Interpretation of Data Questions and Answers

Answer: B

The angle for 'other' is 90°.

The angles of a pie chart add up to 360°:

\[ \begin{aligned} x &= 360 - (108 + 63 + 72 + 27) \\ &= 360 - 270 = 90^\circ \end{aligned} \]

63°, 108° and 72° are the angles already given for clothing, food and house rent, not the remaining sector.

Answer: B

Food accounts for 30% of the expenses.

A sector's share of the total is its angle divided by 360°:

\[ \frac{108}{360} \times 100 = 30\% \]

25% would be a 90° sector, the 'other' expenses, and 20% a 72° sector, house rent. 90% confuses the 'other' sector's angle with a percentage.

Answer: A

The difference is maximum for Engineering and allied.

Subtract district B from district E for each type:

\[ \begin{aligned} \text{Agro} &: 165 - 86 = 79 \\ \text{Forest} &: 120 - 70 = 50 \\ \text{Livestock} &: 176 - 90 = 86 \\ \text{Textile} &: 132 - 92 = 40 \\ \text{Chemical} &: 154 - 86 = 68 \\ \text{Engineering} &: 166 - 76 = 90 \end{aligned} \]

The largest difference, 90, is for engineering and allied. Agro-based (79) and chemical-based (68) are smaller, and textile-based has the smallest difference of all.

Answer: B

The percentage is 17.2%.

District B has 86 agro-based companies out of a total of 500:

\[ \frac{86}{500} \times 100 = 17.2\% \]

86% treats the count as a percentage, 8.6% divides by 1000 instead of 500, and 16.67% would be one-sixth, the share if all six types were equal.

Answer: C

61 players play only one game.

Those in exactly two games are the pairwise figures minus the 3 who play all three:

\[ \begin{aligned} \text{two} &= (8-3) + (10-3) + (7-3) = 16 \\ \text{union} &= 102 - (8 + 10 + 7) + 3 = 80 \\ \text{one} &= 80 - 16 - 3 = 61 \end{aligned} \]

The union equals the 80 players, so everyone plays at least one game. 102 is the plain sum 35 + 42 + 25 without removing overlaps, 80 is everyone, and 63 does not match any count that can be formed from the figures given, such as one game (61) or two games (16).

Answer: D

16 players play exactly two games.

Each pairwise figure includes the 3 players who play all three games, so subtract them from each pair:

\[ \begin{aligned} \text{C and H only} &= 8 - 3 = 5 \\ \text{C and F only} &= 10 - 3 = 7 \\ \text{H and F only} &= 7 - 3 = 4 \\ \text{total} &= 5 + 7 + 4 = 16 \end{aligned} \]

19 is that total plus the 3 who play all three games. 61 is the number who play only one game, and 80 is all the players.

Answer: D

The pass percentage is highest in 2015.

\[ \begin{aligned} 2012: \frac{200}{500} &= 40\% \\ 2013: \frac{400}{700} &\approx 57.1\% \\ 2014: \frac{500}{800} &= 62.5\% \\ 2015: \frac{700}{1000} &= 70\% \end{aligned} \]

The pass rate rises every year, reaching 70% in 2015. 2014 has more passes than 2013 but a lower rate than 2015, and 2012 is the lowest at 40%. Read the hatched part of each bar as the passes, not the whole bar.

Answer: C

The combined pass percentage is 60.

Add the passes and the totals over all four years before dividing:

\[ \begin{aligned} \text{passed} &= 200 + 400 + 500 + 700 = 1800 \\ \text{total} &= 500 + 700 + 800 + 1000 = 3000 \\ \frac{1800}{3000} &= 60\% \end{aligned} \]

Averaging the four yearly rates instead gives about 57.4%, close to the 57.14 option, which is only 2013's rate. 70 is 2015's rate alone, and 50 has no basis in the data.

Answer: B

165 students failed in exactly two languages.

Each 'failed in both' figure includes the 30 who failed in all three, so subtract 30 from each pair:

\[ \begin{aligned} \text{M and E only} &= 125 - 30 = 95 \\ \text{M and H only} &= 70 - 30 = 40 \\ \text{E and H only} &= 60 - 30 = 30 \\ \text{total} &= 95 + 40 + 30 = 165 \end{aligned} \]

255 is the raw sum 125 + 70 + 60, which counts the all-three group three times, and 195 subtracts the 30 only twice instead of three times. 250 is the number who failed in only one language (Q34).

Answer: C

250 students failed in only one language.

For each language subtract the two pairs that include it and add back the 30 who were subtracted twice:

\[ \begin{aligned} \text{Marathi only} &= 220 - 125 - 70 + 30 = 55 \\ \text{English only} &= 300 - 125 - 60 + 30 = 145 \\ \text{Hindi only} &= 150 - 70 - 60 + 30 = 50 \\ \text{total} &= 55 + 145 + 50 = 250 \end{aligned} \]

Check: 250 in one language, 165 in exactly two and 30 in all three make 445 who failed at least once, which agrees with \(670 - 255 + 30 = 445\). 165 is the exactly-two count, and 195 and 255 are wrong pair totals.