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stated measurement."},{"text":"largest","correct":false,"explanation":"This reverses the definitions of the two bounds. The lower bound is the smallest possible original value that could have produced the stated rounded value."},{"text":"average","correct":false,"explanation":"A bound is an endpoint, not a mean of the possible values. The lower bound is the smallest possible value before rounding."},{"text":"nearest","correct":false,"explanation":"This confuses a boundary with the rounded value itself. The lower bound is the smallest possible value in the rounding interval."}]},{"id":"dfe0c2f3-7d49-4576-9862-eab5ef35c53d","cardType":"cloze","points":null,"front":"The lower bound is the smallest possible value before rounding, while the upper bound is the [ ] boundary value.","back":"The lower bound is the smallest possible value before rounding, while the upper bound is the largest boundary value.","mcqOptions":[{"text":"largest","correct":true,"explanation":"The upper bound marks the largest boundary value of the interval containing all original values that round to the stated measurement."},{"text":"smallest","correct":false,"explanation":"This describes the lower boundary rather than the upper boundary. The upper bound is the largest boundary value in the rounding interval."},{"text":"nearest","correct":false,"explanation":"The rounded value may be nearest to the original value, but the upper bound is specifically the largest boundary value permitted."},{"text":"central","correct":false,"explanation":"The centre of the interval is the rounded value, not its upper boundary. The upper bound is the largest boundary value."}]},{"id":"667d4303-7fbf-4370-b926-af78e892b166","cardType":"basic","points":1,"front":"What is the half-gap for a measurement rounded to the nearest $5\\text{ m}$?","back":"$$2.5\\text{ m}$","mcqOptions":[{"text":"$$2.5\\text{ m}$","correct":true,"explanation":"Values rounded to the nearest $5\\text{ m}$ have a rounding gap of $5\\text{ m}$, so the half-gap is $5 \\div 2 = 2.5\\text{ m}$."},{"text":"$$1.25\\text{ m}$","correct":false,"explanation":"This uses one quarter of the rounding interval rather than one half. Half of the $5\\text{ m}$ gap is $2.5\\text{ m}$."},{"text":"$$5\\text{ m}$","correct":false,"explanation":"This uses the full rounding gap instead of halving it. The half-gap for rounding to the nearest $5\\text{ m}$ is $2.5\\text{ m}$."},{"text":"$$10\\text{ m}$","correct":false,"explanation":"This incorrectly doubles the rounding gap. The gap is $5\\text{ m}$ and its half-gap is therefore $2.5\\text{ m}$."}]},{"id":"1e967a84-92e6-46c0-8d34-e5d1f1c0f2eb","cardType":"basic","points":1,"front":"What is the half-gap for a measurement rounded correct to $2$ decimal places?","back":"$$0.005$","mcqOptions":[{"text":"$$0.005$","correct":true,"explanation":"Two decimal places correspond to a rounding increment of $0.01$. The half-gap is therefore $0.01 \\div 2 = 0.005$."},{"text":"$$0.01$","correct":false,"explanation":"This is the full gap between consecutive values to two decimal places. Halving $0.01$ gives the required half-gap of $0.005$."},{"text":"$$0.05$","correct":false,"explanation":"This is the half-gap for rounding to one decimal place. For two decimal places, the gap is $0.01$, giving a half-gap of $0.005$."},{"text":"$$0.0005$","correct":false,"explanation":"This introduces an extra decimal place when halving. Half of the two-decimal-place increment $0.01$ is $0.005$, not $0.0005$."}]},{"id":"3d6ff7d3-13c1-4df0-9085-3c0e24d5cce0","cardType":"basic","points":1,"front":"If a measurement in $\\text{cm}$ is rounded to the nearest $\\text{mm}$, what half-gap value is used?","back":"$$0.05\\text{ cm}$","mcqOptions":[{"text":"$$0.05\\text{ cm}$","correct":true,"explanation":"One millimetre equals $0.1\\text{ cm}$, so rounding to the nearest millimetre gives a half-gap of $0.1 \\div 2 = 0.05\\text{ cm}$."},{"text":"$$0.1\\text{ cm}$","correct":false,"explanation":"This uses the full $1\\text{ mm}$ rounding gap, converted to centimetres, without halving it. The required half-gap is $0.05\\text{ cm}$."