five years ago the father is 2 times the sum of the ages of his three children. after 5 years father's age can be obtained by subtracting 8 get from the sum of the ages of 3 children. fathers present age is
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Let us suppose that Father's age be x years and be y years.
Given : Case I, Five years ago, two times the sum Father's age was of ages of his three _(i) children.
Rightarrow x-5=2(y-15); =>x-5=2y-30; Rightarrow x-2y=-25 \
Case II, After five years, Father's age can 8 from the be obtained by subtracting sum of ages of the three children. = x+5=y+15-8;
Rightarrow*-y=2 \ __(ii) Subtract equation (ii) from equation (i) we get,;
= (x-2y)-(x-y)=-25-2; =>x-2y-x+y=-27 sum of ages of his three son's
Putting the value of y in equation
(i), Rightarrow x-2*27=-25; =x=-25+54=29
Hence,
Present age of father is 29 years.
Step-by-step explanation:
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