Shortcuts & Speed Methods

Combining Ratios

Most ratio questions in Tier 1 come down to a few patterns: two ratios joined through a common term, an amount split in a ratio, a ratio that changes when something is added, and profit shared by capital × months. You get about 36 seconds a question. Each method below says when to use it, works one small question, and lists where it goes wrong, such as joining 3 : 4 and 6 : 5 into 3 : 4 : 5 without matching B.

Make the common term equal

When to use it: you are given A : B and B : C, and the question wants A : B : C or A : C.

Fast way: B has two different values in the two ratios. Make them equal using their LCM, scale each ratio to match, and read off all three.

For example: A : B = 3 : 4 and B : C = 6 : 5. B is 4 in one ratio and 6 in the other, and the LCM is 12. So A : B = 9 : 12 and B : C = 12 : 10, which gives A : B : C = 9 : 12 : 10.

For A : C alone you don't need the LCM at all. Multiply the first terms together and the second terms together: A : C = (3 × 6) : (4 × 5) = 18 : 20 = 9 : 10.

The multiply-down chain for A : B : C : D

When to use it: three links, A : B, B : C and C : D, and you want all four or only A : D.

Fast way: write the ratios one under another as a : b, c : d and e : f. Then

$$A : B : C : D = ace : bce : bde : bdf$$

The pattern is easy to remember. A takes all three first terms and D takes all three second terms. Each term in between swaps one more first term for a second term as you go down the list.

For example: A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7.

• A = 2 × 4 × 6 = 48
• B = 3 × 4 × 6 = 72
• C = 3 × 5 × 6 = 90
• D = 3 × 5 × 7 = 105

Dividing by 3, A : B : C : D = 16 : 24 : 30 : 35, and A : D = 16 : 35. Check one link: B : C = 24 : 30 = 4 : 5. If the question asks only for A : D, skip the full chain: A : D = (product of the first terms) : (product of the second terms), which is 48 : 105 = 16 : 35 here.

Where it fails:

• The links must run in order. If you are given A : B and C : B, flip the second one to B : C before you start.
• Gluing 3 : 4 and 6 : 5 into 3 : 4 : 5 ignores that B is 4 in one ratio and 6 in the other. That trap is in the options more often than you'd think.
• The options usually show the reduced form, so divide out any common factor at the end.

Equal Products and Equal Fractions

2A = 3B = 4C: flip, then clear with the LCM

When to use it: a few quantities multiplied by different numbers come out equal, as in 2A = 3B = 4C, or "5 times A's share equals 6 times B's share".

Fast way: the ratio is made of the reciprocals: A : B : C = $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$. Multiply by the LCM of the bottoms (12) to clear the fractions, and you get 6 : 4 : 3.

With only two quantities it is a plain cross-swap: 5A = 6B means A : B = 6 : 5.

For example: 3A = 5B = 6C. The reciprocals are $\frac{1}{3}$, $\frac{1}{5}$ and $\frac{1}{6}$, and the LCM is 30, so A : B : C = 10 : 6 : 5. Check: 3 × 10 = 5 × 6 = 6 × 5 = 30.

Where it fails: copying the multipliers. 3 : 5 : 6 is the trap, and it is nearly always in the options. The biggest multiplier goes with the smallest share.

A/3 = B/4 = C/5: copy the bottoms

When to use it: the quantities are divided by different numbers instead of multiplied.

Fast way: each one is its bottom number times the same k, so $\frac{A}{3} = \frac{B}{4} = \frac{C}{5}$ gives A : B : C = 3 : 4 : 5 straight away.

If the two forms are mixed, as in $\frac{2A}{3} = \frac{3B}{4} = \frac{4C}{5}$, then A is $\frac{3}{2}k$, B is $\frac{4}{3}k$ and C is $\frac{5}{4}k$. Clearing with 12 gives A : B : C = 18 : 16 : 15.

