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point."},{"text":"That digit appears only once.","correct":false,"explanation":"A digit without recurring notation may appear only once. A dot above the digit instead indicates that it repeats forever."},{"text":"That digit must be rounded up.","correct":false,"explanation":"Rounding is determined by the digit to the right, not by a dot above a digit. The dot shows that the marked digit repeats forever."},{"text":"That digit has a place value of zero.","correct":false,"explanation":"The dot does not alter the digit’s place value or make it zero. It indicates that the marked digit repeats forever."}]},{"id":"68a01069-7f09-4de0-b1f1-08f985f66783","cardType":"basic","points":1,"front":"How is a repeating block of digits (e.g. 0.3636...) shown using recurring notation?","back":"Dots are placed over the first and last digits of the repeating block.","mcqOptions":[{"text":"Dots are placed over the first and last digits of the repeating block.","correct":true,"explanation":"Marking the first and last digits identifies the boundaries of the entire repeating block, including every digit between the two dots."},{"text":"Dots are placed over the first and second digits of the repeating block.","correct":false,"explanation":"Using the first two digits does not show where a longer repeating block ends. Dots belong over the first and last digits of the block."},{"text":"Dots are placed over the middle and last digits of the repeating block.","correct":false,"explanation":"Starting the notation at a middle digit would omit part of the recurring block. The dots must mark its first and last digits."},{"text":"Dots are placed over the last two digits of the repeating block.","correct":false,"explanation":"Marking only the final two digits can falsely suggest that they alone recur. 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Repetition is guaranteed when the same remainder appears again, because the division process then repeats."},{"text":"When a larger remainder appears for a second time.","correct":false,"explanation":"The size of a remainder does not determine whether the decimal recurs. The decimal repeats when exactly the same remainder appears for a second time."},{"text":"When a zero remainder appears for a second time.","correct":false,"explanation":"A zero remainder makes the division terminate rather than recur. A recurring decimal is identified when the same non-zero remainder appears again."}]},{"id":"0955ab14-5dfe-41b3-8979-49a6cf956e7d","cardType":"basic","points":2,"front":"What is the core purpose of multiplying a recurring decimal by powers of 10 in algebra?","back":"To create two equations where the recurring tails are identical.","mcqOptions":[{"text":"To create two equations where the recurring tails are identical.","correct":true,"explanation":"Multiplying by a suitable power of 10 shifts complete recurring blocks so the infinite tails align and can be eliminated by subtraction."},{"text":"To create two equations where the decimal points are identical.","correct":false,"explanation":"Multiplying by 10 moves the decimal point rather than keeping it fixed. Its purpose is to make the recurring tails identical so subtraction removes them."},{"text":"To create two equations where the whole-number parts are identical.","correct":false,"explanation":"The whole-number parts normally become different after multiplication. The important feature is that the recurring tails become identical and therefore cancel under subtraction."},{"text":"To create two equations where the recurring blocks are different.","correct":false,"explanation":"Different recurring blocks would not cancel cleanly when the equations are subtracted. A suitable power of 10 is chosen to make the recurring tails identical."}]},{"id":"9d548039-5314-46e8-a4b0-afbe6e9cf0c2","cardType":"basic","points":2,"front":"Why is subtraction used after lining up the recurring tails of two equations?","back":"To cancel out the infinite repeating part of the decimal.","mcqOptions":[{"text":"To cancel out the infinite repeating part of the decimal.","correct":true,"explanation":"Once the recurring tails are aligned, subtracting one equation from the other removes every repeated decimal digit, leaving a finite equation that can be solved."},{"text":"To round off the infinite repeating part of the decimal.","correct":false,"explanation":"The method gives an exact fraction and does not use rounding. Subtraction cancels the identical infinite repeating tails completely."},{"text":"To double up the infinite repeating part of the decimal.","correct":false,"explanation":"Adding the equations would combine rather than eliminate their recurring tails. Subtraction is used because identical tails cancel out exactly."},{"text":"To shift left the infinite repeating part of the decimal.","correct":false,"explanation":"Multiplication by a power of 10 shifts the decimal digits. Subtraction has the different purpose of cancelling the aligned infinite repeating part."}]},{"id":"b41a1f64-18c1-4764-9fd9-073c4ae0ea2b","cardType":"basic","points":3,"front":"Outline the algebraic steps to prove $0.\\dot{6} = \\frac{2}{3}$.","back":"

