Percentage Change and Successive Change

Percentage Change: The Basic Formula

When a quantity moves from an old value to a new value,

$$\text{Percentage change} = \frac{\text{New} - \text{Old}}{\text{Old}} \times 100\%$$

A positive answer is an increase and a negative one is a decrease. The base is always the old value, the one you started from.

• A price goes from ₹40 to ₹50. The change is 10, and $\frac{10}{40}$ × 100 = 25%, so it is a 25% increase.
• The price comes back from ₹50 to ₹40. The change is still 10, but now $\frac{10}{50}$ × 100 = 20%, so it is a 20% decrease.

The same ₹10 is 25% one way and 20% the other, because the base changed. So before you divide, ask which value is the old one.

The multiplier. An increase of r% multiplies the value by $\left(1 + \frac{r}{100}\right)$, and a decrease of r% multiplies it by $\left(1 - \frac{r}{100}\right)$. Writing the multiplier as a fraction usually makes the arithmetic painless:

So a 20% rise on ₹4,500 gives 4500 × $\frac{6}{5}$ = ₹5,400, and a 12.5% cut on ₹5,600 gives 5600 × $\frac{7}{8}$ = ₹4,900.

Reading the change from a ratio. If you know new ÷ old, the change is right there: new ÷ old = 1.35 means a 35% rise, and new ÷ old = 0.92 means an 8% fall.

When the quantity is itself a percentage. Keep percentage points and per cent apart. A literacy rate rising from 60% to 75% has gone up by 15 percentage points, which is a $\frac{15}{60}$ = 25% increase. The difference is explained in the Percentage Basics section.

[SSC Tip] "By what per cent is 75 more than 60?" and "by what per cent is 60 less than 75?" have different answers, 25% and 20%. In both, divide by the number after "than", because that is the one being compared against.

More Than and Less Than: Switching the Base

If A is x% more than B, then B is less than A by a smaller percentage, because the same difference is now measured against the bigger number A:

$$\text{B is less than A by } \frac{x}{100 + x} \times 100\%$$

If A is x% less than B, then B is more than A by a bigger percentage:

$$\text{B is more than A by } \frac{x}{100 - x} \times 100\%$$

Say A is 25% more than B. Put B = 100, so A = 125. The difference is 25, and measured against A that is $\frac{25}{125}$ = 20%. So B is 20% less than A.

The fraction way. 25% more means A = $\frac{5}{4}$B, so B = $\frac{4}{5}$A. Flip the fraction and read off the change: $\frac{4}{5}$ is one-fifth short of 1, which is 20% less.

"Of" versus "more than". "A is 150% of B" means A is 50% more than B, not 150% more. "A is 250% more than B" means A is 350% of B. SSC uses both wordings, so read them slowly.

Two numbers compared through a third. If A is 20% more than C and B is 50% more than C, take C = 100. Then A = 120 and B = 150, and:

• A is $\frac{30}{150}$ = 20% less than B
• B is $\frac{30}{120}$ = 25% more than A
• A is $\frac{120}{150}$ = 80% of B

"Less than" works the same way. If A and B are 10% and 25% less than C, then A = 90 and B = 75. B is $\frac{15}{90} = 16\frac{2}{3}\%$ less than A, and A is $\frac{15}{75}$ = 20% more than B.

Chains. If A is 25% more than B and B is 20% less than C, then A = $\frac{5}{4} \times \frac{4}{5}$C = C, so A and C are equal. Adding the percentages (25 − 20 = 5) would wrongly make A 5% more than C.

Finding the Original Value

If a value rises by r% and becomes V, then V is (100 + r)% of the original:

$$\text{Original} = V \times \frac{100}{100 + r}$$

After a fall of r%, the original is $V \times \frac{100}{100 - r}$.

A salary rises by 20% to ₹27,000. The new salary is 120% of the old one, so the old salary was 27000 × $\frac{100}{120}$ = ₹22,500.

