Age Related Problems

Age Problems

Question 1:
The sum of the present ages of A and B is 64. Five years ago, the ratio of their ages was 3:2. Find their present ages.
Solution:
Let the present age of A be 3x and B be 2x.
According to the given information, 3x + 2x = 64.
Simplifying the equation, 5x = 64.
Dividing both sides by 5, we get x = 12.8.
Therefore, the present age of A = 3x = 3 * 12.8 = 38.4 years.
And the present age of B = 2x = 2 * 12.8 = 25.6 years.
Requested by: Ahmed

Question 2:
The sum of the present ages of a father and his son is 48 years. Four years ago, the father’s age was five times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 48 – x.
According to the given information, 48 – x – 4 = 5(x – 4).
Simplifying the equation, 48 – x – 4 = 5x – 20.
Combining like terms, 44 – x = 5x – 20.
Bringing like terms to one side, 6x = 64.
Dividing both sides by 6, we get x = 10.67.
Therefore, the present age of the son is approximately 10.67 years.
And the present age of the father is approximately 48 – 10.67 = 37.33 years.
Requested by: Ahmed

Question 3:
The average age of a family of 5 members is 24 years. If the youngest member is 8 years old, what is the average age of the family excluding the youngest member?
Solution:
The sum of the ages of the family members is 24 * 5 = 120 years.
Since the youngest member is 8 years old, the sum of the ages of the remaining 4 members is 120 – 8 = 112 years.
Therefore, the average age of the family excluding the youngest member is 112/4 = 28 years.
Requested by: Inaam

Question 4:
The present age of a father is four times the age of his son. The difference between their ages is 20 years. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 4x.
According to the given information, 4x – x = 20.
Simplifying the equation, 3x = 20.
Dividing both sides by 3, we get x = 6.67.
Therefore, the present age of the son is approximately 6.67 years.
And the present age of the father is approximately 4 * 6.67 = 26.67 years.
Requested by: Faisal

Question 5:
The ratio of the present ages of A and B is 5:3. If the difference between their ages is 8 years, find their present ages.
Solution:
Let the present age of A be 5x and B be 3x.
According to the given information, 5x – 3x = 8.
Simplifying the equation, 2x = 8.
Dividing both sides by 2, we get x = 4.
Therefore, the present age of A = 5x = 5 * 4 = 20 years.
And the present age of B = 3x = 3 * 4 = 12 years.
Requested by: Faisal

Question 6:
The average age of a group of 8 friends is 32 years. If the age of one of the friends is 40 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 8 friends is 32 * 8 = 256 years.
Since one of the friends is 40 years old, the sum of the ages of the remaining 7 friends is 256 – 40 = 216 years.
Therefore, the average age of the remaining friends is 216/7 ≈ 30.86 years.
Requested by: Zohaib

Question 7:
The age of a man is three times the age of his son. After 15 years, the man’s age will be twice that of his son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the man is 3x.
After 15 years, the son’s age will be x + 15, and the man’s age will be 3x + 15.
According to the given information, 3x + 15 = 2(x + 15).
Simplifying the equation, 3x + 15 = 2x + 30.
Bringing like terms to one side, x = 15.
Therefore, the present age of the son is 15 years.
And the present age of the man is 3 * 15 = 45 years.
Requested by: Samad

Question 8:
The sum of the ages of a mother and her daughter is 50 years. Five years ago, the mother was seven times as old as the daughter. Find their present ages.
Solution:
Let the present age of the daughter be x. Therefore, the present age of the mother is 50 – x.
According to the given information, 50 – x – 5 = 7(x – 5).
Simplifying the equation, 45 – x = 7x – 35.
Bringing like terms to one side, 8x = 80.
Dividing both sides by 8, we get x = 10.
Therefore, the present age of the daughter is 10 years.
And the present age of the mother is 50 – 10 = 40 years.
Requested by: Sarang

