Mathematics 0580 | Paper 3 Mock 01 | Calculator (Core)

The glass covers the circular top, so the amount needed is the area of a circle, \( A = \pi r^{2} \), with the radius given directly as \( 0.6 \) m.

Each square metre of glass costs \( \$85 \), so the cost is the area multiplied by the price per square metre:

Correct to the nearest cent, the glass costs \( \$96.13 \) [A1].

Square the radius before multiplying by \( \pi \), since \( \pi \times 0.6^{2} \) is not the same as \( (\pi \times 0.6)^{2} \). The units also work in your favour as a check: an area in square metres multiplied by dollars for each square metre leaves dollars. Keep the unrounded area on the calculator, because rounding it to \( 1.13 \) first would give \( \$96.05 \) and lose the accuracy mark.

When \( BC \) is parallel to \( DE \), the line \( BC \) cuts triangle \( ADE \) so that angle \( ABC \) equals angle \( ADE \) and angle \( ACB \) equals angle \( AED \) (corresponding angles on parallel lines), and angle \( A \) is shared. Three equal pairs of angles means triangle \( ABC \) is similar to triangle \( ADE \).

The critical point is that the enlargement compares \( AB \) with the whole side \( AD \), not with the part \( BD \):

1. \( AD = AB + BD = 5 + 7.5 = 12.5 \) cm.
2. \( \text{scale factor} = \dfrac{AD}{AB} = \dfrac{12.5}{5} = 2.5 \) [M1]
3. \( DE = BC \times 2.5 = 6 \times 2.5 = 15 \)

The length of \( DE \) is \( 15 \) cm. [A1]

The most frequent mistake is using \( 7.5 \div 5 = 1.5 \) and answering \( 9 \) cm. That compares a part of the large triangle with the small triangle, which are not corresponding sides. Always rebuild the full side of the larger triangle first.

Finding a percentage of an amount means finding that many hundredths of it. Here you need \(35\%\) of the \(600\) students shown by the whole pie chart.

A useful check without a calculator: \(10\%\) of \(600\) is \(60\), so \(30\%\) is \(180\); \(5\%\) is half of \(10\%\), which is \(30\); and \(180+30=210\).

So \(210\) students walk to school [B1]

The pie chart itself is not needed for this part, because the percentage is given in words. Do not try to measure the sector angle; use the stated \(35\%\) and the stated total of \(600\).

The number line gives the midnight temperature as \(-7^\circ\)C, a point \(7\) units to the left of zero. A rise in temperature moves to the right along the line, so add the rise:

\[ -7+19=12 \]

The temperature at noon is \(12^\circ\)C. [B1]

Counting it out makes the arithmetic clear. Moving \(7\) units to the right from \(-7\) reaches \(0\), which uses up \(7\) of the \(19\) degrees. The remaining \(19-7=12\) degrees carry the temperature to \(12^\circ\)C above zero.

The common error is to add the digits and ignore the sign, giving \(26^\circ\)C, or to subtract and give \(-26^\circ\)C. A rise always moves the value in the positive direction, and here the rise is larger than the size of the starting temperature, so the result must end up above zero.

The bar chart gives \(42\) goals in season \(1\) and \(60\) goals in season \(5\). Percentage change is always measured against the original value, which here is the season \(1\) figure.

(a) Using \(\text{percentage increase}=\frac{\text{increase}}{\text{original}}\times 100\):

\[ \frac{60-42}{42}\times 100 \] [M1]

\[ =\frac{18}{42}\times 100=42.857\ldots \]

Correct to \(1\) decimal place the increase is \(42.9\%\). [A1]

Dividing by \(60\) instead of \(42\) gives \(30\%\) and is the standard error here. The question asks how much the goals grew relative to where they started, so \(42\) is the denominator.

(b) Scoring \(15\%\) fewer than season \(5\) means keeping \(100\%-15\%=85\%\) of \(60\), so the multiplier is \(0.85\).

\[ 60\times 0.85 \] [M1]

\[ =51 \]

The team scores \(51\) goals in season \(6\). [A1]

Working out \(15\%\) of \(60=9\) and subtracting to get \(60-9=51\) is equally acceptable. The single multiplier is quicker and is essential once repeated changes appear.

The conversion graph is a straight line through the origin, so kilometres and miles are in direct proportion. Reading across the graph converts either way, and once the multiplier is known any distance can be converted by calculation.

