{"id":258,"date":"2026-09-09T06:30:31","date_gmt":"2026-09-09T06:30:31","guid":{"rendered":"https:\/\/repstaq.com\/study-guide-docs\/finance-growth-and-decay-g11-maths-t4p1\/"},"modified":"2026-09-09T06:30:31","modified_gmt":"2026-09-09T06:30:31","slug":"finance-growth-and-decay-g11-maths-t4p1","status":"publish","type":"post","link":"https:\/\/repstaq.com\/study-guide-docs\/finance-growth-and-decay-g11-maths-t4p1\/","title":{"rendered":"Finance, growth and decay-G11-Maths-T4P1"},"content":{"rendered":"\n

Finance, Growth and Decay<\/h2>\n\n\n\n

Finance calculations involve money invested, borrowed or reduced in value over time. Growth occurs when an amount increases, while decay occurs when it decreases. Always convert percentage rates to decimals before substituting into a formula.<\/p>\n\n\n\n

Simple interest<\/h2>\n\n\n\n

Simple interest is calculated only on the original principal amount. The interest earned or charged is the same during every time period.<\/p>\n\n\n\n

The accumulated amount is:<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>n<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n

The simple interest is:<\/p>\n\n\n\n

I<\/mi>=<\/mo>P<\/mi>i<\/mi>n<\/mi><\/mrow><\/math><\/div>\n\n\n\n

In these formulas, P is the principal amount, A is the accumulated amount, i is the interest rate per time period written as a decimal, and n is the number of time periods.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Calculate the accumulated value of an investment of R12 000 at 8% simple interest per annum for 3 years.<\/p>\n\n\n\n

P<\/mi>=<\/mo>12<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>8<\/mn>100<\/mn><\/mfrac>=<\/mo>0.08<\/mn>,<\/mo>n<\/mi>=<\/mo>3<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>n<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>12<\/mn>000<\/mn>(<\/mo>1<\/mn>+<\/mo>0.08<\/mn>\u00d7<\/mo>3<\/mn>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>12<\/mn>000<\/mn>(<\/mo>1.24<\/mn>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>14<\/mn>880<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The interest earned is:<\/p>\n\n\n\n

I<\/mi>=<\/mo>A<\/mi>\u2212<\/mo>P<\/mi><\/mrow><\/math><\/div>\n\n\n\n
I<\/mi>=<\/mo>14<\/mn>880<\/mn>\u2212<\/mo>12<\/mn>000<\/mn>=<\/mo>R<\/mi>2<\/mn>880<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Compound interest<\/h2>\n\n\n\n

Compound interest is calculated on the principal and on interest already added. Interest therefore earns further interest.<\/p>\n\n\n\n

The accumulated amount is:<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

R15 000 is invested at 9% compound interest per annum for 4 years. Calculate the accumulated amount.<\/p>\n\n\n\n

P<\/mi>=<\/mo>15<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.09<\/mn>,<\/mo>n<\/mi>=<\/mo>4<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>15<\/mn>000<\/mn>(<\/mo>1<\/mn>+<\/mo>0.09<\/mn>)<\/mo><\/mrow>4<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>15<\/mn>000<\/mn>(<\/mo>1.09<\/mn>)<\/mo><\/mrow>4<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>21<\/mn>173.72<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>21<\/mn>173.72<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Round money to the nearest cent unless instructed otherwise.<\/p>\n\n\n\n

Compound growth<\/h2>\n\n\n\n

Compound growth occurs when a quantity increases by the same percentage during each time period. The growth factor is greater than one.<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

The original amount can be calculated by rearranging the formula:<\/p>\n\n\n\n

P<\/mi>=<\/mo>A<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

The value of a property increases by 6% per annum. Find its value after 5 years if its current value is R850 000.<\/p>\n\n\n\n

P<\/mi>=<\/mo>850<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.06<\/mn>,<\/mo>n<\/mi>=<\/mo>5<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>850<\/mn>000<\/mn>(<\/mo>1.06<\/mn>)<\/mo><\/mrow>5<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>1<\/mn>137<\/mn>292.39<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>1<\/mn>137<\/mn>292.39<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Compound decay<\/h2>\n\n\n\n

