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Bunuel
A widow received 1/3 of her husband's estate and each of her three sons received 1/3 of the balance. If the widow and one of her sons received a total of $60,000 from the estate, what was the value of the estate?

A. $90000
B. $108000
C. $180000
D. $240000
E. $300000
Let x = the value of the estate

The widow's portion: \(\frac{1}{3}x\)

Portion of the estate that remains: \(\frac{2}{3}x\)

Remainder is divided among three sons. Each son's portion: \(\frac{(\frac{2}{3}x)}{3}=\frac{2}{9}x\)

Widow + one son = \($60,000\)

\((\frac{1}{3}x+\frac{2}{9}x)=\frac{5}{9x}=$60,000\)

\(x=($60,000*\frac{9}{5})=$108,000\)

Answer B
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Let the total value be x
widow received =x/3
each son received = 2x/9
widow+1 son= x/3+2x/3= 5x/9= 60000
Hence x=108000
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Bunuel
A widow received 1/3 of her husband's estate and each of her three sons received 1/3 of the balance. If the widow and one of her sons received a total of $60,000 from the estate, what was the value of the estate?

A. $90000
B. $108000
C. $180000
D. $240000
E. $300000

Since the widow received 1/3, each son received (2/3)/3 = 2/9 of the total. If we let n = the total value of the estate, then we have:

(1/3)n + (2/9)n = 60,000

Multiplying the equation by 9, we have:

3n + 2n = 540,000

5n = 540,000

n = 108,000

Answer: B
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