(a) Read up from \( 60 \) on the kilometres axis to the line, then across to the miles axis.

\( 37.5 \) miles [B1]

(b) Now work in the opposite direction: start at \( 30 \) on the miles axis, read across to the line, then down to the kilometres axis.

\( 48 \) km [B1]

(c) \( 120 \) km is beyond the range of the graph, so use the multiplier instead of extending the line. From part (a), \( 1 \) km \( =\frac{37.5}{60}=0.625 \) miles.

\[ 120\times 0.625 \] [M1]

\[ =75 \text{ miles} \] [A1]

A check without the multiplier: \( 120 \) km is double \( 60 \) km, so it is double \( 37.5 \) miles, which is \( 75 \) miles. Miles are longer than kilometres, so the number of miles must always come out smaller than the number of kilometres.

This question tests reading the type of correlation from a scatter diagram.

Correlation describes how two quantities change together. If the points fall as you move to the right, so that a larger value of one quantity goes with a smaller value of the other, the correlation is negative. If the points rise to the right it is positive, and if there is no clear pattern there is no correlation.

Here the horizontal axis is the age of the car and the vertical axis is its value. Older cars have had more use and wear, so they sell for less, and the points slope downwards from left to right. The correlation is therefore negative [B1]

Notice that "negative" describes the direction of the trend, not the values themselves; the car values are all positive numbers. Also, correlation here shows a strong association between age and value, but you should describe the trend rather than claim that age is the only cause of the fall in price.

A point lies on the curve \(y = \frac{8}{x}\) exactly when its coordinates satisfy that equation, so the value of \(y\) at \(x = 3.2\) can be calculated directly rather than read off the grid. Calculating is more accurate than reading, since a graph can usually only be read to about half a small square.

If the division is not obvious, clear the decimal by multiplying top and bottom by \(10\): \(\frac{8}{3.2} = \frac{80}{32} = \frac{5}{2} = 2.5\).

This is a reciprocal (inverse proportion) curve: as \(x\) increases, \(y\) decreases, and their product is always \(8\). Check the answer with that property: \(3.2 \times 2.5 = 8\). The common error is to multiply instead of divide, giving \(25.6\), which fails the product check immediately.

When powers of the same base are multiplied, the indices are added: \(a^m \times a^n = a^{m+n}\).

Applying this with base \(x\): \[ x^5 \times x^3 = x^{5+3} = x^8 \] [B1]

The rule comes from counting factors. Writing the powers out in full, \(x^5\) is five \(x\) terms multiplied together and \(x^3\) is three more, so altogether there are \(5 + 3 = 8\) factors of \(x\), which is \(x^8\).

The usual error is to multiply the indices and write \(x^{15}\). Multiplying indices belongs to a different rule, raising a power to a power, as in \((x^5)^3 = x^{15}\). Adding is for multiplication of powers, multiplying is for a power of a power, and both rules require the bases to be identical.

(a) Each notebook costs \(\$2.40\), so \(n\) notebooks cost \(2.40n\) dollars. Amir cannot spend more than he has, though he may spend it all exactly, so the cost is at most \(\$20\):

\[ 2.40n\le 20 \] [B1]

(b) Divide both sides by \(2.40\):

\[ n\le\frac{20}{2.40}=8.33\ldots \] [M1]

The number of notebooks must be a whole number, and \(n\) has to stay below \(8.33\ldots\), so the largest possible value is

\[ n=8 \] [A1]

(c) Eight notebooks cost \(8\times 2.40=\$19.20\), so the money left is

\[ 20-19.20=\$0.80 \] [B1]

The important step in part (b) is rounding down rather than to the nearest whole number. Rounding \(8.33\ldots\) to \(8\) happens to agree with normal rounding, but the reason is the context: \(9\) notebooks would cost \(\$21.60\), which Amir cannot afford. Part (c) then confirms that \(8\) is right, since the money left over is less than the price of one more notebook.

Two rays meeting at a point create two angles: the one between them and the reflex angle going the other way round. Together they make one complete turn of \(360^\circ\).

The angle between the rays is \(78.4^\circ\), so the reflex angle is

\[360-78.4=281.6^\circ\]

giving \(281.6^\circ\) [B1].

