47. Calculating present, past, or future ages problems

47. Calculating present, past, or future ages problems

Calculating present, past, or future ages problems involve determining the ages of individuals at different points in time (present, past, or future) based on relationships such as age differences, ratios, or specific conditions. These problems often require setting up equations to represent the relationships and solving for unknown ages, while carefully accounting for time shifts. They are a subset of age-related problems, focusing on the temporal aspect of ages.

Below are 10 calculating present, past, or future ages problems with step-by-step solutions and detailed explanations. Each problem includes the problem statement, solution procedure, verification, and an explanation of the concept. The problems are diverse, covering scenarios with differences, ratios, and time-based conditions, and are distinct from previous age difference and ratio problems. They avoid unrelated contexts (e.g., alcohol or mixing solutions) and focus on family, educational, or social settings.

Problem 1

Five years ago, Anna was twice as old as Ben. Now, Anna is 25 years old. How old is Ben now?

Step-by-Step Solution:

Define variables: Let Ben’s current age be ( b ).
Set up the equation: Five years ago, Anna’s age was
25 - 5 = 20
, and Ben’s age was
b - 5
. Anna was twice as old as Ben:
20 = 2 (b - 5)
Solve the equation:
20 = 2b - 10
20 + 10 = 2b \implies 30 = 2b \implies b = 15
Verify: Current ages: Anna = 25, Ben = 15. Five years ago: Anna =
25 - 5 = 20
, Ben =
15 - 5 = 10
. Check:
20 = 2 \cdot 10
, satisfied.
Answer: Ben is 15 years old now.

Explanation: The problem provides Anna’s current age and a past relationship. The equation relates their ages five years ago, and solving gives Ben’s current age. The key is to account for the time shift consistently.

Problem 2

The sum of the current ages of a father and his son is 50 years. Ten years ago, the father was three times as old as the son. How old are they now?

Step-by-Step Solution:

Define variables: Let the father’s current age be ( f ) and the son’s current age be ( s ).
Set up equations:
- Current sum:
  f + s = 50
- Ten years ago: Father’s age =
  f - 10
  , son’s age =
  s - 10
  . Father was three times as old:
  f - 10 = 3 (s - 10)
Solve the system:
- From the first equation:
  f = 50 - s
- Substitute into the second equation:
  (50 - s) - 10 = 3 (s - 10)
  40 - s = 3s - 30
  40 + 30 = 3s + s \implies 70 = 4s \implies s = 17.5
- Find ( f ):
  f = 50 - 17.5 = 32.5
Verify: Current ages: father = 32.5, son = 17.5. Sum =
32.5 + 17.5 = 50
. Ten years ago: father =
32.5 - 10 = 22.5
, son =
17.5 - 10 = 7.5
. Check:
22.5 = 3 \cdot 7.5
, satisfied.
Answer: The father is 32.5 years old, and the son is 17.5 years old.

Explanation: The sum of current ages and a past relationship (three times as old) form a system of equations. The fractional ages are unusual but mathematically valid, indicating the problem may allow non-integer solutions. The time shift (10 years ago) is critical to the setup.

Problem 3

In 6 years, Clara will be twice as old as she was 4 years ago. How old is Clara now?

Step-by-Step Solution:

Define variables: Let Clara’s current age be ( c ).
Set up the equation: In 6 years, Clara’s age will be
c + 6
. Four years ago, her age was
c - 4
. The future age is twice the past age:
c + 6 = 2 (c - 4)
Solve the equation:
c + 6 = 2c - 8
6 + 8 = 2c - c \implies 14 = c
Verify: Current age: Clara = 14. In 6 years:
14 + 6 = 20
. Four years ago:
14 - 4 = 10
. Check:
20 = 2 \cdot 10
, satisfied.
Answer: Clara is 14 years old now.

