The combined age of husband and wife are three times the combined ages of their children. Three years ago the ratio of their ages was 4 : 1, while 5 years after the ratio will be 20:9. How many children have they?0
Answers
Let p= present sum of the parents ages
Let c= present sum of the children's ages
Let n= number of children
THE SUM OF AGES OF HUSBAND AND HIS WIFE IS 4 TIMES THE SUM OF AGES OF THEIR CHILDREN.
∴ p=4c ....(1)
4 YEARS AGO THE RATIO OF THEIR AGES TO THE AGES OF THEIR CHILDREN WAS 18:1
That will reduce the sum of the parents ages by 8, the children's age by 4n
∴ p−8=18(c−4n)
∴ p−8=18c−72n .....(2)
replace p with 4c (From (1))
∴ 4c−8=18c−72n
subtract 18c from both sides and rearrange to:
∴ −14c+72n=8 .......(3)
2 YEARS HENCE THE RATIO WILL BE 3:1.
that will increase the parents age sum by 4, the children's age by 2n
∴ p+4=3(c+2n)
∴ p+4=3c+6n .....(4)
replace p with 4c (From (1))
∴ 4c+4=3c+6n
subtract 3c from both sides and rearrange to:
c=6n−4 .....(5)
In the (3) replace c with (6n−4)
∴ −14(6n−4)+72n=8
∴ −84n+56+72n=8
∴ −12n=8−56
∴ −12n=−48
∴ n=−48/−12
∴ n=4
Number if children's=4
follow me
Question:--
The combined age of husband and wife are three times the combined ages of their children.
Three years ago the ratio of their ages was 4 : 1, while 5 years after the ratio will be 20:9. How many children have they?
Step-by-step-explanation:
Answer:--
Given:
Let p = present sum of the parents ages
Let c = present sum of the children's ages
Let n= number of children
The sum of ages of husband and his wife is 4 times the sum of ages of their children.
∴ p=4c ....(1)
4 years ago the ratio of their ages to the ages of their children was 18:1.
That will reduce the sum of the parents ages by 8, the children's age by 4n
∴ p−8=18(c−4n)
∴ p−8=18c−72n .....(2)
replace p with 4c (From (1))
∴ 4c−8=18c−72n
subtract 18c from both sides and rearrange to:
∴ −14c+72n=8 .......(3)
2 YEARS HENCE THE RATIO WILL BE 3:1.
that will increase the parents age sum by 4, the children's age by 2n
∴ p+4=3(c+2n)
∴ p+4=3c+6n .....(4)
replace p with 4c (From (1))
∴ 4c+4=3c+6n
subtract 3c from both sides and rearrange to:
c=6n−4 .....(5)
In the (3) replace c with (6n−4)
∴ −14(6n−4)+72n=84
∴ −84n+56+72n=8
∴ −12n=8−56
∴ −12n=−48
∴ n=−48/−12
∴ n=4
Therefore,
Number if children's=4