Fractions, Continued Fractions and Fraction Series

Fractions and Mixed Numbers: The Basic Moves

A fraction $\frac{a}{b}$ has a numerator $a$ on top and a denominator $b$ underneath. These are the names you need:

• Proper ($\frac{3}{7}$): the top is smaller than the bottom, so the value is less than 1.
• Improper ($\frac{11}{4}$): the top is bigger than or equal to the bottom.
• Mixed number ($2\frac{3}{4}$): a whole number and a proper fraction together.
• Equivalent fractions have the same value, like $\frac{2}{3} = \frac{8}{12}$. A fraction is in lowest terms when its top and bottom have HCF 1.

Mixed to improper and back. $2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4}$. To go back, divide: 29 ÷ 6 = 4 remainder 5, so $\frac{29}{6} = 4\frac{5}{6}$.

Adding and subtracting. Bring the fractions to the LCM of the denominators.

$$\frac{5}{6} + \frac{3}{8} - \frac{7}{12} = \frac{20 + 9 - 14}{24} = \frac{15}{24} = \frac{5}{8}$$

With mixed numbers, handle the whole parts and the fraction parts separately. It keeps the numbers small.

$$3\frac{1}{2} + 2\frac{2}{3} - 1\frac{5}{6} = (3 + 2 - 1) + \frac{3 + 4 - 5}{6} = 4 + \frac{2}{6} = 4\frac{1}{3}$$

If the fraction part comes out negative, borrow 1 from the whole part:

$$5\frac{1}{4} - 2\frac{2}{3} = 3 + \frac{3 - 8}{12} = 3 - \frac{5}{12} = 2\frac{7}{12}$$

Multiplying and dividing. Here mixed numbers must become improper fractions first. Cancel before you multiply, and to divide, multiply by the reciprocal (the fraction turned upside down).

$$2\frac{1}{4} \times 1\frac{7}{9} = \frac{9}{4} \times \frac{16}{9} = 4 \qquad 3\frac{3}{4} \div 1\frac{7}{8} = \frac{15}{4} \times \frac{8}{15} = 2$$

Multiplying the whole parts and the fraction parts separately is wrong: $2\frac{1}{4} \times 1\frac{7}{9}$ is not $2 + \frac{7}{36}$.

A fraction of a fraction. "of" means multiply, and the BODMAS order still applies: $\frac{2}{3}$ of $\frac{9}{16}$ ÷ $\frac{3}{4}$ = $\frac{3}{8} \div \frac{3}{4} = \frac{1}{2}$.

Complex fractions. When the top or the bottom has fractions inside it, simplify the top and the bottom on their own, then divide.

$$\frac{\frac{1}{2} + \frac{1}{3}}{\frac{1}{2} - \frac{1}{3}} = \frac{5}{6} \div \frac{1}{6} = 5$$

Negative mixed numbers. A negative mixed number is negative as a whole: $-2\frac{1}{3} = -\left(2 + \frac{1}{3}\right) = -\frac{7}{3}$, not $-2 + \frac{1}{3}$. The same care is needed when you subtract one: $6 - 2\frac{1}{3} = 3\frac{2}{3}$.

Comparing Two Fractions

There are five ways to compare, from the quickest to the most general.

1. Same denominator. The bigger numerator wins: $\frac{7}{11} > \frac{5}{11}$.

2. Same numerator. The smaller denominator wins, because the whole is cut into fewer, bigger pieces: $\frac{5}{7} > \frac{5}{9}$.

3. Cross-multiply. To compare $\frac{a}{b}$ and $\frac{c}{d}$, compare $a \times d$ with $b \times c$. For $\frac{7}{12}$ and $\frac{5}{9}$: 7 × 9 = 63 and 12 × 5 = 60. Since 63 > 60, $\frac{7}{12} > \frac{5}{9}$.

4. Use a half or 1 as a marker. $\frac{6}{13}$ is under a half (half of 13 is 6.5) and $\frac{7}{12}$ is over a half, so $\frac{7}{12}$ is bigger with no working at all.

5. Decimals. Divide to three places. This is the fallback when four options look alike: $\frac{9}{14}$ = 0.643 and $\frac{7}{11}$ = 0.636.

