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17 Sep 2017, 03:24
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Difficulty:
45%
(medium)
Question Stats:
78% (02:43) correct
22%
(02:51)
wrong
based on 85
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Hunter was one fifth as old as Erica, x years ago. In x years, Erica will be twice as old as Hunter. What is the ratio of Hunter's current age to Erica's current age?
A. 11:15
B. 3:7
C. 5:13
D. 9:17
E. 7:23
A. 11:15
B. 3:7
C. 5:13
D. 9:17
E. 7:23
17 Sep 2017, 07:24
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Let age of Erica after x years be E and Hunter's age be H.
From questions stem, E = 2H
Also,\(H - 2x = \frac{1}{5}(E - 2x) => 5H - 10x = E - 2x\)
Substituting the value of E, we will get \(3H = 8x => \frac{H}{x}= \frac{8}{3}\)
If H = 8, E = 16
2x(6) years ago, Hunter's age is 2 and that is one-fifth the age of Erica!
Since, we are asked for their current ages, it must be (8-3:16-3) or 5:13(Option C)
From questions stem, E = 2H
Also,\(H - 2x = \frac{1}{5}(E - 2x) => 5H - 10x = E - 2x\)
Substituting the value of E, we will get \(3H = 8x => \frac{H}{x}= \frac{8}{3}\)
If H = 8, E = 16
2x(6) years ago, Hunter's age is 2 and that is one-fifth the age of Erica!
Since, we are asked for their current ages, it must be (8-3:16-3) or 5:13(Option C)
17 Sep 2017, 11:10
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Hunter was one fifth as old as Erica, x years ago.
=> \((H-x)= \frac{1}{5}(E-x)\)
=> \(5H-5x= E-x\)
=> \(5H =E+4x\)
In x years, Erica will be twice as old as Hunter.
=> \(2(H+x)= E+x\)
=> \(2H+2x=E+x\)
=> \(2H=E-x\)
What is the ratio of Hunter's current age to Erica's current age: \(\frac{H}{E}\)????
=>\(5H =E+4x\)
=> \(x= \frac{5H-E}{4}\)
as \(2H=E-x\) => \(2H=E-\frac{5H-E}{4}\) => \(13H =5E\)
\(\frac{H}{E} = \frac{5}{13}\)
Answer: C
=> \((H-x)= \frac{1}{5}(E-x)\)
=> \(5H-5x= E-x\)
=> \(5H =E+4x\)
In x years, Erica will be twice as old as Hunter.
=> \(2(H+x)= E+x\)
=> \(2H+2x=E+x\)
=> \(2H=E-x\)
What is the ratio of Hunter's current age to Erica's current age: \(\frac{H}{E}\)????
=>\(5H =E+4x\)
=> \(x= \frac{5H-E}{4}\)
as \(2H=E-x\) => \(2H=E-\frac{5H-E}{4}\) => \(13H =5E\)
\(\frac{H}{E} = \frac{5}{13}\)
Answer: C
