Free GRE Course > GRE Quant Questions > Word Problems > Graphs and Data Interpretation

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### 1. Displaying Data

Data can be displayed in many different ways, including tables and graphs. Both tables and graphs can be used in calculations to evaluate and interpret the data.

Graphs show the relationship of numbers and quantities in visual form, so they are the most common way of displaying data. By surveying a graph, you can very quickly learn about the relationship between two or more sets of information and determine if there are any trends.

The data can be about just one set of events, such as wins and losses or percentages of responses in a survey. This takes one graph. Data can also compare two different events, such as sales vs. earnings or production vs. capacity. Displaying this data can take more than one graph.

Data interpretation involves computing and approximating numerical values based on tables and graphs. GRE questions go beyond just reading the data, requiring you to calculate averages or compare changes, for example. This type of question is usually in the Problem Solving format, and can appear in sets of 2 to 3 questions.

### 2. Tables

Tables give values that are organized but not represented visually. You may graph the data from the table to make comparisons and see trends. You may create a table from a graph to make calculations easier.

Examples 1−5 are based on the table below, which is a record of the performance of a baseball team for the first seven weeks of the season.

 Week Games Won Games Lost Total Games Played 1 5 3 8 2 4 4 16 3 5 2 23 4 6 3 32 5 4 2 38 6 3 3 44 7 2 4 50
##### Example 1

How many games did the team win during the first seven weeks?

(A) 2
(B) 21
(C) 29
(D) 50
(E) 58

### Solution

To find the total number of games won, add the number of games won for all the weeks.
5 + 4 + 5 + 6 + 4 + 3 + 2 = 29

The correct answer is choice (C).

##### Example 2

What percent of the games did the team win?

(A) 21%
(B) 29%
(C) 42%
(D) 58%
(E) 72%

### Solution

The team won 29 out of 50 games.
29/50 = 58/100 = 58%

The correct answer is choice (D).

##### Example 3

Which week was the worst for the team?

(A) second week
(B) fourth week
(C) fifth week
(D) sixth week
(E) seventh week

### Solution

Compare the numbers in the won – lost columns.
The seventh week was the only week that the team lost more games than it won.

The correct answer is choice (E).

##### Example 4

Which week was the best week for the team?

(A) first week
(B) third week
(C) fourth week
(D) fifth week
(E) sixth week

### Solution

Compare the numbers in the win – loss columns. Since there is more than 1 week with more wins than losses, you need to do a quick calculation.

Method 1

Compare the values of the ratio wins/losses. The greatest value will be the best week.

5/3 ≈ 1.67
5/2 = 2.5
6/3 = 2
4/2 = 2

The value for the third week is 2.5.

The correct answer is choice (B).

Method 2

Find the percentage of games won using the number of wins and the number of games each week.

5/8 = 62.5%
5/7 ≈ 70%
6/9 ≈ 67%
4/6 ≈ 67%

The team won 70% of its games in the third week.

The correct answer is choice (B).

##### Example 5

If there are 50 more games to play in the season, how many more games must the team win to end up winning 70% of the games?

(A) 18
(B) 21
(C) 35
(D) 41
(E) 42

### Solution

To win 70% of all the games, the team must win 70 out of 100.
Since the team won 29 games out of the first 50 games, it must win 70 – 29 = 41 games out of the next 50 games.

The correct answer is choice (D).

### 3. Circle Graphs

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Circle graphs represent values as “slices of pie.” Each part is labeled with a fraction or a percent, with the values adding to 1 or 100%.

Be careful in computing problems with percentages. Remember that to convert a decimal into a percentage you must multiply it by 100. For example, 0.04 is 4%. To divide by a percentage, you can multiply using the reciprocal of the percentage as a fraction, as shown in the Example below.

Examples 6 and 7 are based on the circle graph below, which shows the results of a survey taken of people who want a career working with animals. ##### Example 6

If 400 people were surveyed, how many people chose animal trainer?

(A) 36
(B) 40
(C) 45
(D) 360
(E) 364

### Solution

Using the graph, 9% of the number surveyed chose animal trainer.

x = 400 × 9%
x = 400 × 0.09 = 400 × 9/100 = 36

The correct answer is choice (A).

##### Example 7

If 55 people surveyed chose pet store owner, how many people were surveyed?

(A) 40
(B) 110
(C) 121
(D) 250
(E) 400

### Solution

Using the graph, 55 people is 22% of the number surveyed.

55 = 0.22x
55 = 22/100 x

x = 55 × 100/22 = 5 × 11 × 100/2 × 11 = 250

The correct answer is choice (D).

Notice that the total number of people surveyed is not the same in both questions, even though the questions use the same graph.