A fraction of A equals a fraction of B

When to use it: "two-fifths of A's money equals three-sevenths of B's money", or "A's $\frac{1}{3}$ = B's $\frac{1}{4}$".

Fast way: cross-swap the fractions. If $\frac{2}{5}$ of A = $\frac{3}{7}$ of B, then A : B = $\frac{3}{7} : \frac{2}{5}$ = 15 : 14. For A's $\frac{1}{3}$ = B's $\frac{1}{4}$, it is simply A : B = 3 : 4.

To check the first one: $\frac{2}{5}$ × 15 = 6 and $\frac{3}{7}$ × 14 = 6. They match, so 15 : 14 is right.

Three people work the same way. If $\frac{1}{2}$ of A = $\frac{2}{3}$ of B = $\frac{3}{4}$ of C, then A : B : C = $2 : \frac{3}{2} : \frac{4}{3}$, and multiplying by 6 gives 12 : 9 : 8.

Where it fails: the same swap trap. Writing A : B = $\frac{2}{5} : \frac{3}{7}$ = 14 : 15 is backwards. The person with the bigger fraction has the smaller amount. "a% of A = b% of B" works the same way with percentages.

The Units Method for Dividing Amounts

Think of a ratio as units. In 4 : 5 : 8, the three people hold 4, 5 and 8 units of the same size. Find what one unit is worth, and every question below takes a single division.

Splitting a total

When to use it: "Divide ₹N among A, B and C in the ratio a : b : c."

Fast way: one unit = N ÷ (a + b + c). Then multiply only the share you need.

For example: ₹7,650 among P, Q and R in the ratio 4 : 5 : 8. There are 17 units, and 7,650 ÷ 17 = 450. R gets 8 × 450 = ₹3,600. You never need P's or Q's share.

"A gets ₹x more than B"

Fast way: the gap in units is worth ₹x.

For example: money is split between Rohan and Sana in the ratio 7 : 4, and Rohan gets ₹1,260 more. The gap is 3 units, so 1 unit = ₹420. Sana gets ₹1,680 and the total is 11 × 420 = ₹4,620.

Ratios given as fractions, or "inversely as"

Fast way: clear the fractions with the LCM first, then split. "Inversely proportional to 3, 4 and 6" means $\frac{1}{3} : \frac{1}{4} : \frac{1}{6}$, which is 4 : 3 : 2 after multiplying by 12.

For example: ₹6,300 divided in inverse proportion to 3, 4 and 6. That is 4 : 3 : 2, so 9 units at ₹700 each, and the shares are ₹2,800, ₹2,100 and ₹1,400. Splitting in 3 : 4 : 6 instead is the trap: it gives the biggest share to the person who should get the least.

Numbers from their ratio and their sum, difference, product or squares

Take the numbers as ak and bk.

• Sum or difference: use units, as above.
• Product: $ab \times k^2$ is the product, so $k^2$ = product ÷ ab.
• Sum of squares: $k^2$ = (sum of squares) ÷ $(a^2 + b^2)$.

For example: two numbers in the ratio 3 : 5 have a product of 960. Then $k^2$ = 960 ÷ 15 = 64, so k = 8 and the numbers are 24 and 40. Two numbers in the ratio 2 : 3 have squares that add up to 1,053. Then $k^2$ = 1,053 ÷ 13 = 81, so k = 9 and the numbers are 18 and 27.

Where it fails:

• With a product or squares you get $k^2$, not k. Stopping at 64 and writing the numbers as 192 and 320 is the trap.
• Split the right total. If a prize is shared after ₹500 goes to charity, split what is left, not the full prize.

Changing Ratios: the Fixed Part and the Fixed Gap

When numbers are added to or taken out of a ratio, ask one question first: what doesn't change? It is either one of the two parts, or the gap between them.

One part stays fixed: match its units

When to use it: something is added to or taken from only one part. New girls join and the boys stay the same, or blue balls are added and the red ones stay the same.

