DRDO Multi tasking staff Model paper 7

Let the three positive numbers be x, y and z
Given,
⇒

x

y

=

5

6

,

y

z

=

9

1

andxz=480 ⇒ 6x = 5y ----(1)
⇒ y = 9z ----(2)
From (1) and (2),
⇒

6x

5

= 9z ⇒ 6x = 45z ⇒ 2x = 15z ⇒ z =

2x

15

----(3)
Substituting the value of z in xz = 480
x×

2x

15

=480⇒x2=3600⇒x=60
Putting x = 60 in (1) and (3):
⇒ y = (6 × 60)/5 = 72
⇒ z = 120/15 = 8
∴ x + y + z = 60 + 72 + 8 = 140