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marks that digit as the recurring block, so the digit continues repeating indefinitely rather than terminating."},{"text":"That digit appears only once.","correct":false,"explanation":"A dot is not used to show that a digit occurs once. It marks the recurring digit, meaning that digit repeats forever."},{"text":"That digit is rounded up.","correct":false,"explanation":"Rounding is usually indicated by an approximation sign or stated degree of accuracy. A dot above the digit instead means that it repeats forever."},{"text":"That digit ends the decimal.","correct":false,"explanation":"The final digit of a terminating decimal has no dot above it. A dot indicates that the marked digit repeats forever."}]},{"id":"44f03563-b1eb-b313-e79c-845cb67a3d09","cardType":"basic","points":1,"front":"In the decimal $0.\\dot{4}0\\dot{5}$, which digits form the repeating block?","back":"405","mcqOptions":[{"text":"405","correct":true,"explanation":"The dots mark the beginning and end of the recurring section. Every digit from 4 through 5 is included, so the block is 405."},{"text":"45","correct":false,"explanation":"This omits the 0 between the two dotted digits. The dots mark the endpoints of the entire repeating block, so the correct block is 405."},{"text":"40","correct":false,"explanation":"This stops before the second dotted digit, but both dotted digits are included as endpoints. The complete repeating block is therefore 405."},{"text":"05","correct":false,"explanation":"This incorrectly begins the block after the first dotted digit. The first dot is above 4, so the full repeating block is 405."}]},{"id":"74234f81-8dfd-5d20-0e4d-cdcb51b18299","cardType":"cloze","points":null,"front":"A recurring decimal is a decimal where a [ ] repeats forever.","back":"A recurring decimal is a decimal where a digit or block of digits repeats forever.","mcqOptions":[{"text":"digit or block of digits","correct":true,"explanation":"A recurring decimal contains either one digit or a longer block of digits that repeats indefinitely, such as 7 in $0.777\\ldots$ or 405 in $0.405405\\ldots$."},{"text":"non-repeating block of digits","correct":false,"explanation":"A non-repeating block does not satisfy the definition of a recurring decimal. The repeating part must be a digit or block of digits that continues forever."},{"text":"decimal point or place-value marker","correct":false,"explanation":"The decimal point separates whole-number and fractional places but does not itself recur. A digit or block of digits is what repeats forever."},{"text":"whole-number part of the number","correct":false,"explanation":"The whole-number part remains fixed rather than repeating through successive decimal places. A recurring decimal has a digit or block of digits that repeats forever."}]},{"id":"09c95dc9-e485-679b-ab19-2109b75aada7","cardType":"basic","points":2,"front":"How is a fraction like $\\frac{5}{11}$ converted into a decimal without a calculator?","back":"Divide the numerator (5) by the denominator (11) using long division.","mcqOptions":[{"text":"Divide the numerator (5) by the denominator (11) using long division.","correct":true,"explanation":"A fraction bar represents division, with the numerator as the dividend and the denominator as the divisor. Therefore, calculate $5 \\div 11$ using long division."},{"text":"Divide the denominator (11) by the numerator (5) using long division.","correct":false,"explanation":"This reverses the fraction and calculates $11 \\div 5$, which corresponds to $\\frac{11}{5}$. For $\\frac{5}{11}$, divide 5 by 11."},{"text":"Multiply the numerator (5) by the denominator (11) using long multiplication.","correct":false,"explanation":"Multiplying 5 by 11 calculates a product, not the value of the fraction. A fraction bar means divide the numerator by the denominator."},{"text":"Add the numerator (5) to the denominator (11) using column addition.","correct":false,"explanation":"Adding the numerator and denominator gives 16, which does not represent the fraction. Convert it by dividing the numerator, 5, by the denominator, 11."}]},{"id":"31dc9cc8-dede-9fb5-738d-749827a7175c","cardType":"basic","points":1,"front":"When converting a fraction to a decimal using division, how do you identify that the decimal is recurring?","back":"A remainder repeats/reappears during the division process.","mcqOptions":[{"text":"A remainder repeats/reappears during the division process.","correct":true,"explanation":"When a remainder reappears, the same subsequent division steps and quotient digits must occur again. This creates a repeating block in the decimal expansion."},{"text":"A remainder becomes zero during the division process.","correct":false,"explanation":"A remainder of zero means the division has finished exactly, producing a terminating decimal. A recurring decimal is identified when a non-zero remainder reappears."},{"text":"The divisor repeats/reappears during the division process.","correct":false,"explanation":"The divisor stays fixed in every division step, so its repeated use does not identify recurrence. Recurrence is detected when a remainder repeats or reappears."