The present ages of three persons are in proportions $4 : 7 : 9$ . Eighty years ago, the sum of their ages was $56$ . Find their present ages (in years).

Question

The present ages of three persons are in proportions 4-7-9- Eighty years ago- the of their ages was 56- Find their present ages -in years-8- 20- 2816- 28- 3620- 35- 4513- 16- 25

Toppr Admin · Accepted Answer

X-y-z-4-7-9x4-y7-z97x-x2212-4y -x2026-x2026-x2026-1-9x-4z -x2026-x2026-x2026-x2026-2-x-x2212-8-y-x2212-8-z-x2212-8-56x-y-z-80x-7x24-9x4-80 from -1- and -2-5x-80x-16y-28z-3616-28-36-

The present ages of three persons are in proportions $4 : 7 : 9$ . Eighty years ago, the sum of their ages was $56$ . Find their present ages (in years).
$8, 20, 28$
$16, 28, 36$
$20, 35, 45$
$13, 16, 25$

$x : y : z = 4 : 7 : 9$
$\frac{x}{4} = \frac{y}{7} = \frac{z}{9}$
$7 x - 4 y$ ……….. $(1)$
$9 x = 4 z$ ………… $(2)$
$(x - 8) + (y - 8) + (z - 8) = 56$
$x + y + z = 80$
$x + \frac{7 x}{24} + \frac{9 x}{4} = 80$ from $(1)$ and $(2)$
$5 x = 80$
$x = 16$
$y = 28$
$z = 36$
$16, 28, 36$ .

The present ages of three persons are in proportions 4:7:9. Eighty years ago, the sum of their ages was 56. Find their present ages (in years).8, 20, 2816, 28, 3620, 35, 4513, 16, 25

x:y:z=4:7:9x4=y7=z97x−4y ………..(1)9x=4z …………(2)(x−8)+(y−8)+(z−8)=56x+y+z=80x+7x24+9x4=80 from (1) and (2)5x=80x=16y=28z=3616,28,36.

The present ages of three persons are in proportions $4 : 7 : 9$ . Eighty years ago, the sum of their ages was $56$ . Find their present ages (in years).
$8, 20, 28$
$16, 28, 36$
$20, 35, 45$
$13, 16, 25$

$x : y : z = 4 : 7 : 9$
$\frac{x}{4} = \frac{y}{7} = \frac{z}{9}$
$7 x - 4 y$ ……….. $(1)$
$9 x = 4 z$ ………… $(2)$
$(x - 8) + (y - 8) + (z - 8) = 56$
$x + y + z = 80$
$x + \frac{7 x}{24} + \frac{9 x}{4} = 80$ from $(1)$ and $(2)$
$5 x = 80$
$x = 16$
$y = 28$
$z = 36$
$16, 28, 36$ .