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16 Jun 2018, 09:23
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (03:36) correct
38%
(03:40)
wrong
based on 99
sessions
History
Date
Time
Result
Not Attempted Yet
Lou has three daughters: Wen, Mildred, and Tyla. Three years ago, when Lou was twice as old as Tyla, he was 30 years older than Mildred. Now, he is 47 years older than Wen. In 4 years, Wen will be half as old as Tyla. What is the sum of the current ages of Lou and his three daughters?
(A) 138
(B) 144
(C) 154
(D) 166
(E) 181
(A) 138
(B) 144
(C) 154
(D) 166
(E) 181
16 Jun 2018, 10:36
Kudos
Bookmarks
L - 3 = 2(t-3) -> L = 2T-3
L - 3 = M + 27
L = 47 + W
W + 4 = (T+4)/2 -> 2W + 4 = T
In 4th equation lets replace the value of w and t with L from 1st and 3rd equation.
We get: (L-47) + 4 = (L+3)/2
Solving for L we get 61. Sub this value in the other equations and we'd get
l = 61
t = 32
m = 31
w = 14
Sum of all these = 138.
OA-A.
L - 3 = M + 27
L = 47 + W
W + 4 = (T+4)/2 -> 2W + 4 = T
In 4th equation lets replace the value of w and t with L from 1st and 3rd equation.
We get: (L-47) + 4 = (L+3)/2
Solving for L we get 61. Sub this value in the other equations and we'd get
l = 61
t = 32
m = 31
w = 14
Sum of all these = 138.
OA-A.
13 Jan 2020, 14:03
Kudos
Bookmarks
Bunuel
We can create the equations, letting L, W, M, and T = the current ages of Lou, Wen, Mildren, and Tyla, respectively:
L - 3 = 2(T - 3)
L - 3 = 2T - 6
L = 2T - 3 → Eq. 1
and
L - 3 = 30 + M - 3
L = 30 + M
L - 30 = M → Eq. 2
and
L = 47 + W → Eq. 3
and
W + 4 = ½(T + 4)
2W + 8 = T + 4
2W + 4 = T → Eq. 4
Substituting 2W + 4 for T in Eq. 1, we have:
L = 2(2W + 4) - 3
L = 4W + 8 - 3
L = 4W + 5
Substituting 4W - 5 for L in Eq. 3, we have:
4W + 5 = 47 + W
3W = 42
W = 14
So T = 2W + 4 = 2(14) + 4 = 32
L = 4W + 5 = 4(14) + 5 = 61
M = L - 30 = 61 - 30 = 31
Thus, the sum of the current ages of the 4 people is 14 + 32 + 61 + 31 = 138.
Answer: A