Let $x = 0.666\\dots$
Multiply by 10 ($10x = 6.666\\dots$)
Subtract ($9x = 6$) and simplify $x = \\frac{6}{9}$

","mcqOptions":[{"text":"

Let $x = 0.666\\dots$
Multiply by 10 ($10x = 6.666\\dots$)
Subtract ($9x = 6$) and simplify $x = \\frac{6}{9}$

","correct":true,"explanation":"Multiplying by 10 aligns the recurring tails. Subtracting the original equation gives $9x=6$, so $x=\\frac{6}{9}=\\frac{2}{3}$."},{"text":"

Let $x = 0.666\\dots$
Multiply by 10 ($10x = 6.666\\dots$)
Subtract ($9x = 6$) and simplify $x = \\frac{6}{10}$

","correct":false,"explanation":"Dividing $9x=6$ by 10 is the slip; the coefficient of $x$ is 9. Dividing both sides by 9 gives $x=\\frac{6}{9}=\\frac{2}{3}$."},{"text":"

Let $x = 0.666\\dots$
Multiply by 10 ($10x = 6.666\\dots$)
Subtract ($10x = 6$) and simplify $x = \\frac{6}{10}$

","correct":false,"explanation":"Subtracting $x$ from $10x$ gives $9x$, not $10x$. The correct equation is $9x=6$, which simplifies to $x=\\frac{2}{3}$."},{"text":"

Let $x = 0.666\\dots$
Multiply by 10 ($10x = 6.666\\dots$)
Subtract ($-9x = -6$) and simplify $x = -\\frac{6}{9}$