},{"text":"A quotient digit becomes zero during the division process.","correct":false,"explanation":"A zero quotient digit can occur within either terminating or recurring decimals, so it is not decisive. A repeated remainder guarantees that the decimal cycle repeats."}]},{"id":"e8c3746e-86e9-6ded-32b3-ca05a843c208","cardType":"basic","points":2,"front":"What is the primary goal of multiplying a recurring decimal $x$ by powers of 10 in an algebraic proof?","back":"To make the infinite recurring tails match so they cancel out when subtracted.","mcqOptions":[{"text":"To make the infinite recurring tails match so they cancel out when subtracted.","correct":true,"explanation":"Multiplying by a suitable power of 10 aligns identical recurring tails. Subtracting the equations then removes those infinite tails and leaves a finite equation for $x$."},{"text":"To turn the recurring decimal into a whole number without any subtraction.","correct":false,"explanation":"No finite power of 10 turns a non-zero recurring decimal into an integer. The multiplication instead aligns the recurring tails so subtraction cancels them."},{"text":"To make the infinite recurring tails different so they combine when added.","correct":false,"explanation":"The recurring tails must be identical and aligned, not different. Once they match, subtracting one equation from the other cancels the infinite decimal parts."},{"text":"To remove the recurring digits by rounding them to a terminating decimal.","correct":false,"explanation":"Rounding would replace the recurring decimal with an approximation and could not prove an exact fraction. Powers of 10 align the recurring tails for exact cancellation."}]},{"id":"041267f8-fde1-d249-594e-3decb965f043","cardType":"cloze","points":null,"front":"In the proof for $0.\\dot{7}$, if $x = 0.777\\ldots$, then $10x = \\text{[...]}$.","back":"In the proof for $0.\\dot{7}$, if $x = 0.777\\ldots$, then $10x = \\boxed{7.777\\ldots}$.","mcqOptions":[{"text":"$$7.777\\ldots$","correct":true,"explanation":"Multiplying by 10 shifts every digit one place to the left, so $0.777\\ldots$ becomes $7.777\\ldots$. The recurring tail remains unchanged and aligned."},{"text":"$$77.777\\ldots$","correct":false,"explanation":"This shifts the decimal point two places, which corresponds to multiplying by 100 rather than 10. A one-place shift gives $10x=7.777\\ldots$."},{"text":"$$0.0777\\ldots$","correct":false,"explanation":"This shifts the digits one place to the right, which corresponds to dividing by 10. Multiplying by 10 shifts them left, giving $7.777\\ldots$."},{"text":"$$7.0777\\ldots$","correct":false,"explanation":"A decimal-point shift moves all digits together and does not insert a zero after the decimal point. Multiplying by 10 gives $7.777\\ldots$."}]},{"id":"95ab520a-b5d3-78f5-1801-7b0e82956fc4","cardType":"basic","points":1,"front":"What is $0.\\dot{7}$ expressed as a fraction?","back":"$$\\frac{7}{9}$","mcqOptions":[{"text":"$$\\frac{7}{9}$","correct":true,"explanation":"Let $x=0.777\\ldots$. Then $10x=7.777\\ldots$, so subtraction gives $9x=7$ and therefore $x=\\frac{7}{9}$."},{"text":"$$\\frac{7}{10}$","correct":false,"explanation":"This treats the decimal as the terminating value $0.7$ and ignores the infinitely repeated 7s. Algebraic subtraction gives $9x=7$, so $x=\\frac{7}{9}$."},{"text":"$$\\frac{7}{90}$","correct":false,"explanation":"This incorrectly uses 90 as the coefficient after shifting and subtracting. Since $10x-x=9x$, the equation is $9x=7$, giving $\\frac{7}{9}$."},{"text":"$$\\frac{1}{7}$","correct":false,"explanation":"This reverses the numerator and denominator. From $9x=7$, division by 9 gives $x=\\frac{7}{9}$, not its reciprocal."}]},{"id":"bbf00c94-72b9-8217-f1a0-b271843112b4","cardType":"basic","points":2,"front":"For $0.5\\dot{3}$, why is subtracting $x$ from $10x$ ineffective for canceling the decimal?","back":"The tails do not match ($5.333\\ldots$ vs $0.533\\ldots$).","mcqOptions":[{"text":"The tails do not match ($5.333\\ldots$ vs $0.533\\ldots$).","correct":true,"explanation":"Subtraction cancels a recurring tail only when the digits after the aligned decimal positions are identical. Here, multiplying by 10 does not align the recurring parts of the two numbers."},{"text":"The decimal points do not align ($5.333\\ldots$ vs $0.533\\ldots$).","correct":false,"explanation":"The decimal points can be aligned for subtraction, but that alone does not cancel an infinite recurring tail. The method fails because the tails do not match: $5.333\\ldots$ and $0.533\\ldots$."},{"text":"The digits do not recur ($5.333\\ldots$ vs $0.533\\ldots$).","correct":false,"explanation":"Both numbers do have recurring digits, so recurrence is not the problem. Subtraction is ineffective because their tails do not match: one has $5.333\\ldots$ while the other has $0.533\\ldots$."},{"text":"The whole-number parts do not match ($5.333\\ldots$ vs $0.533\\ldots$).","correct":false,"explanation":"The whole-number parts do not need to match when subtracting recurring decimals. The issue is that the recurring tails are misaligned, giving $5.333\\ldots$ and $0.533\\ldots$."}]},{"id":"b8de99c8-6605-34c4-980f-04efe978574a","cardType":"basic","points":2,"front":"Which two multiples of $x$ should be subtracted to convert $0.5\\dot{3}$ to a fraction?","back":"