},{"text":"$$0.5\\text{ cm}$","correct":false,"explanation":"This incorrectly treats half a millimetre as half a centimetre. Since $0.5\\text{ mm}=0.05\\text{ cm}$, the half-gap is $0.05\\text{ cm}$."},{"text":"$$0.005\\text{ cm}$","correct":false,"explanation":"This divides by ten once too many during the unit conversion. Half a millimetre is $0.05\\text{ cm}$, not $0.005\\text{ cm}$."}]},{"id":"649e5e97-315e-4bd0-aa73-cd2392c54311","cardType":"cloze","points":null,"front":"A rounded value $l$ is expressed as an interval using the symbols: $\\text{[...]} l < <$.","back":"A rounded value $l$ is expressed as an interval using the symbols: $\\boxed{\\le} l < <$.","mcqOptions":[{"text":"$$\\le$","correct":true,"explanation":"The lower boundary is included in a bounds interval, so the lower bound is less than or equal to the value: $\\text{lower bound} \\le l$."},{"text":"$$<$","correct":false,"explanation":"Using a strict inequality would exclude the lower boundary. Bounds intervals include the lower bound, so the left-hand comparison must use $\\le$."},{"text":"$$\\ge$","correct":false,"explanation":"This reverses the direction of the comparison. The interval is written from the lower bound to $l$, so the required symbol is $\\le$."},{"text":"$$>$","correct":false,"explanation":"This both reverses the comparison and excludes the lower boundary. The correct relationship is $\\text{lower bound} \\le l$."}]},{"id":"4a470f47-348b-4213-96b8-c70bec464a14","cardType":"cloze","points":null,"front":"A rounded value $l$ is expressed as an interval using the symbols: $\\le l < \\text{[...]}$.","back":"A rounded value $l$ is expressed as an interval using the symbols: $\\le l < \\boxed{<}$.","mcqOptions":[{"text":"$$<$","correct":true,"explanation":"The upper boundary is excluded from the interval, so the value must be strictly less than the upper bound. This requires the symbol $<$."},{"text":"$$\\le$","correct":false,"explanation":"The upper boundary is not included because values exactly at it round to the next value. The right-hand comparison must therefore use $<$."},{"text":"$$>$","correct":false,"explanation":"This reverses the comparison, suggesting that $l$ exceeds the upper bound. Instead, $l$ must be strictly less than the upper bound, using $<$."},{"text":"$$\\ge$","correct":false,"explanation":"This reverses the comparison and incorrectly includes the upper boundary. The value must be strictly less than the upper bound, so the correct symbol is $<$."}]},{"id":"37537fac-2004-4146-8d9e-70660c7d6038","cardType":"basic","points":1,"front":"How do you calculate the upper bound for the area of a rectangle?","back":"$$\\text{upper length} \\times \\text{upper width}$","mcqOptions":[{"text":"$$\\text{upper length} \\times \\text{upper width}$","correct":true,"explanation":"For positive rectangle dimensions, the area increases as either dimension increases. Its upper bound is therefore found by multiplying both upper bounds."},{"text":"$$\\text{lower length} \\times \\text{lower width}$","correct":false,"explanation":"Using both lower dimensions gives the lower bound for the area, not the upper bound. The maximum area uses the upper length and upper width."},{"text":"$$\\text{upper length} \\times \\text{lower width}$","correct":false,"explanation":"Using the lower width prevents the area from being as large as possible. The upper bound requires both the upper length and upper width."},{"text":"$$\\text{lower length} \\times \\text{upper width}$","correct":false,"explanation":"Using the lower length prevents the area from being maximised. Both dimensions must take their upper bounds to find the upper bound for area."}]},{"id":"287315df-d0c9-4d8f-bfd5-6e8500360419","cardType":"basic","points":2,"front":"How do you calculate the lower bound for the perimeter of a rectangle?","back":"$$2 \\times (\\text{lower length} + \\text{lower width})$","mcqOptions":[{"text":"$$2 \\times (\\text{lower length} + \\text{lower width})$","correct":true,"explanation":"The perimeter increases as either dimension increases, so using both lower bounds gives the smallest possible perimeter: twice the sum of the lower length and lower width."