Explanation: The problem relates Clara’s future age to her past age, with the present age as the reference point. The equation bridges the time gap (6 years forward, 4 years back), solving for the current age. This is a single-variable problem, making it straightforward.

Problem 4

Eight years ago, David was half as old as Emma. In 4 years, Emma will be 1.5 times as old as David. How old are they now?

Step-by-Step Solution:

Define variables: Let David’s current age be ( d ) and Emma’s current age be ( e ).
Set up equations:
- Eight years ago: David’s age =
  d - 8
  , Emma’s age =
  e - 8
  . David was half as old:
  d - 8 = \frac{1}{2} (e - 8) \implies 2 (d - 8) = e - 8 \implies 2d - 16 = e - 8 \implies e = 2d - 8
- In 4 years: David’s age =
  d + 4
  , Emma’s age =
  e + 4
  . Emma will be 1.5 times as old:
  e + 4 = 1.5 (d + 4)
Solve the system:
- Substitute
  e = 2d - 8
  into the second equation:
  (2d - 8) + 4 = 1.5 (d + 4)
  2d - 4 = 1.5d + 6
  2d - 1.5d = 6 + 4 \implies 0.5d = 10 \implies d = 20
- Find ( e ):
  e = 2 \cdot 20 - 8 = 40 - 8 = 32
Verify: Current ages: David = 20, Emma = 32. Eight years ago: David =
20 - 8 = 12
, Emma =
32 - 8 = 24
. Check:
12 = \frac{1}{2} \cdot 24
. In 4 years: David =
20 + 4 = 24
, Emma =
32 + 4 = 36
. Check:
36 = 1.5 \cdot 24
.
Answer: David is 20 years old, and Emma is 32 years old.

Explanation: The problem uses past and future relationships to determine current ages. The equations account for time shifts (8 years back, 4 years forward), and the system solves for both ages. Verification ensures consistency across all conditions.

Problem 5

The sum of Tom’s and Jerry’s current ages is 36 years. In 5 years, Tom will be as old as Jerry is now. How old are they now?

Step-by-Step Solution:

Define variables: Let Tom’s current age be ( t ) and Jerry’s current age be ( j ).
Set up equations:
- Current sum:
  t + j = 36
- In 5 years, Tom’s age =
  t + 5
  , which equals Jerry’s current age:
  t + 5 = j
Solve the system:
- Substitute
  j = t + 5
  into
  t + j = 36
  :
  t + (t + 5) = 36
  2t + 5 = 36 \implies 2t = 31 \implies t = 15.5
- Find ( j ):
  j = t + 5 = 15.5 + 5 = 20.5
Verify: Current ages: Tom = 15.5, Jerry = 20.5. Sum =
15.5 + 20.5 = 36
. In 5 years: Tom =
15.5 + 5 = 20.5
, which equals Jerry’s current age (20.5).
Answer: Tom is 15.5 years old, and Jerry is 20.5 years old.

Explanation: The sum of ages and the future condition (Tom’s age matching Jerry’s current age) form a simple system. The fractional ages are valid mathematically, though real-world ages are typically integers. The time shift (5 years forward) is key to the relationship.

Problem 6

Twelve years ago, a mother was four times as old as her daughter. In 8 years, the mother will be twice as old as the daughter. How old are they now?

Step-by-Step Solution:

Define variables: Let the mother’s current age be ( m ) and the daughter’s current age be ( d ).
Set up equations:
- Twelve years ago: Mother’s age =
  m - 12
  , daughter’s age =
  d - 12
  . Mother was four times as old:
  m - 12 = 4 (d - 12)
- In 8 years: Mother’s age =
  m + 8
  , daughter’s age =
  d + 8
  . Mother will be twice as old:
  m + 8 = 2 (d + 8)
Solve the system:
- First equation:
  m - 12 = 4d - 48 \implies m = 4d - 36
- Substitute into the second equation:
  (4d - 36) + 8 = 2 (d + 8)
  4d - 28 = 2d + 16
  4d - 2d = 16 + 28 \implies 2d = 44 \implies d = 22
- Find ( m ):
  m = 4 \cdot 22 - 36 = 88 - 36 = 52
Verify: Current ages: mother = 52, daughter = 22. Twelve years ago: mother =
52 - 12 = 40
, daughter =
22 - 12 = 10
. Check:
40 = 4 \cdot 10
. In 8 years: mother =
52 + 8 = 60
, daughter =
22 + 8 = 30
. Check:
60 = 2 \cdot 30
.
Answer: The mother is 52 years old, and the daughter is 22 years old.