Check that the answer really is reflex: \(281.6^\circ\) lies between \(180^\circ\) and \(360^\circ\), so it qualifies. Subtracting from \(180^\circ\) instead would give \(101.6^\circ\), which is obtuse rather than reflex and so cannot be the answer to this question.

A decagon has \( 10 \) sides. In a regular polygon the exterior angles are all equal and add up to \( 360^\circ \), so each one is

The order of rotational symmetry is the number of positions in one full turn in which the shape looks exactly the same as it started. Turning a regular decagon by one exterior angle, \( 36^\circ \), moves each vertex onto the next and leaves the shape looking identical, and \( 360 \div 36 = 10 \) such turns fit into a full revolution. The order of rotational symmetry is therefore \( 10 \). [B1]

This is a general pattern worth remembering: a regular \( n \)-sided polygon has \( n \) lines of symmetry and rotational symmetry of order \( n \), and its exterior angle is \( \frac{360}{n} \) degrees. Do not confuse the exterior angle of \( 36^\circ \) with the interior angle, which here is \( 180 - 36 = 144^\circ \).

Both parts test one idea: find a quantity first, then take a percentage of it. The area of any triangle is half the base times the perpendicular height, and the perpendicular height must be the one measured at right angles to the base you use. Here the stem states the base as \(46\) m and the perpendicular height as \(28\) m, so those two values go straight into the formula.

(a) Substituting into \(A=\frac{1}{2}bh\):

\[ A=\frac{1}{2}\times 46\times 28 \] [M1]

\(46\times 28=1288\), and half of that is \(644\), so the area is \(644\) m\(^2\). [A1]

Note the unit: two lengths in metres multiply to an area in square metres.

(b) "Maize is planted on \(62.5\%\) of the field" means you take \(62.5\%\) of the answer to part (a). Convert the percentage to a decimal multiplier by dividing by \(100\): \(62.5\div 100=0.625\).

\[ 644\times 0.625 \] [M1]

\[ =402.5 \]

The area planted with maize is \(402.5\) m\(^2\). [A1]

An equivalent route is \(\frac{644}{100}\times 62.5\), or noticing that \(62.5\%=\frac{5}{8}\) so \(644\times\frac{5}{8}=402.5\). All three give the same value and earn the same marks. The mark for the method here follows through from your own area, so an arithmetic slip in (a) does not cost you the method mark in (b).

A common error is to use a sloping side of the triangle instead of the perpendicular height. Only the height at right angles to the chosen base works in \(\frac{1}{2}bh\).

The right angle is at \( Y \), and the known angle is at \( X \). Relative to the \( 37^\circ \) angle, \( YZ \) is the opposite side and \( XY = 12.5 \) cm is the adjacent side, so tangent is the correct ratio.

Rearranging to make \( YZ \) the subject:

To \( 3 \) significant figures, \( YZ = 9.42 \) cm. [A1]

Because \( 37^\circ \) is less than \( 45^\circ \), \( \tan 37^\circ \lt 1 \), so \( YZ \) must be shorter than \( XY \), and \( 9.42 \lt 12.5 \) as expected. Dividing by \( \tan 37^\circ \) instead of multiplying would give \( 16.6 \) cm, which fails that check.

A net is the flat shape that folds up to make a solid, so the faces on the net are exactly the faces of the finished solid. Every face here is a rectangle, and a closed solid whose faces are all rectangles has six of them, meeting three at each corner.

The solid is a cuboid. [B1]

A cuboid has \( 6 \) rectangular faces, \( 12 \) edges and \( 8 \) vertices, with opposite faces identical and parallel. A cube is the special case in which all six rectangles happen to be squares, so "cube" is only correct if the net shows squares. "Rectangle" on its own is not accepted here, because that names a flat two-dimensional shape rather than the solid; questions asking for the name of a solid want a three-dimensional name such as cuboid, prism, pyramid, cylinder or cone.

An estimate question asks you to round first and then calculate, so that the arithmetic can be done mentally. Correct to \(1\) significant figure the two measurements of the rectangular sheet become \(7\) cm and \(4\) cm.

\[\text{Area}\approx 7\times 4\] [M1]

\[=28\ \text{cm}^{2}\] [A1]

The area of a rectangle is length \(\times\) width, and multiplying centimetres by centimetres gives square centimetres, so the unit must be cm\(^{2}\).

