CAT Exam Model Paper 4 with solutions for free online practice

Let n units of each item be produced by machine A:
Hence, $\frac{\mathit{n}}{75}+\frac{\mathit{n}}{150}+\frac{\mathit{n}}{125}+\frac{\mathit{n}}{50}=4$
$\Rightarrow \mathit{n}\left(\frac{1}{75}+\frac{1}{150}+\frac{1}{125}+\frac{1}{50}\right)=4$
$\Rightarrow \left(\frac{\mathit{n}}{25}\right)\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{5}+\frac{1}{2}\right)=4$
$\Rightarrow \mathit{n}=\frac{100}{1.2}=83.33$
Since the machine, at maximum, can work for four hours, we will have to discard decimal part of 83.33.
At maximum, 4 × 83 = 332 units can be produced by machine A.
Hence, 4n = 332.
Let k units of each item be produced by machine B:
Hence, $\frac{\mathit{k}}{125}+\frac{\mathit{k}}{75}+\frac{\mathit{k}}{100}+\frac{\mathit{k}}{100}=4$
$\mathit{k}\left(\frac{1}{125}+\frac{1}{75}+\frac{1}{100}+\frac{1}{100}\right)=4$
$\left(\frac{\mathit{k}}{25}\right)\left(\frac{1}{5}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}\right)=4$
$\mathit{k}=\frac{100}{1.033}=96.77$
At maximum, 4 x 96 = 384 units can be produced by machine B.
Hence, 4k = 384
Ratio = 332 : 384 = 83 : 96.