Question

If HCF and LCM of two positive numbers (a,b) is 2 and 24 respectively and the sum of the two numbers is 14, find the positive numbers a and b.​

Answer 1

Answer:

The numbers are 8 and 6.

Step-by-step explanation:

The two numbers = a and b

HCF and LCM of a and b = 2 and 24

Sum of the numbers = 14

According to the Question,

$\longrightarrow$ a + b = 14

$\longrightarrow$ a = 14 - b ----- (Equation I)

We know that,

★ HCF × LCM = Product of numbers

$\longrightarrow$ 2 × 24 = a × b

$\longrightarrow$ 48 = (14 - b) × b

$\longrightarrow$ 48 = -b² + 14b

$\longrightarrow$ b² - 14b + 48 = 0

$\longrightarrow$ b² - 6b - 8b + 48 = 0

$\longrightarrow$ b(b - 6) - 8(b - 6) = 0

$\longrightarrow$ (b - 6)(b - 8) = 0

$\longrightarrow$ b = 6 or b = 8

Taking b as 6, substitute the value of b in equation I.

$\longrightarrow$ a = 14 - 6

$\longrightarrow$ a = 8

The numbers are :

• a = 8
• b = 6

Therefore, the numbers are 8 and 6.

Answer 2

Answer:

HCF and LCM of a and b = 2 and 24

Sum of the numbers = 14

• According to the Question,

$\sf \dashrightarrow \: a + b = 14$

$\sf \dashrightarrow a = 14 - b \: \: \: ----- (Equation I)$

• Formula which we used
• HCF × LCM = Product of numbers

$\sf \: \dashrightarrow 2 × 24 = a × b$

$\sf \dashrightarrow 48 = (14 - b) × b$

$\sf \: \dashrightarrow 48 = -b² + 14b$

$\sf \: \dashrightarrow b² - 14b + 48 = 0$

$\sf \: \dashrightarrow b² - 6b - 8b + 48 = 0$

$\sf \dashrightarrow \: b(b - 6) - 8(b - 6) = 0$

$\sf \dashrightarrow \: (b - 6)(b - 8) = 0$

$\sf \dashrightarrow \: b = 6 \: \: or \: \: b = 8$

Therefore, the numbers are 6 and 8.