ABC26GN3122 · Ratio and Proportion

Subject: General Aptitude · Chapter: Ratio and Proportion · Exam: · Marks: · Difficulty:

What number must be added to each of the numbers 7, 11 and 19 so that the resulting numbers may be in continued proportion?

(a)-3

(b)-4

(c)3

(d)4

Answer

Answer (as printed): A

Explanation

Let the required number be $x$. Then, $$\begin{aligned} & (7+x):(11+x)::(11+x):(19+x) \\ & \Rightarrow \frac{7+x}{11+x}=\frac{11+x}{19+x} \Rightarrow(7+x)(19+x)=(11+x)^{2} \\ & \Rightarrow x^{2}+26 x+133=x^{2}+22 x+121 \\ & \Rightarrow 4 x=-12 \Rightarrow x=-3 \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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