Fast way: make the fixed part's terms equal in both ratios using the LCM. The change in the other part's units is the number added or removed.

For example: boys : girls in a hostel = 7 : 5. After 39 more girls join, the ratio becomes 3 : 4. The boys don't change, so make their terms equal: the LCM of 7 and 3 is 21, and the ratios become 21 : 15 and 21 : 28. The girls went from 15 units to 28 units, so 13 units = 39 and 1 unit = 3. There were 63 boys and 45 girls.

The same number added to both: match the gap

When to use it: the same amount is added to (or taken from) both parts. Ages are the classic case, and so is "if 10 is added to each number".

Fast way: the difference between the two parts can't change. Make the gaps of the old and new ratios equal, and the change in either part's units is the number added.

For example: two numbers are in the ratio 4 : 7. When 10 is added to each, the ratio becomes 3 : 4. The gaps are 3 and 1, so scale the new ratio by 3 to get 9 : 12. The first number went from 4 units to 9 units, so 5 units = 10 and 1 unit = 2. The numbers are 8 and 14. Check: 18 : 24 = 3 : 4.

When one part goes up and the other goes down, the gap changes by the total of both moves. Question 3 below shows this.

The number to add to both terms

When to use it: "What number must be added to each of a and b so that they are in the ratio c : d?"

Fast way: solving (a + x) : (b + x) = c : d once gives

$$x = \frac{ad - bc}{c - d}$$

A negative answer means the number has to be subtracted.

For example: what must be added to 7 and 16 to make the ratio 2 : 3? x = $\frac{7 \times 3 - 16 \times 2}{2 - 3} = \frac{-11}{-1}$ = 11. Check: 18 : 27 = 2 : 3.

For four numbers a, b, c and d that should become proportional, the same idea gives

$$x = \frac{bc - ad}{(a + d) - (b + c)}$$

What must be taken from 14, 17, 26 and 33 to make them proportional? x = $\frac{17 \times 26 - 14 \times 33}{47 - 43} = \frac{-20}{4}$ = −5, so subtract 5. Check: 9 : 12 = 21 : 28.

Where it fails: with four options in front of you, trying them is often quicker than any formula. Add the option to each term and see if the ratio comes out. Each try takes about five seconds, and you can't get a sign wrong. The formula also assumes the order a : b = c : d exactly as the numbers are given.

Mean, Third and Fourth Proportionals

These come up almost every year, and the only thing that really goes wrong is the order of the numbers.

When to use it: any question that uses these names.

Fast way: go straight to the formula. The middle terms multiply and the first term divides.

For example:

• Fourth proportional to 6, 15 and 14: $\frac{15 \times 14}{6}$ = 35.
• Third proportional to 12 and 18: $\frac{18^2}{12} = \frac{324}{12}$ = 27.
• Mean proportional between 8 and 50: $\sqrt{400}$ = 20.

Mixed chains

When to use it: one proportional feeds into another, like "the mean proportional between the third proportional to 4 and 6 and the fourth proportional to 3, 4 and 12".

Fast way: work from the inside out, one formula at a time, and write each result down before the next step.

For example: the third proportional to 4 and 6 is $\frac{36}{4}$ = 9. The fourth proportional to 3, 4 and 12 is $\frac{4 \times 12}{3}$ = 16. The mean proportional between 9 and 16 is $\sqrt{144}$ = 12.

Continued proportion

a, b and c are in continued proportion when a : b = b : c, so $b^2 = ac$ and b is simply the mean proportional of a and c. If 2, x and 18 are in continued proportion, $x^2$ = 36 and x = 6. For four numbers a, b, c and d in continued proportion, every step has the same ratio, so $\frac{a}{d} = \left(\frac{a}{b}\right)^3$.

Where it fails:

• The order. The third proportional to 12 and 18 is 27, but the third proportional to 18 and 12 is $\frac{144}{18}$ = 8. Both turn up in the options.
• The mean proportional is not the average. Between 8 and 50 it is 20, not 29.
• In "fourth proportional to a, b, c", the first number goes at the bottom, not the last one.

