Problem 1 :
Martin is four times as old as his brother Luther at present. After 10 years he will be twice the age of his brother. Find their present ages.
Solution :
Let x and y be the present ages of Martin and Luther respectively.
Given : Martin is four times as old as his brother Luther at present.
x = 4y ----(1)
Given : After 10 years, Martin will be twice the age of his brother Luther.
x + 10 = 2(y + 10)
x + 10 = 2y + 20
Subtract 10 from each side.
x = 2y + 10 ----(2)
From (1) and (2),
4y = 2y + 10
Subtract 2y from each side.
2y = 10
Divide each side by 2.
y = 5
Substitute 5 for y in (1).
x = 4(5)
x = 20
So, the present ages of martin and Luther are 20 years and 5 years respectively.
Problem 2 :
A father is 30 years older than his son,and one year ago he was four times as old as his son. Find the present ages of his father and his son.
Solution :
Let x and y be the present ages of father and son respectively.
Given : A father is 30 years older than his son.
x = y + 30 ----(1)
Given : One year ago, father was four times as old as his son.
x - 1 = 4(y - 1)
x - 1 = 4y - 4
Add 1 to each side.
x = 4y - 3 ----(2)
From (1) and (2),
y + 30 = 4y - 3
Subtract y from each side.
30 = 3y - 3
Add 3 to each side.
33 = 3y
Divide each side by 3.
11 = y
Substitute 5 for y in (1).
(1)----> x = 11 + 30
x = 41
So, the present ages of father and son are 41 years and 11 years respectively.
Problem 3 :
The ages of Abraham and Adam are in the ratio 5 : 7. Four years from now, the ratio of their ages will be 3 : 4. Find the present ages of them.
Solution :
Given : The ages of Abraham and Adam are in the ratio
5 : 7
From the above ratio,
age of Abraham = 5x
age of Adam = 7x
Four years from now,
age of Abraham = 5x + 4
age of Adam = 7x + 4
Given : Four years from now, the ratio of their ages will be
3 : 4
Then, we have
(5x + 4) : (7x + 4) = 3 : 4
4(5x + 4) = 3(7x + 4)
20x + 16 = 21x + 12
Subtract 20x from each side.
16 = x + 12
Subtract 12 from each side.
4 = x
5x = 5(4) = 20
7x = 7(4) = 28
So, the present ages of Abraham and Adam are 20 years and 28 years.
Problem 4 :
Airi's mother is four times as old as Airi. After five years her mother will be three times as old as she will be then. Find their present ages.
Solution :
Let x and y be the present ages of Mother and Airi respectively.
Given : Airi's mother is four times a old as Airi.
x = 4y ----(1)
Given : After five years Airi's mother will be three times as old as Airi will be then.
x + 5 = 3(y + 5)
x + 5 = 3y + 15
Subtract 5 from each side.
x = 3y + 10 ----(2)
From (1) and (2),
4y = 3y + 10
Subtract 3y from each side.
y = 10
Substitute 10 for y in (1).
x = 4(10)
x = 40
So, the present ages of Airi's mother and Airi are 40 years and 10 years respectively.
Problem 5 :
The sum of the present ages of Kiran and Kate is 60 years. If the ratio of their present ages be 7 : 8, find their present age.
Solution :
Let x and y be the present ages of Kiran and Kate respectively.
Given : The ratio of the present ages of Kiran and Kate is
7 : 8
present age of Kiran = 7x
present age of Kate = 8x
Given : The sum of the present ages of Kiran and Kate is 60 years.
7x + 8x = 60
15x = 60
Divide each side by 15.
x = 4
7x = 7(4) = 28
8x = 8(4) = 32
So, the present ages of Kiran and Kate are 28 years and 32 years respectively.
Problem 6 :
Andrea is three times as old as her sister Anu. Three years ago, she was two years less than four times the age of her sister. Find their present ages.
Solution :
Let x and y be the present ages of Andrea and Anu respectively.
Given : Andrea is three times as old as her sister Anu.
x = 3y ----(1)
Given : Three years ago, Andrea was two years less than four times the age of her sister Anu.
x - 3 = 4(y - 3) - 2
x - 3 = 4y - 12 - 2
x - 3 = 4y - 14
Add 3 to each side.
x = 4y - 11 ----(2)
From (1) and (2),
3y = 4y - 11
Subtract 3y from each side.
0 = y - 11
Add 11 to each side.
11 = y
Substitute 11 for y in (1).
x = 3(11)
x = 33
So, the present ages of Andrea and Anu are 33 years and 11 years respectively.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Aug 09, 24 08:39 PM
Aug 09, 24 08:36 PM
Aug 09, 24 06:15 AM