DS Practice Questions

Let there be S Shovels and R rakes. The wordy question language may then be simplified to reiterate the question as saying: What is the value of (14S + 9R) ?
STATEMENT (1) alone: A 2 : 3 ratio does not give us a fix on the total number of shovels and rakes purchased. It merely states the proportion of items that constitutes shovels and rakes. For instance there can 2 Shovels and 3 Rakes or 4 Shovels and 6 Rakes or 8 Shovels and 12 Rakes (and so on) in his purchase. Each possibility yields a different value for (14S + 9R). Which is why,
STATEMENT (1) alone - INSUFFICIENT
STATEMENT (2) alone: This statement says S + R = 50. Although this statement gives us a fix on the total number purchased. It gives no clue as to the individual number of each contained in the sum total of 50. Like the above explanation, we can have multiple sets of (S,R) (say (10,40) or (25,25) for instance) values that add up to 50. All those values again yield different value for (14S + 9R).
STATEMENT (2) alone - INSUFFICIENT
STATEMENT (1) & (2) together: Together we have a fix on both the total number and the distribution (ratio) of the two items within the total. This is enough to yield a unique value for both S & R. Alternatively, mathematically this may be seen as being given two equations: (1) S = (2/3)*R & (2) S + R = 50 to solve for two variables: S & R uniquely. The unique set (S,R) further yields a unique value of (14S + 9R). hence,
STATEMENT (1) & (2) together - SUFFICIENT
ANSWER – (C).

The question introduces two variable sets with the possibility/certainty of an overlap. Such language is typical of two variable sets questions and these questions are best tackled by chalking out the information on a table as follows: $$\begin{array}{|c|c|c|c|} \hline & \text{SET A} & \overline{\text{SET A}} & \text{TOTAL} \\ \hline \text{SET B} & \begin{array}{c} \text{No. of elements} \\ \text{common to both} \end{array} & \begin{array}{c} \text{No. of elements in} \\ \text{B but not in A} \end{array} & \begin{array}{c} \text{Total of SET B} \\ \text{(SUM of left 2)} \end{array} \\ \hline \overline{\text{SET B}} & \begin{array}{c} \text{No. of elements in} \\ \text{A but not in B} \end{array} & \begin{array}{c} \text{No. of elements in} \\ \text{neither A nor B} \end{array} & \begin{array}{c} \text{Total of everything} \\ \text{not of B} \\ \text{(SUM of left 2)} \end{array} \\ \hline \text{TOTAL} & \begin{array}{c} \text{Total of SET A} \\ \text{(SUM of above 2)} \end{array} & \begin{array}{c} \text{Total of everything} \\ \text{not of A} \\ \text{(SUM of above 2)} \end{array} & \begin{array}{c} \text{ENTIRE} \\ \text{TOTAL} \end{array} \\ \hline \end{array}$$ \(\overline{\text{SET A}}\) and \(\overline{\text{SET B}}\) represent the complements of SET A and SET B, which is nothing but the set of elements not belonging to SET A and SET B respectively. Note that: No. of elements(SET A) + No. of elements(\(\overline{\text{SET A}}\)) = Entire Total. ; same for B. Using the information given only in the question we can begin by creating our table and filling in the information and placing a ‘?’ sign at the place that we're required to find. Let us say the Total number of students we're dealing with is \(X\). $$ \begin{array}{|c|c|c|c|} \hline & \text{Likes Lima} & \text{Dislikes Lima} & \text{TOTAL}\\ \hline \text{Likes Sprouts} & & ? & \\ \hline \text{Dislikes Sprouts} & & \frac35\cdot\frac23 X & \\ \hline \text{TOTAL} & \frac13 X & \frac23 X & X \\ \hline \end{array} $$ Filling in the above info allows us to calculate for the required box: The required box has a value $$ \frac23 X - \frac35\!\left(\frac23 X\right)=\left(1-\frac35\right)\frac23 X=\frac25\cdot\frac23 X=\frac4{15}X. $$ However, the above isn't a fixed unique value but a variable in \(X\). Let us consider the statements now. STATEMENT (1) alone: Statement clearly gives out the value of \(X\) which in turn gives us the (unique) value of what is asked. STATEMENT (1) alone – SUFFICIENT. STATEMENT (2) alone:Statement fills in the value for the box \(\frac13 X\). Or in other words says \(\frac13 X = 40\), giving us a (unique) value of \(X\) and hence of what is asked. STATEMENT (2) alone – SUFFICIENT. ANSWER – (D).

Question statement asks for number of distinct prime factors. We should bear in mind the possibility that integer \(n\) can still be a multiple of different prime numbers raised to powers. Remember that the whole idea of the question is to arrive at a unique value of the number of distinct prime factors of the integer. STATEMENT (1) alone: Unaware of whether \(n\) is even or odd, paves way for two possibilities: \(n\) is odd, in which case considering that \(2\) itself is prime and the fact that \(2n\) has four different prime factors, gives us \(3\) prime factors of \(n\). \(n\) is even, in which case considering that \(2\) itself is prime and the fact that \(2n\) has four different prime factors, gives us \(4\) prime factors of \(n\). STATEMENT (1) alone - INSUFFICIENT STATEMENT (2) alone: The squaring operation (performed on any integer) only raises all existing prime factors to a power twice the original (if \(n = 2^1 \cdot 3^2 \cdot 5^3\) then \(n^2 = 2^2 \cdot 3^4 \cdot 5^6\)). Or, squaring can never add another prime factor to an already existing integer. Thus if \(n^2\) has \(4\) distinct prime factors then so does the integer \(n\). STATEMENT (2) alone – SUFFICIENT ANSWER – (B).