Question 9:
The average age of a group of 10 friends is 25 years. If the age of one of the friends is 18 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 10 friends is 25 * 10 = 250 years.
Since one of the friends is 18 years old, the sum of the ages of the remaining 9 friends is 250 – 18 = 232 years.
Therefore, the average age of the remaining friends is 232/9 ≈ 25.78 years.
Requested by: Ali

Question 10:
The sum of the present ages of a father and his son is 72 years. Ten years ago, the father’s age was five times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 72 – x.
According to the given information, 72 – x – 10 = 5(x – 10).
Simplifying the equation, 62 – x = 5x – 50.
Bringing like terms to one side, 6x = 112.
Dividing both sides by 6, we get x = 18.67.
Therefore, the present age of the son is approximately 18.67 years.
And the present age of the father is approximately 72 – 18.67 = 53.33 years.
Requested by: Aakash

Question 11:
The ratio of the present ages of A and B is 7:5. If the difference between their ages is 12 years, find their present ages.
Solution:
Let the present age of A be 7x and B be 5x.
According to the given information, 7x – 5x = 12.
Simplifying the equation, 2x = 12.
Dividing both sides by 2, we get x = 6.
Therefore, the present age of A = 7x = 7 * 6 = 42 years.
And the present age of B = 5x = 5 * 6 = 30 years.
Requested by: Anwar

Question 12:
The average age of a group of 12 friends is 28 years. If the age of one of the friends is 30 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 12 friends is 28 * 12 = 336 years.
Since one of the friends is 30 years old, the sum of the ages of the remaining 11 friends is 336 – 30 = 306 years.
Therefore, the average age of the remaining friends is 306/11 ≈ 27.82 years.
Requested by: Zulfiqar

Question 13:
The age of a man is four times the age of his daughter. After 10 years, the man’s age will be three times that of his daughter. Find their present ages.
Solution:
Let the present age of the daughter be x. Therefore, the present age of the man is 4x.
After 10 years, the daughter’s age will be x + 10, and the man’s age will be 4x + 10.
According to the given information, 4x + 10 = 3(x + 10).
Simplifying the equation, 4x + 10 = 3x + 30.
Bringing like terms to one side, x = 20.
Therefore, the present age of the daughter is 20 years.
And the present age of the man is 4 * 20 = 80 years.
Requested by: Asim

Question 14:
The sum of the ages of a mother and her daughter is 60 years. Five years ago, the mother was four times as old as the daughter. Find their present ages.
Solution:
Let the present age of the daughter be x. Therefore, the present age of the mother is 60 – x.
According to the given information, 60 – x – 5 = 4(x – 5).
Simplifying the equation, 55 – x = 4x – 20.
Bringing like terms to one side, 5x = 75.
Dividing both sides by 5, we get x = 15.
Therefore, the present age of the daughter is 15 years.
And the present age of the mother is 60 – 15 = 45 years.
Requested by: Azhar

Question 15:
The average age of a group of 15 friends is 35 years. If the age of one of the friends is 42 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 15 friends is 35 * 15 = 525 years.
Since one of the friends is 42 years old, the sum of the ages of the remaining 14 friends is 525 – 42 = 483 years.
Therefore, the average age of the remaining friends is 483/14 ≈ 34.5 years.
Requested by: Mazhar

Question 16:
The ratio of the present ages of A and B is 2:3. After 10 years, the ratio of their ages will be 3:4. Find their present ages.
Solution:
Let the present age of A be 2x and B be 3x.
After 10 years, the ages of A and B will be 2x + 10 and 3x + 10, respectively.
According to the given information, (2x + 10)/(3x + 10) = 3/4.
Cross-multiplying, we get 8x + 40 = 9x + 30.
Bringing like terms to one side, x = 10.
Therefore, the present age of A = 2x = 2 * 10 = 20 years.
And the present age of B = 3x = 3 * 10 = 30 years.
Requested by: Faiz

