Data Sufficiency Practice Test 3

Using Statement I:
Lalita’s investment = a
Shalini’s investment $=\frac{120\mathit{a}}{100}=\frac{6\mathit{a}}{5}$
Using Statement II:
Lalita’s investment = $\frac{2}{3}$ of Manisha’s investment
Using Statement III:
Manisha’s investment = $\frac{125}{100}$ Shalini
Using Statements I and II:
Let Lalita’s investment = a
Shalini’s investment = $\frac{120\mathit{a}}{100}=\frac{6}{5}\mathit{a}$
Manisha’s investment $=\frac{5}{4}\times \frac{6}{5}\mathit{a}=\frac{3}{2}\mathit{a}$
Ratio of investment
$\begin{array}{ccccc}\mathit{L}& :& \mathit{S}& :& \mathit{M}\\ \mathit{a}& :& \frac{6}{5}\mathit{a}& :& \frac{3}{2}\mathit{a}\\ 10\mathit{a}& :& 12\mathit{a}& :& 15\mathit{a}\\ 10& :& 12& :& 15\end{array}$
Manisha’s share in profit $=\frac{15}{37}\mathit{a}$
Using Statements II and III:
Let Shalini’s investment = a
Manisha’s investment $=\frac{5}{4}\mathit{a}$
Lalita’s investment $=\frac{2}{3}\times \frac{5}{4}\mathit{a}=\frac{5}{6}\mathit{a}$
Ratio of investment
$\begin{array}{ccccc}\mathit{L}& :& \mathit{S}& :& \mathit{M}\\ \frac{5}{6}\mathit{a}& :& \mathit{a}& :& \frac{5}{4}\mathit{a}\\ 10\mathit{a}& :& 12\mathit{a}& :& 15\mathit{a}\\ 10& :& 12& :& 15\end{array}$
Manisha’s share $=\frac{15}{37}\mathit{a}$
Using Statements III and I:
Let Lalita’s investment = a
Shalini’s investment $=\frac{6}{5}\mathit{a}$
Manisha’s investment $=\frac{6}{5}\times \frac{5}{4}\mathit{a}=\frac{3}{2}\mathit{a}$
Ratio of investment:
$\begin{array}{ccccc}\mathit{L}& :& \mathit{S}& :& \mathit{M}\\ \mathit{a}& :& \frac{6}{5}\mathit{a}& :& \frac{3}{2}\mathit{a}\\ 10& :& 12& :& 15\end{array}$
Manisha’s share $=\frac{15}{37}\mathit{a}$
Hence, solution can be derived by using any two of the three statements.