Rounding, Estimation and Bounds: past paper questions & worked solutions

Correct to \(2\) decimal places means the nearest \(0.01\) kg.

Rounding to the nearest kilogram allows half a kilogram below the stated value.

The rounding interval is \(1177.5\le w<1178.5\); the upper endpoint is the upper bound, not an included weight.

Round each number to one significant figure:

Round each number to one significant figure:

The weight \(475\) g correct to the nearest g has lower bound

The height \(120\) cm correct to the nearest \(10\) cm has upper bound

The weight \(475\) g correct to the nearest g has lower bound

The height \(120\) cm correct to the nearest \(10\) cm has upper bound

The length is \(142.8\) m correct to 1 decimal place.

Half of \(0.1\) is \(0.05\).

The value is correct to 1 decimal place, so the half-unit is \(0.05\).

The length is \(1.4\) m correct to one decimal place.

Half of \(0.1\) is \(0.05\).

The side length is \(18\) mm correct to the nearest \(0.5\) mm.

Half of \(0.5\) mm is \(0.25\) mm.

Use half of each rounding unit to obtain the bounds.

Minimize the positive numerator by using the lower bound of d and upper bound of e; maximize the positive denominator.

Round the boundary value to the requested accuracy.

For the upper bound, use the upper bounds of \(x\) and \(y\), and the lower bound of \(w\).

Correct to 3 significant figures,

For the lower bound, use the lower bound of \(c\) and make the denominator as large as possible.

For the upper bound,

Correct to 3 significant figures,

For the upper bound, use the upper bounds of all three values.

For the upper bound, make the numerator large and the denominator small.

For the lower bound, use the lower bound of \(w\), and make \(\frac{x^2}{y}\) as large as possible.

Using the sine rule,

For the upper bound of \(YZ\):

Correct to one decimal place,

For the upper bound, make the numerator large and the denominator small.

For the upper bound, use the upper bound of \(v\), the lower bound of \(u\), and the lower bound of \(t\).

For the lower bound of \(k\), use the lower bound of \(t\) and make the denominator as large as possible.

For the upper bound of \(P\), make the numerator as large as possible and the denominator as small as possible.

For the upper bound of the shaded area, use the upper bound of the outer rectangle and subtract the lower bound of the cut-out rectangle.

Outer rectangle:

Cut-out rectangle:

Bounds:

Lower bound:

Upper bound:

So

Both bounds round to \(90\) to the nearest \(10\).

For \(17.8\) correct to 1 decimal place, the lower bound is

For \(145\) correct to the nearest \(5\) grams, the upper bound is

For the estimate,

Round to 1 significant figure.

For the upper bound of density, use the upper bound of the mass and the lower bound of the volume.

The radius is \(2.8\) cm correct to 1 decimal place, so the lower bound is

\(r=2.75\)

\(M=260\) correct to 2 significant figures, so the upper bound is

\(M=265\)

The mass is \(M\pi\) and the volume of the sphere is \(\dfrac{4}{3}\pi r^3\).

So

\(\text{density}=\dfrac{M\pi}{\frac{4}{3}\pi r^3}=\dfrac{3M}{4r^3}\)

Upper bound:

\(\dfrac{3(265)}{4(2.75)^3}=9.5567...\)

The fence length \(50\) m correct to the nearest \(5\) m has bounds

The panel length \(1.3\) m correct to the nearest \(10\) cm has bounds

Smallest possible number of panels:

Greatest possible number of panels:

Difference in number of panels:

Greatest difference in cost:

To show Kaidan's average speed could be greater, compare Kaidan's greatest possible speed with Sonja's smallest possible speed.

For Kaidan:

For Sonja:

Since

Kaidan's average speed could have been greater.

For an upper bound for \(b\), use the upper bound of \(c\):

\(c=11.55\)

Use the lower bounds of \(x\) and \(y\), because this makes angle \(B\) as large as possible and makes \(\sin y\) as small as possible.

\(x=79.5^\circ,\quad y=74.5^\circ\)

So

\(\angle B=180^\circ-79.5^\circ-74.5^\circ=26^\circ\)

Using the sine rule,

\(\dfrac{b}{\sin 26^\circ}=\dfrac{11.55}{\sin 74.5^\circ}\)

\(b=\dfrac{11.55\sin 26^\circ}{\sin 74.5^\circ}=5.254...\)

The weight \(4.3\) kg correct to 2 significant figures has bounds

So the upper bound is \(4.35\) kg and the lower bound is \(4.25\) kg.

For

the upper bound uses the upper bound of \(e\) and the lower bound of \(f\).

Martin's greatest possible distance is \(82.5\) km and his least possible time is \(2.65\) hours.

Lucia's least possible distance is \(32.5\) km and her greatest possible time is \(1.05\) hours.

So Martin's speed could have been greater than Lucia's speed.

To get the upper bound for density, use the upper bound for mass and the lower bound for volume.

The mass is \(5\) kg correct to the nearest \(50\) g.

Half of this is \(0.025\) kg, so

The volume is \(1.84\times 10^{-3}=0.00184\) m\(^3\), correct to 3 significant figures.

So the upper bound for the density is

Correct to 1 decimal place,

Use half a rounding unit on each side of the stated value.

For \(v=2s/t-u\), the lower limiting value is \[v_L=\frac{2(99.5)}{1.685}-0.55=117.550890\ldots\text{ m/s}.\]

Decrease the numerator, increase the positive denominator and subtract the largest u.

The upper limiting value is \[v_U=\frac{2(100.5)}{1.675}-0.45=119.55\text{ m/s}.\]

Increase the numerator, decrease the denominator and subtract the smallest u.

Both limiting values round to \(120\text{ m/s}\) to 2 significant figures. They do not round to the same integer, so \[v=120\text{ m/s}\quad(2\text{ s.f.}).\]

This is the greatest common significant-figure accuracy justified throughout the interval.

\(99.5\le s<100.5\), \(1.675\le t<1.685\), \(0.45\le u<0.55\)

Big \(s\), small \(t\), small \(u\): \[v_{\max}=2\left(\frac{100.5}{1.675}\right)-0.45=119.55\]

Small \(s\), big \(t\), big \(u\): \[v_{\min}=2\left(\frac{99.5}{1.685}\right)-0.55=117.55\ldots\]

Both bounds round to \(120\) to 2 s.f. (they differ to 3 s.f.), so \(v=120\).

Bounds:

Lower bound:

Upper bound:

So

Both bounds round to \(3.6\) to 1 decimal place.