# Age Problem Quiz 5

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# Age Problem Quiz 5

### Introduction

Age Problem Quiz 5 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 5. 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

The ratio of Rohan’s age 4years ago and Rahul’s age after 4years is 1:1. If at present, the ratio of their ages is 5:3, then find the ratio between Rohan’s age 4 years hence and Rahul’s age 4years ago.
A. 1 : 3 B. 3 : 1 C. 4 : 3 D. 3 : 4

B
Hint: If ages in the numerical are mentioned in ratio A : B, then A : B will be Ax and Bx
1) At present: Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.
Rohan’s age 4 years ago = 5x – 4
Rahul’s age after 4 years = 3x + 4
2) Ratio of Rohan’s age 4 years ago and Rahul’s age after 4 years is 1 : 1 Therefore,
$\frac {(5x – 4)}{(3x + 4)} = \frac {1}{1}$
Solving, we get x = 4
3) We are asked to find the ratio between Rohan’s age 4 years hence and Rahul’s age 4 years ago.
Rohan’s age : (5x + 4)
Rahul’s age: (3x – 4)
Ratio of Rahul’s age and Rohan’s age
$\frac {(5x + 4)}{(3x - 4)} = \frac {24}{8} = \frac {3}{1} = 3:1$

### Q2

5 years ago, sister’s age was 5 times the age of her brother and the sum of present ages of sister and brother is 34 years. What will be the age of her brother after 6 years?
A. 12 years B. 13.5 years C. 15 years D. 20 years

C
Let present age of brother be x and sister’s age be 34 – x.
Relation Past Age (5 Yrs Ago) Present Age Future Age (After 6 Yrs)
Brother (x – 5) x (x + 6) = ?
Sister (34 – x) - 5 (30 – x)

We are given, 5 years ago sister’s age was 5 times the age of her brother.
Therefore,
(34 – x) – 5 = 5 (x – 5)
34 – x – 5 = 5x – 25
5x + x = 34 – 5 + 25
6x = 54
x = 9
Future age (after 6 yrs) = (x + 6) = (9 + 6) = 15 years

### Q3

Father is 3 times more aged than his daughter. If after 5 years, he would be 3 times of daughter’s age, then further after 5 years, how many times he would be of his daughter’s age?
A. 1 times B. 2 times C. 2.5 times D. 3 times

C
Let daughter’s age be x and father’s age be 3x.
Father’s age is 3 times more aged than his daughter, therefore father’s present age = x + 3x = 4x
After 5 years, father’s age is 3 times more than his daughter age.
(4x + 5) = 3 (x + 5)
x = 10
After 5 years it was (4x + 5), then after further 5 years, father’s age = (4x +10) and daughter’s age = (x + 10)
$\frac {(4x + 10)}{(x + 10)} = ?$
Substitute the value of x, we get
$\frac {(4x + 10) + 10}{(10 + 10)} = \frac {50}{20} = 2.5$
After further 5 years, father will be 2.5 times of daughter’s age.

### Q4

8 years ago, Geeta was half as old as Saina. Saina is now 20 years older than Geeta. How old will Geeta be in 10 years?
A. 45 B. 40 C. 35 D. 38

D
Saina is now 20 years older than Geeta $\rightarrow$ S = G + 20 $\rightarrow$ Saina 8 years
G + 20 - 8 = G + 12 Years old
$(G - 8) \times 2 = G + 12 \rightarrow G = 28 . 28 + 10 = 38 years$

### Q5

Lilly is 5 years older than Martin, and Catherine is twice as old as Lilly. If the sum of their ages is 95, how old is Lilly?
A. 26 B. 27 C. 28 D. 25

D
L = 5 + M and C = 2L and C + M+ L = 95 $\Rightarrow$ 4L = 100 $\Rightarrow$ L = 25