The gap pattern. When the numerator and denominator differ by the same amount in every fraction, look at what is missing from 1.

$$\frac{3}{5} = 1 - \frac{2}{5} \qquad \frac{5}{7} = 1 - \frac{2}{7} \qquad \frac{7}{9} = 1 - \frac{2}{9}$$

The missing piece shrinks as the numbers grow, so $\frac{3}{5} < \frac{5}{7} < \frac{7}{9} < \frac{9}{11}$. For fractions bigger than 1 it flips. $\frac{5}{3} = 1 + \frac{2}{3}$ and $\frac{9}{7} = 1 + \frac{2}{7}$, so now the one with bigger numbers is smaller: $\frac{5}{3} > \frac{7}{5} > \frac{9}{7}$.

The same idea tells you that adding the same positive number to the top and bottom pulls a fraction towards 1. A fraction below 1 goes up ($\frac{2}{5} < \frac{3}{6} < \frac{4}{7}$), and a fraction above 1 comes down.

The step test. [Tier 2] To compare $\frac{a}{b}$ with a fraction whose top and bottom are both bigger, look at the step: (rise in the top) ÷ (rise in the bottom). If the step is bigger than $\frac{a}{b}$, the new fraction is bigger. If it is smaller, the new fraction is smaller. From $\frac{3}{5}$ to $\frac{4}{7}$ the step is $\frac{1}{2}$, which is less than $\frac{3}{5}$, so $\frac{4}{7} < \frac{3}{5}$. From $\frac{3}{5}$ to $\frac{5}{8}$ the step is $\frac{2}{3}$, which is more than $\frac{3}{5}$, so $\frac{5}{8} > \frac{3}{5}$.

Cross-multiplying compares only two fractions at a time. With four options, keep the winner of each pair and compare it with the next one; three comparisons settle it.

Largest, Smallest, Ordering and Fractions In Between

Largest or smallest of four. Convert each to a decimal to three places, or cross-multiply in pairs. For $\frac{5}{8}$, $\frac{3}{5}$, $\frac{7}{11}$ and $\frac{9}{14}$:

So in ascending order, $\frac{3}{5} < \frac{5}{8} < \frac{7}{11} < \frac{9}{14}$.

Sum or difference of the largest and smallest. Once the order is known, go back to the exact fractions. The rounded decimals are only for ordering. Here, $\frac{9}{14} - \frac{3}{5} = \frac{45 - 42}{70} = \frac{3}{70}$.

Mixed lists. When decimals and percentages are mixed in with fractions, change everything to decimals. For 0.63, $\frac{5}{8}$, $\frac{7}{11}$ and 62%, the values are 0.630, 0.625, 0.636 and 0.620. The largest is $\frac{7}{11}$ and the smallest is 62%.

A fraction between two fractions. For two positive fractions $\frac{a}{b} < \frac{c}{d}$, the fraction $\frac{a + c}{b + d}$ (add the tops, add the bottoms) always lies between them. Between $\frac{2}{5}$ and $\frac{3}{7}$ it gives $\frac{5}{12}$, and indeed 0.400 < 0.417 < 0.429. The average, $\frac{1}{2}\left(\frac{2}{5} + \frac{3}{7}\right) = \frac{29}{70}$, also lies between them, exactly halfway.

"Which fraction lies between" with options. Change the two end fractions to decimals and test every option. Two options can look close, so don't stop at the first one that seems right.

Ordering with a pattern. If the fractions follow the gap pattern from Comparing Two Fractions, no working is needed. $\frac{11}{13}$, $\frac{13}{15}$, $\frac{15}{17}$ and $\frac{17}{19}$ are already in ascending order, because each one misses 1 by a smaller piece.

[SSC Tip] When the options are full orderings like "A < B < C < D", you rarely need the whole list. Find the smallest fraction first. Usually only one or two options start with it, and one more comparison finishes the question.

Fraction Word Problems: Servings, Containers and Pieces

"of" means multiply. $\frac{3}{8}$ of 640 = 240.

Part to whole. If $\frac{3}{5}$ of a number is 54, the number is 54 ÷ $\frac{3}{5}$ = 54 × $\frac{5}{3}$ = 90. Divide the part by its fraction.

How many pieces or servings. Divide the total by the size of one piece. The whole-number part of the answer is the number of full pieces, and whatever is left over is the leftover.

• A 25 m rope is cut into pieces of $1\frac{3}{4}$ m. 25 ÷ $\frac{7}{4}$ = $\frac{100}{7}$ = $14\frac{2}{7}$, so there are 14 full pieces. They use 14 × $\frac{7}{4}$ = $24\frac{1}{2}$ m, which leaves $\frac{1}{2}$ m.
• 6 litres of milk are poured into glasses that hold $\frac{3}{8}$ litre each. 6 ÷ $\frac{3}{8}$ = 6 × $\frac{8}{3}$ = 16 glasses.

