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From 1945 to 2008 , one country's per capita gross domestic product (GDPPC) increased about 6% per year, ending at $19,600. For the next few years, the growth slowed to an increase of about $1,300 per year. To the nearest dollar, how much less did the GDPPC grow during the next 6 years than it would have if it had maintained its earlier growth rate? ◻

From 19451945 to 20082008, one country's per capita gross domestic product (GDPPC) increased about 6%6\% per year, ending at $19,600\$19,600. For the next few years, the growth slowed to an increase of about $1,300\$1,300 per year. To the nearest dollar, how much less did the GDPPC grow during the next 66 years than it would have if it had maintained its earlier growth rate?\newline□\square

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Q. From 19451945 to 20082008, one country's per capita gross domestic product (GDPPC) increased about 6%6\% per year, ending at $19,600\$19,600. For the next few years, the growth slowed to an increase of about $1,300\$1,300 per year. To the nearest dollar, how much less did the GDPPC grow during the next 66 years than it would have if it had maintained its earlier growth rate?\newline□\square
  1. Calculate GDPPC Growth: First, we need to calculate the GDPPC growth over the next 66 years if it had maintained the 6%6\% per year growth rate.\newlineThe formula for compound growth is:\newlineGDPPCend=GDPPCstart×(1+growth_rate)number_of_yearsGDPPC_{\text{end}} = GDPPC_{\text{start}} \times (1 + \text{growth\_rate})^{\text{number\_of\_years}}\newlineHere, GDPPCstartGDPPC_{\text{start}} is $19,600\$19,600, growth_rate\text{growth\_rate} is 6%6\% or 0.060.06, and number_of_years\text{number\_of\_years} is 66.
  2. Perform Calculation: Now, we perform the calculation:\newlineGDPPCend=$(19,600)×(1+0.06)6GDPPC_{\text{end}} = \$(19,600) \times (1 + 0.06)^6\newlineUsing a calculator, we find:\newlineGDPPCend≈$(19,600)×(1.418519)GDPPC_{\text{end}} \approx \$(19,600) \times (1.418519)\newlineGDPPCend≈$(27,803.17)GDPPC_{\text{end}} \approx \$(27,803.17)\newlineThis is the GDPPC at the end of 66 years with a 6%6\% growth rate per year.
  3. Calculate Slower Growth: Next, we calculate the GDPPC growth with the slower rate of $1,300\$1,300 per year for 66 years.\newlineThe formula for linear growth is:\newlineGDPPCend=GDPPCstart+(increase_per_year×number_of_years)\text{GDPPC}_{\text{end}} = \text{GDPPC}_{\text{start}} + (\text{increase\_per\_year} \times \text{number\_of\_years})\newlineHere, GDPPCstart\text{GDPPC}_{\text{start}} is $19,600\$19,600, increase_per_year\text{increase\_per\_year} is $1,300\$1,300, and number_of_years\text{number\_of\_years} is 66.
  4. Find Difference: Now, we perform the calculation:\newlineGDPPCend_{\text{end}} = $19,600\$19,600 + $1,300×6\$1,300 \times 6\newlineGDPPCend_{\text{end}} = $19,600\$19,600 + $7,800\$7,800\newlineGDPPCend_{\text{end}} = $27,400\$27,400\newlineThis is the GDPPC at the end of 66 years with a growth of $1,300\$1,300 per year.
  5. Perform Subtraction: Finally, we find the difference between the GDPPC growth at the 6%6\% rate and the slower $1,300\$1,300 per year rate.\newlineDifference = GDPPC\_end (6%(6\% growth) - GDPPC\_end ($1,300/year(\$1,300/year growth)\newlineDifference = $27,803.17\$27,803.17 - $27,400\$27,400
  6. Perform Subtraction: Finally, we find the difference between the GDPPC growth at the 66\% rate and the slower (\$)\(1\),\(300\) per year rate.\(\newline\)Difference = GDPPC\_end (\(6\)\% growth) - GDPPC\_end (\(\$\)\(1\),\(300\)/year growth)\(\newline\)Difference = \(\$27,803.17 - \$27,400\)Now, we perform the subtraction:\(\newline\)Difference \(\approx \$403.17\)\(\newline\)To the nearest dollar, the GDPPC grew \(\$403\) less during the next \(6\) years than it would have if it had maintained its earlier growth rate of \(6\)\% per year.

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