## LCM and HCF

#### LCM and HCF

1. If A and B are the H.C.F. and L.C.M. respectively of two algebraic expressions p and q, and A + B = p + q, then the value of A3 + B3 is

1. Let the numbers are p and q and HCF = A, LCM = B
Using Rule, we have
Product of numbers = HCF × LCM
pq = AB
⇒ p + q = A + B (given) ...(i)
(p – q)2 = (p + q)2 – 4pq
or, (p – q)2 = (A + B)2 – 4AB
or, (p – q)2 = (A – B)2
or, (p – q) = A – B ...(ii)
Using (i) and (ii), we get
p = A and q = B

##### Correct Option: D

Let the numbers are p and q and HCF = A, LCM = B
Using Rule, we have
Product of numbers = HCF × LCM
pq = AB
⇒ p + q = A + B (given) ...(i)
(p – q)2 = (p + q)2 – 4pq
or, (p – q)2 = (A + B)2 – 4AB
or, (p – q)2 = (A – B)2
or, (p – q) = A – B ...(ii)
Using (i) and (ii), we get
p = A and q = B
∴ A3 + B3
= p3 + q3

1. The LCM of two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the other number is

1. Let the HCF of two numbers be H . then
According to question ,
LCM = 44H
∴ 44H + H = 1125
⇒ 45H = 1125

 ∴ H = 1125 = 25 45

∴ LCM = 44 × 25 = 1100
As we know that ,
First number × Second number = LCM × HCF
⇒ 25 × Second number = 1100 × 25

##### Correct Option: A

Let the HCF of two numbers be H . then
According to question ,
LCM = 44H
∴ 44H + H = 1125
⇒ 45H = 1125

 ∴ H = 1125 = 25 45

∴ LCM = 44 × 25 = 1100
As we know that ,
First number × Second number = LCM × HCF
⇒ 25 × Second number = 1100 × 25
 ∴ Second number = 1100 × 25 = 1100 45

1. The L.C.M. of two numbers is 20 times their H.C.F. The sum of H.C.F. and L.C.M. is 2520. If one of the number is 480, the other number is :

1. Let the H.C.F. be H.
According to question ,
∴ L.C.M. = 20H
and H + 20H = 2520
⇒ 21 H = 2520

 ⇒ H = 2520 = 120 21

∴ L.C.M. = 20H = 20 × 120 = 2400
We can find the Second number with the help of given formula ,
First number × Second number = L.C.M. × H.C.F.
⇒ 480 × Second number = 2400 × 120

##### Correct Option: D

Let the H.C.F. be H.
According to question ,
∴ L.C.M. = 20H
and H + 20H = 2520
⇒ 21 H = 2520

 ⇒ H = 2520 = 120 21

∴ L.C.M. = 20H = 20 × 120 = 2400
We can find the Second number with the help of given formula ,
First number × Second number = L.C.M. × H.C.F.
⇒ 480 × Second number = 2400 × 120
 ⇒ Second number = 2400 × 120 = 600 480

1. The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :

1. Let the numbers be pH and qH where H is the HCF and qH > pH.
∴ LCM = pqH
∴ pqH = 2qH ⇒ p = 2
Again, pH – H = 4
⇒ 2H – H = 4 ⇒ H = 4

##### Correct Option: C

Let the numbers be x H and yH where H is the HCF and yH > x H.
∴ LCM = xy H
∴ xyH = 2yH ⇒ x = 2
Again, x H – H = 4
⇒ 2H – H = 4 ⇒ H = 4
∴ Smaller number = pH = 2 × 4 = 8

1. If the HCF and LCM of two consecutive (positive) even numbers be 2 and 84 respectively, then the sum of the numbers is

1. Let the numbers be 2p and 2q , where p and q are prime to each other.
∴ LCM = 2pq
⇒ 2pq = 84
⇒ pq = 42 = 6 × 7
Here p = 6 and q = 7

##### Correct Option: B

Let the numbers be 2p and 2q , where p and q are prime to each other.
∴ LCM = 2pq
⇒ 2pq = 84
⇒ pq = 42 = 6 × 7
Here p = 6 and q = 7
∴ Numbers are 2p = 2 × 6 = 12 and 2q = 2 × 7 = 14.
Hence , Sum of numbers = 12 + 14 = 26