GRE: Data Interpretation

Bar Graph:- These can be used to display the information that would otherwise appear in a table. On a bar Graph, the height of each column shows its value.

Line Graph:- Line graph follow the same principle as bar graphs, except the values are presented as points.

Pie Charts:- A Pie Charts shows how things are distributed; the fraction of a circle occupied by each piece of the “pie” indicates what fraction of the whole it represents.

Pie chart showing John’s Expenditure, 2004

100% = $20,000

To Find the amt. John Paid in federal taxes in 2004, We find the slice labeled “Federal Taxes” and We see that federal taxes represented 20% of this expenditures. Since the whole is $20,000, We find that John’s Federal taxes for 2004 were 20% of $20,000 or 1/5 × $20,000    =    $ 4,000


Problem 1): Number of hours worked per week per employee at Marshville Toy Company

Chart for problem 1 & problem 2

Number of hours worked per week per employee at Marshville toy company

No. of EmployeesHours worked per weak

What is the positive difference between the mode and the range of the numbers of hours worked per week per employee at Marshal Toy Company?

Sol. The chart is a short way of showing a list of values that would begin: 15,15,15,15,25,25,25,25,25,25,25,25,25…..,.,…. (then the number 35 fifteen times, then the number 40 twenty- seven times, then the number 50 five times)

The mode is the number that appears in the list with the greatest frequency. Since 27 people worked 40 hours a weak. Therefore mode=40.

The range is the highest value in the list minus the lowest value in the list: 50-15=35. The positive difference between 40 and 35 = 5 Ans


Problem 2): Approximately What percent of the days with maximum temperature of  90 °F or more in St. Louis Occurred in July?

Ans:- From the grey bars on the “hot” day chart. St. Louis has a total of 1+8+15+12+4+4=44 days, when the temperature reached at least 90°F and 15 of these were in July.

These July days account for ( 15/44×100 )% = 34% Of all the hot days in St. Louis (approximately).

 Use the following graph to answer problem 3-4



Problem 3)  If $720 million was spent on remodeling bedrooms in 2004, how many was spent remodeling kitchens?


Let total amount spent on remodeling be x.

\(\frac { 15 }{ 100 } \times x=$720\\ x=\frac { 720\times 100 }{ 15 } =$4800\)

Amount spent on remodeling kitchen =

\(=\frac { 35 }{ 100 } \times 4800\\ =$1680\)


Problem 4)  On average, a homeowner will see a 75% return on investment for kitchen remodels. One homeowner increased the value of her home by $6000 by remodeling her kitchen. Assuming that gets the same percent return for bathroom remodels & that her spending follows 2004 averages, how much more would her home increase in value (to the nearest dollar) if she also remodeled the bathrooms?

Sol.Let amount spent on remodeling of kitchen be y

\((y\times \frac { 75 }{ 100 } )=$6000\\ y=(\frac { 6000\times 100 }{ 75 } )=8000$\)

Amount spent on remodelling of kitchen is 35% of total amount.

Let total amount spent on remodeling be x.

\((x\times \frac { 35 }{ 100 } )=$8000\\ x=(\frac { 8000\times 100 }{ 35 } )=(\frac { 800000 }{ 35 } )$\)

Amount spent on remodeling bathroom =

\(=(\frac { 25 }{ 100 } \times \frac { 800,000 }{ 35 } )=(\frac { 25\times 8000 }{ 35 } )\)

Increase in home’s value if he/she gets 25% return on remodelling of bathroom

\(=(\frac { 75 }{ 100 } \times \frac { 25\times 8000 }{ 35 } )=(\frac { 75\times 25\times 80 }{ 35 } )\\ =$4285.7\)


Use the following graph to solve problem 5-6

Problem 5) In 2009, a homeowner put his house on the market. If the average drop from listing price to sales price is 6%. What was the minimum price (in whole dollars) that the house could have been listed at so that it sold at the price greater them or equal to the medium price?

Sol. Medium price in 2009 = $240,000

Average drop from listing price to sales price = 6%

Let minimum price be x

\(x-(\frac { x\times 6 }{ 100 } )=240,000\\ x=$255,320\)


Problem 6) A real estate agent says the percent change between 2007 & 2008 is the same as between 2008 & 2009. Is this true or false?

Sol. Percentage change =

\((\frac { Amount\quad of\quad change }{ Original\quad Amount } )\times 100\)

Percentage change from 2007 to 2008 =

\(=(\frac { 230,000-240,000 }{ 240,000 } )\times 100\\ =-4.17%\)

Percentage change from 2008 to 2009 =

\(=(\frac { 240,000-230,000 }{ 240,000 } )\times 100\\ =-4.35%\)

Because 4.35% is not equal to 4.17%, the statement is false.


Number of hours worked per week per employee at Marshville toy company

No. of EmployeesHours worked per weak

Problem 7) What is the average (arithmetic mean) number of hours worked per week per employee at Marshville Toy Company?

a) 32

b) 33

c) 35

d) \(35\frac { 2 }{ 3 }\)

e) \(36\frac { 1 }{ 3 }\)

Sol. Total number of employees = 60

4 employees worked for 15 hours per week. 9 employees worked for 25 hours per week. 15 employees worked for 35 hours per week. 27 employees worked for 40 hours per week. 5 employees worked for 50 hours per week.

