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3 years ago, Henry's age was six years more than twice the age of Mary
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26 Mar 2021, 02:30
26 Mar 2021, 02:30
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3 years ago, Henry's age was six years more than twice the age of Mary. The average of their ages from that time and the present is 30 years. What would Mary’s age in years be in 5 years' time?
A. 25
B. 24
C. 29
D. 30
E. 45
A. 25
B. 24
C. 29
D. 30
E. 45
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Joined: 08 Mar 2021
Posts: 222
Given Kudos: 0
Re: 3 years ago, Henry's age was six years more than twice the age of Mary
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26 Mar 2021, 07:06
26 Mar 2021, 07:06
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Let, current age of Henry = H and current age of Mary = M.
M + 5 =?
Using data:
3 years ago,
H - 3 = 6 + 2 * (M - 3)
=> H - 3 = 6 + 2M - 6
=> H - 3 = 2M
=> H = 2M + 3
The average of their ages three years ago and of present time is 30 years.
\(\frac{M + H + (M - 3) + (H - 3)}{4}\) = 30
=> 2M + 2H - 6 = 120
2M + 2 * (2M + 3) - 6 = 120 (using H = 2M + 3)
=> 2M + 4M + 6 - 6 = 120
=> 6M = 120
=> M = \(\frac{120}{6}\) = 20 years
M + 5 = 20 + 5 = 25 years.
Mary's age after 5 years will be 25 years.
So, correct answer is option A.
M + 5 =?
Using data:
3 years ago,
H - 3 = 6 + 2 * (M - 3)
=> H - 3 = 6 + 2M - 6
=> H - 3 = 2M
=> H = 2M + 3
The average of their ages three years ago and of present time is 30 years.
\(\frac{M + H + (M - 3) + (H - 3)}{4}\) = 30
=> 2M + 2H - 6 = 120
2M + 2 * (2M + 3) - 6 = 120 (using H = 2M + 3)
=> 2M + 4M + 6 - 6 = 120
=> 6M = 120
=> M = \(\frac{120}{6}\) = 20 years
M + 5 = 20 + 5 = 25 years.
Mary's age after 5 years will be 25 years.
So, correct answer is option A.
Re: 3 years ago, Henry's age was six years more than twice the age of Mary
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05 Aug 2024, 01:13
05 Aug 2024, 01:13
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