Compound decay occurs when a quantity decreases by the same percentage during every time period. The decay factor is less than one.<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>\u2212<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A machine is worth R80 000 and loses 12% of its value each year. Calculate its value after 3 years.<\/p>\n\n\n\n

P<\/mi>=<\/mo>80<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.12<\/mn>,<\/mo>n<\/mi>=<\/mo>3<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>\u2212<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>80<\/mn>000<\/mn>(<\/mo>1<\/mn>\u2212<\/mo>0.12<\/mn>)<\/mo><\/mrow>3<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>80<\/mn>000<\/mn>(<\/mo>0.88<\/mn>)<\/mo><\/mrow>3<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>54<\/mn>517.76<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Common Mistake<\/strong><\/p>\n\n\n\n

For decay, subtract the rate from one. Do not use a negative exponent or add the rate.<\/p>\n\n\n\n

Depreciation<\/h2>\n\n\n\n

Depreciation is the decrease in the value of an asset over time. Vehicles, machinery and electronic equipment normally depreciate because of age, use and wear.<\/p>\n\n\n\n

The original purchase price is called the cost price. The value after depreciation is called the book value. Two common methods are straight line depreciation and reducing balance depreciation.<\/p>\n\n\n\n\n

\n\n\"\"\n<\/a>\n<\/figure>\n\n\n\n

Straight line depreciation<\/h2>\n\n\n\n

With straight line depreciation, the same rand amount is deducted every year. Depreciation is calculated on the original cost price.<\/p>\n\n\n\n

The book value is:<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>\u2212<\/mo>i<\/mi>n<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n

The depreciation per year is:<\/p>\n\n\n\n

D<\/mi>=<\/mo>P<\/mi>i<\/mi><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A vehicle costs R240 000 and depreciates at 15% per annum according to the straight line method. Calculate its book value after 4 years.<\/p>\n\n\n\n

P<\/mi>=<\/mo>240<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.15<\/mn>,<\/mo>n<\/mi>=<\/mo>4<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>\u2212<\/mo>i<\/mi>n<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>240<\/mn>000<\/mn>(<\/mo>1<\/mn>\u2212<\/mo>0.15<\/mn>\u00d7<\/mo>4<\/mn>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>240<\/mn>000<\/mn>(<\/mo>0.40<\/mn>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>96<\/mn>000<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The annual depreciation is:<\/p>\n\n\n\n

D<\/mi>=<\/mo>240<\/mn>000<\/mn>(<\/mo>0.15<\/mn>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
D<\/mi>=<\/mo>R<\/mi>36<\/mn>000<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Reducing balance depreciation<\/h2>\n\n\n\n

With reducing balance depreciation, depreciation is calculated on the current book value. The rand amount of depreciation becomes smaller each year.<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>\u2212<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Equipment costing R150 000 depreciates at 20% per annum on the reducing balance method. Find its book value after 3 years.<\/p>\n\n\n\n

P<\/mi>=<\/mo>150<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.20<\/mn>,<\/mo>n<\/mi>=<\/mo>3<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>150<\/mn>000<\/mn>(<\/mo>1<\/mn>\u2212<\/mo>0.20<\/mn>)<\/mo><\/mrow>3<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>150<\/mn>000<\/mn>(<\/mo>0.80<\/mn>)<\/mo><\/mrow>3<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>76<\/mn>800<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Remember<\/strong><\/p>\n\n\n\n

Straight line depreciation is a simple decay model. Reducing balance depreciation is a compound decay model.<\/p>\n\n\n\n

Inflation<\/h2>\n\n\n\n

Inflation is the general increase in the prices of goods and services over time. If inflation remains at a constant percentage, compound growth is used.<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A basket of goods currently costs R2 500. Estimate its cost after 4 years if inflation is 5.5% per annum.<\/p>\n\n\n\n

P<\/mi>=<\/mo>2<\/mn>500<\/mn>,<\/mo>i<\/mi>=<\/mo>0.055<\/mn>,<\/mo>n<\/mi>=<\/mo>4<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>2<\/mn>500<\/mn>(<\/mo>1.055<\/mn>)<\/mo><\/mrow>4<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>3<\/mn>096.94<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>3<\/mn>096.94<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Inflation reduces the purchasing power of money. This means that the same amount of money buys fewer goods in the future.<\/p>\n\n\n\n