The leftover is always total − (pieces × size). A common slip is to give the fractional part, $\frac{2}{7}$, as the leftover length. That $\frac{2}{7}$ is a fraction of one piece: $\frac{2}{7}$ of $1\frac{3}{4}$ m is $\frac{1}{2}$ m, which matches.

Containers at two levels. When a tank goes from one fraction full to another, the difference of the fractions matches the litres added or removed. A tank is $\frac{2}{5}$ full. After 36 litres are added it is $\frac{4}{5}$ full. The 36 litres filled $\frac{4}{5} - \frac{2}{5} = \frac{2}{5}$ of the tank, so the tank holds 36 ÷ $\frac{2}{5}$ = 90 litres.

Full and part-full weights. A bucket full of water weighs 17 kg, and half full it weighs 9.5 kg. The difference, 7.5 kg, is half the water, because the bucket is in both readings. So the water weighs 15 kg and the bucket weighs 17 − 15 = 2 kg.

Parts of one whole. A pole stands with $\frac{1}{6}$ of its length in mud and $\frac{1}{2}$ in water, and 4 m above the water. Mud and water take $\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$, so the 4 m is the other $\frac{1}{3}$ and the pole is 12 m long.

Before adding fractions in a word problem, check whether each one is "of the whole" or "of the remaining". Fractions of the whole can be added. Fractions of a remainder can't, as the next part shows.

Remainder Chains, Mistake Questions and Changing Fractions

Fractions of the remaining. Multiply the fractions that are left, not the fractions that are used. A man spends $\frac{1}{4}$ of his salary on rent and $\frac{2}{3}$ of the remaining on food, and saves ₹ 6,500. After rent, $\frac{3}{4}$ is left. Food takes $\frac{2}{3}$ of that, which leaves $\frac{1}{3}$ of $\frac{3}{4}$, that is $\frac{1}{4}$ of the salary. So the salary is 6,500 × 4 = ₹ 26,000.

Adding $\frac{1}{4} + \frac{2}{3}$ here is the classic mistake, because the $\frac{2}{3}$ is taken from a smaller amount.

Fractions of the whole. When every share is a fraction of the same total, add them. A sum is shared so that A gets $\frac{1}{3}$, B gets $\frac{1}{4}$, C gets $\frac{1}{5}$ and D gets the remaining ₹ 1,300. A, B and C take $\frac{20 + 15 + 12}{60} = \frac{47}{60}$, so D's ₹ 1,300 is $\frac{13}{60}$ of the total, which makes the total ₹ 6,000.

Mistake questions. "A student had to find $\frac{4}{7}$ of a number but found $\frac{7}{4}$ of it. His answer was 165 more than the correct one." The extra 165 is $\frac{7}{4} - \frac{4}{7} = \frac{49 - 16}{28} = \frac{33}{28}$ of the number, so the number is 165 × $\frac{28}{33}$ = 140. If a student divides by $\frac{3}{4}$ instead of multiplying by it, he has really multiplied by $\frac{4}{3}$, so the gap is $\frac{4}{3} - \frac{3}{4} = \frac{7}{12}$ of the number.

Changing the numerator and denominator.

• By the same number. "If 3 is added to the numerator and the denominator, the fraction becomes $\frac{2}{3}$. If 2 is taken from both, it becomes $\frac{1}{2}$." Testing the options is fastest. $\frac{7}{12}$ works, because $\frac{10}{15} = \frac{2}{3}$ and $\frac{5}{10} = \frac{1}{2}$. Always check both conditions, since a wrong option is often built to pass only one.
• By percentages. If the numerator rises by 25% and the denominator falls by 20%, the fraction is multiplied by $\frac{1.25}{0.8} = \frac{25}{16}$. If the new fraction is $\frac{5}{4}$, the original is $\frac{5}{4} \times \frac{16}{25} = \frac{4}{5}$.

A fraction and its reciprocal. "A fraction plus its reciprocal is $2\frac{1}{12}$." Test the options again: $\frac{3}{4} + \frac{4}{3} = \frac{9 + 16}{12} = \frac{25}{12} = 2\frac{1}{12}$.

In a chain question, write one line saying what fraction is left after each step. It takes ten seconds and stops you adding fractions of different amounts.