The trap. The tempting move is to take 20% of ₹27,000 and subtract it: 27000 − 5400 = ₹21,600. That is wrong, because the 20% was worked out on the old salary, not on ₹27,000. Check any answer by pushing it forward: 22500 + 20% of 22500 = 22500 + 4500 = 27000.

The same goes for a fall. After a 20% cut, a price is ₹1,600. The original is 1600 × $\frac{100}{80}$ = ₹2,000, not 1600 + 320 = ₹1,920.

From one changed value to another. "A number increased by 12% becomes 1008. What would it be if it were decreased by 8% instead?" The original is 1008 × $\frac{100}{112}$ = 900, and 92% of 900 = 828. You can do it in one line:

$$1008 \times \frac{92}{112} = 828$$

The gap between two changes. "When a number is increased by 20%, it is 105 more than when it is decreased by 15%." The two results are 120% and 85% of the number, which are 35 percentage points apart. So 35% of the number is 105, and the number is 300.

The change given in rupees. "After a 15% fall, a phone costs ₹2,250 less than before. What is the new price?" The fall of 15% is ₹2,250, so 1% is ₹150 and the old price was ₹15,000. The new price is 15000 − 2250 = ₹12,750.

Getting Back to the Start

A rise followed by an equal fall never brings you back, because the fall is taken on a bigger number. To undo a change exactly:

• After a rise of x%, you need a fall of $\frac{x}{100 + x} \times 100\%$.
• After a fall of x%, you need a rise of $\frac{x}{100 - x} \times 100\%$.

These are the same formulas as in More Than and Less Than above, for the same reason: the changed value is now the base.

The fall-then-rise rows are where marks are lost: a share price that halves has to double to recover.

The fraction way. A 20% fall multiplies by $\frac{4}{5}$, so getting back means multiplying by $\frac{5}{4}$, a 25% rise. A $16\frac{2}{3}\%$ rise multiplies by $\frac{7}{6}$, so getting back means multiplying by $\frac{6}{7}$. That is a fall of $\frac{1}{7}$, or $14\frac{2}{7}\%$.

Reaching a target other than the start. "A number is increased by 20%. By what per cent must the new number be decreased so that it ends up 4% more than the original?" Take the original as 100. After the rise it is 120, and the target is 104. The fall needed is 16 out of 120, which is $13\frac{1}{3}\%$.

"A price is cut by 40%. By what per cent must the new price be raised so that it is 5% above the original?" From 100 the price drops to 60, and the target is 105. The rise needed is 45 out of 60, which is 75%.

[SSC Tip] The fall needed to undo a rise is always smaller than the rise, and the rise needed to undo a fall is always bigger than the fall. That alone knocks out two options in many questions.

Successive Percentage Changes

When one change follows another, the second is taken on the value left after the first. So the two percentages cannot just be added. For a change of a% followed by a change of b%:

$$\text{Net change} = \left(a + b + \frac{ab}{100}\right)\%$$

Use a plus sign for an increase and a minus sign for a decrease.

• 20% rise then 10% rise: 20 + 10 + $\frac{20 \times 10}{100}$ = 32% rise
• 20% rise then 10% fall: 20 − 10 − 2 = 8% rise
• 20% fall then 10% fall: −20 − 10 + 2 = −28%, a 28% fall
• 25% rise then 20% fall: 25 − 20 − 5 = 0, no change at all

The multiplier method. Multiply the multipliers. A 20% rise then a 10% fall is 1.2 × 0.9 = 1.08, an 8% rise. This is the cleanest way when there are three or more changes. A 10% rise, a 20% fall and then a 25% rise give

$$\frac{11}{10} \times \frac{4}{5} \times \frac{5}{4} = \frac{11}{10}$$

which is a 10% rise overall.

Three changes with the formula. Combine the first two, then combine that result with the third. For three 10% rises: 10 and 10 give 21%, then 21 and 10 give 21 + 10 + 2.1 = 33.1%. The full three-change formula is $a + b + c + \frac{ab + bc + ca}{100} + \frac{abc}{10000}$, but doing it in two steps is easier to remember.