[SSC Tip] Use size to strike out options. The mean proportional of two numbers always lies between them. The third proportional to a and b, when b is bigger than a, is bigger than b, because the numbers keep growing by the same ratio.

Expressions from a Ratio, and Componendo and Dividendo

Put the ratio numbers straight in

When to use it: you know a : b (or x : y) and want the value of something like $\frac{5a + 2b}{3b - a}$.

Fast way: if every term in the expression has the same power (all plain a and b, or all $a^2$, ab and $b^2$), just put in the ratio numbers. The common k cancels out.

For example: a : b = 4 : 7. Then $\frac{5a + 2b}{3b - a} = \frac{20 + 14}{21 - 4} = \frac{34}{17}$ = 2. With squares: if x : y = 3 : 2, then $\frac{x^2 + y^2}{x^2 - y^2} = \frac{9 + 4}{9 - 4} = \frac{13}{5}$.

Where it fails: a stand-alone number breaks it. $\frac{a + 2}{b + 2}$ can't be found from a : b alone, because the 2 doesn't grow with k. Mixed powers like $\frac{a^2 + b}{a + b}$ fail the same way.

Going backwards: find x : y

When to use it: "$\frac{4x + 3y}{2x + 5y} = \frac{11}{13}$. Find x : y."

Fast way: cross-multiply, then put the x terms on one side and the y terms on the other. Here 13(4x + 3y) = 11(2x + 5y) gives 52x + 39y = 22x + 55y, so 30x = 16y and x : y = 8 : 15.

Componendo and dividendo in one line

When to use it: the top and bottom are the sum and the difference of the same two things, like $\frac{3x + 2y}{3x - 2y}$.

Fast way:

$$\text{if } \frac{a + b}{a - b} = \frac{p}{q}, \text{ then } \frac{a}{b} = \frac{p + q}{p - q}$$

It works the other way round too.

For example: $\frac{3x + 2y}{3x - 2y} = \frac{5}{3}$. Then $\frac{3x}{2y} = \frac{5 + 3}{5 - 3}$ = 4, so $\frac{x}{y} = 4 \times \frac{2}{3} = \frac{8}{3}$ and x : y = 8 : 3. Check: $\frac{24 + 6}{24 - 6} = \frac{30}{18} = \frac{5}{3}$.

The same step clears square roots. If $\frac{\sqrt{x + 4} + \sqrt{x - 4}}{\sqrt{x + 4} - \sqrt{x - 4}} = 2$, then $\frac{\sqrt{x + 4}}{\sqrt{x - 4}} = \frac{2 + 1}{2 - 1} = 3$. Squaring, x + 4 = 9(x − 4), so 8x = 40 and x = 5.

Where it fails:

• Stopping one step early. In the example above, $\frac{3x}{2y}$ = 4 does not mean x : y = 4 : 1. You still have to undo the 3 and the 2.
• It needs the same pair on top and bottom. $\frac{2x + 3y}{3x - y}$ is not a sum-and-difference pair, so cross-multiply instead.

Variation: Keep the Constant

Every variation question has one quantity that stays fixed. Write it for the old values and the new values, set the two equal, and you are done in one line.

When to use it: "varies as", "varies inversely as", "varies jointly as", or any real-life direct or inverse link like cost and quantity.

Fast way: don't find the constant unless the question asks for it. Scale the old value by the ratio of the changes.

For example:

• y varies as $x^2$, and y = 48 when x = 4. When x = 6, y = 48 × $\left(\frac{6}{4}\right)^2$ = 48 × $\frac{9}{4}$ = 108.
• y varies inversely as x, and y = 15 when x = 8. When x = 12, y = 15 × $\frac{8}{12}$ = 10.
• x varies jointly as y and z, and x = 36 when y = 3 and z = 4. When y = 5 and z = 6, x = 36 × $\frac{30}{12}$ = 90.