Continued Fractions

A continued fraction is a fraction stacked inside fractions, like

$$2 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{2}}}$$

Evaluating: start at the bottom. Work one layer at a time from the lowest line up. The only rule you need is that 1 divided by a fraction is that fraction turned upside down: $\frac{1}{a/b} = \frac{b}{a}$.

• Bottom layer: $4 + \frac{1}{2} = \frac{9}{2}$, and 1 over it is $\frac{2}{9}$.
• Next: $3 + \frac{2}{9} = \frac{29}{9}$, and 1 over it is $\frac{9}{29}$.
• Top: $2 + \frac{9}{29} = \frac{67}{29}$.

When a numerator is not 1, divide by the layer below instead of just flipping: $\cfrac{3}{2 + \cfrac{1}{4}} = 3 \div \frac{9}{4} = \frac{4}{3}$. Subtraction layers work the same way: $2 - \cfrac{1}{2 - \cfrac{1}{2}} = 2 - \frac{2}{3} = \frac{4}{3}$.

Solving for x: peel from the top. Do the steps in reverse. Take away the whole number, flip what is left, and repeat until $x$ is alone.

$$1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{x}}} = \frac{15}{11}$$

• Take away 1 and flip $\frac{4}{11}$: $2 + \cfrac{1}{1 + \cfrac{1}{x}} = \frac{11}{4}$
• Take away 2 and flip $\frac{3}{4}$: $1 + \frac{1}{x} = \frac{4}{3}$
• Take away 1: $\frac{1}{x} = \frac{1}{3}$, so $x = 3$

Writing a fraction as a continued fraction. SSC also asks it the other way round: "If $\frac{43}{30} = 1 + \cfrac{1}{a + \cfrac{1}{b + \cfrac{1}{c}}}$, find $a$, $b$ and $c$." Split off the whole number, flip the remainder, and repeat.

• $\frac{43}{30} = 1 + \frac{13}{30}$, and $\frac{30}{13} = 2 + \frac{4}{13}$
• $\frac{13}{4} = 3 + \frac{1}{4}$

So $a = 2$, $b = 3$ and $c = 4$.

A pattern to recognise. With all 1s, the answers are made of the numbers 1, 2, 3, 5, 8, 13, … where each is the sum of the two before it. For instance, $\cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}}} = \frac{5}{8}$.

[Tier 2] Never-ending continued fractions. If the pattern repeats forever, the part below the first layer equals the whole thing. For $x = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cdots}}$, write $x = 2 + \frac{1}{x}$. Then $x^2 - 2x - 1 = 0$, and the positive root is $x = 1 + \sqrt{2}$. In the same way, $1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}} = \frac{1 + \sqrt{5}}{2}$.

[SSC Tip] The options here are usually the right answer, its flip, and one with a slipped layer, such as $\frac{67}{29}$, $\frac{29}{67}$ and $\frac{67}{9}$. Write every layer down. Doing two flips in your head is where the slip happens.

Telescoping Series

In a telescoping sum each term splits into two parts, and the parts cancel down the line until only the first and the last are left.

The basic split.

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

So

$$\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{9 \times 10} = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{9} - \frac{1}{10}\right) = \frac{1}{2} - \frac{1}{10} = \frac{2}{5}$$

In general, $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

When the gap is not 1. If the two factors differ by $k$,

$$\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right)$$

$$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \cdots + \frac{1}{19 \times 22} = \frac{1}{3}\left(1 - \frac{1}{22}\right) = \frac{7}{22}$$

Don't forget the $\frac{1}{k}$ in front. Leaving it out gives $\frac{21}{22}$, and SSC will have that in the options. If each numerator already equals the gap, as in $\frac{3}{1 \times 4} + \frac{3}{4 \times 7} + \cdots$, the $\frac{1}{k}$ cancels and the sum is simply first minus last.

Disguised terms. The products are often multiplied out, so learn to spot them:

So $\frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110} + \frac{1}{132} = \frac{1}{7} - \frac{1}{12} = \frac{5}{84}$.

Terms like $\frac{1}{n^2 - 1}$ are the gap-2 kind, because $n^2 - 1 = (n - 1)(n + 1)$. So $\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} = \frac{1}{2}\left(1 - \frac{1}{9}\right) = \frac{4}{9}$.

[Tier 2] Three factors. $\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$, so

$$\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \cdots + \frac{1}{8 \times 9 \times 10} = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{90}\right) = \frac{11}{45}$$

Product Series

In a product series each bracket is a fraction, and the top of one bracket cancels the bottom of the next.