The same percentage up and down. A rise of x% followed by a fall of x%, in either order, always ends in a net loss:

$$x - x - \frac{x \times x}{100} = -\frac{x^2}{100}$$

Up 10% and down 10% is a 1% loss. Up 30% and down 30% is a 9% loss. Down 50% and up 50% is a 25% loss.

Order does not matter. Multiplication can be done in any order, so a 20% rise then a 10% fall gives the same result as a 10% fall then a 20% rise. If a question asks which order is better, the answer is that both are the same.

Successive discounts work exactly like this, and so does growth at the same rate year after year.

[SSC Tip] Watch the sign of $\frac{ab}{100}$. Two rises: plus. Two falls: also plus, because minus times minus is plus, so the total fall is smaller than the plain sum. One rise and one fall: minus.

Successive Changes: Working Back from the Result

Once you know the net multiplier, every "find the original" question is a single division.

Final value given. "A price is raised by 25% and then cut by 10%. It is now ₹5,625. What was it at first?" The net change is 25 − 10 − 2.5 = 12.5%, a multiplier of 1.125, which is $\frac{9}{8}$. So the first price was 5625 × $\frac{8}{9}$ = ₹5,000.

The change itself given. "A salary is raised by 10% and then by 20%. Altogether it has gone up by ₹4,800. What was the original salary?" The net rise is 10 + 20 + 2 = 32%, and 32% of the salary is ₹4,800. So the salary was 4800 × $\frac{100}{32}$ = ₹15,000.

The same x up and down. "After a rise of 20% and then a fall of 20%, a price is ₹96 less than at the start." The loss is $\frac{20^2}{100}$ = 4% of the original, and 4% is ₹96, so the original was ₹2,400.

Turned round: "A price is raised by some per cent and then cut by the same per cent. In the end it is 16% lower. Find the per cent." Here $\frac{x^2}{100}$ = 16, so $x^2$ = 1600 and x = 40.

Two people, two changes. "A's salary rises by 25% and B's falls by 20%. Now they earn the same. Find the ratio of their original salaries." 1.25A = 0.8B, so $\frac{A}{B} = \frac{0.8}{1.25} = \frac{16}{25}$, and A : B = 16 : 25.

Up and down to a given result. "A number is first increased by 30% and then decreased by 30%. The result is 182. Find the number." The net multiplier is 1.3 × 0.7 = 0.91, so the number is 182 ÷ 0.91 = 200.

Changes on the same base. Be careful when a question says the changes are both taken on the original value, as in "a price goes up by 20% of the original and then down by 5% of the original". Then the percentages really do add: the net change is 20 − 5 = 15%. This wording is rare, but it does turn up, and it is the one case where plain adding is right.

Errors in Calculation

These questions describe someone who did the wrong operation and ask for the percentage error. The error is measured against the correct answer:

$$\text{Percentage error} = \frac{\left|\text{Wrong} - \text{Correct}\right|}{\text{Correct}} \times 100\%$$

The number being worked on never matters, so pick something convenient, such as 1 or the LCM of the denominators.

Multiplied by the upside-down fraction. A student should multiply by $\frac{4}{3}$ but multiplies by $\frac{3}{4}$. Take the number as 12. The correct answer is 16, the wrong one is 9, and the error is $\frac{7}{16}$ = 43.75%. In general, if the correct multiplier is $\frac{p}{q}$ and the student uses $\frac{q}{p}$:

$$\text{Error} = \frac{\left|p^2 - q^2\right|}{p^2} \times 100\%$$

For $\frac{4}{3}$ that is $\frac{16 - 9}{16}$ = 43.75%, as above.

Divided instead of multiplied. The student should multiply by k but divides by k. With the number 1, the correct answer is k and the wrong one is $\frac{1}{k}$, so the error is $\frac{k^2 - 1}{k^2} \times 100\%$. For k = 8 it is $\frac{63}{64}$ = 98.4375%.

