Specimen Paper 2 Foundation

In 16.35 the digit after the decimal point is the tenths digit.

So the 3 is worth 3 tenths, which is $\frac{3}{10} = 0.3$

Check each number: $24 \div 1, 24 \div 3, 24 \div 6, 24 \div 12, 24 \div 24$ all give whole numbers.

$24 \div 9 = 2.66...$, so 9 is not a factor.

The last digit, 1, is in the hundredths column.

So $0.21 = \frac{21}{100}$

(a) $5f - f + 2f = 4f + 2f = 6f$

(b) Multiply the numbers: $2 \times 8 = 16$, then write the letters: $16mn$

(c) Two lots of $t^2$ make $2t^2$

(a) $18 \times 27 = 486$p $= £4.86$

$£10 - £4.86 = £5.14$

(b) The pens cost more, so she spends more and gets less change.

Take off the extra £72: $458 - 72 = 386$

Split equally: $386 \div 2 = 193$ (Michelle)

Wayne: $193 + 72 = £265$

$10\%$ of £90 is £9

$70\% = 7 \times 9 = £63$

Use a denominator of 24:

$\frac{1}{2} = \frac{12}{24}$, $\frac{3}{4} = \frac{18}{24}$, $\frac{5}{12} = \frac{10}{24}$

Order: $\frac{10}{24}, \frac{12}{24}, \frac{17}{24}, \frac{18}{24}$, i.e. $\frac{5}{12}, \frac{1}{2}, \frac{17}{24}, \frac{3}{4}$

There are 20 squares in the grid.

Shaded: 10 whole squares and 5 halves, so $10 + 2.5 = 12.5$ squares.

$\frac{12.5}{20} \times 100 = 62.5\%$

(i) Dresses only come in even sizes, so a size of 15.3 does not exist.

(ii) The mode is 14. It tells the manager which size sells most often, so which to stock more of.

Week 1: $65 \times 0.8 = £52$

Week 2: $52 - 10 = £42$

£42 is more than £40, so Maria does not have enough.

$3.6$ m $= 360$ cm

$360 \div 30 = 12$ cm

(a) 12 | 3 5 9

13 | 0 3 3 5 7 8

14 | 7 7 8 9

15 | 0 1

Key: 12 | 3 means 123 cm

(b) Heights over 140: 147, 147, 148, 149, 150, 151, which is 6 children.

$P = \frac{6}{15} = \frac{2}{5}$

(a) C is on the $y$-axis at $-1$: $(0, -1)$

(b) From B to A is 3 right, 1 up. Do the same from C: D is at $(3, 0)$

(c) Midpoint of B$(-1, 2)$ and C$(0, -1)$: $\left(\frac{-1+0}{2}, \frac{2+(-1)}{2}\right) = (-0.5, 0.5)$

(a) $210 \div 5 \times 4 = 168$ cm

(b) $(6 - 2.5)^2 = 3.5^2 = 12.25$

$\sqrt{9.34 - 2.58} = \sqrt{6.76} = 2.6$

$12.25 + 2.6 = 14.85$

(a) $4c = 11 - 5 = 6$, so $c = 6 \div 4 = 1.5$

(b) $e + 7 = 20 \div 5 = 4$, so $e = 4 - 7 = -3$

(c) Multiply the powers: $(m^3)^2 = m^{3 \times 2} = m^6$

Angle $A = 180 - 90 - 22 = 68°$ (angles in a triangle add to $180°$)

Triangle APQ is isosceles since AP = AQ, so angles AQP and APQ are equal (base angles of an isosceles triangle).

Angle $AQP = (180 - 68) \div 2 = 56°$

Scale factor: $72 \div 16 = 4.5$

Butter: $240 \times 4.5 = 1080$ g

Flour: $350 \times 4.5 = 1575$ g

Sugar: $100 \times 4.5 = 450$ g

Mincemeat: $280 \times 4.5 = 1260$ g

(a) $\frac{2}{5} = \frac{4}{10}$ and $\frac{1}{2} = \frac{5}{10}$, so the sum is $\frac{9}{10}$. Lethna changed the denominators but left the numerators alone.

(b) $1\frac{1}{2} \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = \frac{48}{6} = 8$. Dave split the mixed numbers apart, which does not work for multiplying.

$w = 3t + 11$

Subtract 11: $w - 11 = 3t$

Divide by 3: $t = \frac{w - 11}{3}$

Paris: $1980 \div 1.34 = £1477.61$

New York: $2250 \div 1.52 = £1480.26$

London: £1480

The cheapest is Jardins of Paris.

Readings: 110, 84, 78, 94, 90, 120

Total $= 576$

Mean $= 576 \div 6 = 96$

The mean equals 96, so the target was met.

(a) The highest frequency is 10, in the class $160 < h \le 170$

(b) The points are plotted at the ends of the classes instead of the middles (135, 145, ...).

The last point is joined back to the first, which a frequency polygon should not do.

(a) 36 minutes $= 0.6$ hours, so distance $= 15 \times 0.6 = 9$ km.

Plot (9 am, 0) to (9.36 am, 9), then on to (10.45 am, 17).

(b) 15 minutes $= 0.25$ hours

Distance $= 18 \times 0.25 = 4.5$ km

Multiplier for 4% growth is 1.04

$7500 \times 1.04^2 = 7500 \times 1.0816 = £8112$

Let Becky have $x$. Chris has $2x$ and Dan has $2x + 7$.

$x + 2x + 2x + 7 = 57$, so $5x = 50$ and $x = 10$

Becky 10, Chris 20, Dan 27

Equal shares would be $57 \div 3 = 19$ each, but Chris keeps his 20, so Dan is wrong.

Rectangle: $19 \times 16 = 304$ cm²

Circle radius 8: half-circle area $= \frac{1}{2} \pi \times 8^2 = 100.53$ cm²

Shaded: $304 - 100.53 = 203.47$ cm²

$\frac{203.47}{304} \times 100 = 66.9\%$

Draw a right-angled triangle: across $9 - 7 = 2$, down 5

$CD = \sqrt{5^2 + 2^2} = \sqrt{29} = 5.385$ cm

Perimeter $= 9 + 7 + 5 + 5.385 = 26.385$ cm

Side of square $= 26.385 \div 4 = 6.596$ cm

Area $= 6.596^2 = 43.5$ cm²