the sum of ages of husband and his wife is four times the sum of ages of their children.four year ago, the ratio of sum of their ages to the sum of sum of ages of their children was 18:1.two year hence the ration will be3:1 how many children do they have?
Answers
wife's age = Y yrs four yrs ago = Y - 4 yrs
Let them have N children.
Let sum of ages of their children = Z yrs
this age four years ago = Z - 4 * N yrs
this age after 2 yrs from now : Z + 2 * N years
Given
X + Y = 4 * Z ---- (1)
Four years ago:
[ (X-4) + (Y-4) ] : (Z - 4* N) = 18 : 1
(X + Y - 8) = 18 (Z - 4 * N)
X + Y = 18 Z - 72 N + 8 --- (2)
from (1) and (2):
14 Z + 8 = 72 * N
N = (7 Z + 4)/ 36 --- (3)
As N is an integer, 7 Z + 4 = 36 or 72 or 108 or 144 or 180 ...
Then Z = 20 for 7 Z + 4 = 144.
Then N = 4
So X + Y = 80 from (1)
Two years from now:
sum of ages of parents = X+2+Y+ 2 = 84 yrs
sum of ages of children = 20 + 2 * 4 = 28 years
Their ratio = 84: 28 = 3 : 1
Answer:
4
Step-by-step explanation:
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