Partly fixed, partly varying

When to use it: "the fare is partly fixed and partly varies with the distance", or "the cost is partly constant and partly varies as the number of guests".

Fast way: subtract the two given cases. The fixed part cancels, and the rate drops out at once.

For example: a taxi charges ₹220 for 12 km and ₹340 for 20 km. The extra 8 km cost ₹120, so the rate is ₹15 a km, and the fixed part is 220 − 180 = ₹40. For 30 km the fare is 40 + 450 = ₹490.

Where it fails:

• Inverse variation goes the other way. More x means less y, so 15 × $\frac{12}{8}$ = 22.5 in the second example is the trap.
• Forgetting to square. In the first example, 48 × $\frac{6}{4}$ = 72 is wrong.
• "x varies as (y + 3)" is not "x varies as y". Keep the + 3: it is $\frac{x}{y + 3}$ that stays fixed.
• A part-fixed fare does not double when the distance doubles, because the fixed part stays put.

Income and Expenditure with Equal Savings

The cross-multiplication shortcut

When to use it: "The incomes of A and B are in the ratio a : b and their expenditures in the ratio c : d. Each saves ₹S."

Fast way: write the two ratios one under the other.

Cross-multiply to get ad − bc, and take the difference of the expenditure terms in the same order, d − c. Then

$$\text{one unit of income} = \frac{S \times (d - c)}{ad - bc}$$

A's income is a units and B's income is b units.

For example: incomes 5 : 4, expenditures 7 : 5, and each saves ₹3,000. Here ad − bc = 25 − 28 = −3 and d − c = 5 − 7 = −2. One unit = $\frac{3000 \times (-2)}{-3}$ = 2,000, so the incomes are ₹10,000 and ₹8,000. Check: they spend ₹7,000 and ₹5,000, which is 7 : 5.

Why it works, and a second fast way

Equal savings means income minus expenditure is the same for both. Moving the terms around, the gap between the two incomes equals the gap between the two expenditures. So you can also just make the two gaps equal, exactly as in age questions.

In the example, the incomes 5 : 4 have a gap of 1 and the expenditures 7 : 5 have a gap of 2. Double the incomes to 10 : 8. Now both gaps are 2 units, and A saves 10 − 7 = 3 units. So 3 units = ₹3,000, 1 unit = ₹1,000, and the incomes are ₹10,000 and ₹8,000. Same answer, and often quicker when the gaps are small.

One person only

If a person's income and expenditure are in the ratio 7 : 5 and he saves ₹6,000, the saving is 2 units. So 1 unit = ₹3,000 and his income is ₹21,000.

Where it fails:

• The shortcut needs equal savings. If A saves ₹4,000 and B saves ₹3,000, write the two equations ak − cm = 4,000 and bk − dm = 3,000 and solve them, or test the options.
• Keep the order the same top and bottom. If you take ad − bc, take d − c, not c − d. Mixing them gives a negative income, which at least tells you something went wrong.

Multiply Each Term: Coins and Percentage Changes

Coin questions and percentage-change questions look different, but both are solved the same way: each term of the ratio gets multiplied by its own number.

Coins and notes: value ratio = count ratio × face value

When to use it: coins or notes of different values are in a ratio by number, and the total value is given, or the other way round.

Fast way: multiply each count term by its coin's face value to get the value ratio. Split the total value in that ratio. Then divide each value by the face value to get the number of coins.

For example: a bag has ₹1, 50-paise and 25-paise coins in the ratio 3 : 4 : 6 by number, worth ₹156 in all. Value ratio = 3 × 1 : 4 × 0.5 : 6 × 0.25 = 3 : 2 : 1.5 = 6 : 4 : 3. That is 13 units, and 156 ÷ 13 = 12. So the ₹1 coins are worth ₹72, the 50-paise coins ₹48 and the 25-paise coins ₹36. The numbers of coins are 72, 96 and 144. Check: 72 : 96 : 144 = 3 : 4 : 6.

