Question

Three numbers are such that the second is as much lesser than the third as the first is lesser than the second. if the product of the two smaller numbers is 85 and the product of two larger numbers is 115 find the middle number

Answer 1

Given :

The three numbers are such that

The second is as much lesser than the third as the first is lesser than the second

The products of two smaller numbers = 85

The product of two larger numbers = 115

To Find :

The middle number

Solution :

Let The smaller number = x - d

     The middle number = x

     The larger number = x + d

According to question

Statement I

The products of two smaller numbers

i.e   ( x ) × ( x - d ) = 85

or,    x² - x d = 85         ...........1

Statement II

The products of two larger numbers

i.e     ( x ) × ( x + d ) = 115

Or,     x² + x d = 115         ...........2

Solving eq 1 and 2

(   x² + x d ) + (   x² - x d ) = 115 + 85

Or,  (   x² +  x² ) + ( x d - x d ) = 200

Or,    2 x² = 200

Or,       x² = $\dfrac{200}{2}$

i.e        x² = 100

∴        x = $\sqrt{100}$

Or,     x = $\pm$ 10

So, The middle number  = x = $\pm$ 10

Now,

When x  = 10

Put the value of x in eq 2

So,   (10)² + 10 × d = 115

or,               10 d = 115 - 100

or,                    d = 1.5

So, The smaller number = 10 - 1.5 = 8.5

     The middle number = x = 10

     The larger number = x + d = 10 + 1.5 = 11.5

And

When x  = - 10

Put the value of x in eq 2

So,   ( - 10)² + ( -  10 ) × d = 115

or,              - 10 d = 115 - 100

or,                    d = - 1.5

So, The larger number = 10 + 1.5 = 11.5

     The middle number = x = 10

     The smaller number = x + d = 10 - 1.5 = 8.5

Hence, The middle number for the given statements is  $\pm$ 10 Answer