$100x$
$10x$

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

$100x$
$10x$

","correct":true,"explanation":"Multiplying by 10 moves the non-recurring digit before the decimal point, while multiplying by 100 shifts one place further. Both then have the same recurring tail, allowing cancellation."},{"text":"

$10x$
$x$

","correct":false,"explanation":"Using $10x$ and $x$ leaves the recurring digits in different decimal positions, so their tails cannot cancel. The aligned multiples are $100x$ and $10x$."},{"text":"

$1000x$
$10x$

","correct":false,"explanation":"Multiplying by 1000 shifts one place too far relative to $10x$, so the recurring tails are not aligned in the required way. Use $100x$ and $10x$."},{"text":"

$100x$
$x$

","correct":false,"explanation":"Although these multiples differ by a factor of 10, they leave the non-recurring digit in the wrong alignment for the intended subtraction. The required multiples are $100x$ and $10x$."}]},{"id":"36ce3d07-7685-01be-065e-06894cb2593e","cardType":"cloze","points":null,"front":"If a decimal has $m$ non-recurring digits and $n$ recurring digits, you use the multiples $\\text{[...]}$ and $10^{m+n}x$.","back":"If a decimal has $m$ non-recurring digits and $n$ recurring digits, you use the multiples $\\boxed{10^m x}$ and $10^{m+n}x$.","mcqOptions":[{"text":"$$10^m x$","correct":true,"explanation":"Multiplication by $10^m$ moves all $m$ non-recurring digits before the decimal point. This positions the start of the recurring block for alignment with the larger multiple."},{"text":"$$10^n x$","correct":false,"explanation":"Using $10^n$ shifts according to the recurring block’s length rather than the number of non-recurring digits. The first required multiple is $10^m x$."},{"text":"$$10^{m+1}x$","correct":false,"explanation":"Adding one to the exponent shifts the decimal one place too far and misaligns the recurring tails. Exactly $m$ non-recurring digits must be moved, so use $10^m x$."},{"text":"$$10^{m-1}x$","correct":false,"explanation":"Subtracting one from the exponent leaves one non-recurring digit after the decimal point. All $m$ such digits must be shifted, requiring $10^m x$."}]},{"id":"a33a704d-2c27-e8dd-f7b8-1faee11845cb","cardType":"cloze","points":null,"front":"If a decimal has $m$ non-recurring digits and $n$ recurring digits, you use the multiples $10^m x$ and $\\text{[...]}$.","back":"If a decimal has $m$ non-recurring digits and $n$ recurring digits, you use the multiples $10^m x$ and $\\boxed{10^{m+n}x}$.","mcqOptions":[{"text":"$$10^{m+n}x$","correct":true,"explanation":"The larger multiple must shift past both the $m$ non-recurring digits and one complete block of $n$ recurring digits, requiring a total shift of $m+n$ places."},{"text":"$$10^n x$","correct":false,"explanation":"Using only $10^n$ ignores the $m$ non-recurring digits before the repeating block. The total shift must cover both sections, so the multiple is $10^{m+n}x$."},{"text":"$$10^{mn}x$","correct":false,"explanation":"Multiplying the exponents confuses the total number of places with a product. The shifts for the non-recurring and recurring sections are added, giving $10^{m+n}x$."},{"text":"$$10^{m-n}x$","correct":false,"explanation":"Subtracting $n$ shifts in the wrong direction conceptually and cannot move through both sections of digits. Their lengths must be added, producing $10^{m+n}x$."}]},{"id":"0ff4e885-5cbe-2179-9102-c6ee4fcf2084","cardType":"basic","points":1,"front":"Why is $1000x$ used when converting $0.