Fractions, Decimals & Percentages

The sum of \(\dfrac{1}{9} + \dfrac{1}{11}\) is \(\dfrac{11}{99} + \dfrac{9}{99}\), that is \(\dfrac{20}{99}\).

Now 20 and 99 do not have any factors in common except 1. So the fraction cannot be simplified.
In option A, we may write 0.10 as the fraction \(\tfrac{10}{100}\), whose denominator is a power of 10. The same is true of options B, C and D. Therefore, when simplified, none of these fractions can have 99 as a denominator.
It is left to the reader to confirm, by division, that \(\dfrac{20}{99} = 0.\dot{2}\dot{0}\).
An alternative argument follows.
Let \(x = 0.\dot{2}\dot{0} = 0.202020....\) Then \(100x = 20.202020...\) .
Subtracting the first equation from the second gives \(99x = 20\).
So \(0.\dot{2}\dot{0} = \dfrac{20}{99}\).

Werner’s salary is 20% of his boss’s salary. Therefore his boss’s salary is \(\tfrac{100}{20} = 5\) times Werner’s salary. Hence his boss’s salary is 500% of Werner’s salary and so is 400% larger.

Since three fifths of the class swim, three fifths of the class dance and no-one does neither, the fraction of the class who do both is \(\tfrac{3}{5} + \tfrac{3}{5} - 1 = \tfrac{1}{5}\). Hence, since 5 pupils do both, the number of pupils in the class is 5 × 5 = 25.

The value of \(\dfrac{2468 \times 2468}{2468 + 2468}\) is \(\dfrac{2468 \times 2468}{2 \times 2468} = \dfrac{2468}{2}\).

Hence the correct answer is 1234.

The triangle is isosceles so has a pair of equal sides. Since the two unknown sides are not equal, they cannot be the pair of equal sides, and hence the 20 cm side must be one of the pair of equal sides. The base is then two-fifths of 20 cm, namely 8 cm. The perimeter is 20 + 20 + 8 = 48 cm.

We choose units so the outer square has side-length 4, and hence area 16.
The unshaded area of the square consists of two congruent triangles of base 3 and height 2.
So the area of each unshaded triangle is \(\tfrac{1}{2} \times 3 \times 2 = 3\).
Hence the area of the shaded part of the square is \(16 - 2 \times 3 = 10\).
Therefore the fraction of the square which is shaded is \(\tfrac{10}{16} = \tfrac{5}{8}\).

The required difference is 25% of (£37 - £17) = 25% of £20.
Hence the difference between 25% of £37 and 25% of £17 is £5.

The shaded square in the lower right corner of the large square is \(\tfrac{1}{4}\) of the large square. The shaded triangle is half of \(\tfrac{1}{4}\) of the large square. Hence it is \(\tfrac{1}{8}\) of the large square. The two small shaded squares in the upper left corner together are half of \(\tfrac{1}{4}\), or \(\tfrac{1}{8}\), of the large square. Therefore the fraction of the large square that is shaded is \(\tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{8} = \tfrac{1}{2}\).

The values of the fractions shown are \(\tfrac{13}{3} = 4\tfrac{1}{3}\), \(\tfrac{8}{8} = 1\), \(\tfrac{8}{8} = 1\), \(\tfrac{11}{5} = 2\tfrac{1}{5}\) and \(\tfrac{3}{13}\). Hence the fraction which has the largest value is \(\tfrac{8+5}{3}\).

Note that \(\dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{10} > \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}\).

Similarly \(\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} > \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6}\). Also \(\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} > \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3} + \dfrac{1}{4}\).

Finally, \(\dfrac{1}{3} + \dfrac{1}{4} > \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}\). Therefore the expression with the largest value is that in option E.

From the equation \(\dfrac{x}{y} = \dfrac{2}{3}\) we deduce that \(3x = 2y\). We are given that \(x + y = \dfrac{2}{3}\), so \(3x + 3y = 2\).

Substituting for \(3x\) gives \(2y + 3y = 2\), so \(y = \dfrac{2}{5}\).

Therefore \(x - y = x + y - 2y = \dfrac{2}{3} - \dfrac{4}{5} = \dfrac{10}{15} - \dfrac{12}{15} = -\dfrac{2}{15}\).