Explanation: The past and future relationships (four times and twice as old) provide two equations. The time shifts (12 years back, 8 years forward) are handled by adjusting ages accordingly, and the solution ensures consistency across both conditions.

Problem 7

In 7 years, the sum of the ages of two brothers will be 50 years. Seven years ago, the older brother was twice as old as the younger. How old are they now?

Step-by-Step Solution:

Define variables: Let the older brother’s current age be ( o ) and the younger brother’s current age be ( y ).
Set up equations:
- In 7 years: Sum of ages =
  (o + 7) + (y + 7) = 50 \implies o + y + 14 = 50 \implies o + y = 36
- Seven years ago: Older =
  o - 7
  , younger =
  y - 7
  . Older was twice as old:
  o - 7 = 2 (y - 7)
Solve the system:
- From the first equation:
  o = 36 - y
- Substitute into the second equation:
  (36 - y) - 7 = 2 (y - 7)
  29 - y = 2y - 14
  29 + 14 = 2y + y \implies 43 = 3y \implies y = \frac{43}{3} \approx 14.33
- Find ( o ):
  o = 36 - \frac{43}{3} = \frac{108 - 43}{3} = \frac{65}{3} \approx 21.67
Verify: Current ages: older =
\frac{65}{3}
, younger =
\frac{43}{3}
. Sum =
\frac{65 + 43}{3} = \frac{108}{3} = 36
. In 7 years: older =
\frac{65}{3} + 7 = \frac{65 + 21}{3} = \frac{86}{3}
, younger =
\frac{43}{3} + 7 = \frac{43 + 21}{3} = \frac{64}{3}
. Sum =
\frac{86 + 64}{3} = \frac{150}{3} = 50
. Seven years ago: older =
\frac{65}{3} - 7 = \frac{65 - 21}{3} = \frac{44}{3}
, younger =
\frac{43}{3} - 7 = \frac{43 - 21}{3} = \frac{22}{3}
. Check:
\frac{44}{3} = 2 \cdot \frac{22}{3}
.
Answer: The older brother is approximately 21.67 years old, and the younger brother is approximately 14.33 years old.

Explanation: The future sum and past relationship form a system. The fractional ages are valid, though integers are more common in real-world scenarios. The time shifts (7 years forward and back) are handled carefully to ensure consistency.

Problem 8

Four years ago, a grandfather was six times as old as his grandson. Now, the grandfather is 60 years old. How old is the grandson now?

Step-by-Step Solution:

Define variables: Let the grandson’s current age be ( g ).
Set up the equation: Four years ago, the grandfather’s age was
60 - 4 = 56
, and the grandson’s age was
g - 4
. The grandfather was six times as old:
56 = 6 (g - 4)
Solve the equation:
56 = 6g - 24
56 + 24 = 6g \implies 80 = 6g \implies g = \frac{80}{6} = \frac{40}{3} \approx 13.33
Verify: Current ages: grandfather = 60, grandson =
\frac{40}{3}
. Four years ago: grandfather =
60 - 4 = 56
, grandson =
\frac{40}{3} - 4 = \frac{40 - 12}{3} = \frac{28}{3}
. Check:
56 = 6 \cdot \frac{28}{3}
, satisfied.
Answer: The grandson is approximately 13.33 years old now.

