## LCM and HCF

#### LCM and HCF

1. The LCM of two numbers is 12 times their HCF. The sum of the HCF and the LCM is 403. If one of the number is 93, then the other number is

1. Let the HCF of numbers = H
∴ Their LCM = 12H
According to the question,
12H +H = 403
⇒ 13H = 403

 ⇒ H = 403 = 31 13

⇒ LCM = 12 × 31 and HCF = 31
Now we know that , First number × second number = HCF × LCM
⇒ 93 × Second Number = 31 × 31 × 12

##### Correct Option: A

Let the HCF of numbers = H
∴ Their LCM = 12H
According to the question,
12H +H = 403
⇒ 13H = 403

 ⇒ H = 403 = 31 13

⇒ LCM = 12 × 31 and HCF = 31
Now we know that , First number × second number = HCF × LCM
⇒ 93 × Second Number = 31 × 31 × 12
 Second number = 31 × 31 × 12 = 124 93

1. The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

1. Given that , HCF of two numbers = 27
∴ Let the numbers be 27p and 27q where p and q are prime to each other.
According to the question ,
27p + 27q = 216
⇒ 27 (p + q) = 216

 ⇒ p + q = 216 = 8 27

∴ Possible pairs of p and q = (1 , 7) and (3 , 5)

##### Correct Option: B

Given that , HCF of two numbers = 27
∴ Let the numbers be 27p and 27q where p and q are prime to each other.
According to the question ,
27p + 27q = 216
⇒ 27 (p + q) = 216

 ⇒ p + q = 216 = 8 27

∴ Possible pairs of p and q = (1 , 7) and (3 , 5)
∴ Numbers = (27 , 189) and (81 , 135)
Hence , required answer is 2 .

1. The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is

1. Let the larger number be p.
We can find the required answer with the help of given formula ,
Product of two numbers = HCF × LCM
∴ Smaller number = p – 2
∴ First number × Second number = HCF × LCM
⇒ p (p – 2) = 24
⇒ p2 – 2p – 24 = 0
⇒ p2 – 6p + 4p – 24 = 0
⇒ p (p – 6) + 4 (p – 6) = 0
⇒ (p – 6) (p + 4) = 0
⇒ p = 6 because p ≠ – 4

##### Correct Option: C

Let the larger number be p.
We can find the required answer with the help of given formula ,
Product of two numbers = HCF × LCM
∴ Smaller number = p – 2
∴ First number × Second number = HCF × LCM
⇒ p (p – 2) = 24
⇒ p2 – 2p – 24 = 0
⇒ p2 – 6p + 4p – 24 = 0
⇒ p (p – 6) + 4 (p – 6) = 0
⇒ (p – 6) (p + 4) = 0
⇒ p = 6 because p ≠ – 4
Hence , the greater number is 6.

1. The H.C.F. of two numbers, each having three digits , is 17 and their L.C.M. is 714. The sum of the numbers will be :

1. Let the numbers be 17p and 17q where p and q are co-prime. LCM of 17p and 17q = 17pq
According to the question ,
17pq = 714

 ⇒ pq = 714 = 42 = 6 × 7 17

⇒ p = 6 and q = 7
or, p = 7 and q =6.

##### Correct Option: C

Let the numbers be 17p and 17q where p and q are co-prime. LCM of 17p and 17q = 17pq
According to the question ,
17pq = 714

 ⇒ pq = 714 = 42 = 6 × 7 17

⇒ p = 6 and q = 7
or, p = 7 and q =6.
∴ First number = 17p = 17 × 6 = 102
Second number = 17q = 17 × 7 = 119
∴ Sum of the numbers = 102 + 119 = 221

1. The sum of two numbers is 45. Their difference is 1/9 of their sum. Their L.C.M. is

1. Let the number be p and q.
According to the question,
∴ p + q = 45 ......... (i)

 Again, p – q = 1 (p + q) 9

 or p – q = 1 × 45 9

or p – q = 5 ..... (ii)
By (i) + (ii) we have,
p + q = 45
 p – q = 5 = 50 2p

##### Correct Option: C

Let the number be p and q.
According to the question,
∴ p + q = 45 ......... (i)

 Again, p – q = 1 (p + q) 9

 or p – q = 1 × 45 9

or p – q = 5 ..... (ii)
By (i) + (ii) we have,
p + q = 45
 p – q = 5 = 50 2p

or, p = 25
∴ q = 45 – 25 = 20.
Now, LCM of 25 and 20 = 100.