Two things lose marks here: working out the exact area with a calculator and then rounding the answer (the rounding must come first, and the question says to write each measurement to \(1\) significant figure), and giving the perimeter instead of the area. Estimation is being tested as a check on calculator work, so the rounded values should be visible in the working.

(a) The points rise from lower left to upper right: more hours of sunshine goes with more visitors. That is positive correlation [B1]. Correlation describes the direction of the trend, so the answer is a type, not a number.

(b) Use the line of best fit, not an individual plotted day. Find \(7\) on the hours axis, go vertically up to the line, then horizontally across to the visitors axis [M1]. The reading is

Any value from \(5.2\) to \(5.6\) is accepted, since a hand-drawn line and a reading between gridlines carry a small tolerance. The axis is already in hundreds, so the answer is left as \(5.4\) hundred rather than converted.

(c) Two separate reasons are needed:

• \(15\) hours is outside the range of the data collected, so using the line there is extrapolation rather than reading within the evidence [B1].
• The trend may not continue beyond the days that were recorded, for example the beach has a limited capacity, so the straight-line pattern cannot be assumed to hold [B1].

Saying "the line does not reach \(15\)" twice in different words scores once only. Give one reason about the data range and one about the trend not continuing.

The table is read as: \(1\) dollar buys \(0.79\) pounds, \(148\) yen, \(0.93\) euros or \(18.6\) rand. That single fact decides both parts. Going from dollars into a foreign currency you multiply by the rate; coming back from a foreign currency into dollars you divide by the rate.

(a) Dollars to pounds, so multiply by \(0.79\):

\[ 340\times 0.79 \] [M1]

\[ =268.6 \]

\(\$340\) is \(268.60\) pounds. [A1] Money is written to two decimal places, so \(268.60\) rather than \(268.6\).

(b) Yen to dollars, so divide by \(148\):

\[ 5550\div 148 \] [M1]

\[ =37.5 \]

\(5550\) yen is \(\$37.50\). [A1]

If you are unsure which way round to go, sanity-check the size of the answer. Since one dollar is worth \(148\) yen, a few thousand yen must be only a few tens of dollars, so the answer has to be much smaller than \(5550\). Multiplying would have given \(821\,400\), which is clearly wrong.

The shape is a rectangle with a semicircle sitting on the top edge. That top edge of the rectangle is inside the shape, so it counts for the area but not for the perimeter.

(a) Going round the outline: the bottom of the rectangle is \(20\) cm, the two vertical sides are \(14\) cm each, and the curved edge is half the circumference of a circle of diameter \(20\) cm.

\[ P=20+2\times 14+\frac{\pi\times 20}{2} \] [M1]

\[ P=20+28+31.415...=79.415... \]

Correct to 3 significant figures, the perimeter is \(79.4\) cm [A1] (\(79.415...\) is accepted).

(b) For the area, add the rectangle and the semicircle. The semicircle has diameter \(20\) cm, so its radius is \(10\) cm:

\[ A=20\times 14+\frac{\pi\times 10^2}{2} \] [M1]

\[ A=280+157.07...=437.07... \]

Correct to 3 significant figures, the area is \(437\) cm\(^2\) [A1] (\(437.07...\) is accepted).

Notice how the \(20\) cm is used differently in the two parts: as a diameter in the arc length \(\frac{\pi d}{2}\), and halved to a radius of \(10\) cm in the area \(\frac{\pi r^2}{2}\). Mixing these up is the usual source of error in composite circle questions.

A bearing is measured clockwise from north and is always written with three figures. The bearing of \(A\) from \(B\) is the back bearing of the bearing of \(B\) from \(A\), and the two differ by \(180^\circ\) because the north lines at \(A\) and at \(B\) are parallel, so the two bearings are co-interior angles that add to a straight turn.

The given bearing is less than \(180^\circ\), so add \(180^\circ\):

The bearing of \(A\) from \(B\) is \(252^\circ\). [B1]

The rule to remember is: if the given bearing is under \(180^\circ\), add \(180^\circ\); if it is \(180^\circ\) or more, subtract \(180^\circ\). Either way the answer must lie between \(000^\circ\) and \(360^\circ\).

