Q. Passage-2
The history of renewable energy suggests there is a steep learning curve, meaning that, as more is produced, costs fall rapidly because of economies of scale and learning by doing. The firms’ green innovation is path-dependent: the more a firm does, the more it is likely to do in the future. The strongest evidence for this is the collapse in the price of solar energy, which became about 90% cheaper during the 2010s, repeatedly beating forecasts. Moving early and gradually gives economies more time to adjust, allowing them to reap the benefits of path-dependent green investment without much disruption. A late, more chaotic transition is costlier.
Q.7) How many possible values of (p + q + r) are there satisfying 1/p+ 1/q+ 1/r=1, where p, q and r are natural numbers (not necessarily distinct)?
Exp) Option c is the correct answer
The problem asks us to find the number of distinct possible values for the sum (p + q + r) given the equation 1/p + 1/q + 1/r = 1, where p, q, and r are natural numbers (positive integers: 1, 2, 3, …). The numbers p, q, and r are not necessarily distinct.
Let’s assume, without loss of generality, that 1 <= p <= q <= r. This ordering helps us systematically find all possible unique sets of (p, q, r). Once we find these sets, we can calculate their sums (p + q + r) and then count the number of distinct sum values.
From the equation 1/p + 1/q + 1/r = 1, since p, q, and r are natural numbers, 1/p, 1/q, and 1/r must be positive.
The possible values for p are given by:-
Since p <= q <= r, it follows that 1/p >= 1/q >= 1/r. Therefore, 1 = 1/p + 1/q + 1/r <= 1/p + 1/p + 1/p = 3/p. So, 1 <= 3/p, which implies p <= 3.
This means p can only be 1, 2, or 3.
Let’s check each case for p:
Case 1: p = 1 If p = 1, the equation becomes 1/1 + 1/q + 1/r = 1. This simplifies to 1 + 1/q + 1/r = 1, which means 1/q + 1/r = 0. Since q and r are natural numbers, 1/q and 1/r are both positive. Their sum cannot be 0. Therefore, p cannot be 1.
Case 2: p = 2 If p = 2, the equation becomes 1/2 + 1/q + 1/r = 1. Subtracting 1/2 from both sides gives 1/q + 1/r = 1 – 1/2 = 1/2.
Now, we need to find natural numbers q and r such that q >= p (so q >= 2) and r >= q. Since q <= r, we have 1/q >= 1/r. So, 1/2 = 1/q + 1/r <= 1/q + 1/q = 2/q. This implies 1/2 <= 2/q, which means q <= 4.
So, when p = 2, the possible values for q are 2, 3, or 4.
- Subcase 2.1: p = 2, q = 2 The equation becomes 1/2 + 1/2 + 1/r = 1. 1 + 1/r = 1, which means 1/r = 0. This is not possible for a natural number r. So, (2, 2, r) is not a solution.
- Subcase 2.2: p = 2, q = 3 The equation becomes 1/2 + 1/3 + 1/r = 1. To find 1/r, subtract 1/2 and 1/3 from 1: 1/r = 1 – 1/2 – 1/3 1/r = (6 – 3 – 2) / 6 1/r = 1/6 So, r = 6. This gives the ordered solution (p, q, r) = (2, 3, 6). Check if this satisfies p <= q <= r: 2 <= 3 <= 6. Yes. Sum for this solution: p + q + r = 2 + 3 + 6 = 11.
- Subcase 2.3: p = 2, q = 4 The equation becomes 1/2 + 1/4 + 1/r = 1. To find 1/r: 1/r = 1 – 1/2 – 1/4 1/r = (4 – 2 – 1) / 4 1/r = 1/4 So, r = 4. This gives the ordered solution (p, q, r) = (2, 4, 4). Check if this satisfies p <= q <= r: 2 <= 4 <= 4. Yes. Sum for this solution: p + q + r = 2 + 4 + 4 = 10.
Case 3: p = 3 If p = 3, the equation becomes 1/3 + 1/q + 1/r = 1. Subtracting 1/3 from both sides gives 1/q + 1/r = 1 – 1/3 = 2/3.
Now, we need to find natural numbers q and r such that q >= p (so q >= 3) and r >= q. Since q <= r, we have 1/q >= 1/r. So, 2/3 = 1/q + 1/r <= 1/q + 1/q = 2/q. This implies 2/3 <= 2/q, which means q <= 3.
So, when p = 3, the only possible value for q (given q >= 3) is q = 3.
- Subcase 3.1: p = 3, q = 3 The equation becomes 1/3 + 1/3 + 1/r = 1. 2/3 + 1/r = 1. To find 1/r: 1/r = 1 – 2/3 1/r = 1/3 So, r = 3. This gives the ordered solution (p, q, r) = (3, 3, 3). Check if this satisfies p <= q <= r: 3 <= 3 <= 3. Yes. Sum for this solution: p + q + r = 3 + 3 + 3 = 9.
Summary of Unique Ordered Solutions (p, q, r) and their sums (p + q + r):
- (2, 3, 6) -> Sum = 11
- (2, 4, 4) -> Sum = 10
- (3, 3, 3) -> Sum = 9
These are the only possible sets of natural numbers (up to reordering) that satisfy the given equation. The problem asks for the number of possible values of (p + q + r).
The distinct sums we found are 9, 10, and 11.
There are three distinct possible values for p + q + r.
Therefore Option (c) is the correct answer.
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