Let the original heights of the first and second candles be \(x\) and \(y\) respectively. Since the first candle lasts six hours, after three hours its height is \(\tfrac{1}{2}x\). Similarly, since the second candle lasts eight hours, after three hours its height is \(\tfrac{5}{8}y\). After three hours the two candles have the same height and hence \(\tfrac{1}{2}x = \tfrac{5}{8}y\). Therefore \(\tfrac{x}{y} = \tfrac{5}{4}\) and hence the ratio of their original heights is 5 : 4.

It may be shown that \(0.\dot{1} = \dfrac{1}{9}\). So \(0.\dot{8} = \dfrac{8}{9}\) and \(0.0\dot{7} = \dfrac{7}{90}\).

Therefore \(0.\dot{8} + 0.0\dot{7} = \dfrac{8}{9} + \dfrac{7}{90} = \dfrac{80}{90} + \dfrac{7}{90} = \dfrac{87}{90} = \dfrac{29}{30}\). Now \(\dfrac{29}{3} = 9 + \dfrac{2}{3} = 9 + 0.\dot{6} = 9.\dot{6}\).
So \(\dfrac{29}{30} = 0.9\dot{6}\). Hence \(0.\dot{8} + 0.0\dot{7} = 0.9\dot{6}\).

(It is left as an exercise to the reader to prove the assertion in the first sentence.)

Michael has 24 pets. Since \(\tfrac{1}{8}\) of them are dogs, he has 3 dogs. Since \(\tfrac{3}{4}\) are not cows, \(\tfrac{1}{4}\) of them are cows and hence he has 6 cows. Similarly, since \(\tfrac{2}{3}\) are not cats, \(\tfrac{1}{3}\) of them are cats and hence he has 8 cats. Therefore, the number of kangaroos he has is 24 − 3 − 6 − 8 = 7.

The cycling and running account for \(\tfrac{3}{4} + \tfrac{1}{5} = \tfrac{15}{20} + \tfrac{4}{20} = \tfrac{19}{20}\) of the distance. So the swimming, which is 2 km, is the remaining \(\tfrac{1}{20}\) of the distance. Hence the distance is 20 × 2 = 40km.

Since Jane had a success rate of 55% after her first 20 shots, the number of times she had scored out of 20 was 0.55 × 20 = 11. Similarly, since her success rate had increased to 56% after 5 more shots, the number of times she had scored out of 25 was 0.56 × 25 = 14. Hence the number of shots she scored out of the last five was 14 − 11 = 3.

To obtain the smallest value, the numerators should be as small as possible, so 1 or 2, and the denominators should be as large as possible, so 9 or 10. The two possible candidates are \(\tfrac{1}{10} + \tfrac{2}{9}\) and \(\tfrac{1}{9} + \tfrac{2}{10}\). The first gives \(\tfrac{9}{90} + \tfrac{20}{90} = \tfrac{29}{90}\), and the second gives \(\tfrac{10}{90} + \tfrac{18}{90} = \tfrac{28}{90} = \tfrac{14}{45}\), which is smaller.

The answers to the three calculations are as follows:

\(X\)  \(0.6 \times 0.5 + 0.4 = 0.3 + 0.4 = 0.7\);
\(Y\)  \(0.6 \times 0.5 \div 0.4 = 0.3 \div 0.4 = 3 \div 4 = 0.75\);
\(Z\)  \(0.6 \times 0.5 \times 0.4 = 0.3 \times 0.4 = 0.12\).
So the correct order is \(YXZ\).

The largest grey square is a quarter of the large square. The smaller grey squares are each one ninth of the size of the largest grey square. Hence the fraction of the large square which is shaded is \(\tfrac{1}{4} + \tfrac{7}{9} \times \tfrac{1}{4} = \tfrac{1}{4} \times \left(1 + \tfrac{7}{9}\right) = \tfrac{1}{4} \times \tfrac{16}{9} = \tfrac{4}{9}\)

First note that three quarters of the UK red squirrels live in Scotland. So one quarter of the UK red squirrels live in the remainder of the UK and the number of such squirrels is therefore one third of the number of red squirrels living in Scotland, which is \(\dfrac{1}{3} \times 120\,000 = 40\,000\).

So the required difference equals 120 000 − 40 000 = 80 000.

