Question Stem:
Let's assume the number of Oranges and Apples to be O and A. respectively.
4 Oranges and 7 Apples are added.
Now the number Oranges = O + 4
Number of Apples = A + 7
And, \(\frac{{O + 4}}{{A + 7}} > \frac{1}{2} \)
Rather than simplifying, we can deduce inference from here.
After adding fruits, the number of oranges is more than half the number of apples.
Statement 1: The total number of oranges and apples after the addition was 25.Total oranges and apples added = 7 + 4 = 11
Before addition, O + A = 14
Since after addition the number of oranges is more than half the number of apples, 9 is the minimum value orange can have.
Since we need to distribute the number 25 between apples and oranges such that oranges > \(\frac{1}{2} \)* orange
Now the number of apples can at least be 7 + 1 = 8. (Since the bucket has some apples initially so the min number is taken as 1, one can argue about 0 but it won't matter).
Maximum value of O+ 4 = 25 - 8 = 17 or, O = 13
O can have multiple values from 5 to 13, inclusive.
Not sufficient. We eliminate A and D.
Statement 2: The product of the initial number of apples and oranges is 33.It can be any of the following,
1. A = 1, O = 33, here \(\frac{{O + 4}}{{A + 7}} = \frac{37}{8} \), valid.
2. A = 3, O = 11, here \(\frac{{O + 4}}{{A + 7}} = \frac{15}{10} \), valid.
3. A = 11, O = 3, here \(\frac{{O + 4}}{{A + 7}} = \frac{7}{18} \), NOT valid.
4. A = 33, O = 1 here \(\frac{{O + 4}}{{A + 7}} = \frac{5}{40} \), NOT valid.
Here, we have two possibilities.
Not sufficient. We eliminate B.
CombineThe only valid scenario is when O = 11 and A = 3.
The answer is C.