Question

Sohan's average for the first six years was:
I. equal to that of the last six years.
II. equal to that of the middle six years.
III. 225

• A. III only
• B. I & III
• C. I, II & III
• D. II & III
• A. 1984
• B. 1985
• C. 1986
• D. never
• A. It increases by 50% every year.
• B. It increases by 25% every year.
• C. It increases by 50 every year.
• D. It increases by 25 every year.
• A. 700
• B. 825
• C. 900
• D. 225
• A. one
• B. two
• C. three
• D. four

Answer 1

The Correct Option is C

Step 1: Read Sohan's score for every year straight off the chart.
The dark blue line on the radar chart is Sohan's score, plotted for each year from 1981 to 1989. Look at where this line touches each of the nine spokes: it lands on the same ring, 225, every single time. So Sohan's nine yearly scores are all 225: 1981 is 225, 1982 is 225, 1983 is 225, 1984 is 225, 1985 is 225, 1986 is 225, 1987 is 225, 1988 is 225, and 1989 is 225.

Step 2: Work out the average for the first six years (1981 to 1986).
Sum of these six scores = 225 + 225 + 225 + 225 + 225 + 225 = 1350.
Average = 1350 / 6 = 225.

Step 3: Work out the average for the last six years (1984 to 1989).
These six years also each score 225, so the sum is again 225 x 6 = 1350, and the average is 1350 / 6 = 225.
This is the same as the first-six-year average, so statement I (first six years equal to last six years) is true.

Step 4: Work out the average for the middle six years (1982 to 1987).
Once again the sum is 225 x 6 = 1350 and the average is 225, matching the first-six-year average. So statement II (first six years equal to middle six years) is also true.

Step 5: Check statement III.
Statement III simply claims the common average is 225. We just calculated that every one of these six-year averages equals exactly 225, so statement III is true as well.

Why the other options fail:
"III only" (option 1) drops I and II, but we showed both hold. "I & III" (option 2) leaves out II, and "II & III" (option 4) leaves out I, but both I and II check out from the working above, so neither partial option can be correct.

Final Answer:
Because Sohan's score never moves away from 225 in any of the nine years, every six-year slice of the data, first, middle, or last, averages to exactly 225. All three statements hold together. \[ \boxed{\text{Option 3: I, II and III}} \]

Answer 2

Step 1: Read both students' scores off the chart.
Sohan's line (dark blue) sits at 225 in every single year from 1981 to 1989, as we can read directly from the chart.
Mohan's line (pink) climbs steadily: 25 in 1981, 50 in 1982, 75 in 1983, 100 in 1984, 125 in 1985, 150 in 1986, 175 in 1987, 200 in 1988, and 225 in 1989.

Step 2: Set up the condition for "exactly half".
Mohan's score is half of Sohan's when \( \text{Mohan} = \frac{\text{Sohan}}{2} \). Since Sohan is always 225, this means we need Mohan's score to equal \( \frac{225}{2} = 112.5 \) in some year.

Step 3: Check whether 112.5 ever appears in Mohan's list.
Mohan's scores are 25, 50, 75, 100, 125, 150, 175, 200, 225. Every one of these is a whole multiple of 25, so none of them can equal 112.5, which sits between 100 and 125 but is not itself a multiple of 25.

Step 4: Check the specific years offered as options.
In 1984, Mohan is 100, and half of Sohan's 225 is 112.5, these do not match.
In 1985, Mohan is 125, again not 112.5.
In 1986, Mohan is 150, still not 112.5.
So none of 1984, 1985 or 1986 actually satisfies the condition, even though 1984 and 1985 bracket the true halfway point of 112.5 the closest.

Final Answer:
Since Sohan's score is fixed at 225 and Mohan's score only ever takes multiples of 25, Mohan's score is never exactly half of Sohan's in any of the nine years. \[ \boxed{\text{never (Option 4)}} \]

Answer 3

Step 1: List Mohan's score for every year from the chart.
Mohan's pink line reads: 25 in 1981, 50 in 1982, 75 in 1983, 100 in 1984, 125 in 1985, 150 in 1986, 175 in 1987, 200 in 1988, and 225 in 1989.