\\dot{1}8\\dot{9}$ to a fraction?","back":"The repeating block contains 3 digits.","mcqOptions":[{"text":"The repeating block contains 3 digits.","correct":true,"explanation":"The dots mark the first and last digits of the repeating block, so the block is $189$. Since it has three digits, multiplying by 1000 shifts one complete repetition."},{"text":"The repeating block contains 2 digits.","correct":false,"explanation":"Counting only the dotted endpoint digits overlooks the $8$ between them. The complete repeating block is $189$, so it contains three digits and requires multiplication by 1000."},{"text":"The repeating block contains 1 digit.","correct":false,"explanation":"The notation does not make the first digit non-recurring; the dots enclose the whole block $189$. Therefore, the repeating block contains three digits, not one."},{"text":"The repeating block contains 4 digits.","correct":false,"explanation":"Repeating the first digit visually can lead to counting an extra place, but one complete block is simply $189$. It contains three digits, which is why $1000x$ is used."}]},{"id":"b2285813-0c63-8e74-af76-17c945ea083b","cardType":"basic","points":2,"front":"Convert $0.\\dot{1}8\\dot{9}$ to its simplest fraction form.","back":"$$\\frac{7}{37}$","mcqOptions":[{"text":"$$\\frac{7}{37}$","correct":true,"explanation":"Letting $x=0.189189\\ldots$ gives $1000x-x=189$, so $999x=189$. Thus $x=\\frac{189}{999}=\\frac{7}{37}$ after dividing by 27."},{"text":"$$\\frac{7}{27}$","correct":false,"explanation":"A denominator of 27 results from reducing the denominator incorrectly. Since $1000x-x=189$, the fraction is $\\frac{189}{999}$, which simplifies to $\\frac{7}{37}$."},{"text":"$$\\frac{7}{36}$","correct":false,"explanation":"Subtracting 1 from 1000 gives 999, not a value that reduces to a denominator of 36. Simplifying $\\frac{189}{999}$ correctly gives $\\frac{7}{37}$."},{"text":"$$\\frac{189}{1000}$","correct":false,"explanation":"This uses 1000 as the denominator without accounting for the recurring subtraction. The correct equation is $999x=189$, giving $\\frac{189}{999}=\\frac{7}{37}$."}]},{"id":"b89677b7-6aec-1a5c-37fa-00c6aec92fb1","cardType":"cloze","points":null,"front":"A fraction is in its simplest form when the numerator and denominator have no [ ] greater than 1.","back":"A fraction is in its simplest form when the numerator and denominator have no common factors greater than 1.","mcqOptions":[{"text":"common factors","correct":true,"explanation":"A fraction can be reduced only when its numerator and denominator share a factor greater than 1. Having no such common factors therefore means it is in simplest form."},{"text":"common multiples","correct":false,"explanation":"Every pair of positive integers has common multiples, so this cannot determine whether a fraction is simplified. Simplest form requires the numerator and denominator to have no common factors greater than 1."},{"text":"prime factors","correct":false,"explanation":"The numerator and denominator may each have prime factors while sharing none of them. The relevant condition is that they have no common factors greater than 1."},{"text":"common denominators","correct":false,"explanation":"A denominator belongs to a fraction rather than being a shared divisor of its two terms. Simplest form is defined by having no common factors greater than 1."}]},{"id":"1a30e5a4-e603-3ed8-7960-15b09c141f9b","cardType":"basic","points":2,"front":"For $2.4\\dot{3}\\dot{6}$, which two multiples of $x$ are needed to align the recurring tails?","back":"