},{"text":"$$2 \\times (\\text{upper length} + \\text{upper width})$","correct":false,"explanation":"Using both upper bounds gives the upper bound, not the lower bound, of the perimeter. The lower bound is $2 \\times (\\text{lower length} + \\text{lower width})$."},{"text":"$$2 \\times (\\text{lower length} + \\text{upper width})$","correct":false,"explanation":"Mixing a lower length with an upper width does not produce the smallest possible perimeter. Both dimensions must use their lower bounds in the perimeter formula."},{"text":"$$2 \\times (\\text{upper length} + \\text{lower width})$","correct":false,"explanation":"Using the upper length makes the calculated perimeter larger than its minimum possible value. The lower bounds of both the length and width are required."}]},{"id":"864da493-6d2a-42f1-aee8-612614c278f7","cardType":"basic","points":1,"front":"To find the lower bound for a circle's circumference ($C=2\\pi r$), which bound of the radius is required?","back":"The lower bound of the radius.","mcqOptions":[{"text":"The lower bound of the radius.","correct":true,"explanation":"Circumference is directly proportional to radius because $C=2\\pi r$. Therefore, the smallest possible radius produces the lower bound of the circumference."},{"text":"The upper bound of the radius.","correct":false,"explanation":"Using the upper radius produces the upper bound of the circumference because circumference increases with radius. The lower bound of the radius is required for the lower circumference bound."},{"text":"The rounded value of the radius.","correct":false,"explanation":"The rounded radius is not the smallest possible radius and therefore cannot guarantee the lower circumference bound. Use the lower bound of the radius in $C=2\\pi r$."},{"text":"The midpoint value of the radius.","correct":false,"explanation":"The midpoint lies between the radius bounds, so it gives neither an upper nor a lower circumference bound. The lower radius bound gives the lower circumference bound."}]},{"id":"5f04ea48-9ad1-4597-8500-0e0e43a20d60","cardType":"basic","points":1,"front":"How do you calculate the upper bound of a product $v = x \\times y$?","back":"$$\\text{upper } x \\times \\text{upper } y$","mcqOptions":[{"text":"$$\\text{upper } x \\times \\text{upper } y$","correct":true,"explanation":"For positive quantities, a product is greatest when both factors take their greatest possible values. Therefore, multiply the upper bound of $x$ by the upper bound of $y$."},{"text":"$$\\text{lower } x \\times \\text{lower } y$","correct":false,"explanation":"Using both lower bounds gives the lower bound of the product for positive quantities. The upper product bound requires $\\text{upper } x \\times \\text{upper } y$."},{"text":"$$\\text{upper } x \\times \\text{lower } y$","correct":false,"explanation":"Mixing an upper bound with a lower bound does not maximise both positive factors. The largest possible product is found by multiplying both upper bounds."},{"text":"$$\\text{lower } x \\times \\text{upper } y$","correct":false,"explanation":"Using the lower bound of $x$ prevents the product from reaching its greatest possible value. For the upper bound, multiply the upper bounds of both variables."}]},{"id":"2e45a5c8-aa12-44b7-aa6c-4d29b9377ca1","cardType":"cloze","points":null,"front":"To find the upper bound of a fraction, use the [ ] numerator and the lower denominator.","back":"To find the upper bound of a fraction, use the upper numerator and the lower denominator.","mcqOptions":[{"text":"upper","correct":true,"explanation":"A fraction becomes larger when its positive numerator increases. Therefore, its upper bound uses the upper numerator together with the lower denominator."},{"text":"lower","correct":false,"explanation":"Using the lower numerator tends to minimise rather than maximise a positive fraction. Its upper bound requires the upper numerator and the lower denominator."},{"text":"central","correct":false,"explanation":"A central numerator value does not represent an extreme and therefore cannot guarantee an upper bound. Use the upper numerator and lower denominator to maximise the fraction."},{"text":"rounded","correct":false,"explanation":"The rounded numerator is only an estimate and need not be the greatest possible value. The upper numerator must be used to obtain the fraction’s upper bound."