Multiplied instead of divided. The student should divide by k but multiplies. Now the correct answer is $\frac{1}{k}$ and the wrong one is k, so the error is $(k^2 - 1) \times 100\%$. For k = 3 it is 800%. Errors above 100% are perfectly possible.

Increased instead of decreased. A price should be cut by x% but is raised by x%. From 100, the correct value is 100 − x and the wrong one is 100 + x, so the error is $\frac{2x}{100 - x} \times 100\%$. For x = 10 that is $\frac{20}{90} = 22\frac{2}{9}\%$. The opposite slip, decreasing instead of increasing, gives $\frac{2x}{100 + x} \times 100\%$, which is $18\frac{2}{11}\%$ for x = 10.

A number read or copied wrongly. Compare the wrong number with the right one directly. If 36 is copied as 63, the difference is 27, and the error is $\frac{27}{36}$ = 75%. When the two digits of a number are swapped, the difference is always 9 times the difference of the digits, which gives the top line quickly.

[SSC Tip] Divide by the correct value, never the wrong one. SSC nearly always puts the answer you get by dividing by the wrong value among the choices.

Population Growth and Depreciation

A population that grows by r% every year is multiplied by $\left(1 + \frac{r}{100}\right)$ each year. After n years:

$$P_n = P\left(1 + \frac{r}{100}\right)^n$$

Something that loses value at r% a year, like a machine, a car or a phone, follows the same rule with a minus sign. This is called depreciation:

$$V_n = V\left(1 - \frac{r}{100}\right)^n$$

Each year's change is taken on the value at the start of that year, which is why the total change is not simply r × n per cent.

Forward. A town of 64,000 grows at 12.5% a year. Since 12.5% is $\frac{1}{8}$, the multiplier is $\frac{9}{8}$:

$$64000 \times \frac{9}{8} \times \frac{9}{8} = 81000$$

After 3 years it would be 81000 × $\frac{9}{8}$ = 91,125.

Backward. To find the value n years earlier, divide instead of multiply:

$$P_{\text{earlier}} = \frac{P}{\left(1 + \frac{r}{100}\right)^n}$$

A car is worth ₹4,05,000 today after losing 10% of its value each year for 2 years. Its price 2 years ago was 405000 ÷ (0.9 × 0.9) = 405000 ÷ 0.81 = ₹5,00,000.

Different rates in different years. Multiply the separate multipliers. A population of 20,000 grows 10% in the first year, falls 5% in the second and grows 20% in the third:

$$20000 \times 1.1 \times 0.95 \times 1.2 = 25080$$

The change in one particular year. "A machine bought for ₹80,000 depreciates 15% a year. How much value does it lose in the second year?" After the first year it is worth 80000 × 0.85 = ₹68,000. In the second year it loses 15% of ₹68,000, which is ₹10,200, not 15% of ₹80,000.

With rates like 10%, 20% or 25%, use fractions: $\left(\frac{11}{10}\right)^2 = \frac{121}{100}$, $\left(\frac{6}{5}\right)^2 = \frac{36}{25}$ and $\left(\frac{4}{5}\right)^3 = \frac{64}{125}$. Option values are usually built so these cancel neatly. The growth formula is the same as the compound interest formula.

Population with Births, Deaths and Migration

Births and deaths. If the birth rate is b% a year and the death rate is d% a year, the population grows at (b − d)% a year. Rates are often given "per thousand": 45 births per 1000 people is 4.5%, and 20 deaths per 1000 is 2%. The net growth is then 2.5% a year, and a town of 32,000 has

$$32000 \times 1.025 \times 1.025 = 33620$$

people after 2 years.

Migration as a percentage. If the population grows by 6% a year but 1% of the people move away each year, the net rate is 5% a year. Setters treat both percentages as taken on the same yearly figure, so they simply combine.