Population growth<\/h2>\n\n\n\n

Population growth is usually modelled using compound growth when the population grows by a fixed percentage each year.<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>)<\/mo><\/mrow>n<\/mi><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A town has a population of 48 000 people. The population grows by 2.4% per annum. Estimate the population after 6 years.<\/p>\n\n\n\n

P<\/mi>=<\/mo>48<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.024<\/mn>,<\/mo>n<\/mi>=<\/mo>6<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>48<\/mn>000<\/mn>(<\/mo>1.024<\/mn>)<\/mo><\/mrow>6<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>55<\/mn>342.79<\/mn><\/mrow><\/math><\/div>\n\n\n\n

A population must be given as a whole number:<\/p>\n\n\n\n

A<\/mi>\u2248<\/mo>55<\/mn>343<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Exchange rates<\/h2>\n\n\n\n

An exchange rate compares the values of two currencies. Decide whether to multiply or divide by considering the direction of the conversion.<\/p>\n\n\n\n

Suppose the exchange rate is:<\/p>\n\n\n\n

1<\/mn>USD<\/mi>=<\/mo>R<\/mi>18.50<\/mn><\/mrow><\/math><\/div>\n\n\n\n

To convert US dollars to rand, multiply by the exchange rate.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Convert 350 US dollars to rand.<\/p>\n\n\n\n

350<\/mn>\u00d7<\/mo>18.50<\/mn>=<\/mo>6<\/mn>475<\/mn><\/mrow><\/math><\/div>\n\n\n\n
350<\/mn>USD<\/mi>=<\/mo>R<\/mi>6<\/mn>475<\/mn><\/mrow><\/math><\/div>\n\n\n\n

To convert rand to US dollars, divide by the exchange rate.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Convert R9 250 to US dollars.<\/p>\n\n\n\n

9<\/mn>250<\/mn><\/mrow>18.50<\/mn><\/mfrac>=<\/mo>500<\/mn><\/mrow><\/math><\/div>\n\n\n\n
R<\/mi>9<\/mn>250<\/mn>=<\/mo>500<\/mn>USD<\/mi><\/mrow><\/math><\/div>\n\n\n\n

Exam Tip<\/strong><\/p>\n\n\n\n

Check whether the answer should be larger or smaller. When converting rand to a stronger currency, the numerical answer will normally be smaller.<\/p>\n\n\n\n

Nominal interest rates<\/h2>\n\n\n\n

A nominal interest rate is the annual rate quoted before considering the effect of compounding during the year. Divide it by the number of compounding periods per year to obtain the rate per period.<\/p>\n\n\n\n

i<\/mi>period<\/mi><\/msub>=<\/mo>i<\/mi>nominal<\/mi><\/msub>m<\/mi><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

Here, m is the number of compounding periods per year.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Find the monthly interest rate for a nominal rate of 12% per annum compounded monthly.<\/p>\n\n\n\n

i<\/mi>monthly<\/mi><\/msub>=<\/mo>0.12<\/mn>12<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
i<\/mi>monthly<\/mi><\/msub>=<\/mo>0.01<\/mn><\/mrow><\/math><\/div>\n\n\n\n
i<\/mi>monthly<\/mi><\/msub>=<\/mo>1<\/mn>%<\/mo><\/mrow><\/math><\/div>\n\n\n\n

Effective interest rates<\/h2>\n\n\n\n

The effective annual interest rate is the actual percentage increase over one year after all compounding has been included.<\/p>\n\n\n\n

i<\/mi>effective<\/mi><\/msub>=<\/mo>(<\/mo>1<\/mn>+<\/mo>i<\/mi>nominal<\/mi><\/msub>m<\/mi><\/mfrac>)<\/mo><\/mrow>m<\/mi><\/msup>\u2212<\/mo>1<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Calculate the effective annual rate corresponding to 12% per annum compounded monthly.<\/p>\n\n\n\n

