Question

Q 25 The sum of the LCM and HCF of two numbers is 176 and their difference is 160. If the difference between the numbers is 32, determine the sum of the
numbers.
Ops: A.
O 60
B.
80
C.
72
D.
076​

Answer 1

Answer:

Since the HCF of the two numbers is 12 the

numbers are 12a and 12b where a and b are

prime to each other

Given 12a−12b=12 ⇒ a−b=1

⇒ a and b are consecutive numbers that are prime

to each other. By inspection we can see that

84, 96 i.e. 12 x 7 and 12 x 8 are the two

numbers satisfying this condition (8−7=1)

Answer 2

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         [Option B]