Percentage Questions and Answers

Answer: D

The original number is 272. After a 25% increase and then a 15% decrease the number is 1.25 × 0.85 = 1.0625 times the original, so the net change is 6.25%.

340, 289 and 110 do not give a 17 difference: 6.25% of 340 is 21.25 and 6.25% of 289 is 18.06.

Check: 1.25 × 272 = 340, and 340 × 0.85 = 289, and 289 - 272 = 17, which agrees with the statement. The net change is not 10% (25 minus 15) but 6.25%, because the 15% decrease is on the larger, increased number.

Answer: B

The population after three years is 30240. Apply the changes in order: \[ 20000 \times 1.2 \times 1.4 \times 0.9 = 30240 \] The first year gives 24000, the second 33600, and the 10% fall in the third gives 30240.

24000 is the figure after year one only, and 33240 and 36940 do not come from these multipliers; 33600 would be the population before the third-year fall.

Because percentage changes multiply and do not add, a 20 per cent rise, a 40 per cent rise and a 10 per cent fall do not equal a net 50 per cent rise.

Answer: D

If 80 percent of the students passed, then the 24 students who failed are the other 20 percent. So 20 percent of the total is 24, and the total is 24 / 0.2 = 120 students. Check: 80 percent of 120 is 96 who passed, and 96 + 24 = 120. The answer is 120, option (D). The values 130, 160 and 100 would not give 24 failures at a 20 percent failure rate; 100 would give 20 and 160 would give 32. As an alternative method, 80 percent is 4/5, so the 24 failures are 1/5 of the total, and the total is 24 x 5 = 120.

Answer: A

T2 got 605 marks. T1 got 45 percent less than T2, so T1 has 55 percent of 605, which is 332.75 marks. T1 also got 21 percent more than T3, which means T1 = 1.21 x T3, so T3 = 332.75 / 1.21 = 275 marks. Check: 21 percent more than 275 is 332.75, and 332.75 is 55 percent of 605. So T3 got 275 marks, option (A). The other values, 255, 285 and 265, would not give T1 = 332.75 when increased by 21 percent. In per cent questions of this kind it is easiest to work with the decimal factors 0.55 and 1.21, as above, and to check the result by applying them back.

Answer: B

His expenditure increases by 12.5 %. Let the income be 100. Savings are 18 and expenditure is 82. The income rises by 6.65 % to 106.65, and the savings fall by 20 % to 14.4.

\[ \text{New expenditure} = 106.65 - 14.4 = 92.25 \]

\[ \frac{92.25 - 82}{82} \times 100 = 12.5\% \]

The options 12 %, 10.5 % and 10 % are not the ratio of the rise of 10.25 to the original expenditure of 82. As a check, the new income of 106.65 equals the new expenditure of 92.25 plus the new savings of 14.4, and the old income of 100 equals 82 plus 18.

Answer: B

The price decreases by 9%.

An increase of 30% followed by a decrease of 30% gives a net factor of 1.3 × 0.7.

\[ 1.3\times0.7=0.91 \]

The final price is 91% of the original, a fall of 9%. 'No change' wrongly assumes that the two percentages cancel because they are applied to different amounts.

The percentages are applied to different bases, since the 30% decrease is taken on the increased price and not on the original. For a price of 100 the price goes to 130 and then falls by 39 to 91. This gives the net fall of 9%, a result that is true for any equal rise and fall of p%, which is p squared over 100 percent.

Answer: C

The total number of valid votes is 30000.

The winner got 62% and the loser 38%, so the margin is 24% of the votes, which equals 7200.

\[ 0.24\times N=7200 \Rightarrow N=30000 \]

273600, 3600 and 27360 do not satisfy this equation.

The 38% candidate lost by 7200 votes, so the difference between the two candidates, which is 62% − 38% = 24% of the votes, is 7200. One percent is therefore 300 votes and the total of 100% is 30000.

Answer: C

The percentage error is 1.67%.

The error is 12.2 − 12 = 0.2 cm on a true length of 12 cm.

\[ \frac{0.2}{12}\times100\approx1.67\% \]

0.16% and 0.67% are not the error on 12 cm, and 16.7% divides by 1.2 instead of 12.

The percentage error is always taken on the true value, here 12 cm, and not on the measured value. Using 12.2 as the base would give 1.64%, which is not an option. The difference 0.2 is the absolute error.

Answer: C

The expenditure increases by 5%. Take the original price as 100. After a 40% rise it is 140, and after a 20% fall it is \(140\times 0.8 = 112\), so the price factor is 1.12. A reduction of \(6\frac{1}{4}\%\) in consumption leaves \(1 - \frac{1}{16} = \frac{15}{16}\) of it. \[ 1.12\times\frac{15}{16} = 1.05 \] The expenditure is 105% of the original, an increase of 5%. A decrease of 5% or 15% would need a factor below 1, and an increase of 15% would need the factor 1.15, neither of which results.

Answer: B

The lowest percentage increase belongs to Q.

Percentage increase = (2010 runs - 2009 runs) / 2009 runs x 100:

\[ \begin{aligned} P &: 206/802 = 25.7\% \\ Q &: 147/765 = 19.2\% \\ R &: 190/429 = 44.3\% \\ S &: 200/501 = 39.9\% \end{aligned} \]

Q has the smallest increase, both in runs (147) and as a percentage. P has the highest totals and might look slowest-growing at a glance, but its increase is 25.7%. R and S, starting from lower bases, grew fastest.