Explanation: The grandfather’s current age and the past relationship simplify the problem to a single equation. The fractional age is mathematically valid, though rounding to 13 or 14 might be considered in practice. The time shift (4 years ago) is key.

Problem 9

In 10 years, Sarah will be three times as old as she was 5 years ago. How old is Sarah now?

Step-by-Step Solution:

Define variables: Let Sarah’s current age be ( s ).
Set up the equation: In 10 years, Sarah’s age will be
s + 10
. Five years ago, her age was
s - 5
. The future age is three times the past age:
s + 10 = 3 (s - 5)
Solve the equation:
s + 10 = 3s - 15
10 + 15 = 3s - s \implies 25 = 2s \implies s = 12.5
Verify: Current age: Sarah = 12.5. In 10 years:
12.5 + 10 = 22.5
. Five years ago:
12.5 - 5 = 7.5
. Check:
22.5 = 3 \cdot 7.5
, satisfied.
Answer: Sarah is 12.5 years old now.

Explanation: The problem relates Sarah’s future age to her past age, with the present as the reference. The single equation accounts for both time shifts (10 years forward, 5 years back), and the fractional age is valid. This is a concise single-variable problem.

Problem 10

The sum of the current ages of two friends is 40 years. Six years ago, one friend was as old as the other is now. How old are they now?

Step-by-Step Solution:

Define variables: Let the first friend’s current age be ( a ) and the second friend’s current age be ( b ).
Set up equations:
- Current sum:
  a + b = 40
- Six years ago, the first friend’s age =
  a - 6
  , which equals the second friend’s current age:
  a - 6 = b
Solve the system:
- From the second equation:
  b = a - 6
- Substitute into the first equation:
  a + (a - 6) = 40
  2a - 6 = 40 \implies 2a = 46 \implies a = 23
- Find ( b ):
  b = a - 6 = 23 - 6 = 17
Verify: Current ages: first friend = 23, second friend = 17. Sum =
23 + 17 = 40
. Six years ago: first friend =
23 - 6 = 17
, which equals the second friend’s current age (17).
Answer: The first friend is 23 years old, and the second friend is 17 years old.

Explanation: The sum of ages and the past condition (one friend’s past age matching the other’s current age) form a system. The time shift (6 years ago) is handled by adjusting the first friend’s age, and the solution ensures both conditions are met.

General Notes on Calculating Present, Past, or Future Ages Problems

Key Concept: These problems focus on determining ages at specific times (present, past, or future) using relationships like sums, multiples, or equalities. They require careful handling of time shifts to set up equations.
Key Steps:
1. Define variables for current ages (or ages at a reference point).
2. Set up equations based on given relationships, adjusting ages for past or future times (e.g., subtract for past, add for future).
3. Solve the equations, typically using substitution or elimination for systems.
4. Verify by checking all conditions at the specified times.
Common Equations:
- Sum:
  a + b = s
- Multiple:
  a = k \cdot b
  (e.g., twice as old)
- Time shift: For ( t ) years ago, ages are
  a - t
  ,
  b - t
  ; for ( t ) years later, ages are
  a + t
  ,
  b + t
  .
Challenges:
- Time shifts must be applied consistently (e.g., subtract the same number of years for all individuals in past conditions).
- Fractional ages may arise, which are mathematically valid but may suggest integer solutions in real-world contexts.
- Ensure ages are positive and realistic (e.g., a person’s age
  a - t \geq 0
  for past conditions).
Applications: These problems model family relationships, educational settings, or social groups, requiring logical reasoning about age progression over time.
Verification: Always verify by substituting solutions back into all conditions (e.g., past relationships, future sums) to ensure consistency.

These problems demonstrate a range of scenarios, from single-variable cases to systems involving past and future conditions, highlighting the importance of time shifts in age calculations. If you need more problems, variations, or further clarification, let me know!

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47. Calculating present, past, or future ages problems

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