The two basic chains.

$$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\cdots\left(1 - \frac{1}{n}\right) = \frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{n - 1}{n} = \frac{1}{n}$$

$$\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\cdots\left(1 + \frac{1}{n}\right) = \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{n + 1}{n} = \frac{n + 1}{2}$$

Watch where it starts. Only the first bottom and the last top survive, so the starting point matters. If the chain starts at $\frac{1}{d}$ instead of $\frac{1}{2}$:

• $\left(1 - \frac{1}{d}\right)\cdots\left(1 - \frac{1}{n}\right) = \frac{d - 1}{n}$. For instance, $\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{6}\right)\cdots\left(1 - \frac{1}{40}\right) = \frac{4}{40} = \frac{1}{10}$.
• $\left(1 + \frac{1}{d}\right)\cdots\left(1 + \frac{1}{n}\right) = \frac{n + 1}{d}$. For instance, $\left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{5}\right)\cdots\left(1 + \frac{1}{99}\right) = \frac{100}{4} = 25$.

Squares in the bottom. Each bracket splits into a minus part and a plus part, $1 - \frac{1}{k^2} = \left(1 - \frac{1}{k}\right)\left(1 + \frac{1}{k}\right)$, so

$$\left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\cdots\left(1 - \frac{1}{n^2}\right) = \frac{1}{n} \times \frac{n + 1}{2} = \frac{n + 1}{2n}$$

Up to $\frac{1}{10^2}$ this gives $\frac{11}{20}$.

Odd-number chains. Watch for the pattern $\left(2 - \frac{1}{3}\right)\left(2 - \frac{3}{5}\right)\left(2 - \frac{5}{7}\right)\cdots$. Each bracket simplifies first: $2 - \frac{1}{3} = \frac{5}{3}$, $2 - \frac{3}{5} = \frac{7}{5}$ and $2 - \frac{5}{7} = \frac{9}{7}$. Now the product cancels down to (last top) ÷ 3. If the last bracket is $2 - \frac{19}{21} = \frac{23}{21}$, the product is $\frac{23}{3}$.

Mixed chains. If a chain alternates or skips, as in $\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{7}\right)$, the tops and bottoms no longer meet, so there is no shortcut. Just multiply: $\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} = \frac{48}{105} = \frac{16}{35}$. Check that the chain really runs through consecutive numbers before you use the formula.

To see what survives, write the first two brackets and the last one as single fractions. The cancelling becomes obvious without writing the middle.

Recurring Decimals in Calculations

Chapter 1 shows how to turn a recurring decimal into a fraction: $0.\overline{7} = \frac{7}{9}$, $0.\overline{45} = \frac{45}{99}$ and $0.1\overline{6} = \frac{16 - 1}{90}$. Here the job is to use them inside a sum. The rule is simple: change every recurring decimal to a fraction, calculate, and change back only if the options need it.

• Adding. $0.\overline{6} + 0.\overline{7} + 0.\overline{8} = \frac{6 + 7 + 8}{9} = \frac{21}{9} = 2.\overline{3}$
• Subtracting. $3.\overline{4} - 1.\overline{7} = \frac{31}{9} - \frac{16}{9} = \frac{15}{9} = 1.\overline{6}$
• Multiplying. $0.\overline{3} \times 0.\overline{6} = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} = 0.\overline{2}$
• Dividing. $0.\overline{36} \div 0.\overline{12} = \frac{36}{99} \div \frac{12}{99} = 3$
• Mixed recurring. $0.2\overline{3} + 0.\overline{4} = \frac{21}{90} + \frac{40}{90} = \frac{61}{90} = 0.6\overline{7}$

Why digit-by-digit adding fails. Added column by column without carries, 0.777… + 0.555… looks like 1.222…. But every column also gets a carry of 1 from the column on its right, so the true answer is $\frac{12}{9} = 1.\overline{3}$. Fractions avoid the problem completely.

Pairs that make 1. Two pure recurring decimals with blocks of the same length whose blocks add to all 9s make exactly 1: $0.\overline{37} + 0.\overline{62} = \frac{99}{99} = 1$. If the blocks add to 101 instead, the answer is $\frac{101}{99} = 1.\overline{02}$, as in $0.\overline{45} + 0.\overline{56}$.