$1000x$
$10x$

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

$1000x$
$10x$

","correct":true,"explanation":"The single non-recurring digit requires a shift by 10, while the three-digit recurring block requires three further places. Thus $1000x$ and $10x$ have matching recurring tails."},{"text":"

$100x$
$10x$

","correct":false,"explanation":"Using $100x$ shifts only two places beyond the original decimal and does not span the complete three-digit recurring block. The required multiples are $1000x$ and $10x$."},{"text":"

$1000x$
$x$

","correct":false,"explanation":"Subtracting $x$ fails to account separately for the one non-recurring digit, leaving the recurring tails misaligned. The correctly aligned multiples are $1000x$ and $10x$."},{"text":"

$1000x$
$100x$

","correct":false,"explanation":"The difference between these shifts is only one decimal place, but the recurring block has three digits. A three-place difference is needed, using $1000x$ and $10x$."}]},{"id":"b34dd752-3baf-2af8-f9a5-de7873737b61","cardType":"basic","points":1,"front":"What is the value of the recurring decimal $0.\\dot{9}$?","back":"1","mcqOptions":[{"text":"1","correct":true,"explanation":"The decimal $0.\\dot{9}$ approaches 1 exactly; there is no positive gap between them. Algebraically, if $x=0.\\dot{9}$, then $10x-x=9$, so $x=1$."},{"text":"0.9","correct":false,"explanation":"Treating the recurring decimal as slightly less than 1 is a common misconception. Its infinitely many 9s make it exactly equal to 1, not 0.9."},{"text":"0.99","correct":false,"explanation":"This value truncates the decimal after two 9s, but the dot means that 9 repeats infinitely. The recurring decimal $0.\\dot{9}$ is exactly equal to 1."},{"text":"1.00","correct":false,"explanation":"This rounds the recurring decimal to two decimal places rather than giving its exact value. Since $0.\\dot{9}=1$, the exact answer is 1."}]},{"id":"0f74c715-f31f-af4c-325e-81c420d2c610","cardType":"cloze","points":null,"front":"Because $0.\\dot{9}=1$, the decimal $1.2\\dot{9}$ is exactly equal to [ ].","back":"Because $0.\\dot{9}=1$, the decimal $1.2\\dot{9}$ is exactly equal to 1.3.","mcqOptions":[{"text":"1.3","correct":true,"explanation":"The recurring 9 increases the preceding digit: $1.2\\dot{9}=1.2999\\ldots$, which is exactly 1.3 because $0.0999\\ldots=0.1$."},{"text":"1.29","correct":false,"explanation":"This truncates the decimal after one recurring digit and ignores the infinitely many following 9s. Since $1.2999\\ldots$ reaches the next tenth exactly, the answer is 1.3."},{"text":"1.23","correct":false,"explanation":"This incorrectly replaces the recurring 9 with a recurring 3-type value. The sequence $1.2999\\ldots$ is exactly equal to 1.3, not 1.23."},{"text":"1.0","correct":false,"explanation":"Rounding to the nearest whole number loses the required exact decimal value. Although 1 is a whole-number approximation, $1.2\\dot{9}$ is exactly 1.3."}]},{"id":"b4f39d24-286d-8bc6-aece-4399b82e3fe5","cardType":"basic","points":2,"front":"How should you approach a calculation (like multiplication or division) involving recurring decimals?","back":"Convert every recurring decimal into a fraction before calculating.","mcqOptions":[{"text":"Convert every recurring decimal into a fraction before calculating.","correct":true,"explanation":"Fractions preserve the exact values of recurring decimals, allowing multiplication or division without rounding errors. The resulting fraction can then be simplified or converted back to a decimal."