# Quantitative Aptitude: Problems on Ages Set 18

1. Ratio of the ages of A and B is 2:3 and the sum of the squares of their ages is 208 years . Find the age of B 4 years ago.
6
4
9
8
none of this
Option D
Let,
their ages be 2x and 3x years .
(2x)^2+(3x)^2 =208
4x^2+9x^2 =208
x^2=16
x=4 Age of A = 8
Age of B= 12
4 years ago age of B = 12-4=8 years.

2. The age of father is 5 times that of his son’s age and sum of their ages is 54 years .Find the son’s age after 5 years.
14
15
16
18
none of this
Option
Let,
son’s age be x.
and father’s age 5x years .
5x+x =54
x =9 years
son’s age after 5 years= 14 years

3. At present the difference of ages of A and B is 20 years. After 3 years their ages are in the ratio 9:19. Find the present age of B.
35
38
40
20
none of this
Option A
Let,
age of A and B after 3 years be 9x and 19x.
Present age of A =9x-3
present age of B = 19x-3
19x-3-(9x-3)=20
19x-3-9x+3 =20
10x=20
x=2
Present age of B = 19x-3 = 38-3 =35 years.

4. At present Ratio of ages of A and B is 8:5. After 2 years it will be 14:9 . Find the age of B 3 years ago.
22
23
25
28
none of this
Option A
Let,
Present age of A and B be 8x and 5x .
ATQ,
(8x+2)/(5x+2) =14/9
72x+18=70x+28
x=5 years.
Present age of B = 5x =25 years
3 years ago=22 years.

5. At present A is 10 years older than B. After 10 years A will be 4 times as old as B was 10 years ago . find the Ratio of their ages.
3:2
4:5
9:4
3:4
none of these
Option A
Let,
B’s present age be x years
A’s present age = x+10 years
after 10 years A’s age=x+20
10 years ago B’s age=x-10
ATQ, x+20=4(x-10)
x+20=4x-40
x=20
ratio of the age of A & B=30:20=3:2

6. The average age of 8 men is increased by 2 years when two of them whose ages are 21 and 23 years replaced by two new men. Find the average age of the two new men.
48
35
40
30
none of these
Option D
Let the average age of 8 men be x
total age of 8 men=8x
these two new men=a and b
(8x-21-23+a+b)/8=x+2
8x-44+a+b=8x+16
x+y=60
average age of two new men=60/2=30 years

7. There are 40 students in a class. The average age of first 10 students is 14.5 years. The average age of the remaining 30 students is 20.5 years. Find the average age of (in years) of the students of the whole class.
19
18
20
30
none of this
Option A
ATQ, total age of 10 students=14.5*10=145
total age of remaining 30 students=20.5 *30=615
average age of students of whole class=145+615/40=19years

8. Present age of B is 50% more than that of A and average of present age of B & C is 30 years. If 4 years ago sum of ages of A & B is 32 years then find the sum of ages of A & C .
54
52
46
45
none of this
Option B
Present age of B 50% more than A.
Ratio of present age of A and B= 2:3
total present age of B and C = 30*2=60
4 years ago some of age of A and B 32 years present age of A and B together =32+8=40 years
present of A =40*2/5=16 years
present age of B =40*3/5= 24 years
present age of C= 60-24= 36 years
sum of present age of A and C =16+36= 52 years

9. The average age of 30 boys in a class is 15 years . One boy aged 20 years left the class but two new boys come in his place whose age differs by 5 years. If the average age of all the boys now in the class still remains 15 years. Find the age of elder newcomer.
15
30
35
20
none of this
Option D
Total age of 30 boys =15*30=450
one boys aged 20 years left the remaining 29 students age =430
after the two newcomer age added then total age of 31 students = 465
the age of two new comer =465-430= 35 years
difference the age of two newcomer =5 years
the age of new elder one = 20 years .

10. The Ratio of the present age of Sima and Lima is 2:1. The ratio of their ages after 30 years will be 7:6. Find the different between their ages after 9 years.
8
10
6
12
none of these
Option
ATQ,
(2x+30)/(x+30)=7/6
12x+180 = 7x+210
5x= 30, x=6
present age of Sima =12
present age of Lima =6
after 9 years their ages 21 and 15 years respectively.
difference between their ages =21-15=6 years

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