Sevenths. $\frac{1}{7} = 0.\overline{142857}$, and every other seventh uses the same six digits in a rotated order: $\frac{2}{7} = 0.\overline{285714}$ and $\frac{3}{7} = 0.\overline{428571}$. So $0.\overline{285714} \times 7 = 2$.

[SSC Tip] Options are often written in bar form, so learn two families. Ninths are a single repeating digit ($\frac{4}{9} = 0.\overline{4}$), and elevenths repeat a two-digit multiple of 9 ($\frac{3}{11} = 0.\overline{27}$, $\frac{5}{11} = 0.\overline{45}$). With these you can turn your fraction back into bar form in a second.

Solved Examples

Question 1

Simplify $4\frac{2}{5} + 2\frac{3}{4} - 3\frac{1}{2}$.

Answer. I do the whole parts and the fraction parts separately. Whole parts: 4 + 2 − 3 = 3. Fraction parts: $\frac{2}{5} + \frac{3}{4} - \frac{1}{2} = \frac{8 + 15 - 10}{20} = \frac{13}{20}$. So the answer is $3\frac{13}{20}$.

Question 2

Simplify $2\frac{1}{3}$ ÷ $1\frac{3}{4}$ of $\frac{2}{3}$ + $\frac{1}{2} \times \frac{3}{4}$. (a) $1\frac{19}{72}$ (b) $2\frac{3}{8}$ (c) $2\frac{1}{4}$ (d) $\frac{7}{8}$

Answer. "of" comes before ÷, so first $1\frac{3}{4}$ of $\frac{2}{3}$ = $\frac{7}{4} \times \frac{2}{3} = \frac{7}{6}$. Then $\frac{7}{3} \div \frac{7}{6} = \frac{7}{3} \times \frac{6}{7} = 2$. Next, $\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$. Adding gives $2\frac{3}{8}$. Reading "of" as an ordinary × gives $\frac{8}{9} + \frac{3}{8} = 1\frac{19}{72}$, which is option (a). The answer is (b) $2\frac{3}{8}$.

Question 3

Simplify $\dfrac{\frac{1}{3} + \frac{1}{4}}{\frac{1}{2} - \frac{1}{5}}$.

Answer. Top: $\frac{1}{3} + \frac{1}{4} = \frac{7}{12}$. Bottom: $\frac{1}{2} - \frac{1}{5} = \frac{3}{10}$. Now I divide: $\frac{7}{12} \div \frac{3}{10} = \frac{7}{12} \times \frac{10}{3} = \frac{70}{36} = \frac{35}{18}$, which is $1\frac{17}{18}$.

Question 4

Which is the largest: (a) $\frac{5}{8}$ (b) $\frac{7}{11}$ (c) $\frac{9}{14}$ (d) $\frac{3}{5}$

Answer. I turn each into a decimal: $\frac{5}{8}$ = 0.625, $\frac{7}{11}$ = 0.636, $\frac{9}{14}$ = 0.643 and $\frac{3}{5}$ = 0.6. To be sure about the top two, I cross-multiply $\frac{9}{14}$ and $\frac{7}{11}$: 9 × 11 = 99 and 14 × 7 = 98, so $\frac{9}{14}$ is bigger. The answer is (c) $\frac{9}{14}$.

Question 5

Arrange $\frac{4}{5}$, $\frac{6}{7}$, $\frac{2}{3}$ and $\frac{7}{9}$ in descending order.

Answer. Three of them follow a pattern: $\frac{2}{3}$, $\frac{4}{5}$ and $\frac{6}{7}$ each miss 1 by $\frac{1}{3}$, $\frac{1}{5}$ and $\frac{1}{7}$, so they go up in that order. $\frac{7}{9}$ misses 1 by $\frac{2}{9}$, which is between $\frac{1}{5}$ and $\frac{1}{3}$. So it sits between $\frac{2}{3}$ and $\frac{4}{5}$. In descending order: $\frac{6}{7} > \frac{4}{5} > \frac{7}{9} > \frac{2}{3}$.

Question 6

Find the difference between the largest and the smallest of $\frac{3}{7}$, $\frac{5}{9}$, $\frac{4}{11}$ and $\frac{7}{12}$.

Answer. In decimals they are 0.429, 0.556, 0.364 and 0.583. So the largest is $\frac{7}{12}$ and the smallest is $\frac{4}{11}$. Now I use the exact fractions: $\frac{7}{12} - \frac{4}{11} = \frac{77 - 48}{132}$ = $\frac{29}{132}$.

