Quantitative Aptitude: Problems on Ages Set 2

Problems on Ages for SBI PO/IBPS PO, NIACL Assisatant, NICL AO and other upcoming exams.

  1. Four times the difference in ages of C and A is one more than the age if B. Percentage of A’s age to C’s age is 75%. If ratio of B’s age 5 years hence to C’s age 1 year ago is 4 : 3. Find the average of ages A and C.
    A) 20
    B) 19
    C) 12
    D) 14
    E) 8
    View Answer
    Option D
    Solution:

    4 (C-A) = B + 1
    A/C * 100 = 75
    (B+5)/(C-1) = 4/3
    Solve
    A = 12, C = 16
  2. 10 years ago daughter’s age was two-fifth of her mother’s age that time. While 10 years hence her age will be three-fifth of her mother’s age then. Find the difference in the ages of the two.
    A) 24
    B) 19
    C) 26
    D) 38
    E) 16
    View Answer
    Option A
    Solution:

    (x-10) = 2/5 (y-10)
    (x+10) = 3/5 (y+10)
    Solve, x = 26 and y = 50
  3. B is as more younger than C as he is elder than A. Ratio of ages of A to C is 5 : 9. If B’s age after 10 years will be 24, find the average of all of their present ages.
    A) 15
    B) 16
    C) 14
    D) 22
    E) 19
    View Answer
    Option C
    Solution:

    B’s present age = 24-10 = 14
    So C’ age = 14+x
    And A’ age = 14-x
    (14+x)/(14-x) = 9/5
    Solve, x = 4
    So average age = (10+14+18)/3 = 14 years
  4. Kaira is 4 years younger to his brother. Her father was 30 years old when her sister was born while her mother was was 30 years old when she was born. If her sister was 4 years old when their brother was born, find the age of her father when her mother was born.
    A) 11
    B) 12
    C) 4
    D) 10
    E) 8
    View Answer
    Option E
    Solution:

    When Kaira was born:
    Mother was 30.
    She is 4 years younger to her brother, so brother was 4 years old.
    Sister was 4 years old when brother was born, so sister is 4 years elder to brother, so sister was 8 years old.
    Father was 30 when sister was born, so father is 30 years elder to sister, so father was 30+8 = 38 years old.
    Now when Kaira was born, mother was 30 and father was 38
    So difference in their ages is 8 years. So when mother was born, father was 8.
  5. 6 years ago, three times the age of B was 2 more than the age if A that time. 6 years hence, twice age of B will be equal to A’s age that time. Find the total of their ages.
    A) 48
    B) 66
    C) 56
    D) 65
    E) 60
    View Answer
    Option B
    Solution:

    3 * (B-6) = 2 + (A-6)
    2 * (B+6) = A + 6
    Solve, A = 46, B = 20
  6. If 6 years are subtracted from the present age of Babita and the remainder is divided by 18, then the present age of her granddaughter Geeta is obtained. If Geeta is 2 years younger to Sita whose age is 5 years, then what is Babita’s present age?
    A) 77
    B) 65
    C) 84
    D) 43
    E) 79
    View Answer
    Option A
    Solution:

    Geeta’s age = (5-2) = 3 years
    Let age of Babita = x years
    So (x-6)/18 = 3
    Solve, x = 60
  7. A’s age is twice C’ age. Ratio of age of B 2 years hence to age of C 2 years ago is 5 : 2. C is 14 years younger than D. Difference in ages of D and A is 4 years. Find the average of their ages.
    A) 36
    B) 25
    C) 27
    D) 13
    E) 18
    View Answer
    Option E
    Solution:

    A = 2C
    (B+2)/(C-2) = 5/2
    C = D – 14
    D – A = 4
    Solve, A = 20, B = 18, C = 10, D = 24
  8. When the couple was married the average of their ages was 25 years. When their first child was born, the average age of family became 18 years. When their second child was born, the average age of the family became 15 years. Find the average age of the couple now.
    A) 31
    B) 27
    C) 28
    D) 29
    E) 30
    View Answer
    Option D
    Solution:

    Sum of ages of couple = 25*2 = 50
    When 1st child born, total age of 3 = 18*3 = 54 years
    At this time the child’s age was 0, so age of father and mother would have increased by same. So increased by 2 years each. So 50 +2 + 2 = 54
    Now when 2nd child born, total age of 4 = 15*4 = 60
    So this time second child’s age = 0 and age of father, mother and first child would have increased by same. So increased by 2 each such that 54 + 2+2+2 = 60
    So now this time (after 4 years from age 50), total age of couple is 50+4+4 = 58
    So average = 29 years
  9. Ratio of age of A to B is 3 : 2 and that of A to C is 1 : 2. Difference in ages of B and C is 24 years. Find the average of their present ages.
    A) 24
    B) 22
    C) 14
    D) 26
    E) 31
    View Answer
    Option B
    Solution:

    B/A = 2/3 and A/C = 1/2
    So B : A : C = 2*1 : 3*1 : 3*2 = 2 : 3 : 6
    So 6x – 2x = 24, 4x = 24, x = 6
    So total of their present ages = (2+3+6)*6. So average = (2+3+6)*6 / 3 = 22 years
  10. Ratio of ages of A 5 years hence to B’s age 3 years ago is 5 : 3. Also ratio of ages of A 4 years ago to B’s age 2 years hence is 4 : 5. Find the age of the elder.
    A) 15
    B) 18
    C) 21
    D) 20
    E) 24
    View Answer
    Option D
    Solution:

    (A+5)/(B-3) = 5/3
    (A-4)/(B+2) = 4/5
    Solve A = 20, B = 18

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12 Thoughts to “Quantitative Aptitude: Problems on Ages Set 2”

  1. Deepak sahu

    Thank you mam

  2. Deepak sahu

    Ap apne studendts ka bahut khyaal rakhe h uske liye dhanyawaad
    Apka matterial acha rahta h

  3. Yogi ji

    Option se ho sakte ye ques

  4. Nishi

    @suraj sir plz check ques 3

      1. jazzzz

        in ques 3 value of X is coming in -ve

  5. Deepak sahu

    mam/sir first question ki equetion nahi solve kar pa raha hu plz ap detail me bata dijiye

    1. 4 (C-A) = B + 1
      4C – 4A = B + 1 ……….(1)
      A/C = 3/4
      4A = 3C …………….(2)
      (B+5)/(C-1) = 4/3
      3B + 15 = 4C – 4
      So 3B – 4C = -19 ………….(3)

      Now hume B ki value nhi chaiye, so (1) me se B ki value (3) me pot karo
      Ab 2 equations bach jaengi – this new obtained and (2)
      ye dono A and C me hai
      solve now

  6. Ati

    So average age = (10+14+18)/3 = 12 years
    it will be 14 yrs not 12 yrs correct it in 3rd ques

      1. Ati

        mam 6th ques ka ans 60 hi aa rha … u have given a option

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