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Age of Dave = D
Age of Triplet = T

D = T+36
D+3T = D+15
3T = 15
T = 5

Therefore D = 5+36
D = 41

Answer D
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Bunuel
Dave is 36 years older than his triplet daughters. If the sum of their four ages is 15 greater than Dave's age, how old is Dave?

A. 38
B. 39
C. 40
D. 41
E. 42

Solution


    • Let us assume that Dave’s age is \(x\) years.
      o This means, the age of each triplet is \(x – 36\) years.
    • The sum of the ages of all four \(= x + 3*(x -36) = x + 15 ⟹3*(x – 36) = 15 ⟹ x - 36= 5 ⟹ x = 41\)
      o Therefore, Dave is 41 years old.
Thus, the correct answer is Option D.
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Bunuel
Dave is 36 years older than his triplet daughters. If the sum of their four ages is 15 greater than Dave's age, how old is Dave?

A. 38
B. 39
C. 40
D. 41
E. 42

Given: Dave is 36 years older than his triplet daughters.
Asked: If the sum of their four ages is 15 greater than Dave's age, how old is Dave?

Let age of Dave be x and age of his triplet daughters be y

Dave is 36 years older than his triplet daughters.
x - y = 36
x = y + 36

The sum of their four ages is 15 greater than Dave's age
x + 3y = x + 15
3y = 15
y = 5

x = y + 36 = 5 + 36 = 41

IMO D
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