Question 17:
The average age of a group of 20 friends is 40 years. If the age of one of the friends is 50 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 20 friends is 40 * 20 = 800 years.
Since one of the friends is 50 years old, the sum of the ages of the remaining 19 friends is 800 – 50 = 750 years.
Therefore, the average age of the remaining friends is 750/19 ≈ 39.47 years.
Requested by: Sahil

Question 18:
The present age of a father is three times the age of his son. The sum of their ages is 48 years. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 3x.
According to the given information, x + 3x = 48.
Simplifying the equation, 4x = 48.
Dividing both sides by 4, we get x = 12.
Therefore, the present age of the son is 12 years.
And the present age of the father is 3 * 12 = 36 years.
Requested by: Zubair

Question 19:
The sum of the ages of a mother and her daughter is 36 years. Four years ago, the mother was five times as old as the daughter. Find their present ages.
Solution:
Let the present age of the daughter be x. Therefore, the present age of the mother is 36 – x.
According to the given information, 36 – x – 4 = 5(x – 4).
Simplifying the equation, 32 – x = 5x – 20.
Bringing like terms to one side, 6x = 52.
Dividing both sides by 6, we get x = 8.67.
Therefore, the present age of the daughter is approximately 8.67 years.
And the present age of the mother is 36 – 8.67 = 27.33 years.
Requested by: Mazhar

Question 20:
The average age of a group of 25 friends is 30 years. If the age of one of the friends is 35 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 25 friends is 30 * 25 = 750 years.
Since one of the friends is 35 years old, the sum of the ages of the remaining 24 friends is 750 – 35 = 715 years.
Therefore, the average age of the remaining friends is 715/24 ≈ 29.79 years.
Requested by: Faizan

Question 21:
The present ages of John and Mary are in the ratio of 5:3. If the sum of their ages is 64 years, what are their present ages?
Solution:
Let the present age of John be 5x and Mary be 3x.
According to the given information, 5x + 3x = 64.
Simplifying the equation, 8x = 64.
Dividing both sides by 8, we get x = 8.
Therefore, the present age of John is 5x = 5 * 8 = 40 years.
And the present age of Mary is 3x = 3 * 8 = 24 years.
Requested by: Azaan

Question 22:
The average age of a family of 6 members is 35 years. If the average age of the parents is 40 years, what is the average age of the children?
Solution:
The sum of the ages of the parents is 40 * 2 = 80 years.
The sum of the ages of the children is the sum of the ages of all family members minus the sum of the ages of the parents.
Therefore, the sum of the ages of the children is (35 * 6) – 80 = 210 – 80 = 130 years.
Since there are 4 children, the average age of the children is 130/4 = 32.5 years.
Requested by: Azaan

Question 23:
The sum of the ages of a father and his son is 48 years. Four years ago, the father’s age was three times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 48 – x.
According to the given information, 48 – x – 4 = 3(x – 4).
Simplifying the equation, 44 – x = 3x – 12.
Bringing like terms to one side, 4x = 56.
Dividing both sides by 4, we get x = 14.
Therefore, the present age of the son is 14 years.
And the present age of the father is 48 – 14 = 34 years.
Requested by: Mustafa

Question 24:
The ratio of the present ages of A and B is 3:5. After 10 years, the ratio of their ages will be 4:7. Find their present ages.
Solution:
Let the present age of A be 3x and B be 5x.
After 10 years, the ages of A and B will be 3x + 10 and 5x + 10, respectively.
According to the given information, (3x + 10)/(5x + 10) = 4/7.
Cross-multiplying, we get 21x + 70 = 20x + 40.
Bringing like terms to one side, x = -30.
Since the age cannot be negative, this is an invalid solution.
Therefore, there is no valid solution for this problem.
Requested by: Mustafa

Question 25:
The average age of a group of 15 friends is 30 years. If the age of one of the friends is 35 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 15 friends is 30 * 15 = 450 years.
Since one of the friends is 35 years old, the sum of the ages of the remaining 14 friends is 450 – 35 = 415 years.
Therefore, the average age of the remaining friends is 415/14 ≈ 29.64 years.
Requested by: Kabeer

