Question

must be added to each of 7, 16, 43 and 79 so that they are in proportion.

• A. 5
• B. 2
• C. 7
• D. 6

Answer 1

The correct option is A 5

Let x be added to each of 7, 16, 43 and 79 so that the resulting number are in proportion. i.e.,
$(7+x):(16+x)::(43+x):(79+x)$

$\Rightarrow \frac{(7+x)}{(16+x)}=\frac{(43+x)}{(79+x)}$
$\Rightarrow (7+x)(79+x)=(16+x)(43+x)$
$\Rightarrow 553+7x+79x+x^2=688+16x+43x+x^2$
$\Rightarrow 553+86x+x^2=688+59x+x^2$
$\Rightarrow 86x–59x+x^2–x^2=688–553$
$\Rightarrow 27x=135$
$\Rightarrow x=\frac{135}{27}=5$
The numbers are: (7 + 5), (16 + 5), (43 + 5) and (79 + 5)
i.e. 12, 21, 48 and 84.