Question

Write 'T' for true and 'F' for false

(i) Mean proportional between 0.4 and 0.9 is 6.
(ii) The third proportional to 9 and 12 is 10.5.
(iii) If 8 : x :: 48 : 18, then x = 3.
(iv) If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = 5 : 2

Answer 1

(i) F
Suppose that the mean proportional is x.
Then, 0.4 : x :: x : 0.9
$\Rightarrow 0.9\times 0.4=x\times x(\mathrm{Product}\mathrm{of}\mathrm{extremes}=\mathrm{Product}\mathrm{of}\mathrm{means})$
$$\begin{gathered} \Rightarrow x^2=0.36 \\ \Rightarrow x=0.6 \end{gathered}$$


(ii) F
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
⇒ 9x = 144 (Product of extremes = Product of means)
⇒ x = 16

(iii) T
8 : x :: 48 : 18
⇒ 144 = 48x (Product of extremes = Product of means)
⇒ x = 3

(iv) T

$$\begin{gathered} \frac{3a+5b}{3a-5b}=\frac{5}{1} \\ \Rightarrow 3a+5b=15a-25b \end{gathered}$$
⇒ 12a = 30b

⇒ a : b = 5 : 2