Shortcuts & Speed Methods

Reading a Mixture as Fractions of the Whole

A ratio tells you what fraction of the whole each part is, and most quick methods for mixtures start from there. In Tier 1 you have about 36 seconds a question, so each method below says when it applies, shows it on a short example, and ends with the cases where it goes wrong, such as alliging on a selling price or subtracting the rounds in a replacement question.

Ratio to fraction to quantity

When to use it: a ratio and a total are given, and you need the amount of one part, its fraction or its percentage.

Fast way: if milk : water = a : b, milk is $\frac{a}{a + b}$ of the mixture and water is $\frac{b}{a + b}$. Divide the total by a + b to get one unit, then multiply.

For example: 91 litres of milk and water are in the ratio 8 : 5. One unit is 91 ÷ 13 = 7 litres, so there are 56 litres of milk and 35 litres of water. Milk is $\frac{8}{13}$ of the mixture, about 61.5%.

A percentage works the other way round. A solution that is 28% acid has acid : water = 28 : 72 = 7 : 18.

Three parts, and one part as a percentage of another

An alloy of copper, zinc and tin in the ratio 7 : 4 : 1 has 12 units. In 54 kg of it one unit is 4.5 kg, so there are 31.5 kg of copper, 18 kg of zinc and 4.5 kg of tin.

Read which whole the question names. Copper as a percentage of zinc is $\frac{7}{4}$ = 175%. Copper as a percentage of the alloy is $\frac{7}{12}$, about 58.3%.

Where it fails:

• In milk : water = 8 : 5, water is $\frac{5}{13}$ of the mixture but $\frac{5}{8}$ of the milk. "Water as a percentage of milk" is 62.5%, while "water as a percentage of the mixture" is about 38.5%. Both tend to sit in the options.
• A ratio is not a quantity. "The ratio is 8 : 5" doesn't mean there are 8 litres of milk. Find the unit first.

Turn every mixture into units at the start and stay in those units. Most of the methods below change the units of one part while the other part stays put, and the answer falls out of the unit count.

The Unchanged Part: Adding Water or One Component

Keep the other part's units fixed

When to use it: water (or only one component) is added, or only one is taken away, and the question gives the old and new ratios.

Fast way: only one part changes, so the other part is the same amount before and after. Scale the two ratios so the unchanged part has the same number in both. The change in the other part, counted in units, is what was added.

For example: 96 litres of milk and water are in the ratio 7 : 5. How much water must be added to make it 4 : 5?

Milk doesn't change, so make the milk numbers equal: 7 : 5 = 28 : 20 and 4 : 5 = 28 : 35. Water goes from 20 units to 35 units, up by 15. The old mixture is 48 units = 96 litres, so one unit is 2 litres and 30 litres of water must go in. Check: 56 litres of milk and 70 litres of water make 4 : 5.

The same idea with percentages

Fast way: the part you don't add is a fixed amount. Find it, then divide by its new share.

For example: 60 litres of a solution are 30% acid. How much pure acid must be added to make it 40% acid? The water, 42 litres, stays. In the new solution water is 60%, so the new total is 42 ÷ 0.6 = 70 litres, and 10 litres of acid go in.

Diluting works the same way. 40 kg of salt water with 15% salt holds 6 kg of salt. To bring the salt down to 12%, the new total must be 6 ÷ 0.12 = 50 kg, so 10 kg of water is added.

Equal amounts of both parts

If the same amount of milk and of water is added, neither part stays fixed, but the difference between them does. Match the difference instead.

For example: 66 litres of milk and water are in the ratio 7 : 4, which is 42 litres of milk and 24 of water, so milk is 18 litres ahead. Equal amounts of both are added and the ratio becomes 5 : 4. In 5 : 4 the gap is 1 unit, and it is still 18 litres, so milk becomes 5 × 18 = 90 litres. It was 42, so 48 litres of each were added.

Where it fails:

• Putting the new ratio on the old total. In the first example, $\frac{5}{9}$ of 96 is about 53.3 litres of water, which would mean adding about 13.3 litres. The total grows when water goes in, so that's wrong.
• In the percentage form, the new percentage is of the new total. Taking 40% of the old 60 litres gives 24 litres of acid and "add 6 litres", which is wrong for the same reason.
• Drying (only water leaves) uses the same idea with the solid fixed.

