All Categories MCQs
Topic Notes: All Categories
General Description
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
7161
A is twice as old as B. 12 years ago, A was five times as old as B. Find the present age of A.
Answer:
32
Present: A = 2B. 12 years ago: A - 12 = 5(B - 12). Substitute A: 2B - 12 = 5B - 60 => 3B = 48 => B = 16. A = 2(16) = 32.
7162
The ratio of the ages of A and B is 3 : 5. The sum of their ages is 40. What is the age of B?
Answer:
25
Total parts = 3 + 5 = 8. 8 parts = 40 years, so 1 part = 5 years. Age of B = 5 parts = 5 * 5 = 25 years.
7163
The sum of the ages of a mother and daughter is 50 years. Also, 5 years ago, the mother's age was 7 times the age of the daughter. The present age of the mother is:
Answer:
40 years
M + D = 50. Five years ago: M - 5 = 7(D - 5). Substitute M = 50 - D: 50 - D - 5 = 7D - 35 => 45 - D = 7D - 35 => 8D = 80 => D = 10. M = 40 years.
7164
My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born?
Answer:
32 years
Father was 28 when sister was born. Sister was 4 when brother was born. Therefore, Father was 28 + 4 = 32 years old when brother was born.
7165
A father said to his son, 'I was as old as you are at the present at the time of your birth.' If the father's age is 38 years now, the son's age five years back was:
Answer:
14 years
Let son's present age be S. Father's age at son's birth was 38 - S. Given that 38 - S = S => 2S = 38 => S = 19. Son's age 5 years back = 19 - 5 = 14 years.
7166
The difference between the ages of two persons is 10 years. Fifteen years ago, the elder one was twice as old as the younger one. The present age of the elder person is:
Answer:
35 years
Let ages be x and x - 10. 15 years ago: (x - 15) = 2(x - 10 - 15) => x - 15 = 2(x - 25) => x - 15 = 2x - 50 => x = 35. Elder person is 35 years old.
7167
The ratio between the school ages of Neelam and Shaan is 5 : 6 respectively. If the ratio between the one-third age of Neelam and half of Shaan's age is 5 : 9, then what is the school age of Shaan?
Answer:
Cannot be determined
Let Neelam's age be 5x and Shaan's be 6x. Condition: (1/3 of 5x) / (1/2 of 6x) = 5/9 => (5x/3) / 3x = 5/9 => 5/9 = 5/9. This is an identity and gives no new information to solve for x. Hence, x cannot be determined.
7168
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Answer:
4 years
Let the ages be x, x+3, x+6, x+9, x+12. Sum = 5x + 30 = 50 => 5x = 20 => x = 4. The youngest child is 4 years old.
7169
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
Answer:
2 times
Three times MORE means Father = S + 3S = 4S. After 8 years: 4S + 8 = 2.5(S + 8) => 4S + 8 = 2.5S + 20 => 1.5S = 12 => S = 8. Father's present age = 32. After further 8 years (16 years from now), Son = 8 + 16 = 24, Father = 32 + 16 = 48. Ratio = 48 / 24 = 2 times.
7170
Ten years ago, A was half of B in age. If the ratio of their present ages is 3 : 4, what will be the total of their present ages?
Answer:
35
Present ages are 3x and 4x. Ten years ago: 3x - 10 = (1/2)(4x - 10) => 6x - 20 = 4x - 10 => 2x = 10 => x = 5. Present ages are 15 and 20. Total = 15 + 20 = 35.