Percentage Questions and Answers

Answer: A

The part present is \(\frac{23}{30}\).

Girls are 3/5 of the class and boys 2/5. Of the girls, \(1 - \tfrac{2}{9} = \tfrac{7}{9}\) are present; of the boys, \(1 - \tfrac{1}{4} = \tfrac{3}{4}\):

\[ \begin{aligned} \tfrac{3}{5} \times \tfrac{7}{9} + \tfrac{2}{5} \times \tfrac{3}{4} &= \tfrac{7}{15} + \tfrac{3}{10} \\ &= \tfrac{14}{30} + \tfrac{9}{30} = \tfrac{23}{30} \end{aligned} \]

7/30 is the part absent, the complement of the answer. 17/25 and 18/49 do not arise from the given fractions.

Answer: A

The voters list has 2705 voters.

Every voter voted, there were no invalid votes and no NOTA, so the votes cast equal the voters on the list, and Y received the other 80%.

\[ \begin{aligned} 80\% - 20\% &= 60\% \text{ of } N = 1623 \\ N &= \frac{1623}{0.6} = 2705 \end{aligned} \]

Check: X got 541 and Y got 2164, a margin of 1623. The larger options cannot be right: 60% of 6905 or of 12705 is far more than 1623.

Answer: D

B scored 75 marks.

Let A score a. Then B = a + 15 and also B = 1.25a.

\[ \begin{aligned} 1.25a &= a + 15 \\ 0.25a &= 15 \Rightarrow a = 60 \\ B &= 60 + 15 = 75 \end{aligned} \]

Check: 75 is 25% more than 60. The option 60 is A's score, the usual trap. 35 and 40 do not satisfy both conditions: for B = 40, A would be 25, and 40 is 60% more than 25, not 25%.

Answer: C

The net change is 8.75%, an increase.

Successive percentage changes multiply, they do not add.

\[ \begin{aligned} 0.75 \times 1.45 &= 1.0875 \\ \text{change} &= +8.75\% \end{aligned} \]

The shortcut formula for changes x and y is \(x + y + xy/100 = -25 + 45 - 11.25 = 8.75\). −11.25 is only the cross term, 20 is the simple sum \(-25 + 45\), and −1.25 is a sign slip.

Answer: B

Only two statements are correct.

Statement 1: 48% of N is 120, so N = 250, and 170% of 250 is 425, not 350. False.

Statement 2: 3.5 × 100% = 350%. True.

Statement 3: passes are \(0.42 \times 1100 = 462\) boys and \(0.30 \times 700 = 210\) girls, 672 in all out of 1800, so 1128 failed.

\[ \frac{1128}{1800} \times 100 = 62\tfrac{2}{3}\% \]

True. So statements 2 and 3 are correct and statement 1 is not.

Answer: A

The percentage is 25%.

Each of the 3 routes from A to B can be followed by any of the 4 routes from B to C, so the number of ways from A to C is the product:

\[ \begin{aligned} A \to C &= 3 \times 4 = 12 \\ \frac{3}{12} \times 100 &= 25\% \end{aligned} \]

300/7% (about 42.9%) comes from adding the routes (3 + 4 = 7) instead of multiplying them, the common error. 74% and 41% do not arise from the given numbers.

Answer: C

About 8.33 kg can be bought now.

Quantity bought with a fixed sum is inversely proportional to price. If the price rises by 20%, the new price is 1.2 times the old:

\[ \frac{10}{1.2} = 8.33\ \text{kg} \]

8 kg is the trap: it assumes a 20% price rise means 20% less coffee. In fact the quantity falls by 1/6, about 16.7%. 9 kg and 7.77 kg do not follow from the figures.

Answer: D

Sudha secured 80 percent of Sunil's marks, so p = 80.

Express both marks as fractions of Radha's marks R and divide.

\[ \begin{aligned} \text{Sudha} &= 0.48R, \quad \text{Sunil} = 0.60R \\ p &= \frac{0.48R}{0.60R} \times 100 = 80 \end{aligned} \]

Radha's marks cancel, so the answer does not depend on them. Taking the difference of the percentages (60 − 48 = 12) or reversing the ratio (60/48 = 125%) leads away from the options; 60 is simply Sunil's own percentage.

Answer: C

He must score 74% in the second paper.

The two papers together carry 180 + 150 = 330 marks, and 50% of that is 165. In the first paper he scored 30% of 180:

\[ \begin{aligned} 0.30 \times 180 &= 54 \\ 165 - 54 &= 111 \\ \frac{111}{150} \times 100 &= 74\% \end{aligned} \]

So he needs 111 out of 150 in the second paper. A common slip is to average percentages, \(2 \times 50 - 30 = 70\%\), which ignores that the papers carry different maximum marks; that gives the 70% option.

Answer: B

The maximum marks are 550.

The student scored 113 marks and fell short of the pass mark by 85, so the pass mark is \(113 + 85 = 198\). That is 36% of the maximum M:

\[ \begin{aligned} 0.36\,M &= 198 \\ M &= \frac{198}{0.36} = 550 \end{aligned} \]

Check: 36% of 550 is 198, and \(198 - 113 = 85\). Taking the 113 marks alone as 36% would give about 314, which is not offered, and 500, 640 and 1000 do not give a pass mark 85 above 113.