Migration as a fixed number. Here the order matters, and the question tells you when people leave. "A town of 50,000 grows 8% a year, and at the end of each year 2,000 people leave."

• Year 1: 50000 × 1.08 = 54000, then 54000 − 2000 = 52000
• Year 2: 52000 × 1.08 = 56160, then 56160 − 2000 = 54160

You cannot use the power formula here, because the 2,000 is not a percentage. Work year by year.

Working backwards with mixed rates. "A population rose 20% in the first year, fell 25% in the second and rose 10% in the third. It is now 39,600. What was it at the start?" The overall multiplier is 1.2 × 0.75 × 1.1 = 0.99, so the starting population was 39600 ÷ 0.99 = 40,000. The order of the three years makes no difference here.

Working backwards from the growth. "Over 2 years at 10% a year, a village grows by 1,680 people. What was its population at the start?" Two years at 10% is a 21% rise (10 + 10 + 1). So 21% of the population is 1,680, and the village had 8,000 people.

[Tier 2] How many years? Sometimes a question asks when a population first crosses a level. Keep multiplying. At 20% a year, $1.2^2 = 1.44$, $1.2^3 = 1.728$ and $1.2^4 = 2.0736$. So a population growing at 20% a year is first more than double after 4 years.

Solved Examples

Question 1

The price of a kilogram of dal rises from ₹96 to ₹120, and later falls back to ₹96. The percentage rise and the percentage fall are, respectively: (a) 25% and 20% (b) 20% and 25% (c) 25% and 25% (d) 24% and 24%

Answer. The change is ₹24 both times. On the way up the base is the old price ₹96, so the rise is $\frac{24}{96}$ = 25%. On the way down the base is ₹120, so the fall is $\frac{24}{120}$ = 20%. The answer is (a) 25% and 20%.

Question 2

Ravi's salary is 40% more than Sunita's. By what per cent is Sunita's salary less than Ravi's? (a) 40% (b) 60% (c) $28\frac{4}{7}\%$ (d) $71\frac{3}{7}\%$

Answer. I take Sunita's salary as 100, so Ravi's is 140. Sunita's is 40 less, and measured against Ravi's 140 that is $\frac{40}{140} = \frac{2}{7}$, which is $28\frac{4}{7}\%$. The answer is (c) $28\frac{4}{7}\%$. Don't pick $71\frac{3}{7}\%$: that is Sunita's salary as a per cent of Ravi's, not how much less it is.

Question 3

The number of students in a school this year is 37.5% less than last year. By what per cent must it rise next year to get back to last year's number? (a) 37.5% (b) 60% (c) 62.5% (d) $27\frac{3}{11}\%$

Answer. I take last year's number as 100, so this year's is 62.5. To get back to 100 it must rise by 37.5, and on a base of 62.5 that is $\frac{37.5}{62.5}$ = 0.6, or 60%. Fractions make it quicker: 37.5% less leaves $\frac{5}{8}$ of the old number, and getting back from $\frac{5}{8}$ to 1 means multiplying by $\frac{8}{5}$, a 60% rise. So the answer is (b) 60%.

Question 4

Two numbers are 20% and 50% more than a third number. By what per cent is the first number less than the second? (a) 30% (b) 25% (c) 80% (d) 20%

Answer. Taking the third number as 100, the first is 120 and the second is 150. The first is 30 less, and against the second (150) that is $\frac{30}{150}$ = 20%, which is option (d) 20%.

Question 5

A is 20% more than B, and B is 25% less than C. A is what per cent of C? (a) 95% (b) 90% (c) 110% (d) 105%

Answer. I multiply the changes. A = 1.2B and B = 0.75C, so A = 1.2 × 0.75 × C = 0.9C. A is 90% of C. The answer is (b) 90%. Adding the percentages, 20 − 25 = −5, gives 95%, which is wrong because the two percentages are taken on different bases.