","correct":false,"explanation":"Dividing two negative quantities produces a positive result. From $-9x=-6$, the correct conclusion is $x=\\frac{6}{9}=\\frac{2}{3}$, not a negative fraction."}]},{"id":"e8276a4d-d226-4431-8649-4c6a9448888d","cardType":"basic","points":1,"front":"What constitutes a fraction in its simplest form?","back":"The numerator and denominator have no common factors other than 1.","mcqOptions":[{"text":"The numerator and denominator have no common factors other than 1.","correct":true,"explanation":"A fraction is fully simplified when no integer greater than 1 divides both its numerator and denominator, so no further cancellation is possible."},{"text":"The numerator and denominator have no common multiples other than 1.","correct":false,"explanation":"Simplification depends on common factors, not common multiples; non-zero integers have many common multiples. The numerator and denominator must share no factor greater than 1."},{"text":"The numerator and denominator are both prime numbers other than 1.","correct":false,"explanation":"The numerator and denominator do not both need to be prime; for example, $8/15$ is simplified. They only need to have no common factor greater than 1."},{"text":"The numerator and denominator have the same number of factors other than 1.","correct":false,"explanation":"The number of factors each value has is irrelevant to simplification. A fraction is simplest when the numerator and denominator have no common factor greater than 1."}]},{"id":"ef935412-6d9d-4177-b09c-de7fc0b96acf","cardType":"basic","points":1,"front":"To convert $0.3\\dot{8}$ to a fraction, which two multiples of $x$ should be compared to align the tails?","back":"$$100x$ and $10x$.","mcqOptions":[{"text":"$$100x$ and $10x$.","correct":true,"explanation":"There is one non-recurring digit and one recurring digit, so multiplying by 10 and 100 produces matching recurring tails that cancel when the equations are subtracted."},{"text":"$$10x$ and $x$.","correct":false,"explanation":"This comparison works when recurrence begins immediately after the decimal point. Here, the non-recurring 3 requires comparing $100x$ and $10x$ to align the recurring tails."},{"text":"$$1000x$ and $100x$.","correct":false,"explanation":"This shifts both decimals one place farther than necessary. Although the tails align, the standard minimal comparison is $100x$ and $10x$, based on one non-recurring and one recurring digit."},{"text":"$$100x$ and $x$.","correct":false,"explanation":"These multipliers differ by two decimal places, but the recurring block contains only one digit. Comparing $100x$ and $10x$ gives identical recurring tails for subtraction."}]},{"id":"11c412e6-8c31-47c8-be73-5fb0e873087c","cardType":"cloze","points":null,"front":"To align tails for a decimal with $n$ non-recurring digits and a block of $r$ recurring digits, compare $\\text{[...]}$ and $10^n x$.","back":"To align tails for a decimal with $n$ non-recurring digits and a block of $r$ recurring digits, compare $\\boxed{10^{n+r}x}$ and $10^n x$.","mcqOptions":[{"text":"$$10^{n+r}x$","correct":true,"explanation":"Multiplying by $10^{n+r}$ shifts past both the $n$ non-recurring digits and one complete recurring block, allowing its tail to align with that of $10^n x$."},{"text":"$$10^r x$","correct":false,"explanation":"This accounts only for the recurring block and ignores the $n$ non-recurring digits. The required larger multiple is $10^{n+r}x$, which shifts past both sections."},{"text":"$$10^{nr}x$","correct":false,"explanation":"Multiplying $n$ and $r$ is a common exponent error. The shifts for the non-recurring digits and recurring block are added, so the correct expression is $10^{n+r}x$."},{"text":"$$10^{n-r}x$","correct":false,"explanation":"Subtracting the block length reduces the shift rather than moving past both sections. Their lengths must be added, making the correct larger multiple $10^{n+r}x$."}]},{"id":"9789f7df-3cea-40a7-b512-f6cde454de08","cardType":"cloze","points":null,"front":"To align tails for a decimal with $n$ non-recurring digits and a block of $r$ recurring digits, compare $10^{n+r}x$ and $\\text{[...]}$.","back":"To align tails for a decimal with $n$ non-recurring digits and a block of $r$ recurring digits, compare $10^{n+r}x$ and $\\boxed{10^n x}$.","mcqOptions":[{"text":"$$10^n x$","correct":true,"explanation":"Multiplying by $10^n$ moves the non-recurring digits before the decimal point. Its recurring tail then aligns with the tail produced by multiplying by $10^{n+r}$."},{"text":"$$10^r x$","correct":false,"explanation":"This uses the recurring block length rather than the number of non-recurring digits. The smaller multiple must be $10^n x$ so that the recurring tails align."