The value of 2019 tenths is \(2019 \times \tfrac{1}{10} = 201.9\)

Let the amount Chloe spent be £\(x\). Therefore Amy spent £\(1.6x\) and Becky spent £\(0.15x\). The total amount spent is £55 and hence \(x + 1.6x + 0.15x = 55\). Therefore \(2.75x = 55\), which has solution \(x = 20\). Hence Amy spent £(1.6 × 20) = £32.

Let \(D\) be the amount that Peter's dad gave, and \(B\) the amount his elder brother gave.
Then \(D = \tfrac{1}{2}(B + 10) \ldots\) (1) and \(B = \tfrac{1}{3}(D + 10) \ldots\) (2).
Substitute (1) into (2) to get \(B = \tfrac{1}{3}\left(\tfrac{1}{2}(B + 10) + 10\right) = \tfrac{1}{6}B + 5\).
Hence \(\tfrac{5}{6}B = 5\), so \(B = 6\) and \(D = \tfrac{1}{2}(6 + 10) = 8\). This gives a total of 6 + 8 + 10 = 24. So the book cost 24 euros.

\(\dfrac{10}{9} + \dfrac{9}{10} = \dfrac{100 + 81}{90} = \dfrac{181}{90} = 2 + \dfrac{1}{90} = 2.0\dot{1}\).

English is taken by 35% of language students, so 65% of language students don’t take English. These are stated as 13% of the overall university population. So 1% of the overall university population is the same as 5% of the language students. Hence 20% of the university population is 100% of the language students.

We are given that \(\dfrac{3}{10} < \dfrac{n}{20} < \dfrac{2}{5}\). Therefore \(\dfrac{6}{20} < \dfrac{n}{20} < \dfrac{8}{20}\). So, as \(n\) is an integer, \(n = 7\).

2018% of 25 is the same as 25% of 2018 (because they are both equal to \(\dfrac{25}{100} \times 2018\)), so their sum is equal to 50% of 2018, which is 1009.

The required fraction is \(\dfrac{26}{206} = \dfrac{13}{103} \approx \dfrac{13}{104} = \dfrac{1}{8}\).

When we simplify the calculation, we obtain \(\dfrac{3 \times 2018}{4 \times 2018}\) which can then be cancelled down to give \(\dfrac{3}{4}\).

The product may be written \(\sqrt{\frac{3}{2}} \times \sqrt{\frac{4}{3}} \times \sqrt{\frac{5}{4}} \times \sqrt{\frac{6}{5}} \times \ldots = \frac{\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{4}}{\sqrt{3}} \times \frac{\sqrt{5}}{\sqrt{4}} \times \frac{\sqrt{6}}{\sqrt{5}} \times \ldots\).
Notice that the numerator of each fraction is cancelled out by the denominator of the following fraction and the only terms which are not cancelled are the denominator of the first fraction and the numerator of the last fraction. So the \(n\)th term of the sequence is \(\frac{\sqrt{n + 2}}{\sqrt{2}} = \sqrt{\frac{n + 2}{2}}\). As \(n \geqslant 1\) the product is not equal to 1.
The product increases with \(n\). So the next possible integer value to consider is 2 and this does occur when \(n = 6\) as \(\sqrt{\frac{6 + 2}{2}} = \sqrt{4} = 2\).
So the smallest number of terms required for the product to be an integer is 6.

Let the capacity of the tank be \(x\) litres. Then \(30 = \frac{5x}{6} - \frac{4x}{5} = \frac{25x - 24x}{30} = \frac{x}{30}\). So \(x = 30 \times 30 = 900\).

Since 35% of the entrants were female, 65% of the entrants were male. Hence, since there were 252 more males than females, 252 people represent (65 − 35)% = 30% of the total number of entrants. Therefore the total number of entrants was (252 ÷ 30) × 100 = 840.

One sixth of the audience are adults. Therefore five sixths of the audience are children. Two fifths of the children are boys and hence three fifths of the children are girls. Therefore three fifths of five sixths of the audience are girls. Now \(\tfrac{3}{5} \times \tfrac{5}{6} = \tfrac{1}{2}\). Hence the fraction of the audience who are girls is \(\tfrac{1}{2}\).

