unanswered question
the sum of ages of parents is twice sum of children's age.5 years ago ,the sum of parent's age is 4 times sum of children's age.in 15 years sum of parent's age will be equal to sum of children's age.so how many children are there in the family?
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- 08 Jun
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strongLet c = sum of the children’s age at presentbr /Let n = no. of childrenbr /Let p = sum of the parent’s age at presentbr /:br /"At present, the sum of the parent's age is twice the sum of the children's ages."br /p = 2cbr /:br /" Five years ago, the sum of the parents' ages was 4 times the sum of the children's ages.br /p-10 = 4(c-5n); (we have to subtract 5 yrs for each person)br /p-10 = 4c - 20nbr /Replace p with 2cbr /2c - 10 = 4c - 20nbr /2c - 4c + 20n = 10br /-2c + 20n = 10br /:br /"fifteen years hence, the sum of the parent's ages will be equal to the sum of the children's ages."br /p + 30 = c + 15n; (we have to add 15 yrs for each person)br /Replace p with 2cbr /2c + 30 = c + 15n/strongbr /strong2c - c - 15n = -30br /c - 15n = -30br /Multiply the above equation by 2, add to the previous equationbr /2c - 30n = -60br /-2c +20n = 10br /-----------------adding eliminates c, find nbr /-10n = -50br /n = +5 childrenbr /:br /:br /Check this by finding c, using c - 15n = -30/strongbr /strongc - 15(5) = - 30br /c = -30 + 75br /c = 45 is the sum of the childrens agebr /thenbr /p = 2(45)br /p = 90 is sum of the parents agebr /:br /Check solution in the statement:br /" Five years ago, the sum of the parents' ages was 4 times the sum of the children's ages.br /90 - 10 = 4(45-5(5))br /80 = 4(45-25)br /80 = 4(20)br /:br /We can say that there are 5 children/strongstrong /strong