### 4. Bar Graphs

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Bar graphs can be used to compare non-number categories by showing the number of times each category occurs.

Examples 8 and 9 are based on the bar graph below, which shows the results of a survey about favorite flavors of ice cream offered in the school cafeteria. ##### Example 8

What percent of the students chose flavors in the top 2 categories?

(A) 25%
(B) 35%
(C) 50%
(D) 55%
(E) 60%

### Solution

The percent of students will be the number of students who chose chocolate and chocolate chip divided by the total number of students in the survey.

Estimate the number of students in each bar, then add.
20 + 5 + 26 + 23 + 13 + 20 = 107

top 2 flavors
26 + 23 = 49

49/107 ≈ 50% Slightly less than half the students chose chocolate or chocolate chip.

The correct answer is choice (C).

##### Example 9

The cafeteria wants to add another flavor to the menu. Based on the survey, which flavor might be the best to add?

(A) Fudge Ripple
(B) Marshmallow
(C) Peanut Butter
(D) Orange Sherbet
(E) Peppermint

### Solution

Look for common characteristics.
The top 2 flavors are chocolate and chocolate chip, with Rocky Road tied for third. The common ingredient is chocolate. So adding another ice cream with chocolate as an ingredient would be the best choice.

The correct answer is choice (A).

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### 5. Double Bar Graphs

Bar graphs can show more than one set of values for each category.

Examples 10−12 are based on the double bar graph below, which shows the results from rolling 2 dice: one white and one red. ##### Example 10

What is the median value rolled by the white die? (Note: Die is the singular of dice.)

(A) 2.5
(B) 3.0
(C) 3.5
(D) 4.0
(E) 4.5

### Solution

Look at just the white bars. Since there were 20 rolls, an even number, the median will be the sum of the 2 middle values divided by 2.

1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6

Since the two values are the same, the median is 3.

The correct answer choice is (B).

##### Example 11

Estimate the arithmetic mean value rolled by the red die.

(A) 2.25
(B) 3.6
(C) 5.85
(D) 7.5
(E) 12

### Solution

Look at just the red bars. The mean will be the sum of each face value multiplied by its frequency, then divided by the number of rolls.

sum of values:
(1 × 7) + (2 × 8) + (3 × 1) + (4 × 2) + (5 × 1) + (6 × 1) = 7 + 16 + 3 + 8 + 5 + 6 = 45

number of rolls:
7 + 8 + 1 + 2 + 1 + 1 = 20 rolls

45 / 20 = 2.25

The correct answer choice is (A).

##### Example 12

Assume that the white die is fair (has an equal chance of landing on each of its faces). What is the probability that after one more roll the mean score will increase?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

### Solution

Look at just the white bars. The mean will be the sum of each face value multiplied by its frequency, then divided by the number of rolls.

sum of values:
(1 × 3) + (2 × 4) + (3 × 4) + (4 × 3) + (5 × 3) + (6 × 3) = 3 + 8 + 12 + 12 + 15 + 18 = 58

number of rolls:
2 + 4 + 4 + 3 + 4 + 3 = 20 rolls

58 / 20 = 2.9

The current mean is 2.9. To increase the mean, the roll will need to be greater than the mean. So the roll will need to be a 3, 4, 5, or 6.

The probability of rolling a 3, 4, 5, or 6 = 4/6 = 2/3

The correct answer choice is (D).

The bars in a double bar graph can be “stacked” to show more than one set of values for each category.

Examples 13−15 are based on the stacked bar graph below, which shows the production vs. the production capacity of a car manufacturing plant. The red portion shows the actual production. ##### Example 13

If the trend of linear growth continued for both the production and the production capacity, in which year would the actual production equal the capacity?

(A) 2007
(B) 2008
(C) 2009
(D) 2010
(E) 2011

### Solution

Each year the gap between production and capacity decreased by 1000. In 2006, the gap is 3000. So in 3 years, in 2009, the gap would disappear.

The correct answer is choice (C).

##### Example 14

The cost of production of one car was \$10,000 in 2004. The cost was \$15,000 in 2006. What was the percent increase in gross production costs from 2004 to 2006?

(A) 290%
(B) 350%
(C) 380%
(D) 410%
(E) 510%

### Solution

Production costs in 2004: \$10,000 × 5,000 = \$50 million

Production costs in 2006: \$15,000 × 17,000 = \$255 million

The percent increase in production costs is the difference in costs divided by the original costs.

difference in costs / original cost = 255 – 50/50 = 205/50 = 410/100 = 410%

The correct answer choice is (D).

##### Example 15

All cars manufactured in 2005 were sold for \$10,000 each and all cars manufactured in 2006 were sold for \$20,000 each. What was the percent increase in sales revenue from 2005 to 2006?