This chart is a short way of showing a list of values that would begin: 15, 15, 15, 15, 25, 25, 25, 25, 25, 25, 25, 25, 25……. (then the number 35 fifteen times, then the number 40 twenty-seven times, then the number 50 five times).

Arithmetic mean of hours worked per week per employees

\(=(\frac { Total\quad no.\quad of\quad hours\quad worked }{ Total\quad no.\quad of\quad employees } )\\ =(\frac { 15\times 4+25\times 9+35\times 15+40\times 27+50\times 5 }{ 60 } )\\ =\frac { 2140 }{ 60 } =\frac { 214 }{ 6 } =\frac { 107 }{ 3 } \\ ={ (35\frac { 2 }{ 3 } ) }^{ Ans }\)


Use the following table to answer exercises 8-9

Number of pets12345
Number of household with this number of pets70246183499
Maximum monthly spending on pet supplies$57.32$75.98$143.57$140.48$170.23
Minimum monthly spending on pet supplies$6.34$35.47$45.84$87.46$111.20
Average monthly spending on pet supplies$31.25$56.42$83.11$127.74$147.38

Problem 8) The household group with which number of pets had the greatest average (arithmetic mean) monthly spending per pet?

Sol. This table tallies the number of households, according to number of pets in the household & each column captures information about these households. For example, the left most columns with numbers indicates that there are 70 households that have one pet & these households spend an average of $31.25 per month on pet supplies. In that group, maximum spending on pet supply was $57.32 and minimum spending on pet supply was $6.34.

We have to find a group whose average monthly spending per pet was maximum.

First group spent an average of $31.25 per pet.

Second group spent an average of \($(\frac { 56.42 }{ 2 } )=$28.21\) per pet.

Third group spent an average of \($(\frac { 83.11 }{ 3 } )=$27.7\) per pet.

Fourth group spent an average of \($(\frac { 56.42 }{ 2 } )=$28.21\) per pet.

Fifth group spent an average of \($(\frac { 56.42 }{ 2 } )=$28.21\) per pet.

Household group with 4 numbers of pets had the greatest average spending per pet.


Problem 9) Approximately what percent of the surveyed households have more than three pets?

a) 10%

b) 20%

c) 30%

d) 40%

e) 50%

Sol. There are 557 households of which 49 have four pets & 9 have five pets.

Percent of the surveyed households having more than three pets =

\(=(\frac { 58 }{ 557 } \times 100)\\ =10.41%\\ \sim 10%\)


Questions 10-12 refer to the following charts.



















Problem 10) Imagine that at the beginning of 2014, Boston & Los Angeles implemented a public health program that reduced deaths of infants less than 1 year old by 20 percent, while the cities of Honolulu & Indianapolis terminated an identical program. What would have been the approximate total impact of these program changes on the number of infant deaths in these cities?

a) These would have been 600 more deaths.

b) These would have been 70 more deaths.

c) These would have been 30 fewer deaths.

d) These would have been 65 fewer deaths.

e) These would have been 150 fewer deaths.


This question asks about deaths of infants under age 1 year in certain cities. That information is found in the bar graph, specifically in the left bar of each cluster. Note that this graph has two y-axes, one on the left and on the right. Deaths that occur before age 1 year are plotted against the y-axes on the left; be sure to read the correct bars against the correct axis.

Boston is represented by the bars on the far left and Los Angeles by the bars on the far right. Honolulu and Indianapolis are represented by the second and fourth pairs of bars, respectively.

Before doing any calculations, you might note that the numbers of infant deaths in both Boston and Los Angeles are significantly higher than in either Honolulu or Indianapolis. Therefore, the effect of introducing public health programs in Boston And Los Angeles will be greater than the effect of terminating programs in the other two cities. In other words, more deaths will be prevented than not prevented, and you can eliminate choices (A) and (B). On Test Day, even if you had no more time to invest in this problem, by using your critical thinking skills to eliminate answer choices, you would improve your odds of guessing the correct answer.

In 2014, Boston had just under 200 infant deaths, and Los Angeles had just over 200. That’ s about 400 infant deaths total for these cities. The question wants you to imagine that a public health program was introduced in these cities, dropping infant deaths by 20%. In that case, there would be (0.2)(400) = 80 fewer infant deaths in those two cities, or 320 infant deaths instead of the actual 400.

Honolulu had about 60 infant deaths, and Indianapolis had not quite 150-call it 140. That’ s 60+140=200 deaths, and you’ re told to imagine what would have happened had a program reducing infant deaths by 20% been terminated. Without such a program acting to reduce infant deaths , the number would go up, but by how much? Call the number of infant deaths in Honolulu and Indianapolis d. The current 200 infant deaths represent 80% of what the number would be without the program, so you can write the following equation and solve: 0.8d = 200; d = 250, There would have been 50 more deaths.