Question 7

Which fraction lies between $\frac{3}{8}$ and $\frac{2}{5}$? (a) $\frac{4}{11}$ (b) $\frac{5}{13}$ (c) $\frac{7}{17}$ (d) $\frac{3}{7}$

Answer. The ends are $\frac{3}{8}$ = 0.375 and $\frac{2}{5}$ = 0.4. Checking each option: $\frac{4}{11}$ = 0.364 is too small, $\frac{5}{13}$ = 0.385 fits, $\frac{7}{17}$ = 0.412 is too big and $\frac{3}{7}$ = 0.429 is too big. In fact $\frac{5}{13}$ is what you get by adding the tops and adding the bottoms of $\frac{3}{8}$ and $\frac{2}{5}$. The answer is (b) $\frac{5}{13}$.

Question 8

A wire 35 m long is cut into pieces of $2\frac{1}{4}$ m each. How many full pieces are there, and how much wire is left over?

Answer. 35 ÷ $\frac{9}{4}$ = 35 × $\frac{4}{9}$ = $\frac{140}{9}$ = $15\frac{5}{9}$, so there are 15 full pieces. They use 15 × $\frac{9}{4}$ = $\frac{135}{4}$ = $33\frac{3}{4}$ m. The leftover is 35 − $33\frac{3}{4}$ = $1\frac{1}{4}$ m. So the answer is 15 pieces, with $1\frac{1}{4}$ m left over.

Question 9

A tank is $\frac{3}{8}$ full. When 33 litres of water are added, it becomes $\frac{5}{6}$ full. What is the capacity of the tank?

Answer. The 33 litres filled $\frac{5}{6} - \frac{3}{8} = \frac{20 - 9}{24} = \frac{11}{24}$ of the tank. So the whole tank is 33 ÷ $\frac{11}{24}$ = 33 × $\frac{24}{11}$ = 72 litres.

Question 10

A bucket full of water weighs 3.6 kg. When it is $\frac{1}{3}$ full, it weighs 1.6 kg. What does the empty bucket weigh?

Answer. The bucket is there in both readings, so the difference of 2 kg is all water. Going from $\frac{1}{3}$ full to full adds $\frac{2}{3}$ of the water, so $\frac{2}{3}$ of the water is 2 kg and the full water is 3 kg. The empty bucket is 3.6 − 3 = 0.6 kg.

Question 11

A man spends $\frac{1}{4}$ of his salary on rent, $\frac{1}{3}$ of the remaining on food and $\frac{2}{5}$ of what is then left on travel. He saves ₹ 9,000. What is his salary?

Answer. I track what is left after each step. After rent, $\frac{3}{4}$ is left. After food, $\frac{2}{3}$ of that is left. After travel, $\frac{3}{5}$ of that is left. So the saving is $\frac{3}{4} \times \frac{2}{3} \times \frac{3}{5} = \frac{3}{10}$ of the salary. The salary is 9,000 × $\frac{10}{3}$ = ₹ 30,000.

Question 12

A student was asked to find $\frac{7}{12}$ of a number. By mistake he found $\frac{7}{4}$ of it, and his answer was 168 more than the correct one. What is the number?

Answer. The extra 168 is $\frac{7}{4} - \frac{7}{12} = \frac{21 - 7}{12} = \frac{14}{12} = \frac{7}{6}$ of the number. So the number is 168 × $\frac{6}{7}$ = 144. Check: $\frac{7}{4}$ of 144 is 252, $\frac{7}{12}$ of 144 is 84, and 252 − 84 = 168.

Question 13

The numerator of a fraction is increased by 20% and its denominator is decreased by 25%. The new fraction is $\frac{4}{5}$. Find the original fraction.

Answer. The fraction gets multiplied by $\frac{1.2}{0.75} = \frac{120}{75} = \frac{8}{5}$. So the original is $\frac{4}{5} \div \frac{8}{5} = \frac{4}{5} \times \frac{5}{8}$ = $\frac{1}{2}$. Check: $\frac{1 \times 1.2}{2 \times 0.75} = \frac{1.2}{1.5} = \frac{4}{5}$.

