Question

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. ​

Answer 1

Answer:

So we have the HCF as 12 and LCM as 2184.

Give that difference of the number is 12.

Let one of the number is n, other number will be n+12.

12 is the common factor, so we can write n as 12m.

So the two numbers are 12m and 12(m+1).

LCM = product of the 2 numbers/ HCF

ie 2184 = 12m x 12(m+1)/12

2184 = 12m^2 + 12m

ie m^2 + m - 182

Solving we get m as 13.

So the numbers are 12x(13) and 12x(13+1)

ie 156 and 168.

Sum of numbers = 156 + 168 = 324

Answer 2

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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