Upper and Lower Bounds (GCSE Maths)

• The value is given to the nearest $10\text{ seconds}$.
• The error interval is half of this value.

Subtract and add the error interval to the original value ($130\text{ seconds}$):

• The value is measured to the "nearest $10\text{ kg}$".
• The error interval is half of this value.

Apply the error interval to the original value ($50\text{ kg}$) to find the bounds:

The value is given to the nearest mile ($1\text{ mile}$). The error interval is half of this value:

The value is correct to the "nearest centimetre" ($1\text{ cm}$). The error interval is half of this value:

The value is measured to the "nearest gram" ($1\text{ g}$). To find the error interval, divide the degree of accuracy by $2$.

Lower bound: Subtract the error interval from the original value.

Upper bound: Add the error interval to the original value.

To find the upper bound (maximum value) of a division, you must divide the largest possible numerator by the smallest possible denominator:

• $s$ is $4.15$ to 2 decimal places.
• The upper bound for $s = 4.155$.

• $t$ is $2.516$ to 3 decimal places.
• The lower bound for $t = 2.5155$.

Rounding to 3 decimal places gives:

Using the area formula $A = \pi r^2$ with the upper bound radius:

Rounding to $3$ significant figures gives $90.4\text{ cm}^2$.

To find the maximum area, use the upper bounds of both measurements identified in Step 1:

• The formula for the area of a circle is $A = \pi r^2$.
• Use the upper bound radius identified in Step 1 to get the maximum area.

To find the minimum area, use the lower bounds of both measurements identified in Step 1.

• $I$ is correct to 2 decimal places (nearest $0.01$).
• The error interval is $0.005$.

• $R$ is correct to 3 decimal places (nearest $0.001$).
• The error interval is $0.0005$.

To maximise the product $V$, multiply the upper bounds of $I$ and $R$ together.

Looking at the fourth decimal digit ($2$), we round down.

• $g = 12.7$ (to 3 s.f., meaning the nearest $0.1$), so the error interval is $0.05$.
• Upper bound $g = 12.75$ and lower bound $g = 12.65$.
• $h = 9.294$ (to 3 d.p., meaning the nearest $0.001$), so the error interval is $0.0005$.
• Upper bound $h = 9.2945$ and lower bound $h = 9.2935$.

To find the maximum value of the fraction, use the upper bound of the numerator ($g$) and the lower bound of the denominator ($h$).

To find the minimum value of the fraction, use the lower bound of the numerator ($g$) and the upper bound of the denominator ($h$).

Compare the upper and lower bounds to see where they agree when rounded:

• $f_{\text{upper}} = 0.384216\dots$
• $f_{\text{lower}} = 0.382665\dots$
• Both numbers round to $0.38$ to 2 decimal places (or 2 significant figures). They do not agree to 3 decimal places ($0.384$ vs $0.383$).
• Therefore, the value of $f$ is $0.38$ because both bounds round to the same value at this degree of accuracy.

Divide both sides by $\pi r$:

• To find the upper bound (maximum value) for a fraction, you must divide the largest possible numerator by the smallest possible denominator.
• Therefore, we need the upper bound of $A$ and the lower bound of $r$.

• $A = 220$ to 3 significant figures (this means it is measured to the nearest $1$). The error interval is $0.5$.
• Upper bound $A = 220 + 0.5 = 220.5$
• $r = 8$ to 1 significant figure (measured to the nearest $1$). The error interval is $0.5$.
• Lower bound $r = 8 - 0.5 = 7.5$

Subtract $2as$ from both sides:

• To find the upper bound (maximum value) for $u$, you need to maximise the positive term ($v^2$) and minimise the term being subtracted ($2as$).
• Therefore, use the upper bound for $v$, and the lower bounds for $a$ and $s$.

• Upper bound of $v = 35.25$
• Lower bound of $a = 9.75$
• Lower bound of $s = 60.345$

The value rounds to $8.11$.

• Square both sides: $T^2 = 4\pi^2 \dfrac{l}{g}$
• Multiply by $g$ and divide by $T^2$:

• $l = 52.0$ (to 3 s.f., meaning nearest $0.1$). The error interval is $0.05$. Therefore, the upper bound is $l_{\text{upper}} = 52.05$ and the lower bound is $l_{\text{lower}} = 51.95$.
• $T = 1.45$ (to 3 s.f., meaning nearest $0.01$). The error interval is $0.005$. Therefore, the upper bound is $T_{\text{upper}} = 1.455$ and the lower bound is $T_{\text{lower}} = 1.445$.

To get the maximum value for a fraction, divide the largest numerator by the smallest denominator.

To get the minimum value for a fraction, divide the smallest numerator by the largest denominator.