},{"text":"$$10^{n-r}x$","correct":false,"explanation":"Subtracting the recurring length from the non-recurring length does not identify the starting point of recurrence. The correct smaller multiple is $10^n x$."},{"text":"$$10^{n+r-1}x$","correct":false,"explanation":"The off-by-one exponent shifts too far unless the recurring block happens to have length one. The general smaller multiple is $10^n x$, determined only by the non-recurring digits."}]},{"id":"d7d547d8-ee04-47a6-830f-ce3d3188b9b6","cardType":"basic","points":2,"front":"For the decimal $0.\\dot{4}\\dot{7}$, why is the decimal multiplied by 100 instead of 10?","back":"The repeating block has 2 digits; multiplying by 100 shifts the tail to match.","mcqOptions":[{"text":"The repeating block has 2 digits; multiplying by 100 shifts the tail to match.","correct":true,"explanation":"The block 47 contains two digits, so multiplication by $10^2=100$ moves one full block before the decimal point and leaves the same recurring tail."},{"text":"The repeating block has 1 digit; multiplying by 100 shifts the tail to match.","correct":false,"explanation":"Treating 47 as a one-digit block misreads the recurrence notation. Both 4 and 7 repeat, so the block has 2 digits and requires multiplication by 100."},{"text":"The repeating block has 2 digits; multiplying by 10 shifts the tail to match.","correct":false,"explanation":"Multiplication by 10 shifts only one digit, reversing the order of the recurring tail. A two-digit block requires multiplication by 100 to reproduce the same tail."},{"text":"The repeating block has 100 digits; multiplying by 100 shifts the tail to match.","correct":false,"explanation":"The multiplier does not equal the number of repeating digits. The block has 2 digits, and $100=10^2$ shifts exactly those two digits to align the tail."}]},{"id":"690a66ad-a2ae-48ad-af66-6c177fa4a630","cardType":"basic","points":1,"front":"Is it necessary to separate the whole number (e.g., the 3 in $3.\\dot{2}$) before using the algebra trick?","back":"No; let $x$ equal the entire decimal and use the standard method.","mcqOptions":[{"text":"No; let $x$ equal the entire decimal and use the standard method.","correct":true,"explanation":"The algebraic subtraction method works with the whole decimal, including its integer part. Setting $x=3.222\\ldots$ and shifting the complete value still cancels the recurring tails correctly."},{"text":"Yes; convert the whole number into a recurring decimal before using the standard method.","correct":false,"explanation":"The integer part does not need to be rewritten as recurring. Let $x$ equal the entire decimal and apply the usual powers-of-ten multiplication and subtraction method directly."},{"text":"Yes; remove the whole number and discard it before using the standard method.","correct":false,"explanation":"Discarding the whole number changes the value and produces only the fractional part. The correct method lets $x$ equal the entire decimal, preserving both integer and recurring parts."},{"text":"No; let $x$ equal only the recurring digit and use the standard method.","correct":false,"explanation":"Setting $x$ equal only to a digit does not represent the original number. Instead, let $x$ equal the entire decimal and perform the standard shifting and subtraction steps."}]},{"id":"4ca3a126-4963-4537-b1a6-1a851ffb1c33","cardType":"cloze","points":null,"front":"The final answer for a decimal greater than 1 (like $3.2\\dot{6}\\dot{3}$) is often an [ ].","back":"The final answer for a decimal greater than 1 (like $3.2\\dot{6}\\dot{3}$) is often an improper fraction.","mcqOptions":[{"text":"improper fraction","correct":true,"explanation":"A decimal greater than 1 has a fractional representation whose numerator is larger than its denominator when written as a single fraction, so it is often an improper fraction."},{"text":"proper fraction","correct":false,"explanation":"A proper fraction is less than 1 because its numerator is smaller than its denominator. A decimal such as $3.2\\dot{6}\\dot{3}$ is greater than 1, so it gives an improper fraction."},{"text":"unit fraction","correct":false,"explanation":"A unit fraction has numerator 1 and is necessarily at most 1. A recurring decimal greater than 1 is instead commonly expressed as an improper fraction."},{"text":"integer","correct":false,"explanation":"A recurring decimal with a non-zero fractional part is not a whole number. When converted into one fraction, a value greater than 1 is generally an improper fraction."