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Let the farmer give Rs x to the 17 years old son & The remaining Rs (2,81,800 - x) to his 19 years old son. Now, [x(1+12/100)]4 = (2,81,800 - x) (1+12/100)2 ⇒ [x(112/100)]2 = (2,81,800 - x) ⇒ [x(28/25)]2 = (2,81,800 - x) ⇒ x(784/625) = (2,81,800 - x) ⇒ (784/625+1) x = 2,81,800 ⇒ ((784 + 625)/625) x = 2,81,800 ⇒ x = (2,81,800× 625)/1409 = 1,25,000 ∴ x = Rs 1,25,000 For 17 years old son = Rs 1,25,000 For 19 years old son = Rs 1,56,800 Alternate shortcut method: They will get the sum in 2nd to 1st child in the ratio of = (1+R/100)^(difference between their age) = (1+12/100)(19 - 17) =(28/25)2 =784/625 So for 17 years old (1st child) , sum = 625/(784+625)×281800=625/1409×281800=125000 & for 19 years old (2nd child) , sum =784/(784+625)×281800=784/1409×281800=156800
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Statements:
B ≥ L = C; N ≤ L < O; Q ≥ N < R
Conclusions:
I. B > N
II. N = B