Adam the Ant has crawled \(\tfrac{2}{3}\) of the length of the pole and so is \(\tfrac{1}{3}\) of the length of the pole from the right-hand end. Benny the Beetle has crawled \(\tfrac{3}{4}\) of the length of the pole and so is \(\tfrac{1}{4}\) of the length of the pole from the left-hand end. Hence the fraction of the length of the pole that Adam and Benny are apart is \(\left(1 - \tfrac{1}{3} - \tfrac{1}{4}\right) = \tfrac{5}{12}\).

Consider options B to E: \(1.354\dot{2} = 1.354222 \ldots\) ; \(1.35\dot{4}\dot{2} = 1.35424242 \ldots\) ; \(1.3\dot{5}4\dot{2} = 1.3542542 \ldots\) ; \(1.\dot{3}54\dot{2} = 1.35423542 \ldots\) . These are all greater than 1.3542, so option A is not correct. The other four options have the same units digit and the first four digits after the decimal point are the same in all. So the largest option is that which has the largest digit five places after the decimal point and this is \(1.3\dot{5}4\dot{2}\).

Seven-eighths of all the guests were adults, of which three-sevenths were men, so the fraction of guests who were adult women equals \(\tfrac{7}{8} \times \tfrac{4}{7} = \tfrac{1}{2}\).

If she wins five more games, then she will have won 14 out of 20, which is equivalent to \(\tfrac{7}{10}\) or 70%.

Let the area of the inner circle be \(A\). The radius of the outer circle is twice that of the inner circle, so its area is four times that of the inner circle , that is \(4A\). Therefore the region between the two circles has area \(3A\). As three of the six equal segments are shaded, the total shaded area is \(3A \div 2\). So the required fraction is \(\frac{3}{2}A \div 4A = \frac{3}{8}\).

When you evaluate correctly the left-hand side of each proposed equality in turn, you obtain 4, 2.5, 2, 1.75 and 1.6. Hence the only true equality is \(\tfrac{5}{2} = 2.5\).

When one third of the circle is shaded, the angle at the centre of the shaded sector is 360° ÷ 3 = 120°. In diagram A, the sector angle is 90°. In diagram C, the sector angle is 90° + 90° ÷ 2 = 135°. So the correct diagram has a sector angle greater than that shown in A, but smaller than that shown in C. The only such sector angle is that in B.

\(\dfrac{2}{5} + \dfrac{2}{50} + \dfrac{2}{500} = \dfrac{200 + 20 + 2}{500} = \dfrac{222}{500} = \dfrac{444}{1000} = 0.444\).

Let \(G\) be the number of grey kangaroos, and \(P\) the number of pink kangaroos, so \(G + P = 2016\). For each grey kangaroo we calculate the fraction as \(\dfrac{P}{G}\); and there are \(G\) of these, so the total of the fractions for grey kangaroos is \(G \times \dfrac{P}{G} = P\).
Similarly the total of the fractions calculated for the pink kangaroos is \(G\). Thus the total of all the fractions is \(P + G = 2016\).

Let the number of boys in the class be \(x\) and let the number of girls be \(y\). Since one third of the boys sit with a girl and one half of the girls sit with a boy, we have \(\tfrac{1}{3}x = \tfrac{1}{2}y\) and hence \(\tfrac{2}{3}x = y\). The total number of students in the class is 20. Therefore \(x + \tfrac{2}{3}x = 20\). Hence \(\tfrac{5}{3}x = 20\), which has solution \(x = 12\). Therefore there are 12 boys in the class.

Let there be \(g\) girls and \(b\) boys in Tegwen's family. Then, as she has the same number of brothers as she does sisters, \(b = g - 1\). Also, each of her brothers has 50% more sisters than brothers. Therefore \(g = \frac{3}{2}(b - 1)\). So \(b + 1 = \frac{3}{2}(b - 1)\) and hence \(2b + 2 = 3b - 3\). Rearranging this equation gives \(b = 5\). So \(g = 5 + 1 = 6\). Therefore there are 5 + 6 = 11 children in Tegwen's family.

For the one red bead to be 10% of the final number of beads, there must be nine blue beads representing 90% of the final number of beads. Therefore the number of beads Petra must remove is 49 − 9 = 40.

