Introduction
Age Problem Quiz 2 are most frequently appearing questions in various competitive exams that include Quantitative Aptitude section. By analyzing the equations from the given data and assuming the unknown values, the age problems are solved.
Algebra is a very powerful branch of Mathematics which can be used to solve the Age Problem Quiz 2. Algebra helps in transforming word problems into mathematical expressions in the form of equations using variables to denote unknown quantities or parameters and thus, providing numerous techniques to solve these mathematical equations and hence, determining the answer to the problem. Identifying key information, organizing information, using mathematical expressions to assume unknown values and thus solving mathematical expressions for the unknown values will help us identify solutions.
Calculate Present age:
If the current age of a person be X, then
- age after n years = X + n
- age n years ago = X – n
- n times the age = nX
- If ages in the numerical are mentioned in ratio A : B, then A : B will be AX and BX
Q1
Father is aged three times more than his son Mohit. After 8 years, he would be two and a half times of Mohit’s age. After further 8 years, how many times would he be of Mohit’s age?
- A. 4
B. 2
C. 2.5
D. 3
Let Monit’s present age be X years. Then, father’s present age = (X + 3X) years = 4X years.
[latex](4X + 8) = \frac {5}{2} x (X + 8)[/latex]
8X + 16 = 5X + 40
3X = 24 so, X = 8
Hence, required ratio = [latex] \frac {(4X + 16)}{(X + 16)} = \frac {48}{24} = 2 [/latex]
Q2
A father said to his son, “I was as old as you are at present at the time of your birth.” If the father’s age is 38 years now, the son’s age five years back was:
- A. 14
B. 19
C. 38
D. 40
Let son’s present age be X years. Then, (38 – X) = X
2X = 38
X = 19
Son’s age 5 years back = (19 – 5) years = 14 years
Q3
One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years?
- A. 14
B. 16
C. 18
D. 12
Let one year ago
Sam’s age be 4X years
And, Ash’s age be 3X years
Present age of Sam = (4X + 1) years
Present age of Ash= (3X + 1) years
One year hence
Sam’s age = (4X + 2) years
Ash’s age = (3X + 2) years
According to question,
[4X + 2] divide by [3X + 2] = [latex]\frac {5}{4}[/latex]
16X + 8 = 15X + 10
or, X = 2
Sum of their present ages = 4X + 1 + 3X + 1
= 7X + 2
= 7 x 2 + 2 = 16 years.
Q4
In 8 years ago P will be twice as old as Q.If A is now 6 years older than Q,then the present age of Q
- A. 10
B. 14
C. 13
D. 12
(P - 8) = 2(Q - 8)
P - 2Q = -8______(1)
P - Q = 6 ___(2)
Solve (1) & (2)
Q = 14
Q5
The difference between the age of the two persons is 6 years. Twelve years ago, the elder was thrice as old as the younger one. The present age of the elder
- A. 16
B. 10
C. 21
D. 13
E - Y = 6________(1)
E - 12 = 3(Y - 12)
E - 3Y = -24_______(2)
Solve (1)& (2)
Y = [latex]\frac {30}{2} = 15[/latex]
E = 15 + 6 = 21