Going the other way, if the values are in a ratio, divide each by its face value. Values of ₹2, ₹1 and 50-paise coins in the ratio 4 : 3 : 2 mean the counts are $\frac{4}{2} : \frac{3}{1} : \frac{2}{0.5}$ = 2 : 3 : 4.

Where it fails: splitting ₹156 in the count ratio 3 : 4 : 6 treats every coin as a rupee, and SSC nearly always puts that answer among the choices. If the decimals bother you, work in paise: 100, 50 and 25.

Ratios after percentage changes

When to use it: the parts of a ratio rise or fall by different percentages, like seats in different courses, men and women in a town, or two salaries.

Fast way: multiply each term by its own multiplier, then clear the decimals.

For example: seats for Science, Commerce and Arts are in the ratio 2 : 3 : 5. They are increased by 75%, 50% and 20%. New ratio = 2 × 1.75 : 3 × 1.5 : 5 × 1.2 = 3.5 : 4.5 : 6 = 7 : 9 : 12. The total went from 10 units to 14 units, a rise of 40%.

Where it fails:

• Using the percentages themselves (75 : 50 : 20) as the new ratio.
• Averaging the percentages for the total change. The total rose 40% here, not the plain average of 75, 50 and 20 (about 48.3%), because the biggest part had the smallest rise.
• If a fixed number is added rather than a percentage, go back to the units method under Changing Ratios above.

Ages: The Gap in Units Stays the Same

Two people's ages grow by the same number of years, so the difference between their ages never changes. That one fact turns most ratio-of-ages questions into a ten-second job.

Match the gaps

When to use it: you know the ratio of two ages now and at another time, some years ago or some years from now.

Fast way: write both ratios and make their gaps (the difference of the terms) equal using the LCM. Now one unit is the same size in both ratios, and the change in either person's units is the number of years that passed.

For example: the ages of P and Q are in the ratio 7 : 4. After 6 years the ratio will be 3 : 2. The gaps are 3 and 1, so scale the second ratio by 3 to get 9 : 6. P goes from 7 units to 9 units, so 2 units = 6 years and 1 unit = 3 years. The ages are 21 and 12. Check: in 6 years, 27 : 18 = 3 : 2.

Past and future in one question

Fast way: count all the years between the two times.

For example: six years ago, the ages of A and B were in the ratio 3 : 5. Four years from now, the ratio will be 5 : 7. The gaps are 2 and 2, already equal. A goes from 3 units to 5 units, so 2 units = 6 + 4 = 10 years and 1 unit = 5. Six years ago they were 15 and 25, so now they are 21 and 31. Check: in four years, 25 : 35 = 5 : 7.

Where it fails:

• Using only 4 or only 6 years in the second example. The units change over the whole 10 years.
• Making the terms equal instead of the gaps. That is right for a fixed part, like boys who stay the same, but in ages both people change.
• If the question gives a sum instead of a second ratio, use the sum. After n years, the sum of two ages goes up by 2n, not n.
• With three people, work with two at a time.

[SSC Tip] As the years pass, a ratio of ages moves closer to 1. A future ratio is always nearer to 1 : 1 than the present one, and a past ratio is further from it. If an option breaks that rule, strike it out.

Partnership: Capital × Months, and Grazing

Profit is shared in the ratio of capital × months

When to use it: every partnership question where the capitals or the times differ.

Fast way: for each partner, multiply each capital by the number of months it stayed in the business, and add up. When a capital changes, draw a quick timeline of the year and mark each change, so no stretch gets missed. Drop the common zeros (work in thousands) and cancel common factors before you split.

For example: A starts a business with ₹42,000. B joins after 3 months with ₹64,000. At the end of the 8th month, A takes out ₹6,000 and B puts in ₹6,000 more. The profit for the year is ₹67,500.