Step 2: Find the year-to-year change in absolute terms.
1982 - 1981: 50 - 25 = 25.
1983 - 1982: 75 - 50 = 25.
1984 - 1983: 100 - 75 = 25.
1985 - 1984: 125 - 100 = 25.
Every single year the increase is exactly 25 marks, all the way through to 1989. This is a constant absolute increase, which matches option 4.

Step 3: Rule out option 3 (increase of 50 every year).
We just found the actual jump each year is 25, not 50. So option 3 doubles the real increase and is wrong.

Step 4: Rule out the percentage options (1 and 2) by checking the percentage change each year.
From 1981 to 1982: increase of 25 on a base of 25 is \( \frac{25}{25} \times 100 = 100\% \).
From 1982 to 1983: increase of 25 on a base of 50 is \( \frac{25}{50} \times 100 = 50\% \).
From 1983 to 1984: increase of 25 on a base of 75 is \( \frac{25}{75} \times 100 \approx 33.3\% \).
The percentage increase keeps shrinking every year (100%, then 50%, then about 33%, and so on), it is never a fixed 25% or 50%. So options 1 and 2 are both wrong, since a genuinely constant percentage growth would need Mohan's scores to multiply by the same factor each year, which they do not, they only add the same amount each year.

Final Answer:
Mohan's score rises by a fixed amount of 25 marks every year, a simple arithmetic progression, not a percentage growth. \[ \boxed{\text{increases by 25 every year (Option 4)}} \]

Answer 4

Step 1: Add up Sohan's total score across the nine years.
Sohan scores 225 in every year on the chart, from 1981 through 1989, that is nine years in total.
Sohan's total = 225 x 9 = 2025.

Step 2: Add up Mohan's total score across the same nine years.
Mohan's scores are 25, 50, 75, 100, 125, 150, 175, 200, 225. This is an arithmetic progression with first term 25, last term 225, and 9 terms.
Sum of an arithmetic progression = \( \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term}) \).
Mohan's total = \( \frac{9}{2} \times (25 + 225) = \frac{9}{2} \times 250 = 9 \times 125 = 1125 \).

Step 3: Subtract to find the difference between the two totals.
Difference = Sohan's total - Mohan's total = 2025 - 1125 = 900.

Step 4: Check why the other options do not fit.
700 and 825 do not match this subtraction at all. 225 is simply Sohan's single-year score, not the difference of the nine-year totals, so picking 225 would mean confusing one year's score with the full nine-year comparison.

Final Answer:
Sohan's nine-year total is 2025 and Mohan's nine-year total is 1125, so their totals differ by exactly 900. \[ \boxed{900} \]

Answer 5

Step 1: List both students' scores year by year from the chart.
Sohan: 225 in every one of the nine years (1981 to 1989).
Mohan: 25 (1981), 50 (1982), 75 (1983), 100 (1984), 125 (1985), 150 (1986), 175 (1987), 200 (1988), 225 (1989).

Step 2: Set up the condition "Sohan is exactly thrice Mohan".
We need to find years where \( \text{Sohan} = 3 \times \text{Mohan} \), that is, where \( 3 \times \text{Mohan} = 225 \), so \( \text{Mohan} = \frac{225}{3} = 75 \).

Step 3: Check each year's Mohan score against this target of 75.
1981: Mohan = 25, thrice this is 75, not 225. No match.
1982: Mohan = 50, thrice this is 150, not 225. No match.
1983: Mohan = 75, thrice this is 225, which equals Sohan's score exactly. Match.
1984: Mohan = 100, thrice this is 300, not 225. No match.
1985: Mohan = 125, thrice this is 375, not 225. No match.
1986: Mohan = 150, thrice this is 450, not 225. No match.
1987: Mohan = 175, thrice this is 525, not 225. No match.
1988: Mohan = 200, thrice this is 600, not 225. No match.
1989: Mohan = 225, thrice this is 675, not 225. No match (this is the year Sohan and Mohan are actually equal to each other, not in the 3:1 ratio).

Step 4: Count the matches.
Only 1983 satisfies the condition, out of all nine years.

Final Answer:
Sohan's score is exactly three times Mohan's score in exactly one year, 1983 (225 is three times 75). \[ \boxed{\text{one (Option 1)}} \]