Age Problem
1. Fathers age is 5 times his son’s age. 4 years back the father was 9 times older than son. Find the fathers’ present age.
Solution:
let. son age=x
fathers age=5x
4 years back, so (x-4)=(5x-4)
then father was 9 times older than son so, 9(x-4)=(5x-4).....(1)
solving eqn (1), x=8
and fathers age =8*5=40 years
2. When Usha was thrice as old as Nisha, her sister Asha was 25, When Nisha was half as old as Asha, then sister Usha was 34. their ages add to 100. How old is Usha?
Solution:
Asha=25 Nisha =n Usha =3n
Asha=2a Nisha =a Usha =34
The difference of all ages must be same
25-2a=n-a=3n-34
We will get two equation
2n+a=34
n+a=25
From the equation we will get n=9 and a=16
The ages are Asha=2a=32 Nisha =16 and Usha =34
Sum of the ages =100
(32+x)+(16+x)+(34+x)=100
82+3x=100
3x=18
x=6
Age of Usha =34+6=40
3. Joe’s father will be twice his age 6 years from now. His mother was twice his age 2 years before. If Joe will be 24 two years from now, what is the difference between his father's and mother's age?
Solution:
(J + 6)*2 = F + 6 ---(1)
(J - 2)*2 = M - 2 ---(2)
J + 2 = 24 => J = 22
put J=22 in (1) we get, (22+6)*2 = F + 6 => F = 50
put J=22 in (2) we get, (22-2)*2 = M - 2 => M = 42
so, difference between his father's and mother's age = F - M =(50-42)= 8 years
4. The difference between the ages oftwo of my three grandchildren is 5. My eldest grandchild's age is four timesthan the age of my youngest grandchild and my eldest grandchild's age is threeyears more than the ages of my two youngest grandchildren added together. Howold is my second grandchild?
Solution:
let, eldest grandchild age=F
second grandchild age= S
youngest grandchild age=T
Given that F-S=5 =>F=S+5
F=4T
F=3+S+T
=> S+5=3+S+T
=>T+3=5 =>T=2
=>F=4T=4(2)=8
=>s=F-5=8-5=3
So, the age of second grandchild is 3 years
5. The ages of two friends are in the ratio 6:5. The sum of their ages is 66.After how many years will the ages be in the ratio 8:7?
Solution:
ratio is 6:5 and sum is 66
6x+5x=66
11x=66
x=6
so x is multiple of 6
x=6(6)=36
x=5(60=30
36+y/30+y=8/7
y=12
6. the difference of two ages is 35. 8 years ago one of the persons age is 6 times of other .then what is the sum of both ages
Solution:
x -y =35 ---(i)
(x-8)= 6(y-8) or x-6y=-40 ---(ii)
On solving (i) & (ii), x=50, y=15
Sum of ages=x+y=50+15=65
7. At the end of 1994 rohit was half an old as his grand mother.The sum of years in which they were born is 3844. How old rohit was at the end of 1999
Solution:
let at the end of 1994 grand mother's age is x
and rohit's age x/2
then we can say....birth year of GM is =(1994-x)
and rohit is = (1994 - x/2)
sum of years is 3844
i.e (1994 -x) + (1994 - x/2) = 3884
=> x = 96
i.e GM age is 96
so rohit age will be 96/2 = 48 years
in 1994 age is 48
1995 49
1996 50
1997 51
1998 52
1999 53
so ans should be 53 years
8. In 4 years, Rajs father will be double Rajs age then. Two years ago, while his mother was twice his age that time. If Raj is going to be 32 years old 8 years from now, then what is the sum of his parents age now?
Solution:
R.F=RAJS FATHER AGE
R=RAJS AGE
R.M=RAJS MOTHER AGE
(R.F+4)=2*(R+4)
(R.M-2)=2*(R-2)
R+8=32
therefore, R=24
therefore, R.M=46
therefore, R.F=52
R.F + R.M = 98
9. 9 years ago I was five times as old as my eldest son. Today I am 3 times his age. How old am I now?