Therefore, had the stated program changes occurred, there would have been about 320+250=570 deaths. There were actually about 400+200 = 600 deaths. That means about 30 fewer deaths would have occurred. Alternatively, you could add the differences in the two cities to get the net difference: -80 + 50 = 30. Either way, the correct answer is (C).


Problem 11) If in 2014, the population of Los Angeles was 75 percent greater than the population of Houston, What is the the ratio of the incidence of pneumonia & influenza related deaths (expressed as a percent of the city’s population) in Houston to the incidence in Los Angeles in that year?

a) \(\frac { 1 }{ 2 }\)

b) \(\frac { 7 }{ 8 }\)

c) \(\frac { 15 }{ 16 }\)

d) \(\frac { 8 }{ 7 }\)

e) \(\frac { 4 }{ 3 }\)


This question concerns deaths from pneumonia/influenza, which you find in the pie chart.

The pie chart represents the 3,532 pneumonia/influenza deaths that occurred in five cities. Of these, 19% occurred in Houston, and 38% occurred in Los Angeles. That’ s a ratio of 19:38 or 1:2 . However, the question is asking for the ratio of the rate of incidence of these deaths.

If Los Angeles’ s population were twice the size of Houston’ s, then the rate of incidence in the two cities would be the same, giving a ratio of 1:1. Because Los Angeles’ s population is less than twice Houston’ s, you know that the rate of incidence is higher in Los Angeles than in Houston. The ratio of Houston to Los Angeles must be less than 1. On the basis of this critical thinking, you can eliminate (D) and (E).

The question does not give you the actual population of the cities, but it does tell you their relative populations. If you call the population of Houston p, then the population of Los Angeles is 1.75p or (7/4)p. As shown in the pie chart, for every 1 death from pneumonia/influenza in Houston, there were 2 in Los Angeles. In Houston, therefore, the rate of incidence of deaths can be represented by the ratio of 1 to p, or (1/p). In Los Angeles, the rate of deaths can be represented as 2 to (7/4)p, or

\(\frac { 2 }{ \frac { 7p }{ 4 } } =2\times \frac { 4 }{ 7p } =\frac { 8 }{ 7p }\)

Now find the ratio of the rates of incidence:

\(\frac { \frac { 1 }{ p } }{ \frac { 8 }{ 7p } } =\frac { 1 }{ p } \times \frac { 7p }{ 8 } =\frac { 7 }{ 8 }\)

The correct answer is (B), \(\frac { 7 }{ 8 }\).


Problem 12) Assuming that 80% of all deaths due to pneumonia or influenza occurs among the elderly, defined as those ages 65 & over, in which city was the least proportion of all deaths among the elderly attributed to pneumonia or influenza?

a) Houston

b) Boston

c) Indianapolis

d) Honolulu

e) Los Angeles


This question asks about deaths due to pneumonia/influenza, so look at the pie chart for those data. It also concerns people who died at the age 65 or over, and that information is in the bar graph, specifically in the right-hand bar of each pair of bars. Remember that the deaths occurring at age 65 and over are plotted against the y-axis on the right side of the bar graph.

Only 80% of the pneumonia/influenza deaths occurred among the elderly. However, because this 80% applies equally to each city, you do not have to take into account as you compare the percentages of deaths in each city attributed to pneumonia/influenza. If a city has the most such deaths when 100% of pneumonia/influenza deaths are counted, then it will also have the most such deaths when the number of pneumonia/influenza deaths is reduced by 20% for all cities.

If you compare cities using estimation, you thinking might go like this: Boston and Houston have about the same number of P/I deaths, making them easy to compare. Houston has far more deaths from all causes, so Houston’ s proportion of P/I deaths is lower than Boston’ s. Houston has about 50% more P/I deaths than Honolulu or Indianapolis, and it has more than four times (Honolulu) and three times (Indianapolis) the deaths from all causes. Therefore, Houston’ s proportion of P/I deaths is lower than Honolulu’ s or Indianapolis’ s. Finally, compare Houston to Los Angeles, Houston has more total deaths and far fewer P/I deaths than Los Angeles has, so its proportion of P/I deaths is definitely lower than Los Angeles’ s. Houston is the winner; choice (A) is correct.

You can also solve using calculation. The pie chart represents a total of 3,532 deaths, and it gives the percentage distribution among the five cities. Determine the number of pneumonia/influenza deaths for each city:

Deaths from P/I

Boston     \(3,532\times 0.18=636\)

Honolulu     \(3,532\times 0.13=459\)

Houston      \(3,532\times 0.19=671\)

Indianapolis    \(3,532\times 0.12=424\)

Los Angeles    \(3,532\times 0.38=1,342\)

Now find the number of deaths among the elderly from the bar graph. Boston is between 4,000 and 5,000 or say 4,500. Honolulu is at about 3,000, Houston is the tallest bar at about 13,000, Indianapolis is at 4,000 and the bar for Los Angeles is at about 9,000.



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