Which of the following numbers must be added to each of the numbers 9, 16, 13 and 23 so that the resulting numbers are in proportion?

Correct option is C

Given:
Numbers are 9, 16, 13, 23.
Formula Used:
If four numbers a, b, c, d are in proportion, then:
​$$\frac{a}{b} = \frac{c}{d} \quad \text{or} \quad a \times d = b \times c$$​​
Solution:

​$$\frac{9+x}{16+x} = \frac{13+x}{23+x} \\(9+x)(23+x) = (16+x)(13+x) \\(9+x)(23+x) = 207 + 32x + x^2 \\(16+x)(13+x) = 208 + 29x + x^2 \\207 + 32x = 208 + 29x \\32x - 29x = 208 - 207 \\3x = 1 \\x = \tfrac{1}{3}\\$$​​