Q.1: The ratio of the present ages of A and B is 4:5. After 6 years, their ages will be in the ratio 6:7. What is the present age of A?
a) 12 years
b) 15 years
c) 18 years
d) 24 years
Ans: a) Solution: Let the present ages be 4x and 5x. After 6 years: (4x + 6) / (5x + 6) = 6 / 7. By cross-multiplication: 7(4x + 6) = 6(5x + 6) => 28x + 42 = 30x + 36 => 2x = 6 => x = 3. Therefore, A’s age = 4 * 3 = 12 years.
Q.2: A father is four times as old as his son. After 5 years, he would be three times as old as his son. Find the son’s present age.
a) 10 years
b) 12 years
c) 15 years
d) 20 years
Ans: a) Solution: Let son = x and father = 4x. After 5 years: (4x + 5) = 3(x + 5) => 4x + 5 = 3x + 15 => x = 10.
Q.3: The sum of the ages of a mother and daughter is 50 years. Five years ago, the mother’s age was 7 times the daughter’s age. What are their present ages?
a) 40, 10
b) 35, 15
c) 45, 5
d) 38, 12
Ans: a)
Solution: Let daughter = d. Mother = 50 – d. Five years ago: (50 – d – 5) = 7(d – 5) => 45 – d = 7d – 35 => 8d = 80 => d = 10. Mother = 40.
Q.4: A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is:
a) 14 years
b) 18 years
c) 20 years
d) 22 years
Ans: d)
Solution: Let son = x, man = x + 24. In 2 years: (x + 24 + 2) = 2(x + 2) => x + 26 = 2x + 4 => x = 22.
Q.5: The average age of 10 students is 15 years. When a teacher’s age is included, the average increases by 1. What is the teacher’s age?
a) 25 years
b) 26 years
c) 30 years
d) 35 years
Ans: b)
Solution: Total age of students = 10 * 15 = 150. Total age with teacher (11 people) = 11 * 16 = 176. Teacher = 176 – 150 = 26.
Q.6: The ratio of present ages of P and Q is 3:4. Five years ago, the ratio was 2:3. What is the sum of their present ages? a) 30 years
b) 35 years
c) 40 years
d) 45 years
Ans: b) Solution: Ages are 3x and 4x. (3x – 5) / (4x – 5) = 2 / 3 => 9x – 15 = 8x – 10 => x = 5. Sum = 3(5) + 4(5) = 35.
Q.7: A is twice as old as B. Ten years ago, A was three times as old as B. What is B’s present age?
a) 15 years
b) 20 years
c) 25 years
d) 30 years
Ans: b) Solution: B = x, A = 2x. (2x – 10) = 3(x – 10) => 2x – 10 = 3x – 30 => x = 20.
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Q.8: The sum of the ages of 5 children born at intervals of 3 years each is 50 years. What is the age of the youngest child? a) 4 years
b) 8 years
c) 10 years
d) 12 years
Ans: a)
Solution: Let ages be x, x+3, x+6, x+9, x+12. Sum = 5x + 30 = 50 => 5x = 20 => x = 4.
Q.9: Present ages of Kiran and Syam are in the ratio 5:4. After 3 years, the ratio becomes 11:9. What is Syam’s present age?
a) 24 years
b) 27 years
c) 30 years
d) 33 years
Ans: a)
Solution: (5x + 3) / (4x + 3) = 11 / 9 => 45x + 27 = 44x + 33 => x = 6. Syam = 4 * 6 = 24.
Q.10: Six years ago, the ratio of the ages of Kunal and Sagar was 6:5. Four years hence, the ratio will be 11:10. What is Sagar’s present age?
a) 16 years
b) 18 years
c) 20 years
d) 22 years
Ans: a)
Solution: Let ages 6 yrs ago be 6x, 5x. After 10 years from that point (6 past + 4 future): (6x + 10) / (5x + 10) = 11 / 10 => 60x + 100 = 55x + 110 => 5x = 10 => x = 2. Sagar’s present age = 5(2) + 6 = 16.
Q.11: The age of a father is twice the age of his elder son. Ten years hence, the father will be three times the age of the younger son. If the difference of the sons’ ages is 15 years, find the father’s age.
a) 50 years
b) 55 years
c) 60 years
d) 70 years
Ans: a)
Solution: Elder son = x, Father = 2x, Younger son = x – 15. In 10 years: 2x + 10 = 3(x – 15 + 10) => 2x + 10 = 3x – 15 => x = 25. Father = 2 * 25 = 50.
Q.12: The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit’s age by 4 years, what is Nikita’s age?
a) 10 years
b) 12 years
c) 15 years
d) 20 years
Ans: b)
Solution: A * N = 240 and 2N – A = 4 => A = 2N – 4. (2N – 4)N = 240 => 2N^2 – 4N – 240 = 0 => N^2 – 2N – 120 = 0. Solving the quadratic gives N = 12.