}]},{"id":"6cb9837b-716e-4c53-b36f-969247797869","cardType":"cloze","points":null,"front":"To find the upper bound of a fraction, use the upper numerator and the [ ] denominator.","back":"To find the upper bound of a fraction, use the upper numerator and the lower denominator.","mcqOptions":[{"text":"lower","correct":true,"explanation":"For a positive fraction, decreasing the denominator increases the quotient. Therefore, the upper bound uses the lower denominator together with the upper numerator."},{"text":"upper","correct":false,"explanation":"Using the upper denominator makes a positive quotient smaller, so it contributes to a lower bound. The upper fraction bound requires the lower denominator."},{"text":"central","correct":false,"explanation":"A central denominator is not an extreme value and cannot guarantee the greatest possible quotient. The lower denominator must be paired with the upper numerator."},{"text":"rounded","correct":false,"explanation":"The rounded denominator is an estimate rather than the smallest possible value. To maximise the fraction, use the denominator’s lower bound."}]},{"id":"ccdad04b-036d-4093-9639-905364104eb1","cardType":"basic","points":2,"front":"In the formula $v = \\frac{s}{t}$, why is the lower bound of $t$ used to find the upper bound of $v$?","back":"A smaller denominator results in a larger overall quotient.","mcqOptions":[{"text":"A smaller denominator results in a larger overall quotient.","correct":true,"explanation":"With a fixed positive numerator, division by a smaller positive number gives a larger result. Thus, the lower bound of $t$ helps produce the upper bound of $v$."},{"text":"A larger denominator results in a larger overall quotient.","correct":false,"explanation":"This reverses the effect of the denominator: increasing a positive denominator decreases the quotient. The upper bound of $v$ therefore uses the lower bound of $t$."},{"text":"A smaller denominator results in a smaller overall quotient.","correct":false,"explanation":"This confuses division with direct proportion. For a fixed positive numerator, a smaller denominator gives a larger quotient, so the lower bound of $t$ is used."},{"text":"A larger denominator results in an unchanged overall quotient.","correct":false,"explanation":"Changing the denominator changes the quotient when the numerator is fixed. A smaller positive denominator creates a larger quotient, which is why the lower bound of $t$ is required."}]},{"id":"193cd522-1f05-441d-aad8-75646c6f3d04","cardType":"basic","points":2,"front":"To find the lower bound of a fraction, which bounds for the numerator and denominator are used?","back":"

Lower bound of the numerator
Upper bound of the denominator

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

Lower bound of the numerator
Upper bound of the denominator

","correct":true,"explanation":"A positive fraction is minimised by making its numerator as small as possible and its denominator as large as possible. These bounds therefore give the lower quotient bound."},{"text":"

Upper bound of the numerator
Lower bound of the denominator

","correct":false,"explanation":"This combination maximises a positive fraction, so it gives the upper bound. The lower bound uses the lower numerator and the upper denominator."},{"text":"

Lower bound of the numerator
Lower bound of the denominator

","correct":false,"explanation":"Although the lower numerator reduces the fraction, the lower denominator increases it. To guarantee the smallest quotient, use the lower numerator and upper denominator."},{"text":"

Upper bound of the numerator
Upper bound of the denominator