[SSC Tip] In one line, new total = old total × $\frac{\text{old share of the fixed part}}{\text{new share of the fixed part}}$. For the acid example that is 60 × $\frac{70}{60}$ = 70 litres, and for the salt water it is 40 × $\frac{15}{12}$ = 50 kg.

Removal, Then Addition

Taking out mixture keeps its ratio

When to use it: some of the mixture is taken out, and then one or more parts are poured in.

Fast way: when mixture (not one liquid) is taken out, every part leaves in the same ratio, so what stays has the same ratio as before. Cut down the units, then add.

For example: a can holds 90 litres of milk and water in the ratio 7 : 3. 20 litres of the mixture are taken out and 10 litres of water are poured in. The 70 litres left are still 7 : 3, which is 49 litres of milk and 21 of water. Add 10 litres of water and the ratio is 49 : 31.

Three liquids: work back from the gap

When to use it: three liquids in a ratio, some mixture drawn off, some of two liquids added, and you're told how much more of one liquid there now is than another. The question wants the starting total.

Fast way: the gap between two liquids before the additions is a fixed number of units of the leftover mixture. Undo the additions to get that gap in litres, and you have the unit.

For example: a tank holds liquids A, B and C in the ratio 4 : 7 : 5. 32 litres are drawn off, then 6 litres of A and 2 litres of B are added. Now there is 14 litres more of B than of A. How much did the tank hold at first?

The additions closed the gap by 6 − 2 = 4 litres, so before them B was 18 litres ahead of A. In the leftover, B − A = 7 − 4 = 3 units, so one unit is 6 litres. The leftover is 16 units = 96 litres, and the tank first held 96 + 32 = 128 litres. Check: from 128 litres down to 96 gives A = 24 and B = 42; after the additions A = 30 and B = 44, a gap of 14.

Where it fails:

• Taking the 20 litres out of the milk alone. Each part loses its own share: 14 litres of milk and 6 of water in the first example.
• Forgetting to add back the litres drawn off. The unit gives the leftover, so 96 litres is a trap answer in the second example.
• If the same amount is taken out and water put back several times, that is removal and replacement. A power formula (later in this section) is much faster than tracking it step by step.

Mixing Two Mixtures: Equal Totals or Fractions

Equal quantities: make the ratio sums equal

When to use it: equal amounts of two mixtures (or two full vessels of the same size) are poured together.

Fast way: make both ratio sums the same with their LCM, then add part by part.

For example: equal quantities of two mixtures with milk : water = 5 : 1 and 7 : 5 are mixed. The sums are 6 and 12, so write the first one as 10 : 2. Now 10 : 2 plus 7 : 5 gives 17 : 7.

Given quantities or a mixing ratio: use fractions of one part

When to use it: the two mixtures are taken in given litres or in a given ratio.

Fast way: write the fraction of one component in each mixture, multiply by the quantity taken, and add. The other component is whatever is left of the total.

For example: acid : water is 9 : 1 in jar A and 1 : 4 in jar B, and A and B are mixed in the ratio 2 : 3. Acid is $\frac{9}{10}$ of A and $\frac{2}{10}$ of B. In 5 parts of the mix, acid = 2 × $\frac{9}{10}$ + 3 × $\frac{2}{10}$ = $\frac{24}{10}$, and water = 5 − $\frac{24}{10}$ = $\frac{26}{10}$. So acid : water = 12 : 13.

Writing both fractions over the same denominator (tenths here) means you only add whole numbers. Alloys work the same way, with copper and zinc (or gold and silver) in place of acid and water.

Where it fails:

• Adding the ratios as they stand. 5 : 1 plus 7 : 5 gives 12 : 6 = 2 : 1, which is wrong because the two ratios use units of different sizes. It is right only when the sums already match.
• Equal sums are for equal quantities. With a mixing ratio of 2 : 3, the first mixture counts twice and the second three times.
• Taking the fraction of acid from one jar and the fraction of water from the other. Stick to one component.