Solution:
Father's age is x years
eldest son age is y years
nine years ago--->x-9 = 5(y-9)
x-9 = 5y-45
now-->x=3y
substitute this in above
3y-9=5y-45
45-9=2y
y=18
x=54(father's age)
10. Bhavna is presently three times as old as her daughter Anushka. Ten years ago Bhavna was five times as old as her daughter Anushka was. After how many years, sum of their ages will be 100.
Solution:
Initially , present age of anushka= x;
present age of bhavna= 3x;
before 10 yrs...
3x-10=5(x-10)
x=20;
3*20=60=bhavya
20 =anushka
if 10 yrs is increased........
bhavya=60+10=70
anushka=20+10=30
total=100
11. 2 years ago of A is x times that of B. 3 Years hence the age of A is 4/3 times of B. What is the present age of B in binary form?
Solution:
let age of A be a and age of B be b.
a-2=x*b--1st eq.
a+3=4/3b--2nd eq.
so (2nd eq)-(1st eq):
5=(4/3-x)*b---3rd eq.
here we can see that 5 is a prime no.
so b shud be 5 becoz if we will take (4/3-x)=5 then x will be negative which is not possible.
so b=5yrs.
x=1/3
a=11/3 yrs.
now changing 5 in binary it will be 101.
12. The age of the two friends were in the ration of 6:5. If the sum of their ages is 66. Then after how many years their ratio will become 7:6?
Solution:
6x+5x=66
11x=66
X=6 then
36+x/30+x= 7/6
By solving it we get x=6
13. roy is now 4 years older then erik and half of that amount older than iris.If in 2 years, roy will be twice as old as erik,then in 2 years what would be roy's age multiplied by iris's age?
Solution:
R=E+4;
R=I+(4/2);
(R+2)=2(E+2) solving
R=6;I=4;E=2;
so after 2 years
(R+2)(I+2)
=(8)(6)
=48
14. The difference between the ages of two of my three granchildren is 3.my eldest grandchild is three times older than the age of my youngest grandchild and my eldest granchild's age is two years more than the ages of my two youngest grandchildren added together.how old is my eldest grandchild?
Solution:
x,y,z be ages where x is of eldest,z is of youngest
y-z=3
x=3z
x=y+z+2
solving x=15
15. When Usha was thrice as old as Misha,her sister Asha was 25. When misha was half as old as Asha, then her sister Usha was 34. Their ages add to 100. How old is Usha?
Solution:
Asha=25 Nisha =n Usha =3n
Asha=2a Nisha =a Usha =34
The difference of all ages must be same
=> 25-2a=n-a=3n-34
We will get two equation
=> 2n+a=34
=> n+a=25
From the equation we will get n=9 and a=16
The ages are Asha=2a=32 Nisha =16 and Usha =34
Sum of the ages =100
=> (32+x)+(16+x)+(34+x)=100
=> 82+3x=100
=> 3x=18
=> x=6
=> Age of Usha =34+6
=40
16. Monisha was seven times as old as her daughter Mercy eight years ago. Now three times Mercy’s age is her mother age. Before how many years the ratio of their ages (ratio of mother's age to Mercy's age) will be increased by 1 from the current ratio
Solution:
Eight years ago -- let Monisha’s age be M and Mercy’s age be D.
M = 7 D
Now M + 8 = 3 (D + 8)
7 D + 8 = 3D + 24
7D - 3 D = 24 - 8 = 16
D = 4 = age of Mercy before 8 years
M = 28 = age of Mother before 8 years
Current age of Mercy = D + 8 = 12
Current age of Monisha = M + 8 = 36
Current ratio of mother's age to Mercy's age = 36/12 = 3.
Let before x years this ratio becomes 1 more than the current ratio. i.e x denotes the number of years before which the ratio becomes 3 + 1 = 4.
Then 36 - x / 12 - x = 4
36 - x = 48 - 4x
3x = 12
Or x = 4
i.e before 4 years, the ratio of their ages will be 4.
17. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
answer is:92
father's age=50
mother's age=42
18. Ten years ago X was half of Y. If the ratio of their present ages is 3:4, what will be the total of their present ages?
Solution:
let the present age of X and Y be x,y
x:y=3:4
x-10=1/2(y-10)
on solving the equations we get
x=15,y=20
so total =35
19. Jalia is twice older than qurban . If jalia was 4 years younger, qurban was 3 years older ther diff. between their ages is 12 years what is the sum of thier ages?
Solution:
1. jalia's age=qurban's age*2
2. jalia's age - 4 -(qurban's age + 3)=12
so by solving this two eqa.
we get jalia's age=38 and qurban's age = 19
and sum=38+19=57
20. Old women said that number of days of my grand daughter is equal to the number of months of my daughter and my age in number of years is equal to the number of weeks of my grand daughter. Sum of their ages is 115. Find the age of the old woman.
Solution:
age of grand daughter is d days d/365 years
and age of daughter is d months d/12yrs
and age of old women is d/7 years
d/7 + d/12 + d/365 = 115years
4380d +2555d + 84d = 115*365*12*7
7019d = 3525900
d = 3525900/7019 = 502
age of old women is 502/7
71.7 years
21. 6 persons standing in queue with different age group, after two years their average age will be 43 and seventh person joined with them. Hence the current average age has become 45. Find the age of seventh person?
Solution:
et the total age of 6 people be x yrs
after 2 yrs it becomes (x+12)
therefore; (x+12)/6=43
=>x= 246
again; (246+ 7th person's age)/7=45
=>7th person's age
= 69 yrs
22. At the end of 1894 Suresh was half as old as his grandmother. The sum of the years were born in 3644. How old Suresh was at the end of 1899?
Solution:
suresh-x, grandmother-2x
(1894-x)=(1894-2x)=3644
3788-3x=3644
3x=144
x=48.........suresh 48 and grandmothe-96...in 1894
in 1899 suresh is 53
23. After 6 years Raju's fathers age will twice of the Raju's age 2 years ago.His mothers age was twice that of Raju's age. Sum of the age of their parents.
Solution:
F+6=2(R+6)
F= 2R+6
M-2=2(R-2)
M= 2R-2
Therefore the sum of Raju’s Parent’s age is
F+M=2R+6+2R-2
F+M=4R+4
4 more than four times Raju’s age
24. 9 years ago I was five times as old as my eldest son. Today I am 3 times his age. How old am I now?
Solution:
x-9=5(y-9)
& x=3y
x-9=5x/3-45
x=54
25. Two years ago ,a man's age was 3 times the sqr of his son's age.In 3 yrs ,his age will be 4 times his son's age.find the present age.
Solution:
man = x , son = y
(x-2)= 3(y-3)^2 ------ 1
(x+3)= 4(y+3) ----- 2
solving 1 & 2
y = 5, y= 1/3(neglect)
x = 29
26. Peter and paul are 2 friends.Sum of their ages is 35 years.Peter is twice as old as paul was when peter was as old as paul is now.What is the present age of peter?
Solution:
present age of 2 friends is 35 years.
peter paul
2x y = 35
y x = 35 - 2(2x-y)
solving this x=20 and y =15
so peter current age is 20
27. A man is 37 yrs old and his two sons are 8 yrs and 3 yrs old. After how many years will he be twice as old as sum of the ages of his sons?
Solution:
Let the ans be x, then
after x years,
2*(8+x +3+x) = 37+x
22+4x = 37+x
3x = 15
x = 5
28. the sum of anitha and sunitha age is 43.hence 11 years anitha age will be 7/6 times of sunitha age.find the present sunitha age?
Solution:
present age of sunitha be x. anitha=43-x..
hence 11years after sunitha=x+11, anitha=43-x+11..
anitha age will be 7/6 times of sunitha..
so 43-x+11=(7/6)x+11
solve we get x=19..
present age of sunitha is 19yr.
29. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
f+6=2(R+6)
m-2=2(r-2)
r+3=25;
r=22;
so,
f+6=56;f=50
m-2=2*20;m=42
there fore.. sum of his parents is 92
30. Roy is now 4 years older than Erik and 2 years older than Iris. If in 2 years, Roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?
