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3 years ago, Henry's age was six years more than twice the

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Asif123
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3 years ago, Henry's age was six years more than twice the [#permalink] New post   23 Feb 2020, 06:37
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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
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A
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Re: 3 years ago, Henry's age was six years more than twice the [#permalink] New post   28 Feb 2020, 10:16
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Asif123 wrote:
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


3 years ago, Henry's age was six years more than twice the age of Mary.
Let M = Mary's age THREE YEARS AGO
This means 2M + 6 = Henry's age THREE YEARS AGO

To find their PRESENT ages, we add three years to each age from THREE YEARS AGO
So M + 3 = Mary's PRESENT age
And 2M + 6 + 3 = Henry's PRESENT age

The average of their ages from that time and the present is 30 years
This means we need to add their ages from three years ago and their present ages

We get: \(\frac{M+(2M+6)+(M+3)+(2M+6+3)}{4 = 30}\)

Simplify: \(\frac{6M+18}{4 = 30}\)

Multiply both sides of the equation by 4 to get: \(6M+18=120\)

Subtract 18 from both sides: \(6M=102\)

Solve: \(M=17\)

So Mary was 17 THREE YEARS AGO

What would Mary’s age in years be in 5 years?
From our above work we see that Mary's age THREE YEARS AGO was 17
This means her PRESENT age is 20
So, in five years, her age will be 25

Answer: A

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Re: 3 years ago, Henry's age was six years more than twice the [#permalink] New post   01 Mar 2020, 02:25
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3 years ago, Henry's age was six years more than twice the age of Mary.
Let M = Mary's age THREE YEARS AGO
This means 2M + 6 = Henry's age THREE YEARS AGO

To find their PRESENT ages, we add three years to each age from THREE YEARS AGO
So M + 3 = Mary's PRESENT age
And 2M + 6 + 3 = Henry's PRESENT age

The average of their ages from that time and the present is 30 years
This means we need to add their ages from three years ago and their present ages

We get: M+(2M+6)+(M+3)+(2M+6+3)4=30M+(2M+6)+(M+3)+(2M+6+3)4=30

Simplify: 6M+184=306M+184=30

Multiply both sides of the equation by 4 to get: 6M+18=1206M+18=120

Subtract 18 from both sides: 6M=1026M=102

Solve: M=17M=17

So Mary was 17 THREE YEARS AGO

What would Mary’s age in years be in 5 years?
From our above work we see that Mary's age THREE YEARS AGO was 17
This means her PRESENT age is 20
So, in five years, her age will be 25

Answer: A
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3 years ago, Henry's age was six years more than twice the [#permalink] New post   03 Mar 2021, 04:29
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Asif123 wrote:
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


Given: \((H - 3) = 2(M - 3) + 6\) ........ (1)

\(\frac{{(H - 3) + (M - 3) + H + M}}{4} = 30\) ........ (2)
\((H - 3) + (M - 3) + H + M = 120\)
\(2H + 2M = 126\)
\(H + M = 63\)

Put \(H = 63 - M\)\(\) in (1);
\(63 - M - 3 = 2M - 6 + 6\)
\(3M = 60\)
\(M = 20\)

So, \(M + 5 = 25\)

Hence, option A
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3 years ago, Henry's age was six years more than twice the [#permalink]
03 Mar 2021, 04:29
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