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Question
The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers
Solution
Let the two numbers be a and b.
Their A.M. = `("a" + "b")/2`
G.M. = `sqrt("ab")` and H.M. = `(2"ab")/("a" + "b")`
We are given A.M. = G.M. + 10 = H.M. + 16
(i.e) `("a" + "b")/2 = sqrt("ab") + 10` .....(1)
and `("a" + "b")/2 = (2"ab")/("a" + "b") + 16` ......(2)
From (2) `("a" + "b")/2 - (2"ab")/("a" + "b")` = 16
⇒ (a + b)2 – 4ab = 16(2)(a + b)
(i.e) (a – b)2 = 32(a + b) .....(3)
(1) ⇒ `("a" + "b")/2 = sqrt('ab") + 10`
⇒ a + b = `2sqrt("ab") + 20`
⇒ a + b – 20 `2sqrt("ab")`
So, (a + b – 20)2 = 4ab
(i.e.) (a + b)2 + 400 – 40(a + b) = 4ab
(a + b)2 – 4ab = 40(a + b) – 400
From (3) (a + b)2 – 4ab = 32(a + b)
⇒ 32(a + b) = 40(a + b) – 400
(÷ by 8) 4(a + b) = 5(a + b) – 50
4a + 4b = 5a + 5b – 50
a + b = 50
a = 50 – b
Substituting a = 50 – b in (3) we get
(50 – b – b)2 = 32(50)
(50 – 2b)2 = 32 × 50
[2(25 – b)]2 = 32 × 50
(i.e.) 4(25 – b)2 = 32 × 50
⇒ (25 – b)2 = `(32 xx 50)/4`
= 8 × 50 = 400
= 202
⇒ 25 – b = `+- 20`
25 – b = 20
⇒ b = 25 – 20 = 5
25 – b = – 20
b = 25 + 20 = 45
When b = 5, a = 50 – 5 = 45
When b = 45, a = 50 – 45 = 5
So the two numbers are 5 and 45.
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