2.017 × 2016 − 10.16 × 201.7 = 201.7 × 20.16 − 10.16 × 201.7
= 201.7(20.16 − 10.16) = 201.7 × 10 = 2017.

Let \(g\) and \(s\) be the number of goats and sheep respectively. Then \(s = 2g\) and \(12 = s - g = 2g - g = g\). Hence the number of animals is \(s + g = 3g = 36\).

By train, the distance in miles of the second sign from Edinburgh is \(200 - 3\frac{1}{2}\). This sign is halfway between London and Edinburgh, so the distance in miles between the two cities is \(2(200 - 3\frac{1}{2}) = 400 - 7 = 393\).

The percentage of teachers coming to school by car is one fifth of the percentage of teachers coming to school by bicycle. Therefore, the number of teachers coming to school by car is \(\tfrac{1}{5} \times 45 = 9\).

The difference between the given options and 1 is \(\dfrac{1}{8}\), \(\dfrac{1}{7}\), \(\dfrac{1}{10}\), \(\dfrac{1}{11}\) and \(\dfrac{1}{10}\) respectively. As \(\dfrac{1}{11}\) is the smallest of these fractions, \(\dfrac{10}{11}\) is closest to 1.

Let the area of each rectangle be \(Y\). Then the total shaded area is \(2(Y - X) + X = 2Y - X\). Therefore \(X = \tfrac{1}{8}(2Y - X)\). So \(8X = 2Y - X\), that is \(9X = 2Y\). Therefore \(\tfrac{X}{Y} = \tfrac{2}{9}\).

The remaining kangaroos weigh (100 − 25 − 60)% = 15% of the total weight. However, this cannot be made up of the weights of more than one kangaroo since the information in the question tells us that the lightest two weigh 25% of the total. Hence there are 2 + 1 + 3 = 6 kangaroos in the mob.

Half of a third, plus a third of a quarter, plus a quarter of a fifth equals
\(\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} = \dfrac{10 + 5 + 3}{60} = \dfrac{18}{60} = \dfrac{3}{10}\).

1 − 0.2 + 0.03 − 0.004 = 0.8 + 0.026 = 0.826.

If the areas of the original and new triangles are the same then the product of the base and the perpendicular height must be the same for the two triangles. When the base of the original triangle is increased by 25%, its value is multiplied by \(\tfrac{5}{4}\). So if the area is to remain unchanged then the perpendicular height must be multiplied by \(\tfrac{4}{5}\), which means that its new value is 80% of its previous value. So it is decreased by 20%.

Let the width of each strip be 1 unit. Then the triangle has base 8 and perpendicular height 8. So its area is equal to \(\tfrac{1}{2} \times 8 \times 8 = 32\). Looking from the right, the area of the first shaded strip is 1 unit of area less than the first unshaded strip. This difference of 1 unit also applies to the other three pairs of strips in the triangle, which means that the shaded area is 4 less than the unshaded area. So the total shaded area is \(\tfrac{1}{2}(32 - 4) = 14\). Therefore the required fraction is \(\tfrac{14}{32} = \tfrac{7}{16}\).

25% of \(\tfrac{3}{4} = \tfrac{1}{4} \times \tfrac{3}{4} = \tfrac{3}{16}\).

From the information given, we may deduce that the number of coins is a multiple of each of 3, 5, 7. Since these are distinct primes, their lowest common multiple is 3 × 5 × 7 = 105. So the number of coins in the bag is a multiple of 105. So there are 105 coins in the bag since 105 is the only positive multiple of 105 less than or equal to 200.

As all the fractions are raised to the power 3, the expression which has the largest value is that with the largest fraction in the brackets.
Each of these fractions is a little larger than \(1\tfrac{1}{2}\). Subtracting \(1\tfrac{1}{2}\) from each in turn, we get the fractions \(\tfrac{1}{14}\), \(\tfrac{1}{6}\), \(\tfrac{1}{4}\), \(\tfrac{3}{10}\), 0, the largest of which is \(\tfrac{3}{10}\) (because \(0 < \tfrac{1}{14} < \tfrac{1}{6} < \tfrac{1}{4} = \tfrac{2\frac{1}{2}}{10} < \tfrac{3}{10}\)). Hence \(\left(\tfrac{9}{5}\right)^3\) is the largest.