Q.13: A father is 28 years old when his daughter is born. His son is 4 years younger than the daughter. What was the father’s age when the son was born?
a) 32 years
b) 34 years
c) 35 years
d) 30 years
Ans: a)
Solution: Daughter born => Father = 28. Son born 4 years later => Father = 28 + 4 = 32.
Q.14: A person’s age is 2/5 of his mother’s age. After 8 years, he will be 1/2 of her age. How old is the mother now?
a) 32 years
b) 36 years
c) 40 years
d) 48 years
Ans: c)
Solution: P = 0.4M. In 8 years: (0.4M + 8) = 0.5(M + 8) => 0.4M + 8 = 0.5M + 4 => 0.1M = 4 => M = 40.
Q.15: Ratio of ages of A and B is 3:5 and the sum is 80 years. What is the ratio after 10 years?
a) 2:3
b) 1:2
c) 3:5
d) 4:5
Ans: a)
Solution: 3x + 5x = 80 => x = 10. Present: 30, 50. After 10 years: 40, 60. Ratio = 40:60 = 2:3.
Q.16: In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, find B’s age.
a) 19 years
b) 29 years
c) 39 years
d) 49 years
Ans: c)
Solution: A = B + 9. (A + 10) = 2(B – 10) => B + 9 + 10 = 2B – 20 => B + 19 = 2B – 20 => B = 39.
Q.17: A + B is 12 years more than B + C. How many years is C younger than A?
a) 12 years
b) 10 years
c) 24 years
d) Data insufficient
Ans: a)
Solution: A + B = (B + C) + 12 => A = C + 12.
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Q.18: A father says: “I was as old as you are now at the time of your birth.” If the father is 38 now, how old was the son 5 years ago?
a) 14 years
b) 19 years
c) 33 years
d) 38 years
Ans: a)
Solution: Father = 38, Son = x. At birth, Father was (38-x). 38-x = x => 2x = 38 => x = 19. 5 years ago: 19 – 5 = 14.
Q.19: F + S = 45. Five years ago, the product of their ages was 4 times the father’s age then. Find their ages.
a) 36, 9
b) 39, 6
c) 40, 5
d) 42, 3
Ans: a) Solution: (F-5)(S-5) = 4(F-5) => S-5 = 4 => S = 9. F = 45 – 9 = 36.
Q.20: A is 3 years older than B and 3 years younger than C. B and D are twins. How much older is C than D?
a) 2 years
b) 3 years
c) 6 years
d) 9 years
Ans: c)
Solution: A = B + 3, A = C – 3. Thus B + 3 = C – 3 => C – B = 6. Since B = D, C – D = 6.
Q.21: Ratio of Father:Son is 17:7. Six years ago, it was 3:1. Father’s present age is:
a) 48 years
b) 51 years
c) 61 years
d) 64 years
Ans: b)
Solution: (17x – 6) / (7x – 6) = 3 / 1 => 17x – 6 = 21x – 18 => 4x = 12 => x = 3. 17 * 3 = 51.
Q.22: Father’s age is 3 times the sum of his 2 children’s ages. In 5 years, he will be twice the sum. Father’s age is:
a) 40 years
b) 45 years
c) 50 years
d) 55 years
Ans: b) Solution: F = 3S. (F + 5) = 2(S + 10) [Note: 2 children both age 5 yrs] => 3S + 5 = 2S + 20 => S = 15. F = 45.
Q.23: E – Y = 10. Fifteen years ago, Elder was twice as old as Younger. Elder’s age is:
a) 25 years
b) 35 years
c) 45 years
d) 55 years
Ans: b)
Solution: E = Y + 10. (E – 15) = 2(Y – 15) => (Y + 10 – 15) = 2Y – 30 => Y – 5 = 2Y – 30 => Y = 25. E = 35.
Q.24: A is 1/6 of B. B will be twice C in 10 years. C was 8 two years ago. A is:
a) 5 years
b) 7.5 years
c) 10 years
d) 15 years
Ans: a)
Solution: C now = 10. C in 10 yrs = 20. B in 10 yrs = 40. B now = 30. A = 30 / 6 = 5.
Q.25: Sum of ages of 3 persons is 150 years. 10 years ago, ratio was 7:8:9. Find current age of the eldest.
a) 40 years
b) 50 years
c) 55 years
d) 70 years
Ans: c)
Solution: Sum 10 yrs ago = 150 – 30 = 120. 7x + 8x + 9x = 120 => 24x = 120 => x = 5. Eldest then = 9 * 5 = 45. Now = 45 + 10 = 55. (Correcting calculation: 9x+10 = 9(5)+10 = 55). Wait, option check: 70 is 9(x)+10 if x was higher. Let’s re-verify: 24x=120, x=5. Eldest=55.
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