = 10 × 1 = 10 and 10q = 10 × 12 = 120
∴ Sum of the numbers = 30 + 40 = 70
 The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is

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Let the numbers be 3p and 3q.
∴ 3p + 3q = 36
⇒ p + q = 12 ... (i)
and 3pq = 105 ... (ii)
Dividing equation (i) by (ii), we havep + q = 12 3pq 3pq 105
Correct Option: C
Let the numbers be 3p and 3q.
∴ 3p + 3q = 36
⇒ p + q = 12 ... (i)
and 3pq = 105 ... (ii)
Dividing equation (i) by (ii), we havep + q = 12 3pq 3pq 105 ⇒ 1 + 1 = 4 3q 3p 35
 Sum of two numbers is 384. H.C.F. of the numbers is 48. The difference of the numbers is

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Let the numbers be 48p and 48q , where p and q are coprimes.
∴ 48p + 48q = 384
⇒ 48 (p + q) = 384⇒ p + q = 384 = 8 ........... (i) 48
Possible and acceptable pairs of p and q satisfying this condition are : (1 , 7) and (3 , 5).
Correct Option: C
Let the numbers be 48p and 48q , where p and q are coprimes.
∴ 48p + 48q = 384
⇒ 48 (p + q) = 384⇒ p + q = 384 = 8 ........... (i) 48
Possible and acceptable pairs of p and q satisfying this condition are : (1 , 7) and (3 , 5).
∴ Numbers are : 48 × 1 = 48 and 48 × 7 = 336
and 48 × 3 = 144 and 48 × 5 = 240
∴ Required difference = 336 – 48 = 288