GRE Data Interpretation Practice

To decide which year makes one combined quantity closest to another, compute the absolute difference $\lvert (\text{coffee} + \text{ice cream}) - \text{poultry} \rvert$ for each candidate year and choose the smallest.

The question asks about the first five years shown: $1920$, $1930$, $1940$, $1950$, and $1960$. Approximate per capita pounds from the graph:

Compare year by year

Choice A ($1920$): Coffee $+$ ice cream $\approx 12 + 7 = 19$; poultry $\approx 13$; difference $\approx 6$.

Choice B ($1930$): $\approx 12 + 10 = 22$ versus $\approx 17$; difference $\approx 5$.

Choice C ($1940$): $\approx 16 + 11 = 27$ versus $\approx 17$; difference $\approx 10$.

Choice D ($1950$): $\approx 16 + 16 = 32$ versus $\approx 24$; difference $\approx 8$.

Choice E ($1960$): $\approx 16 + 18 = 34$ versus $\approx 34$; difference $\approx 0$.

The smallest difference is in $1960$.

Things to Remember:

• “Closest” means the smallest absolute difference.
• Add the two series first, then compare to the third.
• Graph-reading approximations are fine when the ranking of differences is clear.

A percent decrease from an old value to a new value is $\dfrac{\text{old} - \text{new}}{\text{old}} \times 100\%$; a percent increase uses $\dfrac{\text{new} - \text{old}}{\text{old}} \times 100\%$. The arithmetic mean of three numbers is their sum divided by $3$. The standard deviation measures spread about the mean; among data sets of the same size, a larger spread from the center generally means a larger standard deviation.

From the table, City A has rainfall $1.3$, $1.2$, $1.3$; City C has $8.9$, $9.2$, $9.3$; City E has $5.8$, $6.5$, $6.6$ (June, July, August).

Evaluate each statement

Choice A: In City A, percent decrease June to July $= \dfrac{0.1}{1.3} \times 100\% \approx 7.7\%$. Percent increase July to August $= \dfrac{0.1}{1.2} \times 100\% \approx 8.3\%$. These are not equal, so A is false. (Equal absolute changes from different bases are not equal percents.)

Choice B: City C mean $= \dfrac{8.9 + 9.2 + 9.3}{3} = \dfrac{27.4}{3} \approx 9.133$. July’s $9.2$ is greater than this mean, so B is true.

Choice C: City C’s values $8.9$, $9.2$, $9.3$ cluster tightly (range $0.4$). City E’s values $5.8$, $6.5$, $6.6$ spread farther from their mean (range $0.8$, with a larger gap from June to July). So the standard deviation for City E is greater than for City C, and C is true.

The true statements are B and C.

Things to Remember:

• Percent change always divides by the starting value of that interval.
• A value can exceed the mean of its set even when all values are close together.
• Greater spread about the mean implies a greater standard deviation.

Percent increase from an old value to a new value is $\frac{\text{new} - \text{old}}{\text{old}} \times 100\%$. Read the needed years from the bar chart, then compare the computed percent with the choices.

The bar chart gives the number of temporary employees (in thousands): $121$ in $1993$ and $286$ in $1999$.

Compute the percent increase

$\frac{286 - 121}{121} \times 100\% = \frac{165}{121} \times 100\% \approx 136.4\%$

The closest choice is $136\%$.

Choice A: $36\%$ is far too small (near the fractional part of $0.364$ without the full increase).

Choice B: $58\%$ understates the relative change.

Choice C: $136\%$ matches the computed increase.

Choice D: $158\%$ confuses percent increase with a related but larger ratio.

Choice E: $236\%$ is roughly $\frac{286}{121}$ as a percent, which is the new value as a percent of the old value, not the increase.

Things to Remember:

• Percent increase $= \frac{\text{new} - \text{old}}{\text{old}} \times 100\%$.
• Do not confuse "new as a percent of old" with percent increase.
• When choices are rounded, compute exactly first, then pick the nearest option.