Question

Two numbers a,b are such that their gm is less than 40% less than am the ratio between the numbers

Answer 1

Given numbers are a, b

$\sf \: Arithmetic \: \: mean = (a + b) / 2 \\ \\ \sf \: Geometric \: \: mean = \sqrt{ab}$

According to the Question,

G. M is 40% less than AM.

We can say,

G. M + 40% of AM = AM

Further,

G. M = 60% of AM.

Now,

$\sqrt{ab} = 60 \% \: \times \frac{a + b}{2} \\ \\ \sqrt{ab} = \frac{6}{10} \times \frac{a + b}{2} \\ \\ \sqrt{ab} = \frac{3}{10} \times {a + b}$

Squaring on both sides,

$ab \: = ( { \frac{3}{10} })^{2} ( {a + b})^{2} \\ \\ 100ab = 9( {a}^{2} + {b}^{2} + 2ab) \\ \\ 9a ^{2} + 9 {b}^{2} - 82ab = 0$

Divide by ab,

$9( \frac{a}{b} ) + 9( \frac{b}{a} ) = 82$

Let a/b = x ( This is the required ratio)

$9x + \frac{9}{x} = 82 \\ \\ 9 {x}^{2} + 9 = 82x \\ \\ 9 {x}^{2} - 82x + 9 = 0 \\ \\( 9x - 1)(x - 9) = 0$

Now, x = 1/9 or x = 9

So, The ratio of numbers can be a : b = 1 : 9 if a < b And 9 : 1 if a > b

Answer 2

Given: a and b are the numbers given

To find: ratio between the numbers

Solution:

• We know that arithmetic mean is a + b/ 2 and geometric mean is √ab

• So in the question, it says G. M is 40% less than AM,

• So GM + 40% of AM = AM,

• Also, it is given that  G. M = 60% of AM.

            √ab = 60%  X a+b/2 = 60/100 x a+b/2

            √ab = 3/10 x (a+b)

• squaring both sides we get:

             ab = 9/100(a+b)^2

             100ab = 9 (a^2 + b^2 + 2ab)

             100ab = 9a^2 + 9b^2 + 18ab

             9a^2 + 9b^2 - 82ab=0

• Now lets divide the whole equation by ab, we get

             9(a/b) + 9 (b/a) = 82

• So lets consider a/b = z

             9z + 9/z = 82

             9z^2 + 9 = 82z

             9z^2-82z+ 9 =0

             (9z-1)(z-9)=0

             So z=1/9 or z= 9

Answer:

                 So the ratio is a:b= 1:9 or 9:1