Question

find two numbers whose AM exceed GM by 7and their HM by 63/5​

Answer 1

If AM exceeds GM by 7 and their HM by 63/5 then the two numbers are 56 and 14.

Step-by-step explanation:

Let the two numbers be "x" and "y".

Step 1:

It is given that,

A.M. = G.M. + 7  

⇒ G.M. = A.M. – 7 …… (i)

And  

A.M. = H.M. + 63/5 = H.M. + 12.6

⇒ H.M. = A.M. – 12.6 ….. (ii)  

We have the formula,

G.M.² = A.M. * H.M.

Substituting from (i) & (ii), we get

[A.M. – 7]² = A.M. * [A.M. – 12.6]

⇒ A.M.² – 14 A.M. + 49 = A.M.² - 12.6A.M.

⇒ 1.4A.M. = 49

⇒ A.M. = 49/1.4  

⇒ A.M. = 35 …… (iii)

∴ G.M. = 35 – 7 = 28 ….. (iv)

And, H.M. = 35 – 12.6 = 22.4

Step 2:

Also, we know

Arithmetic mean, A.M. of x and y = $\frac{x+y}{2}$ ….. (v)  

Geometric mean, G.M of x and y = $\sqrt{xy}$ ….. (vi)

So, From (iii) & (v), we get

$\frac{x+y}{2}$  = 35

⇒ x + y = 70 …. (vii)

And,

From (iv) & (vi), we get

$\sqrt{xy}$ = 28

⇒ xy = 28² …. [squaring both sides]

⇒ xy = 784  

⇒ y = 784/x …. (viii)

Step 3:

Substituting (viii) in (vii), we get

x + [784/x] = 70

⇒ x² + 784 = 70x

⇒ x² – 70x + 784 = 0

⇒ x(x-56) – 14(x-56) = 0

⇒ (x-56)(x-14) = 0

⇒ x = 56 or 14

∴ if x = 56 then y = 784/56 = 14 and if x = 14 then y = 784/14 = 56

Thus, the two numbers are 56 and 14.

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