Question 6

After a 15% rise, a scooter costs ₹74,750. What was its price before the rise? (a) ₹63,537.50 (b) ₹62,500 (c) ₹64,000 (d) ₹65,000

Answer. The new price is 115% of the old one. So the old price is 74750 × $\frac{100}{115}$ = ₹65,000. Check: 15% of 65,000 is 9,750, and 65,000 + 9,750 = 74,750. The answer is (d) ₹65,000. Taking 15% off ₹74,750 gives ₹63,537.50, option (a), but the 15% was added to the old price, not to the new one.

Question 7

A number increased by 15% becomes 920. What would the number become if it were decreased by 15% instead? (a) 782 (b) 690 (c) 680 (d) 800

Answer. The number is 920 × $\frac{100}{115}$ = 800. Decreasing 800 by 15% gives 800 × 0.85 = 680. The answer is (c) 680. In one step it is 920 × $\frac{85}{115}$ = 680.

Question 8

A number increased by 18% exceeds the same number decreased by 12% by 75. The number is: (a) 250 (b) 125 (c) 500 (d) 375

Answer. The first result is 118% of the number and the second is 88%. The gap is 30% of the number, and that gap is 75. So the number is 75 × $\frac{100}{30}$ = 250. Check: 118% of 250 = 295 and 88% of 250 = 220, and 295 − 220 = 75. The answer is (a) 250.

Question 9

A company's share price falls by 44% in a month. By what per cent must it now rise to get back to where it started? (a) 44% (b) 56% (c) $78\frac{4}{7}\%$ (d) $30\frac{5}{9}\%$

Answer. From 100 the share drops to 56. To get back to 100 it must rise by 44 on a base of 56, which is $\frac{44}{56} = \frac{11}{14}$, or $78\frac{4}{7}\%$. The answer is (c) $78\frac{4}{7}\%$.

Question 10

A number is increased by 25%. By what per cent must the new number be decreased so that the result is 10% less than the original number? (a) 28% (b) 35% (c) 30% (d) 25%

Answer. I take the original as 100. After the 25% rise it is 125. The target is 10% less than the original, which is 90. So the fall is 35 out of 125, and $\frac{35}{125}$ = 28%. The answer is (a) 28%. 35% would be right only if the base were 100, but the fall is taken on 125.

Question 11

The price of an article is increased by 20% and then by 15%. The total increase is: (a) 35% (b) 38% (c) 37% (d) 32%

Answer. Two rises of a% and b% give a + b + $\frac{ab}{100}$. Here that is 20 + 15 + $\frac{20 \times 15}{100}$ = 20 + 15 + 3 = 38%, so the answer is (b) 38%.

Question 12

A salary is first increased by 30% and then decreased by 30%. The net effect on the salary is: (a) no change (b) 9% increase (c) 9% decrease (d) 3% decrease

Answer. The same percentage up and then down always ends in a loss of $\frac{x^2}{100}\%$. Here that is $\frac{30 \times 30}{100}$ = 9%. With multipliers: 1.3 × 0.7 = 0.91, so only 91% of the salary is left. The answer is (c) 9% decrease.

Question 13

A quantity is increased by 10%, then increased by 20%, and then decreased by 25%. The net change is: (a) 1% decrease (b) 5% increase (c) 1% increase (d) 5% decrease

Answer. I multiply the three multipliers: 1.1 × 1.2 × 0.75. First 1.1 × 1.2 = 1.32, and then 1.32 × 0.75 = 0.99. So 99% is left, which is a 1% decrease. The answer is (a) 1% decrease.

Question 14

A number is first decreased by 25% and then increased by 40%. The result is 1260. The original number is: (a) 1200 (b) 1071 (c) 1323 (d) 1260

Answer. The net multiplier is 0.75 × 1.4 = 1.05. So the number × 1.05 = 1260, and the number is 1260 ÷ 1.05 = 1200. The answer is (a) 1200. Check: 1200 falls by 25% to 900, and 900 rises by 40% to 1260.

