Percentage Questions and Answers

Answer: C

The present female population is 5,000.

Let the numbers of males and females be M and F. Then \(M + F = 11000\), and after the increases:

\[ \begin{aligned} 1.11M + 1.2F &= 12660 \\ 1.11(11000) + 0.09F &= 12660 \\ 12210 + 0.09F &= 12660 \\ F &= 5000 \end{aligned} \]

Check: 6,000 males rise by 660 and 5,000 females by 1,000, a total rise of 1,660, from 11,000 to 12,660. 6,000 is the number of males, which is the trap.

Answer: A

The greatest is 14.28% of 700.

Each percentage is close to a simple fraction:

\[ \begin{aligned} 14.28\% \times 700 &\approx \tfrac{1}{7} \times 700 = 100 \\ 11.11\% \times 819 &\approx \tfrac{1}{9} \times 819 = 91 \\ 9.09\% \times 704 &\approx \tfrac{1}{11} \times 704 = 64 \\ 6.25\% \times 960 &= \tfrac{1}{16} \times 960 = 60 \end{aligned} \]

Knowing these fraction equivalents makes the comparison quick: 100 is clearly the largest.

Answer: A

The other candidate got 6,380 valid votes.

20% of the votes were invalid, so 80% were valid. The first candidate got 45% of the valid votes, so the other got 55%:

\[ \begin{aligned} \text{valid} &= 0.8 \times 14500 = 11600 \\ \text{other} &= 0.55 \times 11600 = 6380 \end{aligned} \]

5,220 is the first candidate's share, and 6,525 and 7,975 come from taking 45% or 55% of all 14,500 votes instead of the valid ones.

Answer: B

The net decrease is 1%.

Take the number as 100. After a 10% increase it becomes 110; a 10% decrease of 110 removes 11, leaving 99. The net change is 100 to 99, a decrease of 1%.

\[ \begin{aligned} &\left(1 + \frac{10}{100}\right)\left(1 - \frac{10}{100}\right) \\ &= 1.1 \times 0.9 = 0.99 \end{aligned} \]

The shortcut for an equal rise and fall of \(x\)% is a net fall of \(\frac{x^{2}}{100}\)%. 0% is the trap of assuming the two changes cancel; they do not, because the decrease is taken on a larger base.

Answer: B

A town growing at 6% a year from 238765 has in 2020 a population of \(238765\times1.06\times1.06=268276\) (about), option B.

286726 and 286722 reverse the digits, and 212500 is far below even the 2018 figure plus growth.

Compound growth uses the multiplier 1.06 each year. The first year gives 253,091.9 and the second year 268,277.4, which the question rounds to 268,276 in the options. The two options 286,726 and 286,722 are transpositions that a careless calculation could produce.

Answer: D

A rise of 5% and then a fall of 5% leaves the population at 1.05 x 0.95 = 0.9975 of the start. So the start is 9975 / 0.9975 = 10,000, option D.

9,975 is the final figure itself, and 9,500 and 10,500 do not give 9,975 after the two changes.

Checking the answer: 10,000 increased by 5 per cent is 10,500, and a fall of 5 per cent on 10,500 is 525, which leaves 9,975, exactly the figure in the question. The common error is to think that a rise and a fall of the same percentage cancel out, which would give 9,975 as the start.

Answer: A

The number of bacteria after three hours is \(20000\times1.05\times0.88\times1.10=20328\), option A.

The other values come from using wrong signs, such as adding the 12% instead of subtracting.

The changes multiply: 20,000 x 1.05 = 21,000; x 0.88 = 18,480; x 1.10 = 20,328. A student who adds the percentages, +5 - 12 + 10 = +3 per cent, would get 20,600, which is not an option, and the other values come from using the wrong factors.

Answer: B

The price including 18% GST is 118% of the base price, so the base price is 10,502 / 1.18 = Rs 8,900.

The GST amount is 10,502 - 8,900 = Rs 1,602, option B. Rs 1,620 would be 18% of 9,000.

The price with GST is 118 per cent of the price without it, so the base is 10,502 x 100 / 118 = 8,900. The common error is to take 18 per cent of 10,502, which gives 1,890, while another is 18 per cent of 9,000 (1,620), and neither is the answer.

Answer: C

Let the maximum marks be M. Peter's 30% is 10 short of the pass mark, so pass marks = 0.3M + 10. Paul's 40% is 15 more than the pass mark, so \[0.4M=0.3M+10+15,\ \ 0.1M=25,\ \ M=250.\]

Pass marks = 0.3 x 250 + 10 = 85, option C. Check: 40% of 250 is 100, which is 15 above 85.

The pass mark is the same for both candidates, so the difference between 40% and 30%, which is 10% of M, equals the 25 marks that separate the two results.

Answer: A

Akila's score of 576 is 80% of the maximum marks M, so \[M=\tfrac{576}{0.8}=720.\]

Check: 80% of 720 is 576. The other options give a different 80%: 620 gives 496, 820 gives 656 and 710 gives 568. So the maximum marks are 720, option A.

Equivalently, 576 corresponds to 80 parts out of 100, so one part is 7.2 and 100 parts are 720.