Problem Solving

Given that the sum of two numbers is ab and their difference is 1/7 of their sum. Find their HCF.

A)a+b

B)(a-b)/ab

C)(12/7)ab

D)ab

E)(6/7)ab

OA[spoiler]E)(6/7)ab[/spoiler]

None of the answer choices is correct.
I believe that the problem should read as follows:

Vikas Mishra wrote:Given that the sum of two numbers is ab and their difference is 1/7 of their sum. Find their HCF.

A)a+b

B)(a-b)/ab

C)(12/7)ab

D)ab

E)(1/7)ab

Since the difference of x and y is equal to 1/7 the sum of x and y, we get:
x-y = (1/7)(x+y)
7x-7y = x+y
6x=8y
3x=4y
x=(4/3)y.

Let y=3, implying that x=4.
In this case, ab = the sum of 4 and 3 = 7.
Let a=1 and b=7, with the result that ab=7.
The HCF (highest common factor) of 4 and 3 is 1. This is our target.

Now plug a=1 and b=7 into the answers to see which yields our target of 1.
Only E works:
(1/7)ab = (1/7) = (1/7)(7) = 1.

The correct answer is E.

If we test any case in which x = (4/3)y, the HCF will always be (1/7)(x+y).
If y=6 and x=8, then the HCF = (1/7)(8+6) = 2.
If y=9 and x=12, then the HCF = (1/7)(12+9) = 3.
If y=12 and x=16, then the HCF = (1/7)(16+12) = 4.
Since ab = the sum of x and y, the OA must be [spoiler](1/7)ab[/spoiler].

Here's what's happening algebraically.

If we know that

x - y = (x + y) / 7

then we know

7x - 7y = x + y

6x = 8y

3x = 4y

x = (4/3)y

The GCF of x and y is thus the GCF of (4/3)y and y. The GCF of any number is also a factor of the difference of the numbers, so it must be a factor of (1/3)y.

Since (1/3)y = (1/7) * ((4/3)y + y) = (1/7) * (x + y), we're left with (1/7) * sum, or (1/7) * ab.