The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4√ 3 and their harmonic mean is 6, then what is the father's age?

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Accepted Answer

Given, F = 2E + 4Y ...(I)
√EY = 4√3
⇒ EY = 48 ...(II)
and 2EY/ E + Y = 6 ⇒ E + Y = 16 ...(III)
Now, (E -Y)² = (E + Y)² - 4EY
= (16)² - 4 x 48
= 256 - 192 = 64
∴ E - Y = 8 ...(iv)
From Eqs. (iii) and (iv), we get
E = 12 and Y = 4
From Eq. (i) F = 2 x 12 + 4 x 4 = 40 yr

The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4√ 3 and their harmonic mean is 6, then what is the father's age?

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