Question

Ramesh bought a total of 6 fruits (apples and oranges) from the market. He found that he required one orange less to extract the same quantity of juice as extracted from apples. If Ramesh had used the same number of apples and oranges to make the blend, then which of the following correctly represents the percentage of apple juice in the blend?

• A. 25%
• B. 33.3%
• C. 60%
• D. 66.6%
• E.

Hint:

- When equal counts of two items are mixed, the percentage share depends only on the per-item yield ratio: if \(J_A:J_O = r:1\), then apple percentage \(= \dfracr1+r\).
- Translating “one less is needed to match” often gives a relation like \(a\,J_A=(a-1)\,J_O\), which can be solved for the yield ratio.

Answer 1

Answer: (E)

Step 1: Let the number of apples be \(a\) and oranges be \(o\), with \(a+o=6\). Let the juice per apple be \(J_A\) and per orange be \(J_O\). \[ \text{Total apple-juice} = a\,J_A, \qquad \text{Total orange-juice (needed to match)} = (a-1)\,J_O. \] The statement “one orange less is required to match the apples’ juice” gives \[ a\,J_A=(a-1)\,J_O \;⇒ \frac{J_A}{J_O}=\frac{a-1}{a}. \] Step 2: If he used the same number \(n\) of apples and oranges in a blend, the fraction (and hence percentage) of apple juice in the blend is \[ \frac{nJ_A}{nJ_A+nJ_O}=\frac{J_A}{J_A+J_O}=\frac{\tfrac{a-1}{a}}{1+\tfrac{a-1}{a}}=\frac{a-1}{2a-1}. \] Step 3: From \(a+o=6\) and the feasibility condition that he must have at least \(a-1\) oranges to “require one less orange,” we need \(o\ge a-1⇒ 6-a\ge a-1⇒ a\le 3\). Also \(a\ge 2\) (otherwise \(a-1=0\) is trivial). Hence \(a\in\{2,3\}\).
If \(a=2\), the percentage would be \(\dfrac{1}{3}=33.3%\) (one of the options).
However, the usual non-trivial balanced case with \(a=3\) (and \(o=3\)) satisfies the condition and yields \[ \text{Apple-juice percentage }=\frac{a-1}{2a-1}=\frac{2}{5}=40%. \] Since \(40%\) is not among options (A)–(D), the correct choice is (E) None of the above.