The number of seeds in a special packet is 1.25 × 40 = 50. So the number of seeds which germinate is 0.7 × 50 = 35.

\(\dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{2} + \dfrac{1}{12} = \dfrac{7}{12}\); \(\dfrac{1}{2} + \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{1}{2} + \dfrac{4}{3} = \dfrac{11}{6}\); \(\dfrac{1}{2} \times \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{2}{3}\); \(\dfrac{1}{2} - \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{1}{2} - \dfrac{4}{3} = -\dfrac{5}{6}\); \(\dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{2} - \dfrac{1}{12} = \dfrac{5}{12}\).
Of the fractions \(\dfrac{7}{12}\), \(\dfrac{11}{6}\), \(\dfrac{2}{3}\), \(-\dfrac{5}{6}\), \(\dfrac{5}{12}\), the closest to 0 is \(\dfrac{5}{12}\).

The values of the expressions are £5.40, £6.00, £5.40, £5.40, £5.40 respectively.

1% of 1 000 000 = 1 000 000 ÷ 100 = 10 000. So the least number of fleas which will be eradicated is 1 000 000 − 10 000 = 990 000.

The reduction of 15% off sale prices is equal to a reduction of 7.5% off the original prices. Therefore the total reduction on the original prices is (50 + 7.5)% = 57.5%.

After the first mouse has eaten, \(\tfrac{2}{3}\) of the cheese remains. After the second mouse has eaten, \(\tfrac{2}{3}\) of \(\tfrac{2}{3}\), that is \(\tfrac{4}{9}\), of the cheese remains. Finally, after the third mouse has eaten, \(\tfrac{2}{3}\) of \(\tfrac{4}{9}\), that is \(\tfrac{8}{27}\), of the cheese remains. So the mice ate \(\tfrac{19}{27}\) of the cheese.

20 ÷ 11 = \(1\tfrac{9}{11}\) = 1.818181… . So the first 2011 digits are 1006 ‘1’s and 1005 ‘8’s. Therefore the required total is 1006 × 1 + 1005 × 8 = 1006 + 8040 = 9046.

\(19\tfrac{1}{2} \times 20\tfrac{1}{2} = (20 - \tfrac{1}{2}) \times (20 + \tfrac{1}{2}) = 20^2 - (\tfrac{1}{2})^2 = 400 - \tfrac{1}{4} = 399\tfrac{3}{4}\).

When the cuboid is cut away, the surface area of the solid ‘loses’ two rectangles measuring 10 cm × 5 cm and two squares of side 5 cm. However, it also ‘gains’ two rectangles measuring 10 cm × 5 cm. So the surface area decreases by an area equal to one half of the area of one of the faces of the original cube, that is one twelfth of its original surface area.

The values are A \(\dfrac{1}{50}\), B \(\dfrac{1}{60}\), C \(\dfrac{1}{60}\), D \(\dfrac{1}{50}\), E \(\dfrac{1}{30}\).

As \(\dfrac{1}{x} = 3.5 = \dfrac{7}{2}\), \(x = \dfrac{2}{7}\). So \(x + 2 = \dfrac{16}{7}\). Hence \(\dfrac{1}{x + 2} = \dfrac{7}{16}\).

The ratio of goats to sheep is 100:155 = 20:31.

\(\dfrac{4}{5} = \dfrac{12}{15}\) and \(-\dfrac{2}{3} = -\dfrac{10}{15}\), so the number half way between these is \(\dfrac{1}{2}\left(\dfrac{-10}{15} + \dfrac{12}{15}\right)\), that is \(\dfrac{1}{15}\).

The values of these expressions are 5, 8, 9, 8, 5 respectively.

The five options are: A 0.20088888… B 0.20080000… C 0.20080808… D 0.20080080… E 0.20082008….
So in ascending order they are B D C E A.

The problem may be solved by firstly calculating the third and fourth terms of the sequence, but an algebraic method does reduce the amount of calculation involved. Let the first two terms be \(x\) and \(y\) respectively. Then the third term is \(\tfrac{1}{2}(x + y)\), whilst the fourth term is \(\tfrac{1}{4}(x + 3y)\). So the fifth term is \(\tfrac{1}{8}(3x + 5y)\). Putting \(x = \dfrac{2}{3}\) and \(y = \dfrac{4}{5}\), we obtain \(\dfrac{2 + 4}{8} = \dfrac{3}{4}\).