The share of a component in the mix always lies between its shares in the two mixtures. Acid is $\frac{2}{10}$ and $\frac{9}{10}$ of the two jars, so in the mix it must be in between, and $\frac{12}{25}$ is. Any option outside that range is out at once.

Transfers Between Containers

Some from each into a third

When to use it: given amounts are taken from two containers and poured into a third.

Fast way: track one component only. From each container, multiply the litres taken by that container's fraction of the component. The other component is the rest.

For example: in container A alcohol : water is 3 : 5, and in B it is 4 : 1. 24 litres from A and 10 litres from B go into C. Alcohol = 24 × $\frac{3}{8}$ + 10 × $\frac{4}{5}$ = 9 + 8 = 17 litres out of 34, so water is also 17 litres and C has alcohol : water = 1 : 1.

There and back

When to use it: some liquid moves from A to B, and then the same amount moves from B back to A.

Fast way: after the return trip each container holds what it started with in volume, so every litre of milk that left A has been replaced by a litre of water. The water now in A equals the milk now in B.

For example: jug A has 40 litres of pure milk and jug B has 40 litres of water. 10 litres go from A to B and, after stirring, 10 litres come back. B had 10 litres of milk in 50, so the returning 10 litres carry $\frac{1}{5}$ of that milk, 2 litres, and 8 litres of water. Now A has 32 litres of milk and 8 litres of water, and B has 8 litres of milk and 32 of water.

Where it fails:

• Treating "24 litres from A" as 24 units of A's ratio. Litres taken are a real quantity, so multiply by the fraction.
• On the return trip the jug carries mixture, not pure water. Sending back 10 litres of water would give 10 litres of water in A instead of 8.
• If the containers end with different volumes from the ones they started with, "water in A = milk in B" no longer holds. Track the litres.

[SSC Tip] When liquid goes back and forth two or more times, keep a small table with two columns (milk, water) for each container and update it after every pour. It takes 20 seconds and never loses a litre.

Average Price Forward, Alligation Backward

Which way is the question going?

Questions on mixing varieties at different prices go in one of two directions.

• Forward: the quantities (or their ratio) and the prices are given, and the price of the mix is asked. That's a weighted average.
• Backward: the price of the mix is given, and the ratio or a quantity is asked. That's alligation, the cross shown next.

Read the last line of the question first. If it asks for a price, go forward. If it asks for a ratio or for "how many kg", go backward.

Forward: the base method

Fast way: pick one price as the base, multiply each variety's difference from it by its quantity, add, and divide by the total quantity.

For example: 10 kg of dal at ₹110, 15 kg at ₹125 and 25 kg at ₹145 per kg are mixed. With ₹125 as the base: 10 × (−15) + 25 × (+20) = −150 + 500 = 350. Over 50 kg that is +7, so the mix costs ₹132 per kg.

With a ratio in place of quantities, the mix price = cheaper price + gap × dearer share. For ₹56 and ₹66 rice in the ratio 2 : 3, it is 56 + 10 × $\frac{3}{5}$ = ₹62.

Equal quantities or equal money?

• Equal quantities of each: the mix price is the plain average.
• Equal money spent on each: you get more of the cheaper one, so the mix price is lower. For two varieties it is $\frac{2ab}{a + b}$.

For example: ₹1,080 is spent on each of two teas at ₹90 and ₹135 per kg. That buys 12 kg and 8 kg, so 20 kg cost ₹2,160 and the mix is ₹108 per kg, not the plain average of ₹112.50.

Water is a variety that costs ₹0. 15 litres of milk at ₹56 a litre with 5 litres of water cost ₹840 for 20 litres, which is ₹42 a litre.

Where it fails:

• The plain average when quantities differ. In the dal example it gives about ₹126.67, which is too low because there is more of the ₹145 dal.
• Equal money is not equal quantity. ₹112.50 is the trap in the tea example.

The mix price always sits nearer the variety there is more of. With 3 parts of ₹66 rice to 2 of ₹56, the mix must be above the midpoint ₹61, so ₹62 makes sense at a glance.

The Alligation Cross: Prices, Percentages and Fractions

The cross

When to use it: the mean (price, strength or fraction) is known, and the ratio of the two varieties, or one quantity, is asked.