(A) 166%
(B) 178%
(C) 198%
(D) 209%
(E) 255%

### Solution

In 2005, the revenue was \$10,000 × 11,000 cars = \$110 million.

In 2004 the revenue was \$20,000 × 17,000 cars = \$340 million.

The percent increase in revenue is the difference in revenue divided by the original revenue.

difference in revenue / original revenue = 340 – 110/110 = 230 /110
This is slightly greater than 200%.

The correct answer choice is (D).

### 6. Histograms

A histogram is very similar to a bar graph, but has no spaces between the bars. The main difference is that histograms use continuous grouped data to show frequency trends.

Examples 16 and 17 are based on the histogram below, which shows the results of a survey on the frequency of eating out for various age groups. ##### Example 16

Which age group eats out for a third of a week’s meals?

(A) 25−34
(B) 35−44
(C) 45−54
(D) 55−64
(E) 65−74

### Solution

A week has about 21 meals, so a third of a week’s meals is 7.

The age interval of 35−44 eats about 7 meals out per week.

The correct answer choice is (B).

##### Example 17

What is the arithmetic mean number of meals eaten out per week?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

### Solution

5.5 + 7 + 3.2 + 1.4 + 2.7/5 = 19.8/5 ≈ 4

The mean number of meals eaten out per week by adults is 4.

The correct answer is choice (B).

Note: To do calculations based on data in a histogram, make sure the intervals are all equal. In the above examples, the intervals all are 10 years, so no adjustments need to be made to the calculations.

Examples 18−21 are based on the histograms below, which show the sales and earnings of a company from 2005 to 2010.  ##### Example 18

During what two-year period did the company’s earnings increase the most in absolute terms?

(A) 2005−2007
(B) 2006−2007
(C) 2006−2008
(D) 2007−2009
(E) 2008−2010

### Solution

Using the second graph, you are looking for the greatest change between 3 adjacent bars.

The biggest increase between 2 adjacent bars is from \$5 million in 2006 to \$10 million in 2007, which is a \$5 million increase. Then the change from \$10 million in 2007 to \$12 million in 2008 is a \$2 million increase.

The two-year increase from 2006 to 2008 was \$7 million — clearly the largest on the graph.

The correct answer choice is (C).

##### Example 19

During the years 2006 through 2008, what were the arithmetic mean earnings per year?

(A) \$6 million
(B) \$7.5 million
(C) \$9 million
(D) \$10 million
(E) \$27 million

### Solution

You can make a quick table to organize the data from the graph.

Year   Earnings

2006   \$5 million

2007   \$10 million

2008   \$12 million

5 + 10 + 12/3 = 27/3 = 9

The mean earnings were \$9 million per year.

The correct answer is choice (C).

##### Example 20

In which year did earnings decrease by the greatest percentage from the previous year?

(A) 2006
(B) 2007
(C) 2008
(D) 2009
(E) 2010

### Solution

Use the second graph.

To find the percent decrease, divide the change by the original amount.
Note: Percent decrease here means the proportional change, not the numerical change. You can compare the change as a percent, fraction, or decimal.

Make a quick table to organize the data and calculations.

 Year Earn Decrease Calculation 2005 8 n/a 2006 5 37.5% (8 – 5)/8 = 3/8 = 0.375 = 37.5% 2007 10 increase 2008 12 increase 2009 11 8% (12– 11)/12 = 1/12 ≈ 8% 2010 8 27% (11– 8)/11 = 3/11 ≈ 27%

The correct answer choice is (A).

##### Example 21

If the company’s earnings are less than 10 percent of sales during a year, then the Chief Operating Officer gets a 5% pay cut. How many times between 2005 and 2010 did the Chief Operating Officer take this pay cut?

(A) none
(B) one
(C) two
(D) three
(E) four

### Solution

Use the second graph.

To find the percent decrease, divide the change by the original amount.
Note: Percent decrease here means the proportional change, not the numerical change. You can compare the change as percent, fraction, or decimal.

Make a quick table to organize the data and calculations.

Year
2005
2006
2007
2008
2009
2010

10% of Sales
0.1 × 80 = 8
0.1 × 70 = 7
0.1 × 50 = 5
0.1 × 80 = 8
0.1 × 90 = 9
0.1 × 100 = 10

Earnings
8
5
10
12
11
8

Earnings < 10%?
equal: no cut
5 < 7: cut
10 > 5: no cut
12 > 8: no cut
11 > 9: no cut
8 < 10: cut

Comparing the right columns shows that earnings were less than 10 percent of sales in 2006 and 2010.

The correct answer is choice (C).

### 7. Line Graphs

One type of line graph is a broken line graph. Segments are used to join the values but the graph has multiple slopes.