Solution:
R=4+E-----(1)
R=2+I-----(3)
R+2=2(E+2)
SO, R=2E+2-----(2)
EQUATE 1 AND 2
2E+2=4+E
SUB IN 1 AND 3
E=2
IMPLIES R=6 I=4
AFTER 2 YEARS
R=8 AND I=6;
8*6=48
31. If A's age is 4 yrs more than B's age and 2 yrs more than C's age. After 2 yrs A's age will be twice of B's age. Then find the current ages of A,B & C
Solution:
A=(B+4)=(C+2) ----(1)
after two years
(A+2)=2*(B+2) [put,A=(B+4)]
=> B=2
so, from(1)
A=(B+4)=6 ,
C=(A-2)=4
A=6 yrs, B=2 yrs, C=4 yrs
32. Roy's is now 7 yrs older than Erik and & 2 yrs older than Iris.If in 2 yrs,Roy will be twice as old as Erik, then in 2 yrs ,what would be Roy's age multiplied by Iris age?
Solution:
R=E+7=I+2
R+2=2(E+2)
(E+7)+2=2(E+2)
E=5 SO,R=5+7=12,I=12-2=10
SO,(R+2)(I+2)=14*12=168
33. Ten years ago,the average of family of four members was 24 years.Three children having been born,the average age of family is same today.What are the present ages of children if two children are identically twins and differ by two years from the younger one?
Solution:
en years ago,
age of four members=24*4=96
present total age of those four people=96+40=136
present average of four people and three new children=24
hence total sum=7*24=168
sum of three children=168-136=32let younger member's age be 'z'
z+(z-2)2=32
=>z=12
age of twins=10 each and younger member is of 12 years
34. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
2016 raj's age is 25
2019 he'll be 28,then his father'll be 56
2011 he was 20,then his mother was 40
so,in 2013 there sum of age is
=50+42
=92
Solution:
let. son age=x
fathers age=5x
4 years back, so (x-4)=(5x-4)
then father was 9 times older than son so, 9(x-4)=(5x-4).....(1)
solving eqn (1), x=8
and fathers age =8*5=40 years
2. When Usha was thrice as old as Nisha, her sister Asha was 25, When Nisha was half as old as Asha, then sister Usha was 34. their ages add to 100. How old is Usha?
Solution:
Asha=25 Nisha =n Usha =3n
Asha=2a Nisha =a Usha =34
The difference of all ages must be same
25-2a=n-a=3n-34
We will get two equation
2n+a=34
n+a=25
From the equation we will get n=9 and a=16
The ages are Asha=2a=32 Nisha =16 and Usha =34
Sum of the ages =100
(32+x)+(16+x)+(34+x)=100
82+3x=100
3x=18
x=6
Age of Usha =34+6=40
3. Joe’s father will be twice his age 6 years from now. His mother was twice his age 2 years before. If Joe will be 24 two years from now, what is the difference between his father's and mother's age?
Solution:
(J + 6)*2 = F + 6 ---(1)
(J - 2)*2 = M - 2 ---(2)
J + 2 = 24 => J = 22
put J=22 in (1) we get, (22+6)*2 = F + 6 => F = 50
put J=22 in (2) we get, (22-2)*2 = M - 2 => M = 42
so, difference between his father's and mother's age = F - M =(50-42)= 8 years
4. The difference between the ages oftwo of my three grandchildren is 5. My eldest grandchild's age is four timesthan the age of my youngest grandchild and my eldest grandchild's age is threeyears more than the ages of my two youngest grandchildren added together. Howold is my second grandchild?
Solution:
let, eldest grandchild age=F
second grandchild age= S
youngest grandchild age=T
Given that F-S=5 =>F=S+5
F=4T
F=3+S+T
=> S+5=3+S+T
=>T+3=5 =>T=2
=>F=4T=4(2)=8
=>s=F-5=8-5=3
So, the age of second grandchild is 3 years
5. The ages of two friends are in the ratio 6:5. The sum of their ages is 66.After how many years will the ages be in the ratio 8:7?
