Comprehension

College and University enrollment

Question: 1

The total enrollment in 1985 was approximately how much greater than the total enrollment in 1960?

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When reading from a bar graph, you often have to estimate the values. Read them as carefully as you can, but don't worry about extreme precision. The answer choices are usually spread out enough to accommodate small estimation errors.
Updated On: Oct 4, 2025
  • 4 million
  • 5 million
  • 7 million
  • 9 million
  • 11 million
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This question requires us to read data from a bar graph for two different years, calculate the total for each year, and then find the difference between these totals.
Step 2: Key Formula or Approach:
1. Read the male and female enrollment numbers for 1960 and 1985. 2. Calculate the total enrollment for each year by adding the male and female values. 3. Subtract the 1960 total from the 1985 total to find the difference.
Step 3: Detailed Explanation:
First, let's read the approximate values from the bar graph (in millions). For 1960: - Males (dark bar) \(\approx\) 2.2 million - Females (light bar) \(\approx\) 1.3 million - Total enrollment in 1960 \(\approx\) 2.2 + 1.3 = 3.5 million For 1985: - Males (dark bar) \(\approx\) 6.0 million - Females (light bar) \(\approx\) 6.5 million - Total enrollment in 1985 \(\approx\) 6.0 + 6.5 = 12.5 million Now, find the difference: \[ \text{Difference} = \text{Total in 1985} - \text{Total in 1960} \] \[ \text{Difference} \approx 12.5 \text{ million} - 3.5 \text{ million} = 9.0 \text{ million} \] The total enrollment in 1985 was approximately 9 million greater than in 1960.
Step 4: Final Answer:
The difference in total enrollment is approximately 9 million.
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Question: 2

For which of the years shown was the ratio of male to female enrollment greatest?

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For ratio questions on bar graphs, you can often solve them by visual inspection. Look for the largest proportional difference between the two bars being compared. You don't always need to calculate the exact numbers if the visual difference is clear.
Updated On: Oct 4, 2025
  • 1980
  • 1975
  • 1970
  • 1965
  • 1960
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The Correct Option is

Solution and Explanation

Step 1: Understanding the Concept:
We need to find the year where the ratio of male enrollment to female enrollment was the highest. This means we are looking for the year where the number of males was largest \textit{in comparison to} the number of females.
Step 2: Key Formula or Approach:
The ratio is \( \frac{\text{Males}}{\text{Females}} \). We can either estimate this ratio for each year or visually inspect the graph. A larger ratio means the male bar is much taller than the female bar.
Step 3: Detailed Explanation:
Let's visually inspect the graph for each year. We are looking for the year where the dark bar (Males) is proportionally the largest compared to the light bar (Females).
- 1960: Male bar (\(\approx\) 2.2M) is significantly taller than the female bar (\(\approx\) 1.3M). The ratio is clearly greater than 1. \( \frac{2.2}{1.3} \approx 1.69 \) - 1965: Male bar (\(\approx\) 3.5M) is taller than the female bar (\(\approx\) 2.2M). The ratio is greater than 1, but visually, the proportional difference seems smaller than in 1960. \( \frac{3.5}{2.2} \approx 1.59 \) - 1970: Male bar (\(\approx\) 4.5M) is taller than the female bar (\(\approx\) 3.2M). The proportional difference continues to shrink. \( \frac{4.5}{3.2} \approx 1.41 \) - 1975: Male bar (\(\approx\) 5.8M) is taller than the female bar (\(\approx\) 5.0M). They are getting very close in height. \( \frac{5.8}{5.0} = 1.16 \) - 1980: The male bar (\(\approx\) 5.8M) and female bar (\(\approx\) 6.2M) are very close, and the female bar is now slightly taller. The ratio is less than 1. - 1985: The female bar (\(\approx\) 6.5M) is clearly taller than the male bar (\(\approx\) 6.0M). The ratio is less than 1. By both visual inspection and approximate calculation, the ratio of males to females was highest in 1960, when the male bar was proportionally much taller than the female bar. The ratio has consistently decreased over time.
Step 4: Final Answer:
The greatest ratio of male to female enrollment occurred in 1960.
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Question: 3

For which of the following periods was the percent increase in female enrollment the greatest?

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Don't confuse the largest absolute increase with the largest percent increase. The largest percent increase often happens when the initial value (the denominator) is small. In this case, the absolute increase from 1970-1975 was largest, but the percent increase was highest from 1960-1965 because the starting base was much smaller.
Updated On: Oct 4, 2025
  • 1960 to 1965
  • 1965 to 1970
  • 1970 to 1975
  • 1975 to 1980
  • 1980 to 1985
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
We need to calculate the percent increase in female enrollment for each 5-year period and identify the period with the highest percent increase.
Step 2: Key Formula or Approach:
The formula for percent increase is \( \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100% \). We need to calculate this for the female enrollment (light bars) for each period. A large percent increase can happen either from a large absolute increase or a small original value.
Step 3: Detailed Explanation:
Let's first read the approximate female enrollment values for each year: - 1960: 1.3 M - 1965: 2.2 M - 1970: 3.2 M - 1975: 5.0 M - 1980: 6.2 M - 1985: 6.5 M Now, let's calculate the percent increase for each period:

(A) 1960 to 1965: Increase = 2.2 - 1.3 = 0.9 M. Percent Increase = \( \frac{0.9}{1.3} \times 100% \approx 0.69 \times 100% = 69% \)
(B) 1965 to 1970: Increase = 3.2 - 2.2 = 1.0 M. Percent Increase = \( \frac{1.0}{2.2} \times 100% \approx 0.45 \times 100% = 45% \)
(C) 1970 to 1975: Increase = 5.0 - 3.2 = 1.8 M. (This is the largest absolute increase). Percent Increase = \( \frac{1.8}{3.2} \times 100% \approx 0.56 \times 100% = 56% \)
(D) 1975 to 1980: Increase = 6.2 - 5.0 = 1.2 M. Percent Increase = \( \frac{1.2}{5.0} \times 100% = 0.24 \times 100% = 24% \)
(E) 1980 to 1985: Increase = 6.5 - 6.2 = 0.3 M. Percent Increase = \( \frac{0.3}{6.2} \times 100% \approx 0.05 \times 100% = 5% \)
Comparing the percent increases: 69%, 45%, 56%, 24%, 5%. The greatest percent increase occurred during the period from 1960 to 1965.
Step 4: Final Answer:
The greatest percent increase in female enrollment was from 1960 to 1965, at approximately 69%.
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