Data Interpretation Questions and Answers

Answer: D

The answer is Cannot be determined.

Antarctica now has 8% of the wild animals. The question says its population will grow 2·5% more than that of the rest of the world, but it never says how fast the rest of the world grows.

Antarctica's future share depends on both growth rates. If the rest of the world stayed level, Antarctica's 8 units would grow to 8.2 and its share to about 8.2%; with other growth rates the share changes again. None of 9·5%, 10·5% or 15·7% can be fixed without the missing rate.

Answer: B

On 27th June, 50 dollars would be worth 5,400 Yen.

The Dollar/Yen chart reads 107 on the 21st and 107.5 on the 24th, a rise of 0.5 over those three days. The same trend from the 24th to the 27th takes it to 108.

\[ 50 \times 108 = 5400 \text{ Yen} \]

The question asks about the Dollar/Yen ratio, so the two rupee options are beside the point, and at Dollar/Rupee rates near 45 they would not equal 50 dollars in any case. 6,000 Yen would need a rate of 120, far above the chart.

Answer: A

On 20th June Mohan could buy the jacket only with 18,200 Rupees.

Read each chart on 20th June and convert the 400-dollar price:

\[ \begin{aligned} \text{Rupees} &: 400 \times 45.5 = 18200 \\ \text{Pounds} &: 400 / 1.6 = 250 \\ \text{Yen} &: 400 \times 106.3 \approx 42500 \end{aligned} \]

His ₹ 18,200 exactly meets the rupee price. His 225 Pounds fall short of 250, and 35,000 Yen is far below about 42,500, so neither of those currencies works. Since the answer can be read from the charts, 'Cannot be determined' is wrong.

Answer: D

Among the options, P6 remained not out the most times.

Since average = runs / times out, the number of times each batsman got out is runs divided by average:

P6 was not out in 6 innings, more than P1 (3), P2 (5) or P5 (3). P4 also has 6 not-outs but is not among the options.

Choosing P5 because of the highest average, or P2 because of the most runs, confuses scoring with not-outs.

Answer: B

P3's average would change the least.

Since average = runs / times out, the number of times each batsman got out is runs divided by average:

P3 got out in all 46 innings, so dividing his runs by innings gives the same 36·5: no change at all. The others fall: P1 to 1800 / 48 = 37.5, P5 to 1806 / 38 = 47.5 and P6 to 1445 / 40 = 36.1.

The batsman with more not-outs sees the bigger change, which is why P6 changes the most among the options.

Answer: B

Only 1 batsman got out in every innings he played.

Since average = runs / times out, the number of times each batsman got out is runs divided by average:

Only P3 has as many dismissals as innings (46 of 46). Every other batsman has at least 3 not-outs.

'Cannot be determined' is wrong, because the definition of the average lets you recover the number of dismissals exactly; 0 and 2 miscount P3.

Answer: B

On day 6 Shakuntala plucked 35 apples.

Combine a five-day sum with a three-day sum that starts on the same day:

\[ \begin{aligned} d_7 + d_8 &= (d_4 + \dots + d_8) - (d_4 + d_5 + d_6) \\ &= 270 - 145 = 125 \\ d_6 &= (d_6 + d_7 + d_8) - 125 \\ &= 160 - 125 = 35 \end{aligned} \]

The charts are therefore enough, so 'Cannot be determined' is wrong. 45 and 25 are the counts for other days (day 10 and day 2) when every day is worked out the same way.

Answer: D

The charts fix the apples of all ten days, so none of the options 0, 5 or 3 is right and the answer is None of the above.

Let the days be d₁ to d₁₀. The first question found d₆ = 35 from d₄ + d₅ + d₆ = 145 and the two sums that give d₇ + d₈ = 125. The rest follows step by step:

\[ \begin{aligned} d_4 + d_5 &= 145 - 35 = 110 \\ d_3 &= 165 - 110 = 55 \\ d_3 + \dots + d_7 = 265 &\Rightarrow d_7 = 265 - 55 - 110 - 35 = 65 \\ d_5 &= 140 - 35 - 65 = 40, \quad d_4 = 70 \\ d_8 &= 125 - 65 = 60 \\ d_9 &= 175 - 65 - 60 = 50 \\ d_{10} &= 155 - 60 - 50 = 45 \\ d_2 &= 150 - 55 - 70 = 25, \quad d_1 = 130 - 25 - 55 = 50 \end{aligned} \]

So the ten days are 50, 25, 55, 70, 40, 35, 65, 60, 50 and 45, and every one of the eight three-day sums and six five-day sums in the charts is reproduced by them. Solving all fourteen equations leaves no free unknown, so the exact count is determined for all 10 days. Zero, 5 and 3 are all too few.

Answer: A

The trip costs ₹ 26,650.

Destination C has two flights, at 11.00 for Rs. 6,300 and at 22.00 for Rs. 7,200. The 11.00 flight is the morning one, so the paper's afternoon flight is the later one at Rs. 7,200 per passenger. The extras are Rs. 2,000 per item of hold luggage and Rs. 350 per passenger for the environmental offset.

\[ \begin{aligned} \text{fares} &= 3 \times 7200 = 21600 \\ \text{luggage} &= 2 \times 2000 = 4000 \\ \text{offset} &= 3 \times 350 = 1050 \\ \text{total} &= 26650 \end{aligned} \]

Leaving out the offset gives 25,600; charging the offset for only two people gives 26,300; and charging luggage for three items gives 27,600. Using the 11.00 fare instead gives 23,950, which is not offered.

Answer: A

The ratio is \(\frac{4}{7} : \frac{2}{3}\).

Read the two second-stage diagrams.

• Of the 70 candidates who cleared E₁ first, those who cleared E₂ in the second stage are 10 + 30 = 40, a fraction of 40/70 = 4/7.
• Of the 60 who cleared E₂ first, those who cleared E₁ in the second stage are 10 + 30 = 40, a fraction of 40/60 = 2/3.

The two percentages are therefore in the ratio 4/7 : 2/3.

The options beginning with 3/7 count only the 30 candidates who cleared both E₂ and the viva, leaving out the 10 who cleared E₂ but failed the viva. None of 3/7, 2/7 or 1/5 matches the 40 out of 60 in the other diagram.