i<\/mi>effective<\/mi><\/msub>=<\/mo>(<\/mo>1<\/mn>+<\/mo>0.12<\/mn>12<\/mn><\/mfrac>)<\/mo><\/mrow>12<\/mn><\/msup>\u2212<\/mo>1<\/mn><\/mrow><\/math><\/div>\n\n\n\n
i<\/mi>effective<\/mi><\/msub>=<\/mo>(<\/mo>1.01<\/mn>)<\/mo><\/mrow>12<\/mn><\/msup>\u2212<\/mo>1<\/mn><\/mrow><\/math><\/div>\n\n\n\n
i<\/mi>effective<\/mi><\/msub>=<\/mo>0.126825<\/mn><\/mrow><\/math><\/div>\n\n\n\n
i<\/mi>effective<\/mi><\/msub>\u2248<\/mo>12.68<\/mn>%<\/mo><\/mrow><\/math><\/div>\n\n\n\n

The effective rate is higher than the nominal rate because interest is compounded during the year.<\/p>\n\n\n\n

Different compounding periods<\/h2>\n\n\n\n

Interest may be compounded annually, half-yearly, quarterly, monthly or daily.<\/p>\n\n\n\n

The common values of m are:<\/p>\n\n\n\n

Annually: one compounding period per year.<\/p>\n\n\n\n

Half-yearly: two compounding periods per year.<\/p>\n\n\n\n

Quarterly: four compounding periods per year.<\/p>\n\n\n\n

Monthly: twelve compounding periods per year.<\/p>\n\n\n\n

Daily: usually 365 compounding periods per year.<\/p>\n\n\n\n

For a nominal annual rate compounded m times per year:<\/p>\n\n\n\n

A<\/mi>=<\/mo>P<\/mi>(<\/mo>1<\/mn>+<\/mo>i<\/mi>m<\/mi><\/mfrac>)<\/mo><\/mrow>m<\/mi>n<\/mi><\/mrow><\/msup><\/mrow><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

R20 000 is invested at 10% per annum compounded quarterly for 3 years. Calculate the accumulated amount.<\/p>\n\n\n\n

P<\/mi>=<\/mo>20<\/mn>000<\/mn>,<\/mo>i<\/mi>=<\/mo>0.10<\/mn>,<\/mo>m<\/mi>=<\/mo>4<\/mn>,<\/mo>n<\/mi>=<\/mo>3<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The rate per quarter is:<\/p>\n\n\n\n

i<\/mi>m<\/mi><\/mfrac>=<\/mo>0.10<\/mn>4<\/mn><\/mfrac>=<\/mo>0.025<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The number of compounding periods is:<\/p>\n\n\n\n

m<\/mi>n<\/mi>=<\/mo>4<\/mn>\u00d7<\/mo>3<\/mn>=<\/mo>12<\/mn><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>20<\/mn>000<\/mn>(<\/mo>1<\/mn>+<\/mo>0.10<\/mn>4<\/mn><\/mfrac>)<\/mo><\/mrow>4<\/mn>\u00d7<\/mo>3<\/mn><\/mrow><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>20<\/mn>000<\/mn>(<\/mo>1.025<\/mn>)<\/mo><\/mrow>12<\/mn><\/msup><\/mrow><\/math><\/div>\n\n\n\n
A<\/mi>=<\/mo>R<\/mi>26<\/mn>897.78<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Exam Tip<\/strong><\/p>\n\n\n\n

The interest rate and the number of periods must refer to the same time interval. If interest is compounded monthly, use a monthly rate and the total number of months. Avoid rounding intermediate answers; round only the final monetary answer to the nearest cent.<\/p>\n\n\n\n\n

\n\n\"\"\n<\/a>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"

Finance, Growth and Decay Finance calculations involve money invested, borrowed or reduced in value over time. Growth occurs when an amount increases, while decay occurs when it decreases. Always convert percentage rates to decimals before substituting into a formula. Simple interest Simple interest is calculated only on the original principal amount. The interest earned or […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-258","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":{"document_name":"Finance, growth and decay","grade":11,"subject":"Mathematics","term":4,"paper":1},"_links":{"self":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/258","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/comments?post=258"}],"version-history":[{"count":0,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/258\/revisions"}],"wp:attachment":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/media?parent=258"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/categories?post=258"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/tags?post=258"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}