So A : B = 480 : 600 = 4 : 5. That is 9 units, and 67,500 ÷ 9 = 7,500. A gets ₹30,000 and B gets ₹37,500.

If everyone stays for the same time, the months cancel and the profit goes in the ratio of the capitals alone. That is simple partnership.

Where it fails:

• "B joins after 3 months" means B is in for 9 months, not 3. "B leaves 3 months before the end" also means 9 months.
• The business may not run for 12 months. If the profit is for 18 months, the timeline runs to 18.
• Adding the capitals ("A put in 42 and then 36, so 78") ignores how long each amount was in.
• A loss is shared in exactly the same ratio as a profit.

Grazing: animals × months

When to use it: several people rent one pasture and graze different numbers of animals for different numbers of months.

Fast way: the rent is shared in the ratio of animals × months. If different animals are mixed, change them all into one kind first, using the equivalence given.

For example: A grazes 10 cows for 6 months, B grazes 12 cows for 5 months and C grazes 8 horses for 4 months. 2 horses eat as much as 3 cows, and the rent is ₹4,200. C's 8 horses count as 12 cows, so the shares are 60 : 60 : 48 = 5 : 5 : 4. One unit is 4,200 ÷ 14 = 300, so A and B pay ₹1,500 each and C pays ₹1,200.

Where it fails: converting the wrong way. 2 horses = 3 cows means 1 horse = 1.5 cows, not $\frac{2}{3}$ of a cow.

Take Out the Fixed Parts First

Many partnership questions add a twist: a working partner gets a salary or a commission, partners get interest on their capital, or part of the profit is kept aside. The method never changes.

1. Take out every fixed part first: salary, commission, interest on capital, charity or reserve.
2. Divide what is left in the ratio of capital × months.
3. Add each partner's fixed parts back to get his total.

Salary or commission

For example: A and B invest ₹30,000 and ₹50,000. A runs the business and takes 10% of the profit as commission, and the rest is shared in the ratio of capitals. The profit is ₹48,000. A's commission is ₹4,800, leaving ₹43,200 to split in 3 : 5. One unit is 5,400, so A gets 16,200 + 4,800 = ₹21,000 and B gets ₹27,000.

A monthly salary works the same way. ₹2,000 a month is ₹24,000 for the year, taken out before anything is split.

Interest on capital

For example: P and Q invest ₹80,000 and ₹50,000. Each gets 12% a year as interest on capital, and the rest of the profit is shared equally. The profit is ₹35,600. The interest is ₹9,600 for P and ₹6,000 for Q, which leaves ₹20,000, or ₹10,000 each. So P gets ₹19,600 and Q gets ₹16,000.

Charity or reserve first

If 10% of the profit goes to charity, only 90% is shared. When a question gives one partner's share and asks for the whole profit, run the same steps backwards: his part of the remainder, then the remainder, then the full profit.

Where it fails:

• Sharing the whole profit first and then taking off the salary. The salary comes out before the split, not after.
• "10% of the profit" and "10% of the profit left after his commission" are different. [Tier 2] In the second case his commission is $\frac{10}{110}$ of the profit, not $\frac{10}{100}$.
• If the rest is shared in the same ratio as the capitals, interest on capital at one common rate changes nothing: everyone's total stays in the capital ratio. Interest only matters when the rest is split some other way, like equally, or when someone also draws a salary.
• Forgetting to add the fixed part back. The trap answer is the share from step 2 alone.

Option Elimination

Whole numbers, multiples and direction

When to use it: before the full working, or instead of it, especially in long word problems.

Fast way: test the options against three quick checks.

• Multiples. Anything counted in whole units (people, coins, animals) must be a multiple of the sum of the ratio terms. If boys : girls = 7 : 9, the class size is a multiple of 16, and the number of boys is a multiple of 7.
• Round money. Setters keep money totals tidy. A total of ₹67,500 shared 4 : 5 divides exactly by 9, which is a good sign your ratio is right. If your units come out as ugly decimals, recheck the ratio before you go on.
• Direction. A future age ratio is nearer to 1 than the present one. A partner with more capital × months always gets more. A mean proportional lies between the two numbers.