(a) The frequencies in the table total \(200\), which matches the number of people surveyed, so a probability is a frequency divided by \(200\).

\[ P(\text{juice}) = \frac{35}{200} = \frac{7}{40} \] [B1]

(b) Tea and water are mutually exclusive, since each person names only one favourite, so their frequencies add.

\[ P(\text{tea or water}) = \frac{62 + 25}{200} = \frac{87}{200} \] [B1]

(c) The proportion choosing coffee in the survey is used as an estimate of the proportion in the new group, then applied to \(500\) people.

\[ \frac{78}{200} \times 500 \] [M1]

\[ = 0.39 \times 500 = 195 \text{ people} \] [A1]

Equivalent forms such as \(0.175\) in (a) and \(0.435\) in (b) are accepted. Part (c) is an estimate: it assumes the second group has similar tastes to the first, which is exactly the assumption that makes a sample useful for prediction.

Work through the two stages in order: convert the currency first, then subtract the spending.

The rate says that \(1\) euro is worth \(\$1.08\), so each euro becomes \(1.08\) dollars and \(450\) euros become \(450\) lots of \(\$1.08\):

\[ 450\times 1.08=486 \] [M1]

Nina therefore starts with \(\$486\). Subtracting what she spends:

\[ 486-312.50=173.50 \]

She has \(\$173.50\) left. [A1]

Decide between multiplying and dividing by checking which currency is worth more. One euro buys more than one dollar, so the number of dollars must be larger than the number of euros. An answer of \(450\div 1.08=416.67\) fails that test and would convert dollars into euros instead.

Keep the working in dollars once the conversion is done. Subtracting \(\$312.50\) from the \(450\) euros mixes two different currencies and is not a valid calculation.

A scale of \(1:50000\) means every \(1\) cm on the map represents \(50\,000\) cm on the ground, so multiply the map length by \(50\,000\) and then convert to kilometres.

\[ 3.6 \times 50000 = 180000 \text{ cm} \] [M1]

Since \(1\) km \(=100\,000\) cm:

\[ \frac{180000}{100000} = 1.8 \]

The real length of the lake is \(1.8\) km. [A1]

Do the conversion in one clear step rather than converting to metres and losing track of a factor of \(10\). If you prefer metres as a staging post, \(180\,000\) cm is \(1800\) m, and \(1800\) m is \(1.8\) km, which agrees. Answers such as \(180\) km or \(0.018\) km come from mishandling that conversion, so always ask whether the size is realistic for a lake.

The area of a rectangle is length times width, so when one dimension is written as an algebraic expression the area statement becomes an equation you can solve for the unknown.

The length is \((x+5)\) cm and the width is \(8\) cm, so the area is \(8(x+5)\) cm\(^2\). Setting that equal to the given area of \(96\) cm\(^2\):

\[8(x+5)=96\]

which earns the method mark [M1]. Dividing both sides by \(8\) gives \(x+5=12\), so \(x=7\) [A1].

The perimeter needs the actual dimensions, not \(x\). With \(x=7\) the length is \(7+5=12\) cm and the width is \(8\) cm, so

\[\text{perimeter}=2(12+8)=40\text{ cm}\]

giving the final answer of \(40\) cm [A1]. A quick check: \(12\times 8=96\) cm\(^2\), which matches the stated area.

The common slip is to give the perimeter as \(2(x+5+8)=2x+26\) or to substitute \(x=7\) directly as a side length. Always convert \(x\) back into the real lengths before working out a perimeter or an area.

When dividing powers of the same base, subtract the indices: \( a^{m} \div a^{n} = a^{m-n} \).

Both terms have base \( t \), so \[ \frac{t^{15}}{t^{6}} = t^{15-6} = t^{9} \] [B1]

The rule follows from what an index means. The numerator is 15 factors of \( t \) multiplied together and the denominator is 6 factors, so 6 of the factors cancel in pairs, leaving \( 15 - 6 = 9 \) factors of \( t \), which is \( t^{9} \).

The usual errors are dividing the indices to get \( t^{2.5} \), or subtracting the wrong way round to get \( t^{-9} \). Subtraction of indices applies only when the bases are identical, as they are here.

The scale \( 1 : 12 \) means that \( 1 \) cm on the model represents \( 12 \) cm on the real statue, so the real object is found by multiplying the model measurement by \( 12 \).

1. Multiply by the scale factor: \[ 27.5 \times 12 = 330 \text{ cm} \] [M1]
2. Convert to metres, as the question requires: \( 330 \div 100 = 3.3 \) m.

