The shortest side is \(9\) cm and the longest side is \(12\) cm.

\(\frac{ 9 }{ 12 }\)

Divide the top and bottom by \(3\):

\(\frac{ 3 }{ 4 }\)

The angles in a triangle add up to \(180^\circ\), and \(x = 3y\), so

\(3y + y + 100 = 180\)

\(4y + 100 = 180\)

\(4y = 180 - 100 = 80\)

\(y = 80 \div 4 = 20\)

The ratio has \(2 + 7 + 8 = 17\) parts altogether, and 7 of those parts are gold.

So \(\frac{ 7 }{ 17 }\) of the beads are gold.

8 of the 17 parts are black, so every other part is not black:

\(2 + 7 = 9\) parts out of 17

So \(\frac{ 9 }{ 17 }\) of the beads are not black.

The ratio \(v : u = 1 : 4\) means u is 4 times v.

Take a pair of values in the ratio: \(v = 1\) when \(u = 4\).

The equation \(v = 4u\) gives \(v = 16\) for \(u = 4\), not \(1\).

The equation \(v = u - 3\) does fit this pair, so try the double pair: \(v = 2\) when \(u = 8\) is still in the ratio. But \(u - 3\) gives \(5\) for \(u = 8\), not \(2\).

The equation \(v = u + 3\) fails at the first pair: \(u = 4\) gives \(v = 7\), not \(1\).

The equation \(v = \frac{ u }{ 4 }\) gives \(v = 1\) when \(u = 4\) and \(v = 2\) when \(u = 8\), so it is the correct one.

Write the ratio with \(n\) first:

\(n : m = 5 : 8\)

"\(n\) as a fraction of \(m\)" puts the part for \(n\) on the top and the part for \(m\) on the bottom:

\(\frac{ 5 }{ 8 }\)

So the correct statement is: \(n\) is \(\frac{ 5 }{ 8 }\) of \(m\)

\(\frac{ 8 }{ 5 }\) is the other way up, and \(\frac{ 5 }{ 13 }\) and \(\frac{ 8 }{ 13 }\) compare one part with all 13 parts, not with the other letter.

Think of the pencil case as 13 equal parts. 9 of those parts are green, so the other \(13 - 9 = 4\) parts are black.

The ratio of green pencils to black pencils is \(9 : 4\), and 9 and 4 have no common factor other than 1, so that is already its simplest form.

The green pencils are 9 of the 13 parts, so one part is

\(63 \div 9 = 7\) pencils

There are 13 parts altogether:

\(13 \times 7 = 91\)

The ratio says there are 3 vans for every 11 cars, so the number of vans divided by the number of cars is

\(\frac{ 3 }{ 11 }\)

This time the vans go on the bottom, so the fraction is turned over:

\(\frac{ 11 }{ 3 }\)

It is bigger than 1 because there are more cars than vans.

This one compares the vans with all of the vehicles, not just with the cars.

There are \(3 + 11 = 14\) parts altogether, and 3 of them are vans:

\(\frac{ 3 }{ 14 }\)

The second ratio is given the other way round, so turn it round first: the ratio of vans to lorries is \(8 : 3\).

The two ratios now meet at the vans, and the number of parts for vans has to be made the same before they can be joined.

The lowest common multiple of 2 and 8 is 8.

Written with 8 parts for vans, the first ratio is \(20 : 8\) and the second is \(8 : 3\).

Joining them at vans, the ratio of hatchbacks to vans to lorries is

\(20 : 8 : 3\)

Pick a value of \(a\) that makes the arithmetic easy. Taking \(a = 2\),

\(b = \frac{ 5 }{ 2 } \times 2 = 5\)

So \(a = 2\) goes with \(b = 5\), and

\(a : b = 2 : 5\)

From part (a), \(a\) and \(b\) are in the ratio \(2 : 5\), so \(b\) is \(5 - 2 = 3\) parts more than \(a\).

Those 3 parts are worth 21, so one part is

\(21 \div 3 = 7\)

\(a\) is 2 parts:

\(2 \times 7 = 14\)

Written with x first, the ratio is \(x : y = 5 : 7\), so \(y = \frac{ 7 }{ 5 }x\) and the line goes through the origin.

When \(x = 5\), \(y = 7\). When \(x = 10\), \(y = 14\).

Plot \((0, 0)\), \((5, 7)\) and \((10, 14)\) and join them with a straight line.

The line passes through \((0, 0)\) and \((5, 7)\).

gradient \(= \frac{ \text{change in } y }{ \text{change in } x } = \frac{ 7 }{ 5 }\)

The gradient is \(y\) divided by \(x\), so it is the number of buckets of sand for each 1 bucket of cement.

It is the amount of sand needed for each 1 bucket of cement.

The ratio \(6 : 7\) says that \(\frac{ G }{ S } = \frac{ 7 }{ 6 }\).

Multiplying both sides by \(S\) gives

\(G = \frac{ 7 }{ 6 }S\)

66 is \(66 \div 6 = 11\) lots of 6, so there are 11 lots of 7 as well:

\(G = 11 \times 7 = 77\)

The formula gives the same thing: \(G = \frac{ 7 }{ 6 } \times 66 = 77\)