Bunuel
Tough and Tricky questions: Word Problems.
Roberto has three children: two girls and a boy. All were born on the same date in different years. The sum of the ages of the two girls today is smaller than the age of the boy today, but a year from now the sum of the ages of the girls will equal the age of the boy. Three years from today, the difference between the age of the boy and the combined ages of the girls will be
A. 1
B. 2
C. 3
D. –2
E. –1
Kudos for a correct solution. Approach I (Plugin's)Girl I ............... Girl II ................ Boy1 ....................... 1 ........................ 3 (Assume the current ages)
1 + 1 < 3 .......... (Satisfies the given condition)
1 Year later there ages are
2 ....................... 2 ......................... 4
2 + 2 = 4 ............ (Satisfies the given condition)
After 3 years there ages are
4 ....................... 4 ............................ 6
Difference = 6 - (4+4) = 6 - 8 = -2
Answer = D
Approach II (Arithmetic)Girl I ................. Girl II ............... Boy
a ........................ b ...................... c (Assume there current ages; a+b < c)
a+1 ................... b+1 ................... c+1 (Ages after 1 year)
Given that a+1 + b+1 = c+1
a+b = c - 1 ................... (1)a+3 .................... b+3 .................. c+3 (Ages 3 years later from today)
Difference = c+3 - (a+3+b+3)
= c + 3 - (c-1+6) ............
Substitute from (1) = 3 - 5 = -2
Answer = D