},{"text":"Round every recurring decimal to a whole number before calculating.","correct":false,"explanation":"Rounding first replaces exact recurring values with approximations, so the final result may be inaccurate. Convert every recurring decimal into a fraction before calculating."},{"text":"Remove every recurring decimal’s repeated digits before calculating.","correct":false,"explanation":"Removing the recurring digits truncates the numbers and changes their values, producing an inexact calculation. Convert every recurring decimal into a fraction before calculating."},{"text":"Treat every recurring decimal as a terminating decimal before calculating.","correct":false,"explanation":"Treating recurring decimals as terminating ignores that their repeated digits continue indefinitely. Convert every recurring decimal into a fraction before calculating to retain exact values."}]},{"id":"cf53b39a-34ee-a4e0-5b10-39f19736ab42","cardType":"basic","points":1,"front":"When dividing fractions, what is the 'reciprocal'?","back":"The second fraction flipped upside down.","mcqOptions":[{"text":"The second fraction flipped upside down.","correct":true,"explanation":"In fraction division, the reciprocal is formed by interchanging the numerator and denominator of the second fraction. You then multiply the first fraction by this flipped fraction."},{"text":"The first fraction flipped upside down.","correct":false,"explanation":"A common division error is to flip the first fraction. The reciprocal used when dividing fractions is the second fraction flipped upside down, after which the fractions are multiplied."},{"text":"The two fractions swapped around.","correct":false,"explanation":"Swapping the fractions’ order is not taking a reciprocal and changes the calculation incorrectly. The reciprocal is the second fraction flipped upside down."},{"text":"The second fraction made negative.","correct":false,"explanation":"Changing both signs produces negatives rather than reciprocals. For fraction division, keep the signs and flip the numerator and denominator of the second fraction only."}]},{"id":"c0a64aed-17ba-0ec6-f41e-43c2093f3398","cardType":"cloze","points":null,"front":"To show the pattern to an examiner, write out $x$ with [ ] instead of just the dot notation.","back":"To show the pattern to an examiner, write out $x$ with a few repeated digits instead of just the dot notation.","mcqOptions":[{"text":"a few repeated digits","correct":true,"explanation":"Writing several repeated digits makes the recurring pattern explicit and shows which digits continue indefinitely. This helps justify the algebra used to convert the decimal into a fraction."},{"text":"only non-recurring digits","correct":false,"explanation":"Writing only the non-recurring digits hides the repeating pattern needed for the conversion method. Instead, write a few repeated digits so the recurrence is clear to the examiner."},{"text":"a rounded decimal value","correct":false,"explanation":"Rounding changes the recurring decimal into an approximation and obscures its exact pattern. Write a few repeated digits instead, preserving the value needed for an exact fraction."},{"text":"one repeated digit","correct":false,"explanation":"Writing a single repeated digit may not clearly display a multi-digit recurring pattern. Write a few repeated digits so the examiner can see exactly what repeats."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Unit 1"},"heading":"Recurring Decimals to Fractions","title":"Recurring Decimals to Fractions Flashcards","topbar":{"backHref":"/gcse/maths/wjec/flashcards"},"topicSidebar":"$L39"}]}],"topicId":"17f4b209-b6b2-b7cc-60f6-20ac9a8d5150","walled":"$L3a"}],"$L3b","$L3c"]