","correct":false,"explanation":"The upper denominator reduces the fraction, but the upper numerator increases it, so this does not guarantee the minimum. Use the lower numerator and upper denominator."}]},{"id":"f88b6652-ca1a-4a8f-b35f-06094986717f","cardType":"basic","points":1,"front":"To find the lower bound of the hypotenuse $c$ in $c^2 = a^2 + b^2$, which bounds of $a$ and $b$ are used?","back":"The lower bounds of both $a$ and $b$.","mcqOptions":[{"text":"The lower bounds of both $a$ and $b$.","correct":true,"explanation":"For positive lengths, increasing either shorter side increases $a^2+b^2$ and hence $c$. The lower hypotenuse bound therefore uses the lower bounds of both $a$ and $b$."},{"text":"The upper bounds of both $a$ and $b$.","correct":false,"explanation":"Using both upper bounds makes $a^2+b^2$ as large as possible, producing the upper bound of $c$. The lower bound requires both lower bounds."},{"text":"The lower bound of $a$ and upper bound of $b$.","correct":false,"explanation":"Using the upper bound of $b$ does not minimise $a^2+b^2$, so the result is not the lower hypotenuse bound. Both $a$ and $b$ require lower bounds."},{"text":"The upper bound of $a$ and lower bound of $b$.","correct":false,"explanation":"Using the upper bound of $a$ prevents $a^2+b^2$ from taking its smallest possible value. The lower bounds of both sides are needed."}]},{"id":"02bd653e-61f7-40ed-a633-04426af5959e","cardType":"basic","points":1,"front":"To find the lower bound of side $b = \\sqrt{c^2 - a^2}$, which bound of the hypotenuse $c$ is used?","back":"The lower bound of $c$.","mcqOptions":[{"text":"The lower bound of $c$.","correct":true,"explanation":"A smaller hypotenuse gives a smaller value of $c^2$, so it minimises $c^2-a^2$ and therefore produces the lower bound of $b$."},{"text":"The upper bound of $c$.","correct":false,"explanation":"Using the upper bound of $c$ increases $c^2$ and therefore increases $b$. The lower bound of $c$ is needed to minimise the result."},{"text":"The rounded value of $c$.","correct":false,"explanation":"The stated rounded value is not an extreme of the possible interval. The lower bound of $c$ must be used because it produces the smallest possible $c^2$."},{"text":"The midpoint value of $c$.","correct":false,"explanation":"The midpoint does not give the smallest possible result and is usually just the stated rounded value. The lower bound of $c$ is required to minimise $b$."}]},{"id":"6bce6b84-0568-4b16-884b-e9032beda5a3","cardType":"basic","points":2,"front":"To find the lower bound of side $b = \\sqrt{c^2 - a^2}$, which bound of side $a$ is used?","back":"The upper bound of $a$ (subtracting a larger value minimizes the result).","mcqOptions":[{"text":"The upper bound of $a$ (subtracting a larger value minimizes the result).","correct":true,"explanation":"Because $a^2$ is subtracted, its largest possible value makes $c^2-a^2$ as small as possible. This gives the lower bound of $b$."},{"text":"The lower bound of $a$ (subtracting a smaller value minimizes the result).","correct":false,"explanation":"Subtracting a smaller value actually leaves a larger result, so this would help maximise $b$. The upper bound of $a$ is used to find the lower bound."},{"text":"The rounded value of $a$ (subtracting the stated value minimizes the result).","correct":false,"explanation":"The rounded value is not the largest possible value of $a$, so it cannot guarantee the minimum. The upper bound of $a$ must be subtracted."},{"text":"The midpoint value of $a$ (subtracting the central value minimizes the result).","correct":false,"explanation":"Using the midpoint ignores the extreme values allowed by the bounds. Since $a^2$ is subtracted, its upper bound gives the smallest possible result."}]},{"id":"40aab98e-12be-41b9-b30e-8a91cd79092b","cardType":"basic","points":1,"front":"Define a suitable degree of accuracy in the context of bounds.","back":"The rounding level where the lower bound and upper bound round to the same value.","mcqOptions":[{"text":"The rounding level where the lower bound and upper bound round to the same value.","correct":true,"explanation":"If both endpoints round to the same value at a particular accuracy, every possible value between them also rounds to that value, making the accuracy justified."},{"text":"The rounding level where the lower bound and upper bound round to different values.","correct":false,"explanation":"Different rounded endpoint values mean the interval does not determine one answer at that accuracy. A suitable accuracy is one where both bounds round to the same value."