}]},{"id":"919770c2-46fa-421c-b2da-e2565c665f3a","cardType":"basic","points":1,"front":"What is the resulting fraction for the recurring digit $a$ in the decimal $0.\\dot{a}$?","back":"$$\\frac{a}{9}$","mcqOptions":[{"text":"$$\\frac{a}{9}$","correct":true,"explanation":"If $x=0.aaa\\ldots$, then $10x=a.aaa\\ldots$. Subtracting gives $9x=a$, so $x=\\frac{a}{9}$."},{"text":"$$\\frac{a}{10}$","correct":false,"explanation":"Dividing by 10 represents the terminating decimal $0.a$, not the recurring decimal $0.aaa\\ldots$. Subtraction of $10x-x$ gives $9x=a$, so the fraction is $\\frac{a}{9}$."},{"text":"$$\\frac{a}{99}$","correct":false,"explanation":"A denominator of 99 corresponds to a two-digit repeating block. Here only one digit repeats, so subtracting $x$ from $10x$ gives the correct result $\\frac{a}{9}$."},{"text":"$$\\frac{9}{a}$","correct":false,"explanation":"This reverses the numerator and denominator after solving $9x=a$. Dividing both sides by 9 gives $x=\\frac{a}{9}$, not its reciprocal."}]},{"id":"1bf9627f-428b-4915-b6ce-8f5ad4aa4d34","cardType":"basic","points":1,"front":"What is the resulting fraction for the repeating block in $0.\\dot{0}\\dot{a}$?","back":"$$\\frac{a}{99}$","mcqOptions":[{"text":"$$\\frac{a}{99}$","correct":true,"explanation":"The repeating block is $0a$, which has two digits. If $x=0.0a0a\\ldots$, then $100x=0a.0a\\ldots$, giving $99x=a$ and therefore $x=\\frac{a}{99}$."},{"text":"$$\\frac{a}{9}$","correct":false,"explanation":"A denominator of 9 applies to a one-digit repeating block. Here the repeating block is $0a$, so multiplying by 100 and subtracting gives $99x=a$ and $x=\\frac{a}{99}$."},{"text":"$$\\frac{a}{90}$","correct":false,"explanation":"This would fit a non-recurring zero followed by recurring $a$, written with only $a$ recurring. Both 0 and $a$ are marked as recurring here, so the fraction is $\\frac{a}{99}$."},{"text":"$$\\frac{a}{100}$","correct":false,"explanation":"A denominator of 100 represents the terminating decimal $0.0a$. Because the two-digit block $0a$ repeats indefinitely, subtraction produces a denominator of 99, giving $\\frac{a}{99}$."}]},{"id":"130a1377-4f52-4ae9-ba86-e83738d372c3","cardType":"basic","points":1,"front":"How must recurring decimals be handled before performing multiplication or division?","back":"Convert every recurring decimal into a fraction first.","mcqOptions":[{"text":"Convert every recurring decimal into a fraction first.","correct":true,"explanation":"Fractions represent recurring decimals exactly, so multiplication or division can be completed without introducing rounding errors from an approximate decimal value."},{"text":"Round every recurring decimal to a fixed number of places first.","correct":false,"explanation":"Rounding replaces the recurring decimal with an approximation, which can make the final result inaccurate. Convert every recurring decimal into a fraction first to preserve its exact value."},{"text":"Convert every recurring decimal into a percentage first.","correct":false,"explanation":"Changing a recurring decimal into a percentage does not remove its recurring nature. It should be converted into an exact fraction before multiplication or division."},{"text":"Treat every recurring decimal as a terminating decimal first.","correct":false,"explanation":"A recurring decimal does not terminate, so cutting it off changes its value. Convert it into a fraction first so the calculation remains exact."}]},{"id":"d0b0716a-eb9d-4879-816e-4eb3a7ef75c8","cardType":"cloze","points":null,"front":"When dividing fractions converted from recurring decimals, multiply by the [ ] of the second fraction.","back":"When dividing fractions converted from recurring decimals, multiply by the reciprocal of the second fraction.","mcqOptions":[{"text":"reciprocal","correct":true,"explanation":"Dividing by a fraction is equivalent to multiplying by its reciprocal, found by interchanging its numerator and denominator."},{"text":"numerator","correct":false,"explanation":"Using only the numerator ignores the denominator and changes the fraction’s value. Division requires multiplication by the reciprocal of the entire second fraction."},{"text":"denominator","correct":false,"explanation":"Using only the denominator is a common incomplete inversion error. The correct method is to swap both parts and multiply by the reciprocal."},{"text":"additive inverse","correct":false,"explanation":"The additive inverse changes a number’s sign and is used when dealing with subtraction. Fraction division instead requires the reciprocal of the second fraction."}]},{"id":"b796a081-ec17-4783-aa8c-cb84afc5b759","cardType":"basic","points":2,"front":"What distinguishes a terminating decimal from a recurring decimal?","back":"