Fast way: put the cheaper value c and the dearer value d at the top, the mean m in the middle, and cross the differences.

$$\text{cheaper} : \text{dearer} = (d - m) : (m - c)$$

Each variety gets the difference from the other side.

For example: rice at ₹42 and ₹57 per kg is mixed so that the mix costs ₹51 per kg. Cheaper : dearer = (57 − 51) : (51 − 42) = 6 : 9 = 2 : 3. Check forward: 42 + 15 × $\frac{3}{5}$ = 51.

Percentages and strengths

The cross works on any "per" value: % of acid, % of milk, % of gold. Water is a 0% solution and a pure liquid is 100%.

For example: how much 22% acid must be added to 15 litres of 58% acid to get 31% acid? The 22% acid : 58% acid ratio is (58 − 31) : (31 − 22) = 27 : 9 = 3 : 1, so you need three times as much 22% acid, 45 litres. Check: 9.9 + 8.7 = 18.6 litres of acid in 60 litres is 31%.

Ratios of two mixtures: turn them into fractions

Fast way: write the fraction of the same component in each mixture and in the target, all over one denominator, and cross the numerators.

For example: milk : water is 4 : 1 in vessel A and 1 : 3 in vessel B. In what ratio should they be mixed to get 3 : 2? Over 20ths, milk is $\frac{16}{20}$, $\frac{5}{20}$ and $\frac{12}{20}$. Cross 16, 5 and 12: A gets 12 − 5 = 7 and B gets 16 − 12 = 4, so A : B = 7 : 4.

Where it fails:

• Flipping the answer. The number written against the cheaper variety comes from the dearer price. A quick forward check (the mean should be nearer the bigger share) catches a flip.
• Mixing fractions of different parts, such as milk from A and water from B. Use the same component throughout.
• Ratio of what? The cross gives the ratio of whatever the value is "per". Prices per kg give a ratio by weight, and a strength in % by volume gives a ratio by volume.

[SSC Tip] Water makes the cross even quicker. To bring 21 litres of 64% acid down to 48%, cross 64 and 0 around 48: water : acid solution = 16 : 48 = 1 : 3, so add 7 litres of water.

Alligation for Averages That Aren't Mixtures

When to use it: a whole is split into two parts that earn two different rates, and the overall rate is given. Interest rates on two parts of a sum, profit % on two parts of a stock, speeds over two parts of a journey, pass rates of two schools.

Fast way: the same cross. The ratio it gives is of whatever the rate is "per".

• An interest rate is per ₹100 of principal, so the cross gives the ratio of the sums lent.
• Profit % is per ₹100 of cost, so it gives the ratio of the cost prices (or of the quantities, if the cost per kg is the same).
• Speed is distance per hour, so it gives the ratio of times, not of distances.

For example (interest): ₹30,000 is lent in two parts, at 7% and at 12% simple interest a year. The yearly interest is ₹2,700, which is 9% of the whole. Cross: 7% part : 12% part = (12 − 9) : (9 − 7) = 3 : 2, so ₹18,000 is at 7% and ₹12,000 at 12%.

Another one (profit and loss): a trader sells part of his stock at a 10% loss and the rest at a 15% profit, and makes 5% overall. Write the loss as −10: loss part : profit part = (15 − 5) : (5 + 10) = 10 : 15 = 2 : 3.

One more (speeds): a car covers 230 km in 5 hours, partly at 40 km/h and partly at 55 km/h. The average is 46 km/h. Time at 40 : time at 55 = (55 − 46) : (46 − 40) = 9 : 6 = 3 : 2, so 3 hours at 40 km/h (120 km) and 2 hours at 55 km/h (110 km).

Where it fails:

• With speeds, using 3 : 2 as the ratio of distances gives 138 km at 40 km/h. That's wrong, because speed is per hour, not per km.
• If the two distances are equal, the average speed is $\frac{2ab}{a + b}$, the same rule as equal money spent on two prices.
• Compound interest over several years doesn't allige cleanly. Keep the cross for simple interest (or for a single year).

The cross also handles two groups' averages. If boys average 64 marks, girls 72 and the whole class 67, then boys : girls = (72 − 67) : (67 − 64) = 5 : 3.

