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The friends whose ages from a G.P. divide a certain sum of money in proportion to their ages. If they do that three years later, when the youngest is halfg the age of the oldest, then he will receive 105 rupees more that the he gets now and the middle friends will get 15 reupees more that he gets now, then ages of the friends are |
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Answer» Let their ages be `a,ar,ar^(2)`, After 3 yr, their ages will be `a+3,ar+3,ar^(2)+3`. Given, `2(a+3)=ar^(2)+3" " "……..(i)"` Let x rupee be the sum of the money divided. And Let `y=a+ar+ar^(2)" " "………(ii)"` Then, `y+9=a+3+(ar+3)+(ar^(2)+3)` We have, `(x(a+3))/((y+9))=(xa)/(y)+105` `impliesx((a+3)/(y+9)-(a)/(y))=105" " ".........(iii)"` Also, `(x(ar+3))/((y+9))=(xar)/(y)+15` `impliesx[(ar+3)/(y+9)-(ar)/(y)]=15" " ".........(iv)"` On dividing Eq. (iii)" by Eq. "(iv), we get `implies(y(a+3)-a(y+9))/(y(ar+3)-ar(y+9))=7 implies (y-3a)/(y-3ar)=7` ` implies 6y=21ar-3a implies y=(a(7r-1))/(2)` From Eq. (ii), `(a(7r-1))/(2)=a+ar+ar^(2)` `implies 5ar=3a+2ar^(2)" " ".........(v)"` From Eqs. (i) and (ii), `a=12,r=(3)/(2)` Let ages of these friends are `12,12xx(3)/(2),12xx((3)/(2))^(2)i.e.12,18,27`. |
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