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The respective ratio of the present ages of a mother and daughter is 7 : 1. Four years ago the respective ratio of their ages was 19:1. What will be the mother's age four years from now ?
- A42 years
- B38 years
- C46 years
- D36 years
Answer not set
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0
CCommunity
Solution
Let the present ages of mother and daughter be x and y respectively.
$\frac{y}{x} = \frac{1}{7}$ => x = 7y
$\frac{y-4}{x-4} = \frac{1}{19}$ => 19y - 76 = x - 4 => x = 19y - 72
=> 7y = 19y - 72 => y = 6 and x = 42
Age of mother 4 years from now = 42 + 4 = 46
Let the present ages of mother and daughter be x and y respectively.
$\frac{y}{x} = \frac{1}{7}$ => x = 7y
$\frac{y-4}{x-4} = \frac{1}{19}$ => 19y - 76 = x - 4 => x = 19y - 72
=> 7y = 19y - 72 => y = 6 and x = 42
Age of mother 4 years from now = 42 + 4 = 46
0
CCommunity
Solution
Let the present ages of mother and daughter be x and y respectively.
$\frac{y}{x} = \frac{1}{7}$ => x = 7y
$\frac{y-4}{x-4} = \frac{1}{19}$ => 19y - 76 = x - 4 => x = 19y - 72
=> 7y = 19y - 72 => y = 6 and x = 42
Age of mother 4 years from now = 42 + 4 = 46
Let the present ages of mother and daughter be x and y respectively.
$\frac{y}{x} = \frac{1}{7}$ => x = 7y
$\frac{y-4}{x-4} = \frac{1}{19}$ => 19y - 76 = x - 4 => x = 19y - 72
=> 7y = 19y - 72 => y = 6 and x = 42
Age of mother 4 years from now = 42 + 4 = 46