},{"text":"The rounding level where the lower bound and upper bound have the same digits.","correct":false,"explanation":"Bounds may share several digits without rounding to the same final value. Suitability depends specifically on both bounds rounding to exactly the same value."},{"text":"The rounding level where the lower bound and upper bound differ by the rounding unit.","correct":false,"explanation":"Comparing the interval width with the rounding unit is not sufficient, especially near a rounding boundary. Both bounds must round to the same value."}]},{"id":"dcb02e4d-1784-4e54-9205-80b1e2d8568e","cardType":"basic","points":2,"front":"If $LB = 5.359$ and $UB = 5.371$, why is \"2 decimal places\" NOT a suitable degree of accuracy?","back":"They round to different values ($5.36$ and $5.37$).","mcqOptions":[{"text":"They round to different values ($5.36$ and $5.37$).","correct":true,"explanation":"At two decimal places, $5.359$ rounds to $5.36$, while $5.371$ rounds to $5.37$. Therefore the interval does not guarantee one rounded answer."},{"text":"They round to the same value ($5.36$ and $5.36$).","correct":false,"explanation":"This incorrectly rounds $5.371$ down; its third decimal digit is $1$, so it rounds to $5.37$. The bounds therefore give different two-decimal-place values."},{"text":"They round to different values ($5.35$ and $5.38$).","correct":false,"explanation":"These results come from moving each endpoint outward rather than applying standard rounding. Correctly rounding gives $5.36$ and $5.37$, which are still different."},{"text":"They round to the same value ($5.37$ and $5.37$).","correct":false,"explanation":"This incorrectly rounds $5.359$ up to $5.37$; its hundredths rounding is determined by the third decimal digit, giving $5.36$. The endpoints round differently."}]},{"id":"a2e49925-85b8-4b33-96b1-24b0c3ddd5e7","cardType":"basic","points":2,"front":"What are the bounds for $58.0\\text{ cm}$ correct to the nearest millimetre?","back":"$$57.95 \\le l < 58.05$","mcqOptions":[{"text":"$$57.95 \\le l < 58.05$","correct":true,"explanation":"One millimetre is $0.1$ cm, so the rounding interval extends half of this, $0.05$ cm, on either side of $58.0$ cm."},{"text":"$$57.5 \\le l < 58.5$","correct":false,"explanation":"These are the bounds for a measurement rounded to the nearest centimetre, not millimetre. Nearest millimetre means half of $0.1$ cm, giving $57.95 \\le l < 58.05$."},{"text":"$$57.9 \\le l < 58.1$","correct":false,"explanation":"This subtracts and adds a whole millimetre rather than half a millimetre. Rounding bounds use half the rounding unit, so the correct endpoints are $57.95$ and $58.05$."},{"text":"$$57.99 \\le l < 58.01$","correct":false,"explanation":"This interval uses $0.01$ cm on either side, equivalent to only $0.1$ mm. Half of the $1$ mm rounding unit is $0.5$ mm, or $0.05$ cm."}]},{"id":"6409b478-8bd4-46e6-899d-6b71cafd4fdc","cardType":"cloze","points":null,"front":"When solving bounds problems, you should calculate the bounds for each value [ ] substituting them into a formula.","back":"When solving bounds problems, you should calculate the bounds for each value before substituting them into a formula.","mcqOptions":[{"text":"before","correct":true,"explanation":"Each measured quantity has its own interval of possible values. Finding those bounds before substitution ensures the correct extreme values are selected for the required result."},{"text":"after","correct":false,"explanation":"Substituting the rounded measurements first loses the extreme values needed for a bounds calculation. The lower and upper bounds of every value must be found before substitution."},{"text":"while","correct":false,"explanation":"Working out bounds during substitution encourages values to be bounded inconsistently or overlooked. Determine each quantity’s lower and upper bounds before applying the formula."},{"text":"without","correct":false,"explanation":"Bounds must eventually be substituted into the formula to calculate the result’s limits. They should be calculated before substitution, not used as a replacement for substitution."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Bounds","title":"Bounds Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"4ae2a8e7-ac9e-456a-b383-0e1d7df56def","walled":"$L39"}],"$L3a","$L3b"]