\(S\) = 25% of 60 = 15. \(U = \dfrac{60}{0{\cdot}8} = 75\). \(M = \dfrac{80}{0{\cdot}25} = 320\). So \(S + U + M = 410\).

Let the smaller cubes have side of length 1 unit. So the original cube had side of length 3 units and hence a surface area of 54 units², all of which was painted blue. The total surface area of the 27 small cubes is 27 × 6 units², that is 162 units². So the required fraction is \(\dfrac{54}{162} = \dfrac{1}{3}\).

Let the original number be \(x\). Then \(x^2 = 0{\cdot}7x\), that is \(x(x - 0{\cdot}7) = 0\). So \(x = 0\) or \(x = 0{\cdot}7\), but as \(x\) is non-zero it is 0·7.

The large square is made up of 25 small squares, 15 of which are shaded. So \(\dfrac{15}{25}\) of the large square is shaded, corresponding to \(\tfrac{15}{25}\) × 100% = 60%.

0·75 ÷ \(\tfrac{3}{4}\) = \(\tfrac{3}{4}\) ÷ \(\tfrac{3}{4}\) = 1.

The area of the page inside the margins is 26 cm × 36 cm, that is 936 cm². So the percentage of the page occupied by the margins is \(\dfrac{264}{1200}\) × 100%, that is 22%.

The fraction of the cake which Victoria eats is \(\dfrac{2}{3} - \dfrac{1}{4}\), that is \(\dfrac{5}{12}\).

There are forty dots in the 5 × 8 array and twenty of these are illuminated to form the letter ‘t’.

The profit made would be £14 999.50, which is 29 999 times the original price. So that gives a profit of 29 999 × 100%, that is 2 999 900%.

The mean of \(1.\dot{2}\) and \(2.\dot{1}\) is \((1.\dot{2} + 2.\dot{1}) \div 2 = 3.\dot{3} \div 2 = 1.\dot{6}\).

Note that \(x\)% of \(y\) = \(y\)% of \(x\) = \(xy\)/100. So 2006% of 50 = 50% of 2006 = 1003.

Note that \(\dfrac{1}{4} = \dfrac{6}{24}\) and \(\dfrac{1}{6} = \dfrac{4}{24}\). So \(\dfrac{5}{24}\) is half-way between them.

Let the number of cases solved in 2003 be \(x\). Then, as this was 80% of the number of cases, there were \(5x/4\) cases to solve in 2003. So the number of cases to solve in 2004 was \(5x/4 \times 6/5\), which is \(3x/2\). Inspector Remorse solved 60% of these cases, that is \(3x/2 \times 3/5\), which is \(9x/10\). So the change in the number of cases solved was a 10% decrease.

Four bags of porridge contain one-fifth of a bag of wheatbran. So the proportion of wheatbran in the porridge is 1/20, that is 5%.

The required fraction = \(\dfrac{6\frac{3}{4}}{24} = \dfrac{27}{96} = \dfrac{9}{32}\).

The offer gives the purchaser two items for the price of one and a half items. So the average cost per item is the same as four items for the price of three.

10% of one million is 100 000; 10% of one thousand is 100; 100 000 − 100 = 99 900.

In increasing order, the numbers are 4.004, 4.04, 4.044, 4.4, 4.44.

Consider the cuboid to be made up of 60 unit cubes. The front and side views show that top and bottom layers consist of the same number of cubes and from the top view we see that this number is 14. The front and side views indicate that only the 4 corner cubes remain in the middle layer, so the total number of cubes remaining is 2 × 14 + 4 = 32. The required fraction, therefore, is \(\dfrac{32}{60} = \dfrac{8}{15}\).

As \(P\) and \(Q\) both lie between 0 and 1, their product will be greater than 0 but smaller than \(P\) and smaller than \(Q\). Of the options available, only \(B\) satisfies these conditions. (Furthermore, its position is correct since \(P\) is approximately equal to \(\tfrac{1}{2}\), which means that the product of \(P\) and \(Q\) lies approximately half way between 0 and \(Q\).)

The four fractions total \(\dfrac{5}{4}\), so their mean = \(\dfrac{5}{4} \div 4 = \dfrac{5}{16}\).