Question 15

Anil's salary was raised by 25% and then by 20%. It is now ₹4,500 more than it was before both raises. His original salary was: (a) ₹10,000 (b) ₹6,000 (c) ₹11,250 (d) ₹9,000

Answer. Rises of 25% and 20% make 25 + 20 + 5 = 50% in all. So 50% of the salary is ₹4,500, and the salary was (d) ₹9,000.

Question 16

A's salary is increased by 10% and B's salary is decreased by 10%. After this their salaries are equal. The ratio of A's original salary to B's original salary is: (a) 11 : 9 (b) 9 : 11 (c) 1 : 1 (d) 10 : 11

Answer. After the changes, 1.1 × A = 0.9 × B. So $\frac{A}{B} = \frac{0.9}{1.1} = \frac{9}{11}$, and A : B = 9 : 11. The answer is (b) 9 : 11. It makes sense that A started lower, because A's salary had to rise to meet B's falling one.

Question 17

A student was asked to multiply a number by $\frac{5}{3}$, but he multiplied it by $\frac{3}{5}$ instead. The percentage error in his answer is: (a) 64% (b) 36% (c) 40% (d) $177\frac{7}{9}\%$

Answer. I take the number as 15. The correct answer is 15 × $\frac{5}{3}$ = 25 and the wrong one is 15 × $\frac{3}{5}$ = 9. The error is 16, and against the correct 25 that is $\frac{16}{25}$ = 64%. The answer is (a) 64%. Dividing by the wrong answer 9 instead gives $177\frac{7}{9}\%$.

Question 18

Instead of multiplying a number by 4, a student divided it by 4. The percentage error is: (a) 1500% (b) 75% (c) 6.25% (d) 93.75%

Answer. I take the number as 4. The correct answer is 16 and the wrong one is 1. The error is 15 out of the correct 16, which is $\frac{15}{16}$ = 93.75%. The answer is (d) 93.75%.

Question 19

A teacher entered a student's marks as 54 instead of 45. The percentage error is: (a) $16\frac{2}{3}\%$ (b) 9% (c) 20% (d) 25%

Answer. The marks were wrong by 54 − 45 = 9. Against the correct 45 that is $\frac{9}{45}$ = 20%. The answer is (c) 20%.

Question 20

The population of a town is 1,76,400 and it grows at 5% a year. Its population after 2 years and its population 2 years ago are, respectively: (a) 1,94,481 and 1,60,000 (b) 1,94,040 and 1,58,760 (c) 1,94,481 and 1,58,760 (d) 1,94,040 and 1,60,000

Answer. 5% a year means multiplying by $\frac{21}{20}$ each year. After 2 years: 176400 × $\frac{21}{20} \times \frac{21}{20}$ = 441 × 441 = 1,94,481. Two years ago: 176400 × $\frac{20}{21} \times \frac{20}{21}$ = 400 × 400 = 1,60,000. The answer is (a) 1,94,481 and 1,60,000. A flat 10% for the two years ignores the growth on growth.

Question 21

A machine worth ₹2,50,000 loses 20% of its value in the first year and 10% in each later year. Its value after 3 years is: (a) ₹1,50,000 (b) ₹1,62,000 (c) ₹1,60,000 (d) ₹1,44,000

Answer. The value after 3 years is 250000 × 0.8 × 0.9 × 0.9. 250000 × 0.8 = 2,00,000, then × 0.9 gives 1,80,000, and × 0.9 again gives 1,62,000. The answer is (b) ₹1,62,000.

Question 22

A town of 40,000 people grows by 10% a year, but 1,000 people move away at the end of every year. Its population after 2 years is: (a) 48,400 (b) 46,400 (c) 46,090 (d) 46,300

Answer. I go year by year, because the 1,000 is a fixed number. Year 1: 40000 × 1.1 = 44000, then 44000 − 1000 = 43000. Year 2: 43000 × 1.1 = 47300, then 47300 − 1000 = 46300. The answer is (d) 46,300. Taking away all 2,000 people at the very end gives 46,400, which is close but wrong.