Solution:
ratio is 6:5 and sum is 66
6x+5x=66
11x=66
x=6
so x is multiple of 6
x=6(6)=36
x=5(60=30
36+y/30+y=8/7
y=12
6. the difference of two ages is 35. 8 years ago one of the persons age is 6 times of other .then what is the sum of both ages
Solution:
x -y =35 ---(i)
(x-8)= 6(y-8) or x-6y=-40 ---(ii)
On solving (i) & (ii), x=50, y=15
Sum of ages=x+y=50+15=65
7. At the end of 1994 rohit was half an old as his grand mother.The sum of years in which they were born is 3844. How old rohit was at the end of 1999
Solution:
let at the end of 1994 grand mother's age is x
and rohit's age x/2
then we can say....birth year of GM is =(1994-x)
and rohit is = (1994 - x/2)
sum of years is 3844
i.e (1994 -x) + (1994 - x/2) = 3884
=> x = 96
i.e GM age is 96
so rohit age will be 96/2 = 48 years
in 1994 age is 48
1995 49
1996 50
1997 51
1998 52
1999 53
so ans should be 53 years
8. In 4 years, Rajs father will be double Rajs age then. Two years ago, while his mother was twice his age that time. If Raj is going to be 32 years old 8 years from now, then what is the sum of his parents age now?
Solution:
R.F=RAJS FATHER AGE
R=RAJS AGE
R.M=RAJS MOTHER AGE
(R.F+4)=2*(R+4)
(R.M-2)=2*(R-2)
R+8=32
therefore, R=24
therefore, R.M=46
therefore, R.F=52
R.F + R.M = 98
9. 9 years ago I was five times as old as my eldest son. Today I am 3 times his age. How old am I now?
Solution:
Father's age is x years
eldest son age is y years
nine years ago--->x-9 = 5(y-9)
x-9 = 5y-45
now-->x=3y
substitute this in above
3y-9=5y-45
45-9=2y
y=18
x=54(father's age)
10. Bhavna is presently three times as old as her daughter Anushka. Ten years ago Bhavna was five times as old as her daughter Anushka was. After how many years, sum of their ages will be 100.
Solution:
Initially , present age of anushka= x;
present age of bhavna= 3x;
before 10 yrs...
3x-10=5(x-10)
x=20;
3*20=60=bhavya
20 =anushka
if 10 yrs is increased........
bhavya=60+10=70
anushka=20+10=30
total=100
11. 2 years ago of A is x times that of B. 3 Years hence the age of A is 4/3 times of B. What is the present age of B in binary form?
Solution:
let age of A be a and age of B be b.
a-2=x*b--1st eq.
a+3=4/3b--2nd eq.
so (2nd eq)-(1st eq):
5=(4/3-x)*b---3rd eq.
here we can see that 5 is a prime no.
so b shud be 5 becoz if we will take (4/3-x)=5 then x will be negative which is not possible.
so b=5yrs.
x=1/3
a=11/3 yrs.
now changing 5 in binary it will be 101.
12. The age of the two friends were in the ration of 6:5. If the sum of their ages is 66. Then after how many years their ratio will become 7:6?
Solution:
6x+5x=66
11x=66
X=6 then
36+x/30+x= 7/6
By solving it we get x=6
13. roy is now 4 years older then erik and half of that amount older than iris.If in 2 years, roy will be twice as old as erik,then in 2 years what would be roy's age multiplied by iris's age?
Solution:
R=E+4;
R=I+(4/2);
(R+2)=2(E+2) solving
R=6;I=4;E=2;
so after 2 years
(R+2)(I+2)
=(8)(6)
=48
14. The difference between the ages of two of my three granchildren is 3.my eldest grandchild is three times older than the age of my youngest grandchild and my eldest granchild's age is two years more than the ages of my two youngest grandchildren added together.how old is my eldest grandchild?
