adkikani wrote:

niks18 amanvermagmatWhat is wrong in below approach for St 2:

A = R * T

200 = \(\frac{200}{4}\) * 4 (A + B)

4 = \(\frac{4}{x}\) * x (Only A)

1 = \(\frac{1}{x}\) * x (Only B) we are given same time x for each A and B

adding total rates:

\(\frac{4}{x}\) + \(\frac{1}{x}\) = 50 or x = \(\frac{1}{10}\)

time = Amount of work / rate ie 200 / (1/10) ie 2000

Hi

adkikaniSorry but i could not comprehend your approach

This is a very simple question that do not require any calculation. For DS you need to simplify the question stem first and then take a shot at the statements

Let A's rate be \(r_1\) widget/hr and B's rate be \(r_2\) widget/hr. So in \(4\) hours number of widgets produced by them -

\(4r_1+\)\(4r_2=200=>\) \(r_1+\)\(r_2=50\)-----------(1) we need to find \(r_1\). so we need either the value of \(r_2\) or relationship between \(r_1\) & \(r_2\)

Statement 1: Directly provides the value of \(r_2=\frac{50}{5}=10\). Hence

sufficientStatement 2: Directly gives the relationship between \(r_1\) & \(r_2\). Hence

sufficient.

For the sake of calculation, let B produces \(1\) widget in \(t\) hours so \(r_2=\frac{1}{t}=>t=\frac{1}{r_2}\)

also \(r_1=\frac{4}{t}=>r_1=4r_2\) (substituting the value of \(t\))

Hence we have \(r_1\)\(+r_2=50=>4r_2+r_2=50=>r_2=10\)

so \(r_1=40\). Hence time taken by A to produce \(200\) widgets \(= \frac{200}{40}=5\) hrs