Alligation with Profit, and Free Water

Find the cost of the mix first

When to use it: the mix is sold at a given price with a given gain (or loss) %, and the mixing ratio or a quantity is asked.

Fast way: turn the selling price into the cost price of the mix, then allige the cost prices.

$$\text{CP of mix} = \text{SP} \times \frac{100}{100 + \text{gain}}$$

For a loss, put 100 − loss in the bottom.

For example: teas at ₹290 and ₹350 per kg are mixed, and the mix is sold at ₹396 per kg for a 20% gain. CP of the mix = 396 × $\frac{100}{120}$ = ₹330. Cross: (350 − 330) : (330 − 290) = 20 : 40 = 1 : 2.

Another one, with a quantity: how many kg of rice at ₹56 must be mixed with 36 kg at ₹38 so that selling the mix at ₹55.20 per kg gives a 15% profit? CP of the mix = 55.20 ÷ 1.15 = ₹48. Cross: ₹38 rice : ₹56 rice = (56 − 48) : (48 − 38) = 8 : 10 = 4 : 5. The 36 kg are 4 parts, so 45 kg of the ₹56 rice.

Free water

Water costs ₹0, so it is the cheaper variety at zero.

• Sold at the cost price of milk with an x% gain: all the gain comes from the water, so water : milk = x : 100. A 10% gain means water : milk = 1 : 10.
• Sold at some other price: find the CP of the mix, then cross milk and ₹0. Milk costs ₹77 a litre, and the mix sells at ₹73.50 a litre, below the price of milk, yet for a 5% gain. CP of the mix = 73.50 ÷ 1.05 = ₹70, so milk : water = 70 : (77 − 70) = 70 : 7 = 10 : 1.

Where it fails:

• Alliging with the selling price. In the tea example ₹396 is above both prices, so no cross is possible. If the "mean" lies outside the two prices, you forgot to take out the profit.
• The gain is on cost, so divide by 1.15. Taking 15% off ₹55.20 gives ₹46.92, which is wrong.
• Water : milk = x : 100 needs the mix sold at the cost price of pure milk. At any other price, use the cross.

Three or More Varieties

Pair them across the mean

When to use it: three or more varieties make a mix of known price, and the ratio is asked.

Fast way: pair each variety below the mean with one above it, allige each pair, and add up what each variety gets. There is more than one correct answer, so in the exam you test the options rather than build your own.

For example: rice at ₹29, ₹35 and ₹51 per kg is mixed to cost ₹41 per kg. Pair (29, 51): 29 gets 10 and 51 gets 12. Pair (35, 51): 35 gets 10 and 51 gets 6. Together, 29 : 35 : 51 = 10 : 10 : 18 = 5 : 5 : 9. Check: 145 + 175 + 459 = 779, and 779 ÷ 19 = 41.

Test the options in one line

Fast way: take the mean as the base. A ratio x : y : z works if x × (first difference) + y × (second) + z × (third) = 0.

With the same rice: the differences from ₹41 are −12, −6 and +10.

• 1 : 3 : 3 gives −12 − 18 + 30 = 0, so it works.
• 2 : 1 : 3 gives −24 − 6 + 30 = 0, so it works too.
• 1 : 1 : 2 gives −12 − 6 + 20 = 2, so it doesn't.

That's why a question with three varieties usually asks "which of these can be the ratio".

Two of them in a fixed ratio

If two varieties are mixed in a fixed ratio, combine them into one variety first, then allige with the third.

For example: teas at ₹62 and ₹71 per kg and a third tea are mixed in the ratio 1 : 1 : 2, and the mix is worth ₹80 per kg. The first two together are 2 kg averaging ₹66.50, and the third is also 2 kg, so ₹80 is exactly halfway. The third tea costs 80 + (80 − 66.50) = ₹93.50 per kg.

Where it fails:

• All the varieties on one side of the mean. Then no mix works, and any option that claims one is wrong.
• Treating the ratio you built as the only answer. If it isn't in the options, test each option, because another valid ratio will be there.

[SSC Tip] A variety priced exactly at the mean can go in any amount without changing the price. When testing an option, you only need to balance the other varieties.

Removal and Replacement: The Fraction Left Multiplies

The formula

When to use it: x litres are drawn out of V litres of a liquid and replaced with water (or another liquid), and this is done n times.

