Question

If AM and GM pf two numbers is 10 and 8 find the numbers

Answer 1

Step-by-step explanation:

Given:-

AM and GM pf two numbers is 10 and 8

To find:-

Find the numbers ?

Solution:-

Let the required numbers be X and Y

We know that

The AM of the two numbers X and Y =(X+Y)/2

According to the given problem

The AM of two numbers = 10

=> (X+Y)/2 = 10

=> X+Y = 10×2

=>X+Y = 20

X = 20-Y---------(1)

We know that

The GM of the two numbers X and Y = √(XY)

According to the given problem

The GM of two numbers = 8

=> √(XY) = 8

On squaring both sides then

=>[√(XY)]^2 = 8^2

=>XY = 64

=>(20-Y)Y = 64

(from (1))

=> 20Y-Y^2=64

=>20Y-Y^2-64=0

=>Y^2-20Y+64 = 0

=> Y^2-4Y-16Y+64 = 0

=> Y(Y-4)-16(Y-4)=0

=>(Y-4)(Y-16)=0

=>Y-4 = 0 or Y-16 = 0

=>Y = 4 or 16

If Y = 4 then

X=20-4 = 16

If Y=16 then

X= 20-4 = 16

Therefore, The numbers = 16 and 4

Answer:-

The required two numbers are 16 and 4

Used formulae:-

• The AM of the two numbers X and Y =(X+Y)/2
• The GM of the two numbers X and Y = √(XY)

Answer 2

Answer:

$\huge{\mathcal{\underline{\green{AnSwER}}}}$

⇒AM = $\frac{a + b}{2}$

⇒ GM = √ab

Given AM = 10, GM = 8.

⇒ $\frac{a + b}{2}$ = 10

⇒ a + b = 20

⇒ a = 20–b

⇒ $\sqrt{(20 - b)b}$

⇒ 20b – b²= 64

⇒ b² – 20b + 64 = 0

⇒ b²– 16b – 4b + 64 = 0

⇒ b(b – 16) – 4(b – 16) = 0

⇒ b = 4 or b = 16

⇒ If b = 4

then a = 16

⇒ If b = 16

then a = 4.

Hence, the numbers are 4 and 16