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The sum of the LCM and HCF of two numbers is 176 and their difference is 160. Difference between the numbers is 32. Determine the sum of the  numbers.

Answers

Answered by ItzMiracle
33

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Given :

The sum of the LCM and HCF of two numbers is 176 and their difference is 160. Difference between the numbers is 32.

To find :

Determine the sum of the  numbers.

Solution :

LCM + HCF = 176 (1)

LCM - HCF = 160 (2)

Now adding both equations :

⇒ LCM + HCF + LCM - HCF = 176 + 160

⇒ 2LCM = 336

⇒ LCM = 336/2

⇒ LCM = 168

Now putting this value in (1) :

⇒ 168 + HCF = 176

⇒ HCF = 176 - 168

⇒ HCF = 8

Now we know,

HCF × LCM = Product of 2 numbers.

Now, difference b/w numbers = 32

Let the numbers be 'a' and 'b'     [Where, a > b]

⇒ a - b = 32

⇒ a = b + 32

Now,

⇒ HCF × LCM = ab

⇒ 168 × 8 = (b + 32)b

⇒ b² + 32b = 1344

⇒ b² + 32b - 1344 = 0

⇒ b² + 56b - 24b - 1344 = 0

⇒ b(b + 56) - 24(b + 56) = 0

⇒ (b - 24)(b + 56) = 0

⇒ b = 24   or,   b = - 56

∵ We will neglect negative values here.

∴ b = 24

Now putting value,

⇒ a = 24 + 32

⇒ a = 56

∴ Sum of the numbers = a + b

                                      = 56 + 24

                                      = 80

∴ Sum of the numbers = 80        

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Answered by Anonymous
4

Answer:

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Now,

⇒ HCF × LCM = ab

⇒ 168 × 8 = (b + 32)b

⇒ b² + 32b = 1344

⇒ b² + 32b - 1344 = 0

⇒ b² + 56b - 24b - 1344 = 0

⇒ b(b + 56) - 24(b + 56) = 0

⇒ (b - 24)(b + 56) = 0

⇒ b = 24   or,   b = - 56

∵ We will neglect negative values here.

∴ b = 24

Now putting value,

⇒ a = 24 + 32

⇒ a = 56

∴ Sum of the numbers = a + b

                                      = 56 + 24

                                      = 80

∴ Sum of the numbers = 80        

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