Fast way: each round keeps $\left(1 - \frac{x}{V}\right)$ of the liquid that was there. After n rounds,

$$\text{liquid left} = V\left(1 - \frac{x}{V}\right)^n$$

For example: from 75 litres of pure milk, 15 litres are drawn out and replaced with water, three times. Each round keeps $\frac{4}{5}$, so milk left = 75 × $\frac{64}{125}$ = 38.4 litres. Water is 36.6 litres, and milk : water = 64 : 61. The ratio needs no volume at all: milk is $\frac{64}{125}$ of the vessel.

Finding x or n

Finding x: take the n-th root of the fraction left. 112 litres of pure acid go through two rounds and end with acid : water = 49 : 15. Acid is $\frac{49}{64} = \left(\frac{7}{8}\right)^2$ of the vessel, so each round keeps $\frac{7}{8}$ and x = $\frac{1}{8}$ of 112 = 14 litres.

Finding n: 54 litres are drawn from 162 litres of pure juice and replaced with water each time. Each round keeps $\frac{2}{3}$, and 162 × $\left(\frac{2}{3}\right)^4$ = 32, so it takes 4 rounds to bring the juice down to 32 litres.

Different amounts each round: multiply each round's own fraction. Replacing 15 litres and then 20 litres of 60 litres of milk leaves 60 × $\frac{45}{60}$ × $\frac{40}{60}$ = 30 litres.

Hitting a target in one step

When to use it: "What part of the mixture must be drawn off and replaced with water to make it …?"

Fast way: the part that isn't put back only shrinks, by the fraction kept. So the fraction to replace = 1 − $\frac{\text{new share}}{\text{old share}}$ of that part.

For example: milk : water is 8 : 1, so milk is $\frac{8}{9}$. For 2 : 1, milk must be $\frac{2}{3}$. The fraction kept is $\frac{2}{3} \div \frac{8}{9} = \frac{3}{4}$, so $\frac{1}{4}$ of the mixture must be drawn off and replaced with water. If the refill is the other liquid (say pure milk), do the same with the water.

Where it fails:

• Adding the changes. Three rounds that each take out 20% leave $(0.8)^3$ = 51.2% of the milk, not 40%. In the first example, subtracting 15 litres three times gives 30 litres, a common wrong option.
• Using the formula when the refill is a mixture rather than water or the other pure liquid. Then track both parts round by round.
• If the vessel starts with a mixture, apply the formula only to the part that isn't put back, such as the milk in milk and water topped up with water.

Option Elimination

When to use it: first, before the full working. Sometimes it settles the question on its own.

Fast way: test the options against these four checks.

• The mean lies between. A mix price (or strength) is always between the cheapest and the dearest variety, and nearer the one there is more of.
• The ratio's sum divides the total. If a mixture in the ratio 7 : 5 holds a whole number of litres of each part, its total is a multiple of 12, and each part is a multiple of the unit.
• Direction. Adding water lowers the strength, adding pure liquid raises it, and every round of replacement with water lowers what is left.
• Replacement range. After n rounds, the liquid left is more than V − nx, because every round after the first takes out some water too.

For example: rice at ₹44 and ₹62 per kg is mixed so that the mix costs ₹56 per kg. How much of the ₹62 rice is there in 54 kg of the mix: (a) 18 kg (b) 36 kg (c) 30 kg (d) 40 kg?

₹56 is nearer ₹62, so there is more of the dearer rice, more than half of 54 kg. That rules out (a). The cross gives cheaper : dearer = (62 − 56) : (56 − 44) = 6 : 12 = 1 : 2, so the dearer rice is $\frac{2}{3}$ of 54 = (b) 36 kg. Notice that (a) 18 kg is what a flipped ratio gives.

Another one: 75 litres of milk go through three rounds of 15 litres replaced with water. If an option says 30 litres are left, it fails the range check at once: 75 − 3 × 15 = 30 is the floor that the answer must stay above.

Where it fails:

• The whole-number check needs whole-number parts. Prices, percentages and some quantities come in fractions, so use it only when the options are whole numbers of litres or kg.
• When all four options pass the checks, you still need the method.