Question 26:
The age of a man is four times the age of his son. Five years ago, the man’s age was three times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the man is 4x.
Five years ago, the son’s age was x – 5, and the man’s age was 4x – 5.
According to the given information, 4x – 5 = 3(x – 5).
Simplifying the equation, 4x – 5 = 3x – 15.
Bringing like terms to one side, x = 10.
Therefore, the present age of the son is 10 years.
And the present age of the man is 4 * 10 = 40 years.
Requested by: Ali

Question 27:
The sum of the ages of a brother and sister is 40 years. Four years ago, the brother’s age was twice the age of the sister. Find their present ages.
Solution:
Let the present age of the sister be x. Therefore, the present age of the brother is 40 – x.
Four years ago, the sister’s age was x – 4, and the brother’s age was 40 – x – 4.
According to the given information, 40 – x – 4 = 2(x – 4).
Simplifying the equation, 36 – x = 2x – 8.
Bringing like terms to one side, 3x = 44.
Dividing both sides by 3, we get x = 14.67.
Therefore, the present age of the sister is approximately 14.67 years.
And the present age of the brother is 40 – 14.67 = 25.33 years.
Requested by: Ilyas

Question 28:
The average age of a group of 20 friends is 35 years. If the age of one of the friends is 40 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 20 friends is 35 * 20 = 700 years.
Since one of the friends is 40 years old, the sum of the ages of the remaining 19 friends is 700 – 40 = 660 years.
Therefore, the average age of the remaining friends is 660/19 ≈ 34.74 years.
Requested by: Ilyas

Question 29:
The present ages of two friends are in the ratio of 2:3. Four years ago, the ratio of their ages was 1:2. Find their present ages.
Solution:
Let the present age of the first friend be 2x and the second friend be 3x.
Four years ago, the ages of the friends were 2x – 4 and 3x – 4.
According to the given information, (2x – 4)/(3x – 4) = 1/2.
Cross-multiplying, we get 4x – 8 = 3x – 4.
Bringing like terms to one side, x = 4.
Therefore, the present age of the first friend is 2x = 2 * 4 = 8 years.
And the present age of the second friend is 3x = 3 * 4 = 12 years.
Requested by: Azlan

Question 30:
The average age of a group of 30 friends is 25 years. If the age of one of the friends is 30 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 30 friends is 25 * 30 = 750 years.
Since one of the friends is 30 years old, the sum of the ages of the remaining 29 friends is 750 – 30 = 720 years.
Therefore, the average age of the remaining friends is 720/29 ≈ 24.83 years.
Requested by: Azlan

Question 31:
The ratio of the present ages of a father and his son is 5:2. Four years ago, the ratio of their ages was 11:4. Find their present ages.
Solution:
Let the present age of the father be 5x and the son be 2x.
Four years ago, the ages of the father and son were 5x – 4 and 2x – 4, respectively.
According to the given information, (5x – 4)/(2x – 4) = 11/4.
Cross-multiplying, we get 20x – 16 = 22x – 44.
Bringing like terms to one side, x = 14.
Therefore, the present age of the father is 5x = 5 * 14 = 70 years.
And the present age of the son is 2x = 2 * 14 = 28 years.
Requested by: Azlan

Question 32:
The average age of a group of 40 students is 18 years. If the age of one of the students is 20 years, what is the average age of the remaining students?
Solution:
The sum of the ages of the 40 students is 18 * 40 = 720 years.
Since one of the students is 20 years old, the sum of the ages of the remaining 39 students is 720 – 20 = 700 years.
Therefore, the average age of the remaining students is 700/39 ≈ 17.95 years.
Requested by: Zain