Solution:
x,y,z be ages where x is of eldest,z is of youngest
y-z=3
x=3z
x=y+z+2
solving x=15
15. When Usha was thrice as old as Misha,her sister Asha was 25. When misha was half as old as Asha, then her sister Usha was 34. Their ages add to 100. How old is Usha?
Solution:
Asha=25 Nisha =n Usha =3n
Asha=2a Nisha =a Usha =34
The difference of all ages must be same
=> 25-2a=n-a=3n-34
We will get two equation
=> 2n+a=34
=> n+a=25
From the equation we will get n=9 and a=16
The ages are Asha=2a=32 Nisha =16 and Usha =34
Sum of the ages =100
=> (32+x)+(16+x)+(34+x)=100
=> 82+3x=100
=> 3x=18
=> x=6
=> Age of Usha =34+6
=40
16. Monisha was seven times as old as her daughter Mercy eight years ago. Now three times Mercy’s age is her mother age. Before how many years the ratio of their ages (ratio of mother's age to Mercy's age) will be increased by 1 from the current ratio
Solution:
Eight years ago -- let Monisha’s age be M and Mercy’s age be D.
M = 7 D
Now M + 8 = 3 (D + 8)
7 D + 8 = 3D + 24
7D - 3 D = 24 - 8 = 16
D = 4 = age of Mercy before 8 years
M = 28 = age of Mother before 8 years
Current age of Mercy = D + 8 = 12
Current age of Monisha = M + 8 = 36
Current ratio of mother's age to Mercy's age = 36/12 = 3.
Let before x years this ratio becomes 1 more than the current ratio. i.e x denotes the number of years before which the ratio becomes 3 + 1 = 4.
Then 36 - x / 12 - x = 4
36 - x = 48 - 4x
3x = 12
Or x = 4
i.e before 4 years, the ratio of their ages will be 4.
17. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
answer is:92
father's age=50
mother's age=42
18. Ten years ago X was half of Y. If the ratio of their present ages is 3:4, what will be the total of their present ages?
Solution:
let the present age of X and Y be x,y
x:y=3:4
x-10=1/2(y-10)
on solving the equations we get
x=15,y=20
so total =35
19. Jalia is twice older than qurban . If jalia was 4 years younger, qurban was 3 years older ther diff. between their ages is 12 years what is the sum of thier ages?
Solution:
1. jalia's age=qurban's age*2
2. jalia's age - 4 -(qurban's age + 3)=12
so by solving this two eqa.
we get jalia's age=38 and qurban's age = 19
and sum=38+19=57
20. Old women said that number of days of my grand daughter is equal to the number of months of my daughter and my age in number of years is equal to the number of weeks of my grand daughter. Sum of their ages is 115. Find the age of the old woman.
Solution:
age of grand daughter is d days d/365 years
and age of daughter is d months d/12yrs
and age of old women is d/7 years
d/7 + d/12 + d/365 = 115years
4380d +2555d + 84d = 115*365*12*7
7019d = 3525900
d = 3525900/7019 = 502
age of old women is 502/7
71.7 years
21. 6 persons standing in queue with different age group, after two years their average age will be 43 and seventh person joined with them. Hence the current average age has become 45. Find the age of seventh person?
Solution:
et the total age of 6 people be x yrs
after 2 yrs it becomes (x+12)
therefore; (x+12)/6=43
=>x= 246
again; (246+ 7th person's age)/7=45
=>7th person's age
= 69 yrs
22. At the end of 1894 Suresh was half as old as his grandmother. The sum of the years were born in 3644. How old Suresh was at the end of 1899?
Solution:
suresh-x, grandmother-2x
(1894-x)=(1894-2x)=3644
3788-3x=3644
3x=144
x=48.........suresh 48 and grandmothe-96...in 1894
in 1899 suresh is 53
23. After 6 years Raju's fathers age will twice of the Raju's age 2 years ago.His mothers age was twice that of Raju's age. Sum of the age of their parents.