Terminating decimals stop after a fixed number of places.
Recurring decimals have digits that repeat forever.

","mcqOptions":[{"text":"

Terminating decimals stop after a fixed number of places.
Recurring decimals have digits that repeat forever.

","correct":true,"explanation":"A terminating decimal has finitely many decimal places, whereas a recurring decimal contains a digit or block of digits that continues repeating indefinitely."},{"text":"

Terminating decimals repeat after a fixed number of places.
Recurring decimals have digits that eventually stop.

","correct":false,"explanation":"These definitions have been reversed. Terminating decimals stop after finitely many places, while recurring decimals contain digits that repeat forever."},{"text":"

Terminating decimals stop after a fixed number of places.
Recurring decimals have digits that repeat once.

","correct":false,"explanation":"The first statement is accurate, but recurring digits do not repeat only once. Their repeating pattern continues forever."},{"text":"

Terminating decimals continue after a fixed number of places.
Recurring decimals have digits that repeat forever.

","correct":false,"explanation":"The second statement is accurate, but a terminating decimal does not continue indefinitely. It stops after a finite number of decimal places."}]},{"id":"845612bc-f8d1-4dce-99d9-09240778407f","cardType":"basic","points":1,"front":"Convert the fraction $\\frac{7}{11}$ to recurring decimal notation.","back":"$$0.\\dot{6}\\dot{3}$","mcqOptions":[{"text":"$$0.\\dot{6}\\dot{3}$","correct":true,"explanation":"Dividing 7 by 11 gives $0.636363\\ldots$, so the repeating block is 63. Dots above 6 and 3 mark the beginning and end of that block."},{"text":"$$0.\\dot{3}\\dot{6}$","correct":false,"explanation":"This represents $0.363636\\ldots$, which equals $\\frac{4}{11}$. Dividing 7 by 11 instead produces the repeating block 63, giving $0.\\dot{6}\\dot{3}$."},{"text":"$$0.6\\dot{3}$","correct":false,"explanation":"This represents $0.63333\\ldots$, where only 3 recurs. For $\\frac{7}{11}$, both digits in the block 63 repeat, so both endpoints need dots."},{"text":"$$0.\\dot{7}\\dot{0}$","correct":false,"explanation":"This represents $0.707070\\ldots$, not the result of dividing 7 by 11. Long division gives $0.636363\\ldots$, written $0.\\dot{6}\\dot{3}$."}]},{"id":"f49c74b9-99b4-4def-baa7-7a3b093547bb","cardType":"basic","points":2,"front":"Why is it a common mistake to compare $100x$ and $x$ for the decimal $0.3\\dot{8}$?","back":"The recurring tails will not match (one starts at the first decimal place, the other at the second).","mcqOptions":[{"text":"The recurring tails will not match (one starts at the first decimal place, the other at the second).","correct":true,"explanation":"The multiplier should be chosen so the repeating sections begin in corresponding positions before subtraction. Here, the recurring 8 begins after the non-recurring 3 in $x$."},{"text":"The recurring tails will match immediately (both start at the first decimal place).","correct":false,"explanation":"In $x=0.3888\\ldots$, the first decimal digit is the non-recurring 3, while after multiplying by 100 the recurring tail starts immediately. Their starting positions therefore differ."},{"text":"Multiplying by 100 removes the recurring part (so $100x$ is a whole number).","correct":false,"explanation":"Multiplication by 100 shifts the decimal point two places but does not remove an infinite recurring tail. The issue is that the recurring tails begin in different positions."},{"text":"The non-recurring digit also repeats forever (so the entire decimal forms one recurring block).","correct":false,"explanation":"Only the 8 repeats in $0.3\\dot{8}$; the 3 occurs once. Consequently, the repeating tail begins at the second decimal place in $x$, not the first."}]}]}],"examCountdown":null,"heading":"Recurring Decimals to Fractions","title":"Recurring Decimals to Fractions Flashcards","topbar":{"backHref":"/igcse/maths/cie/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"b6e0f4fc-2ddd-4f27-b84a-fb962c972ee3","walled":"$L39"}],"$L3a","$L3b"]