Question 33:
The sum of the ages of a grandfather and his grandson is 80 years. The grandfather is 4 times as old as the grandson. Find their ages.
Solution:
Let the present age of the grandson be x. Therefore, the present age of the grandfather is 4x.
According to the given information, x + 4x = 80.
Simplifying the equation, 5x = 80.
Dividing both sides by 5, we get x = 16.
Therefore, the present age of the grandson is 16 years.
And the present age of the grandfather is 4 * 16 = 64 years.
Requested by: Rehman

Question 34:
The ratio of the present ages of A and B is 3:7. Five years ago, the ratio of their ages was 2:5. Find their present ages.
Solution:
Let the present age of A be 3x and B be 7x.
Five years ago, the ages of A and B were 3x – 5 and 7x – 5, respectively.
According to the given information, (3x – 5)/(7x – 5) = 2/5.
Cross-multiplying, we get 15x – 25 = 14x – 10.
Bringing like terms to one side, x = 15.
Therefore, the present age of A is 3x = 3 * 15 = 45 years.
And the present age of B is 7x = 7 * 15 = 105 years.
Requested by: Zohaib

Question 35:
The average age of a group of 25 friends is 32 years. If the age of one of the friends is 40 years, what is the average age of the remaining friends?
Solution:
The sum of the ages of the 25 friends is 32 * 25 = 800 years.
Since one of the friends is 40 years old, the sum of the ages of the remaining 24 friends is 800 – 40 = 760 years.
Therefore, the average age of the remaining friends is 760/24 = 31.67 years.
Requested by: Akram

Question 36:
The age of a father is 4 times the age of his son. Four years later, the age of the father will be three times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x. Therefore, the present age of the father is 4x.
Four years later, the ages of the father and son will be 4x + 4 and x + 4, respectively.
According to the given information, 4x + 4 = 3(x + 4).
Simplifying the equation, 4x + 4 = 3x + 12.
Bringing like terms to one side, x = 8.
Therefore, the present age of the son is 8 years.
And the present age of the father is 4 * 8 = 32 years.
Requested by: Aakash

Question 37:
The sum of the present ages of a husband and wife is 60 years. The husband is twice as old as the wife. Find their ages.
Solution:
Let the present age of the wife be x. Therefore, the present age of the husband is 2x.
According to the given information, x + 2x = 60.
Simplifying the equation, 3x = 60.
Dividing both sides by 3, we get x = 20.
Therefore, the present age of the wife is 20 years.
And the present age of the husband is 2 * 20 = 40 years.
Requested by: Zain

Question 38:
The average age of a group of 50 employees is 35 years. If the age of one of the employees is 45 years, what is the average age of the remaining employees?
Solution:
The sum of the ages of the 50 employees is 35 * 50 = 1750 years.
Since one of the employees is 45 years old, the sum of the ages of the remaining 49 employees is 1750 – 45 = 1705 years.
Therefore, the average age of the remaining employees is 1705/49 ≈ 34.80 years.
Requested by: Zohaib

Question 39:
The present ages of two friends are in the ratio of 4:7. Seven years ago, the ratio of their ages was 3:5. Find their present ages.
Solution:
Let the present age of the first friend be 4x and the second friend be 7x.
Seven years ago, the ages of the friends were 4x – 7 and 7x – 7, respectively.
According to the given information, (4x – 7)/(7x – 7) = 3/5.
Cross-multiplying, we get 20x – 35 = 21x – 21.
Bringing like terms to one side, x = 14.
Therefore, the present age of the first friend is 4x = 4 * 14 = 56 years.
And the present age of the second friend is 7x = 7 * 14 = 98 years.
Requested by: Faizan

Question 40:
The average age of a group of 60 students is 21 years. If the age of one of the students is 25 years, what is the average age of the remaining students?
Solution:
The sum of the ages of the 60 students is 21 * 60 = 1260 years.
Since one of the students is 25 years old, the sum of the ages of the remaining 59 students is 1260 – 25 = 1235 years.
Therefore, the average age of the remaining students is 1235/59 ≈ 20.93 years.
Requested by: Hasnain

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