[SSC Tip] Trap options usually come from one slip: a flipped ratio, alliging with the selling price, putting the new ratio on the old total, or subtracting replacement rounds instead of multiplying. If your answer matches an option that one of those slips gives, recheck that one step before you mark it.

Solved Examples

Every answer below uses the quick method and gives a rough time.

Question 1

A 108-litre mixture has milk and water in the ratio 7 : 5. How much milk must be added to make the ratio 3 : 2?

Answer. Only milk goes in, so the water stays fixed. I match the water numbers: 7 : 5 = 14 : 10 and 3 : 2 = 15 : 10. Milk goes up by 1 unit, and the old mixture is 24 of these units. 108 ÷ 24 = 4.5, so 4.5 litres of milk must be added. Check: 67.5 litres of milk against 45 of water is 3 : 2.

Time check: about 20 seconds.

Question 2

75 litres of a solution are 36% alcohol. How much water must be added to bring the alcohol down to 27%?

Answer. The alcohol stays fixed, so the new total = 75 × $\frac{36}{27}$ = 100 litres. That means 25 litres of water. Check: 27 litres of alcohol in 100 litres is 27%.

Time check: about 15 seconds.

Question 3

A drum holds three oils P, Q and R in the ratio 5 : 3 : 4. 36 litres of the mixture are drawn off, and then 3 litres of P and 9 litres of Q are poured in. Now there is 12 litres more of P than of Q. How much oil was in the drum at first?

Answer. The additions gave Q 6 litres more than P (9 − 3), so before them P was 12 + 6 = 18 litres ahead. In the leftover that gap is 5 − 3 = 2 units, so one unit is 9 litres. The leftover is 12 units = 108 litres, and the drum first held 108 + 36 = 144 litres. Check: 144 down to 108 gives P = 45 and Q = 27; after the additions P = 48 and Q = 36.

Time check: about 30 seconds. Equations in x take a minute or more.

Question 4

Two vessels of the same size are full of milk and water, one in the ratio 4 : 1 and the other in the ratio 5 : 7. Both are emptied into a big tub. Find the ratio of milk to water in the tub.

Answer. Equal quantities, so I make the ratio sums equal. The LCM of 5 and 12 is 60: 4 : 1 = 48 : 12 and 5 : 7 = 25 : 35. Adding part by part gives 73 : 47.

Time check: about 20 seconds. Adding 4 : 1 and 5 : 7 as they stand gives 9 : 8, which is wrong.

Question 5

Jar A has acid and water in the ratio 2 : 3, and jar B has them in the ratio 7 : 1. 30 litres from A and 8 litres from B are poured into an empty jar C. What percentage of the liquid in C is acid?

Answer. I track only the acid. From A: 30 × $\frac{2}{5}$ = 12 litres. From B: 8 × $\frac{7}{8}$ = 7 litres. That is 19 litres of acid in 38 litres, which is 50%.

Time check: about 20 seconds.

Question 6

A trader spends ₹2,160 on each of three kinds of pulses, priced at ₹36, ₹54 and ₹108 per kg, and mixes them. What does the mix cost per kg?

Answer. Equal money, not equal quantity. ₹2,160 buys 60 kg, 40 kg and 20 kg, so ₹6,480 buys 120 kg, and the mix costs ₹54 per kg. The plain average of the prices, ₹66, is the trap.

Time check: about 20 seconds.

Question 7

How many kg of wheat at ₹34 per kg must be mixed with 42 kg of wheat at ₹27 per kg so that the mix costs ₹31 per kg?

Answer. Cross: ₹27 wheat : ₹34 wheat = (34 − 31) : (31 − 27) = 3 : 4. The 42 kg are 3 units, so one unit is 14 kg and the ₹34 wheat is 56 kg. Check: 1,134 + 1,904 = 3,038, and 3,038 ÷ 98 = 31.

Time check: about 15 seconds.

Question 8

How many litres of 8% acid must be added to 30 litres of 36% acid to get 16% acid?

Answer. Cross 8 and 36 around 16: 8% acid : 36% acid = (36 − 16) : (16 − 8) = 20 : 8 = 5 : 2. The 30 litres are 2 units, so one unit is 15 litres and I need 75 litres of the 8% acid. Check: 6 + 10.8 = 16.8 litres of acid in 105 litres is 16%.

