Question

Dinalal divides his property among his four sons after donating rs.20,000 and 10% of his remaining property. The amounts received by the last three sons are in arithmetic progression and the amount received by the fourth son is equal to the total amount donated. The first son receives as his share rs.20,000 more than the share of the second son. The last son received rs.1 lakh less than the eldest son. Find the share of the third son

Answer 1

Answer:

The share of the third son is Rs. 1,20,000.

Step-by-step explanation:

Let the total property of Dinalal be “Rs. x”.

Amount donated by Dinalal is [20,000 + (1/10)(x−20,000)]

According to the question, as the amount received by the second, third & forth son are in A.P. Therefore, let the amount received by 2nd, 3rd & 4th son be a+d, a, a-d.

Amount received by 1st son = a+d+20,000

Amount received by the last son is  

(a+d+20,000) – (a-d) = 1,00,000

Or, a+d+20000-a+d=1,00,000

Or, 2 d = 1,00,000 – 20,000

Or, d = 80,000 / 2 = 40,000

∴ the amount received by all four sons are a+60,000, a+40,000, a and a-40,000.

It is also given that the forth son receives an amount equal to the total amount donated.

∴ [20,000 + (1/10)(x−20,000)] = (a - 40,000)  

Or, 20,000 + (x/10) - 2,000 = a-40,000

Or, 18000 + (x/10) = a-40000

Or, 58000 + (x/10) = a

Or, 580000 + x = 10 a …….. (i)

Remaining property of Dinalal = (90/100)(x-20,000)

(Sum of amount received by all three sons) = (Remaining property of Dinalal)

∴(90/100)(x-20,000) = (a + 60,000 + a + 40,000 + a + a-40,000)

Or, (9x/10 - 18000) = 4 a + 60,000

Or, (9x/10) = 4 a + 78000

Or, 9 x = 40 a + 780000

Or, 9 x – 780000 = 40 a …… (ii)

Subtracting equation (i) from (ii), we get

x = Rs. 620000

putting the value of x in eq. (i)

580000 + x = 10 a

Or, 580000 + 620000 = 10 a

Or, 10 a = 1200000

Or, a = 1,20,000

∴ The amount received by 3rd son = a = Rs. 1,20,000