Solution:
F+6=2(R+6)
F= 2R+6
M-2=2(R-2)
M= 2R-2
Therefore the sum of Raju’s Parent’s age is
F+M=2R+6+2R-2
F+M=4R+4
4 more than four times Raju’s age
24. 9 years ago I was five times as old as my eldest son. Today I am 3 times his age. How old am I now?
Solution:
x-9=5(y-9)
& x=3y
x-9=5x/3-45
x=54
25. Two years ago ,a man's age was 3 times the sqr of his son's age.In 3 yrs ,his age will be 4 times his son's age.find the present age.
Solution:
man = x , son = y
(x-2)= 3(y-3)^2 ------ 1
(x+3)= 4(y+3) ----- 2
solving 1 & 2
y = 5, y= 1/3(neglect)
x = 29
26. Peter and paul are 2 friends.Sum of their ages is 35 years.Peter is twice as old as paul was when peter was as old as paul is now.What is the present age of peter?
Solution:
present age of 2 friends is 35 years.
peter paul
2x y = 35
y x = 35 - 2(2x-y)
solving this x=20 and y =15
so peter current age is 20
27. A man is 37 yrs old and his two sons are 8 yrs and 3 yrs old. After how many years will he be twice as old as sum of the ages of his sons?
Solution:
Let the ans be x, then
after x years,
2*(8+x +3+x) = 37+x
22+4x = 37+x
3x = 15
x = 5
28. the sum of anitha and sunitha age is 43.hence 11 years anitha age will be 7/6 times of sunitha age.find the present sunitha age?
Solution:
present age of sunitha be x. anitha=43-x..
hence 11years after sunitha=x+11, anitha=43-x+11..
anitha age will be 7/6 times of sunitha..
so 43-x+11=(7/6)x+11
solve we get x=19..
present age of sunitha is 19yr.
29. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
f+6=2(R+6)
m-2=2(r-2)
r+3=25;
r=22;
so,
f+6=56;f=50
m-2=2*20;m=42
there fore.. sum of his parents is 92
30. Roy is now 4 years older than Erik and 2 years older than Iris. If in 2 years, Roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?
Solution:
R=4+E-----(1)
R=2+I-----(3)
R+2=2(E+2)
SO, R=2E+2-----(2)
EQUATE 1 AND 2
2E+2=4+E
SUB IN 1 AND 3
E=2
IMPLIES R=6 I=4
AFTER 2 YEARS
R=8 AND I=6;
8*6=48
31. If A's age is 4 yrs more than B's age and 2 yrs more than C's age. After 2 yrs A's age will be twice of B's age. Then find the current ages of A,B & C
Solution:
A=(B+4)=(C+2) ----(1)
after two years
(A+2)=2*(B+2) [put,A=(B+4)]
=> B=2
so, from(1)
A=(B+4)=6 ,
C=(A-2)=4
A=6 yrs, B=2 yrs, C=4 yrs
32. Roy's is now 7 yrs older than Erik and & 2 yrs older than Iris.If in 2 yrs,Roy will be twice as old as Erik, then in 2 yrs ,what would be Roy's age multiplied by Iris age?
Solution:
R=E+7=I+2
R+2=2(E+2)
(E+7)+2=2(E+2)
E=5 SO,R=5+7=12,I=12-2=10
SO,(R+2)(I+2)=14*12=168
33. Ten years ago,the average of family of four members was 24 years.Three children having been born,the average age of family is same today.What are the present ages of children if two children are identically twins and differ by two years from the younger one?
Solution:
en years ago,
age of four members=24*4=96
present total age of those four people=96+40=136
present average of four people and three new children=24
hence total sum=7*24=168
sum of three children=168-136=32let younger member's age be 'z'
z+(z-2)2=32
=>z=12
age of twins=10 each and younger member is of 12 years
34. In 6 years, Raj's father will be twice Raj's age then. Two years ago, while his mother was twice his age then. If Raj is going to be 25 years old 3 years from now, then what is the sum of his parent's age now?
Solution:
2016 raj's age is 25
2019 he'll be 28,then his father'll be 56
2011 he was 20,then his mother was 40
so,in 2013 there sum of age is
=50+42
=92
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