Examples 22−24 are based on the line graphs below. The graph on the left shows a company’s earnings for each month between April and August. The graph on the right shows the number of outstanding shares.  ##### Example 22

How much greater were the company’s earnings in June than in April?

(A) 150%
(B) 175%
(C) 200%
(D) 250%
(E) 300%

### Solution

The percentage increase in earnings is the difference in earnings divided by the original earnings.

difference in earnings/original earnings = June earnings – April earnings/April earnings = 5 – 2/2 = 3/2 = 1.5 = 150%

Since both values are in millions, you don’t need trailing zeros in the calculations.

The correct answer is choice (A).

##### Example 23

How much did the earnings per share increase by from May to June?

(A) \$2.50
(B) \$3.33
(C) \$5.00
(D) \$6.67
(E) \$10.00

### Solution

Earnings per share in May:
3,000,000/180,000 = 300/18 = 6 × 50/6 × 3 = 50/3 ≈ 16.67

Earnings per share in June:
5,000,000/250,000 = 500/25 = 20

20 – 16.67 = 3.33
The earnings per share grew by \$3.33.

The correct answer is choice (B).

##### Example 24

Management predicts that earnings will decline by 5% from August to September. How many shares does the company need to buy back so that the earnings per share ratio does not change?

(A) 3,000
(B) 4,000
(C) 5,500
(D) 6,000
(E) 7,500

### Solution

Earnings per share in August:
3,000,000 / 150,000 = 300 / 15 = 15 × 20 / 15= 20

The expected earnings in September are 5% less than August, which means the earnings will be 95% of the August earnings.

Let x be the number of outstanding shares in September.
So the earnings per share ratio in September will be:

2,850,000/x = 20
20x = 2,850,000
x = 142,500

The number of shares in August is 150,000. The number of shares in September needs to be 142,500.

150,000 – 142,500 = 7,500

The company has to buy back 7,500 shares.

The correct answer is choice (E).

Note: Capital stock (buy backs) is not counted in the basic EPS, but may be counted in the diluted EPS. Many companies publish both numbers in their annual reports.

Examples 25−27 are based on the line graphs below, which show exchange rate dynamics during one week.  ##### Example 25

On Tuesday Mr. Smith bought 200 British Pounds using US Dollars. On Wednesday he exchanged these 200 British Pounds back into US Dollars. What profit did Mr. Smith make? (Assume zero transaction cost.)

(A) \$5
(B) \$8
(C) \$10
(D) \$15
(E) \$20

### Solution

Method 1

On Tuesday, 1 pound cost 1.7 dollars. Mr. Smith bought 200 pounds for \$1.70 × 200 = \$340.

On Wednesday, Mr. Smith sold 200 pounds × \$1.75 = \$350 .

So Mr. Smith made a profit of \$350 − \$340 = \$10.

The correct answer is choice (C).

Method 2

Compare the exchange rates. Since 1.75 − 1.70 = 0.05, the profit is 5 cents on each pound he bought and sold.
So 200 × 0.05 = \$10.

The correct answer is choice (C).

##### Example 26

What was the price of one US Dollar in terms of Euro on the Sunday at the end of the week?

(A) 0.67
(B) 0.75
(C) 0.80
(D) 1.20
(E) 1.33

### Solution

At the end of the week, 1 pound cost 1.6 dollars and 1 pound cost 1.2 euros. This means that 1.6 dollars were equivalent to 1.2 euros.

1.6 dollars/1.2 euros = 1 dollar/x euros

x = 1.2/1.6 = 3 × 0.4/4 × 0.4 = 0.75

So, 1 dollar was equivalent to 0.75 euros.

The correct answer is choice (B).

##### Example 26

On the basis of these graphs alone, how much did the dollar depreciate against the euro during this seven-day period?

(A) 5.00%
(B) 5.50%
(C) 6.25%
(D) 7.50%
(E) 8.25%

### Solution

At the end of last week, the first Sunday on the graph, 1 pound cost 1.75 dollars and 1 pound cost 1.4 euros. This means that 1.75 dollars were equivalent to 1.4 euros.

1.75 dollars/1.4 euros = 1 dollar/x euros

x = 1.4/1.75 = 2 × 0.7 /2.5 × 0.7= 4/5 = 0.8

So, 1 dollar was equivalent to 0.8 euros.

From the Example above, at the end of the week, on the second Sunday on the graph, 1 dollar was equivalent to 0.75 euros.

So the dollar depreciated by 0.05 euros.
To find the percentage decrease, divide the change by the original amount.

0.05/0.8 = 5/80 = 5/5 × 16 = 1/16 = 0.0625 = 6.25%

The correct answer is choice (C).