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