The real statue is \( 3.3 \) m tall. [A1]

Because the model was measured in centimetres, the product is also in centimetres; the ratio itself has no unit and never changes the unit you are working in. Converting only at the very end keeps the arithmetic simple and avoids dividing by \( 100 \) at the wrong point.

(a) On a bar chart the height of each bar, read against the vertical scale, gives the number of goals for that team. Ordering the teams by number of goals therefore means ordering the bars by height, so read every bar against the scale rather than judging by eye alone.

The two tallest bars belong to Denby and Barton, so those two teams come first [M1]

Completing the ranking from the tallest bar down to the shortest, the teams in order starting with the most goals are Denby, Barton, Ashford, Chelford, Elmore [A1]

(b) The difference between the greatest and the least is the reading for the tallest bar minus the reading for the shortest bar, which are the bars for Denby and Elmore. Carrying out that subtraction gives

\[ 22 \]

so the difference is \( 22 \) goals [B1]

Two points decide the marks here. First, check the value of one small square on the vertical axis before reading anything; if each square is worth \( 2 \) goals rather than \( 1 \), every reading doubles. Second, the difference is a subtraction of two readings, not a count of bars or a reading of the gap between the tops of the bars by eye.

For a regular polygon the number of lines of symmetry equals the number of sides, so 6 lines of symmetry means this is a regular hexagon.

There are two standard routes to one interior angle, and either earns the method mark.

• Exterior angle first: the exterior angles sum to \(360^\circ\), so each is \(360\div 6 = 60^\circ\), and the interior angle is its supplement, \(180-360\div 6\).
• Angle sum first: the interior angles of an \(n\)-sided polygon sum to \((n-2)\times 180^\circ\), so here \((6-2)\times 180 = 720^\circ\), shared equally as \((6-2)\times 180\div 6\).

\(180-360\div 6\) or \((6-2)\times 180\div 6\) [M1]

Each interior angle is \(120^\circ\). A useful check is that \(6\times 120 = 720\), which matches the angle sum, and that \(120^\circ\) is obtuse, as it must be for any regular polygon with more than four sides.

To read a gradient off an equation, it must first be arranged with \(y\) alone on the left, because only then does the number multiplying \(x\) equal the gradient.

1. Subtract \(6x\) from both sides: \[ 2y = 10 - 6x \]
2. Divide every term by \(2\): \[ y = 5 - 3x \] [M1]

Written in the standard order this is \(y = -3x + 5\), so

\[ \text{Gradient} = -3 \] [A1]

Two points are worth stressing. Every term must be divided by \(2\), not just the \(x\) term. And the gradient is negative, meaning the line falls as \(x\) increases; reading \(3\) from the original \(6x\) without rearranging is the usual mistake.

(a) The four numbers are given as fractions, a decimal and a percentage, so convert them all to decimals before comparing.

• \( \frac{13}{20} = 13 \div 20 = 0.65 \)
• \( 0.66 \) is already a decimal
• \( 64.5\% = 64.5 \div 100 = 0.645 \)
• \( \frac{2}{3} = 2 \div 3 = 0.6666\ldots \)

Writing each number as a decimal is the method step [M1]

Padding to three decimal places gives \( 0.650 \), \( 0.660 \), \( 0.645 \) and \( 0.666\ldots \), so the order from smallest is \( 0.645 \lt 0.650 \lt 0.660 \lt 0.666\ldots \).

Back in the original forms, starting with the smallest: \( 64.5\% \), \( \frac{13}{20} \), \( 0.66 \), \( \frac{2}{3} \) [A1]

(b) \( \frac{2}{3} = 0.6666\ldots \). The first three significant figures are \( 6, 6, 6 \) and the next digit is \( 6 \), which is \( 5 \) or more, so the last one rounds up: \( 0.667 \) [B1]

(c) From the conversions, \( \frac{13}{20} = 0.65 \) and \( 0.65 \lt 0.66 \), so

\[ \frac{13}{20} \lt 0.66 \] [B1]

A frequent error in part (b) is writing \( 0.666 \) by truncating the recurring decimal instead of rounding it. Rounding looks at the first digit discarded, and here that digit is a \( 6 \), so the final \( 6 \) becomes a \( 7 \). A second error is converting \( 64.5\% \) to \( 0.645 \) but then reading it as larger than \( 0.65 \) because it has more digits; compare \( 0.645 \) with \( 0.650 \) at equal length.