Time check: about 20 seconds.

Question 9

Vessel A has milk and water in the ratio 5 : 3, and vessel B has them in the ratio 3 : 5. In what ratio must A and B be mixed to get milk and water in the ratio 9 : 7?

Answer. Milk is $\frac{5}{8}$ of A, $\frac{3}{8}$ of B and $\frac{9}{16}$ of the target. Over 16ths these are 10, 6 and 9. Cross: A gets 9 − 6 = 3 and B gets 10 − 9 = 1, so A : B = 3 : 1.

Time check: about 20 seconds.

Question 10

A dealer bought 80 kg of onions. He sold part of them at a 12% profit and the rest at a 4% loss, and made a 7% profit on the whole lot. How many kg did he sell at a profit?

Answer. I write the loss as −4 and cross around 7: profit part : loss part = (7 + 4) : (12 − 7) = 11 : 5. That's 16 units for 80 kg, so 5 kg a unit and 55 kg went at a profit. Check: 55 × 12 − 25 × 4 = 560 = 80 × 7.

Time check: about 20 seconds.

Question 11

Two kinds of rice at ₹49 and ₹64 per kg are mixed, and the mix is sold at ₹63.25 per kg for a 15% profit. In what ratio were they mixed?

Answer. First the cost of the mix: 63.25 ÷ 1.15 = ₹55. Cross: (64 − 55) : (55 − 49) = 9 : 6 = 3 : 2 (cheaper : dearer). Check: (147 + 128) ÷ 5 = 55.

Time check: about 25 seconds. Crossing ₹63.25 directly gives 0.75 : 14.25, a lopsided ratio that tells you the profit wasn't taken out.

Question 12

A milkman buys milk at ₹52 a litre, mixes in water, and sells the mixture at ₹54.60 a litre, making a 40% profit. Find the ratio of water to milk in the mixture.

Answer. Cost of the mix = 54.60 ÷ 1.4 = ₹39 a litre. Cross milk (₹52) and water (₹0) around ₹39: milk : water = 39 : 13 = 3 : 1, so water : milk = 1 : 3. Check: 3 litres of milk cost ₹156, the 4 litres sell for ₹218.40, and the ₹62.40 profit is 40%.

Time check: about 20 seconds.

Question 13

Tea at ₹150, ₹185 and ₹240 per kg is mixed to get tea worth ₹200 per kg. Which of these can be the mixing ratio: 2 : 1 : 3, 1 : 1 : 1, 1 : 2 : 2 or 2 : 2 : 3?

Answer. The differences from ₹200 are −50, −15 and +40. Testing each: 2 : 1 : 3 gives +5, 1 : 1 : 1 gives −25, 2 : 2 : 3 gives −10, and 1 : 2 : 2 gives −50 − 30 + 80 = 0. So 1 : 2 : 2 works. Check: 150 + 370 + 480 = 1,000 for 5 kg, which is ₹200 a kg.

Time check: about 25 seconds.

Question 14

A can holds 140 litres of pure milk. 35 litres are taken out and replaced with water, and this is done three times in all. Find the ratio of milk to water at the end.

Answer. Each round keeps $\frac{3}{4}$ of the milk, so after three rounds milk is $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ of the can. Water is the other $\frac{37}{64}$, so milk : water = 27 : 37. The 140 litres don't even matter for the ratio.

Time check: about 15 seconds.

Question 15

A 180-litre mixture has alcohol and water in the ratio 7 : 3. How many litres of it must be drawn off and replaced with pure alcohol to make the ratio 5 : 1?

Answer. Pure alcohol goes back in, so I track the water, which only goes down. Water is $\frac{3}{10}$ now and must become $\frac{1}{6}$. The fraction kept is $\frac{1}{6} \div \frac{3}{10} = \frac{5}{9}$, so $\frac{4}{9}$ of the mixture, 80 litres, must be replaced. Check: 126 and 54 litres become 70 and 30 after 80 litres come out, and adding 80 litres of alcohol gives 150 : 30 = 5 : 1.

Time check: about 25 seconds.