Question 14

If 1 is added to both the numerator and the denominator of a fraction, it becomes $\frac{3}{4}$. If 3 is taken away from both, it becomes $\frac{1}{2}$. The fraction is: (a) $\frac{2}{3}$ (b) $\frac{3}{5}$ (c) $\frac{5}{7}$ (d) $\frac{7}{9}$

Answer. I test the options. $\frac{2}{3}$ passes the first condition ($\frac{3}{4}$) but taking 3 from 3 leaves 0 at the bottom, so it fails. $\frac{3}{5}$ gives $\frac{4}{6} = \frac{2}{3}$, so it fails. $\frac{5}{7}$ gives $\frac{6}{8} = \frac{3}{4}$ and $\frac{2}{4} = \frac{1}{2}$, so it passes both. The answer is (c) $\frac{5}{7}$.

Question 15

Find the value of $2 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{4}}}$.

Answer. I start at the bottom. $1 + \frac{1}{4} = \frac{5}{4}$, and 1 over it is $\frac{4}{5}$. Next, $3 + \frac{4}{5} = \frac{19}{5}$, and 1 over it is $\frac{5}{19}$. Top: $2 + \frac{5}{19} = \frac{43}{19}$. The answer is $\frac{43}{19}$, or $2\frac{5}{19}$.

Question 16

Find $x$ if $\cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{x}}} = \frac{5}{13}$.

Answer. I peel it from the top. Flipping both sides: $2 + \cfrac{1}{1 + \cfrac{1}{x}} = \frac{13}{5}$. Take away 2: $\cfrac{1}{1 + \cfrac{1}{x}} = \frac{3}{5}$. Flip: $1 + \frac{1}{x} = \frac{5}{3}$. Take away 1: $\frac{1}{x} = \frac{2}{3}$. So $x = \frac{3}{2}$.

Question 17

If $\frac{61}{27} = 2 + \cfrac{1}{x + \cfrac{1}{y + \cfrac{1}{z}}}$, where $x$, $y$ and $z$ are natural numbers, find $x + y + z$.

Answer. $\frac{61}{27} = 2 + \frac{7}{27}$. Flip the $\frac{7}{27}$: $\frac{27}{7} = 3 + \frac{6}{7}$, so $x = 3$. Flip the $\frac{6}{7}$: $\frac{7}{6} = 1 + \frac{1}{6}$, so $y = 1$ and $z = 6$. So $x + y + z$ = 3 + 1 + 6 = 10.

Question 18

Find $\frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \cdots + \frac{1}{380}$.

Answer. 12 = 3 × 4, 20 = 4 × 5, 30 = 5 × 6, and 380 = 19 × 20. So every term is $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, from $n = 3$ to $n = 19$. Everything in the middle cancels and I am left with $\frac{1}{3} - \frac{1}{20} = \frac{20 - 3}{60}$ = $\frac{17}{60}$.

Question 19

Find $\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \cdots + \frac{1}{28 \times 31}$.

Answer. The two factors differ by 3 each time, so every term is $\frac{1}{3}\left(\frac{1}{n} - \frac{1}{n+3}\right)$. The middle cancels and I get $\frac{1}{3}\left(1 - \frac{1}{31}\right) = \frac{1}{3} \times \frac{30}{31}$ = $\frac{10}{31}$. Forgetting the $\frac{1}{3}$ would give $\frac{30}{31}$.

Question 20

Find $\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right)\cdots\left(1 - \frac{1}{50}\right)$.

Answer. Each bracket is a fraction: $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \cdots \times \frac{49}{50}$. Each top cancels the bottom before it, so only the first top (2) and the last bottom (50) are left. The value is $\frac{2}{50}$ = $\frac{1}{25}$.

Question 21

Find $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\cdots\left(1 - \frac{1}{100}\right)$.

Answer. The bottoms are the squares $2^2$ up to $10^2$, so this is the chain $\left(1 - \frac{1}{k^2}\right)$ from $k = 2$ to $k = 10$. Its value is $\frac{n + 1}{2n}$ with $n = 10$, which is $\frac{11}{20}$. Writing it out: the minus parts give $\frac{1}{10}$ and the plus parts give $\frac{11}{2}$, and $\frac{1}{10} \times \frac{11}{2} = \frac{11}{20}$.

Question 22

Find $0.\overline{36} \times 2.\overline{2}$.

Answer. $0.\overline{36} = \frac{36}{99} = \frac{4}{11}$, and $2.\overline{2} = 2 + \frac{2}{9} = \frac{20}{9}$. Multiplying: $\frac{4}{11} \times \frac{20}{9} = \frac{80}{99}$. A two-digit block over 99 is a two-digit recurring decimal, so the answer is $0.\overline{80}$.