In a medium-size township, the trend of annual immigration is an addition of 20% of the population as it was at the beginning; also 15% of the population as it was at the beginning is estimated to relocate elsewhere every year. If the current population is 80000, what is the likely population three years hence ?

Correct — D, (d) 92610.

Read the two rates first. Immigration adds 20% of the population 'as it was at the beginning'; relocation away removes 15% of the population 'as it was at the beginning'. Both are computed on the SAME base — the population standing at the start of that year — so within any one year they combine into a single net addition of

20% - 15% = 5% of the opening population.

The phrase 'as it was at the beginning' is what makes the two rates additive. If the 15% were taken on the population after the immigrants had arrived, the two would not net out and the arithmetic would be different.

Now the step that decides the question. Each year begins with the population that the previous year ended with, so the 5% is applied to a fresh, larger base every year. Growth is therefore COMPOUND, not simple:

P(3) = 80000 x (1.05)^3

Take it year by year rather than reaching for the cube: Year 1: 80000 + 5% of 80000 = 80000 + 4000 = 84000 Year 2: 84000 + 5% of 84000 = 84000 + 4200 = 88200 Year 3: 88200 + 5% of 88200 = 88200 + 4410 = 92610

The likely population three years hence is 92610, which is option (d).

Notice that the annual addition itself grows — 4000, then 4200, then 4410 — and that the extra over simple growth is only 610 on a base of 80000. That small gap is exactly what the option set is built around: the difference between the compound answer and the simple-growth answer is 610, and both figures are printed on the page. The insight the question is testing is not the percentage arithmetic, which is easy, but the recognition that a population growing at a constant RATE grows by a rising AMOUNT.

Why the others are wrong

• (a)90000 — A round figure that does not come out of the data by any correct route. Reaching 90000 from 80000 in three years needs a total growth of 12.5%, or about 4% a year compounded, and neither number appears anywhere in the stem. It is the lowest of the four options and is placed as an anchor for a candidate who estimates rather than computes: a rough sense that 'a bit of growth for three years' gets you to a round 90000 is exactly the habit the item punishes, since the correct answer is over 2600 higher.
• (b)91200 — This is what you get by applying the 15% to the wrong base, and for one year only: 80000 increased by 20% is 96000, and 96000 reduced by 15% is 91200. That treats the departures as 15% of the population AFTER the immigrants have been added, when the stem says twice — once for each flow — that the base is the population 'as it was at the beginning'. It is the sharpest trap in the set, because the misreading is a natural one: in real life people leave a town after others have arrived in it. Here the paper has specified the base, and specifying it is what makes the two rates net to 5%.
• (c)92000 — This is the simple-growth answer: 5% of the ORIGINAL 80000 is 4000, taken three times gives 12000, and 80000 + 12000 = 92000. It applies the correct net rate to the correct starting figure and then makes the single mistake the question exists to catch — holding the base fixed instead of letting each year start from the previous year's closing population. It sits just 610 below the answer, close enough that a candidate who has computed it will believe it, which is why the difference between simple and compound growth is worth recognising by sight rather than by arithmetic.

This is a compound growth problem wearing demographic clothes, and it has two separable steps.

STEP ONE — NET THE FLOWS. Where an inflow and an outflow are both quoted on the same base, they subtract to a single net rate. Here in-migration of 20% and out-migration of 15%, both on the opening population, give net growth of 5% a year. The common base is a condition, not a formality: rates quoted on different bases cannot be added or subtracted, and a great many percentage errors come from doing it anyway.

STEP TWO — COMPOUND, DO NOT MULTIPLY. A constant rate applied to a base that itself changes gives

P(n) = P(0) x (1 + r)^n

and not P(0) x (1 + nr). The two agree only in the first year and diverge afterwards, always with the compound figure higher for positive growth.

Two practical aids are worth carrying. For small rates and few periods, (1+r)^n is approximately 1 + nr + n(n-1)/2 x r^2, so the excess of compound over simple growth for three years at 5% is about 3 x 0.0025 = 0.0075 of the base, which on 80000 is 600 — near enough to the true 610 to identify the right option without finishing the multiplication. And the annual increments of a compounding series are themselves in geometric progression: 4000, 4200, 4410, each 5% above the last.

The same structure governs compound interest, depreciation at a fixed percentage, inflation over several years and any population question with birth and death rates on a common base. Decline works identically with (1 - r)^n, where the compound figure is always ABOVE the simple one, because each year's reduction is taken on a smaller base.

Key facts

• Rates of inflow and outflow quoted on the SAME base subtract to a single net rate: 20% in and 15% out on the opening population is 5% net growth a year.
• Constant-rate growth over n periods is P(0) x (1+r)^n, not P(0) x (1+nr); the second is simple growth and always understates.
• 80000 x 1.05^3 = 92610, reached year by year as 84000, then 88200, then 92610.
• The annual increments of a compounding series themselves grow at the same rate: 4000, 4200, 4410.
• Over three years at 5%, compound growth exceeds simple growth by about 3r^2 of the base — roughly 600 on 80000, against the exact 610.
• Taking the 15% on the post-immigration figure instead of the opening figure gives 91200 for a single year, which is why the stem specifies the base twice.
• For a declining quantity the formula is P(0) x (1-r)^n, and there the compound figure lies ABOVE the simple one.

Common traps

• Applying the 5% to the original 80000 every year. That is simple growth and gives 92000, which is printed as an option.
• Deducting the 15% from the population after the 20% has been added. The stem fixes both bases at the opening population; that misreading gives 91200.
• Counting only two years because the current population is treated as the end of year one. 'Three years hence' means three full applications of the rate.
• Estimating instead of computing. The four options span only 2610, so a rough figure cannot separate them.

• practice — not a real PYQ

  The population of a town rises by 10% in the first year and falls by 10% in the second year. Compared with its population at the start, the population at the end of the two years is

  • (a)unchanged
  • (b)1% higher
  • (c)1% lower
  • (d)2% lower

  Answer(c) 1% lower — the rise and the fall are taken on different bases, so they do not net out. Taking 100 as the start, 100 becomes 110, and 10% of 110 is 11, leaving 99. The shortcut a + b + ab/100 gives 10 - 10 - 100/100 = -1 per cent.

• practice — not a real PYQ

  A machine costing 50000 rupees depreciates by 20% of its value at the beginning of each year. Its value at the end of three years is

  • (a)20000 rupees
  • (b)25600 rupees
  • (c)30000 rupees
  • (d)32000 rupees

  Answer(b) 25600 rupees — depreciation on the opening value each year compounds downward: 50000 becomes 40000, then 32000, then 25600, which is 50000 x (0.8)^3. Taking 20% of the original 50000 three times would give 30000, the simple-depreciation trap.