For example: boys and girls in a class are in the ratio 4 : 5. Exactly $\frac{3}{8}$ of the boys play hockey and $\frac{2}{5}$ of the girls play basketball. Which of these could be the number of students: (a) 75 (b) 81 (c) 99 (d) 108?

The total must be a multiple of 9, so 75 is out. The boys are $\frac{4}{9}$ of the class, and their number must divide by 8. With 81 students there are 36 boys, and $\frac{3}{8}$ of 36 is 13.5. With 99 students there are 44 boys, giving 16.5. Only (d) 108 works: 48 boys (18 play hockey) and 60 girls (24 play basketball).

Where it fails: when all four options pass the checks, you still need the method. Money can also be split into paise, so the multiple check is strongest for people, animals and coins.

[SSC Tip] Once you have the final ratio, glance at the options before you work out every share. If ₹10,400 is split 4 : 5 : 4, one unit is ₹800, so every share must be a multiple of ₹800. Options that aren't can go straight away.

Solved Examples

These answers use the quick methods from this section, and each one ends with roughly how long it should take.

Question 1

A : B = 3 : 5, B : C = 4 : 7 and C : D = 2 : 3. Find A : D.

Answer. I only need A and D, so I multiply all the first terms and all the second terms: A : D = (3 × 4 × 2) : (5 × 7 × 3) = 24 : 105 = 8 : 35.

Time check: about 10 seconds. Building the whole chain with LCMs takes close to a minute.

Question 2

₹7,600 is divided among A, B and C so that 4 times A's share, 6 times B's share and 9 times C's share are all equal. Find each share.

Answer. 4A = 6B = 9C, so A : B : C = $\frac{1}{4} : \frac{1}{6} : \frac{1}{9}$. The LCM of 4, 6 and 9 is 36, which gives 9 : 6 : 4. That is 19 units, and 7,600 ÷ 19 = 400. So A gets ₹3,600, B gets ₹2,400 and C gets ₹1,600. Check: 4 × 3,600 = 6 × 2,400 = 9 × 1,600 = 14,400.

Time check: about 20 seconds.

Question 3

Men and women in a club are in the ratio 5 : 3. Then 16 men leave and 8 women join, and the numbers become equal. How many men were there at first?

Answer. The men were 2 units ahead. After the change they are level, so the gap closed by 16 + 8 = 24 people. So 2 units = 24, 1 unit = 12, and there were 5 × 12 = 60 men (and 36 women). Check: now there are 44 men and 44 women.

Time check: about 15 seconds.

Question 4

What number must be added to each of 13 and 41 so that they are in the ratio 3 : 7?

Answer. With x = $\frac{ad - bc}{c - d}$, where a = 13, b = 41, c = 3 and d = 7: x = $\frac{91 - 123}{3 - 7} = \frac{-32}{-4}$ = 8. Check: 21 : 49 = 3 : 7.

Time check: about 15 seconds. With options in front of you, adding each one and testing is just as quick.

Question 5

x is the third proportional to 9 and 12, and y is the fourth proportional to 3, 4 and 27. Find the mean proportional between x and y.

Answer. x = $\frac{12^2}{9} = \frac{144}{9}$ = 16. y = $\frac{4 \times 27}{3}$ = 36. Mean proportional = $\sqrt{16 \times 36}$ = 4 × 6 = 24.

Time check: about 20 seconds.

Question 6

If $\frac{7a + 4b}{7a - 4b} = \frac{9}{5}$, find a : b.

Answer. Componendo and dividendo: $\frac{7a}{4b} = \frac{9 + 5}{9 - 5} = \frac{14}{4} = \frac{7}{2}$. So $\frac{a}{b} = \frac{7}{2} \times \frac{4}{7}$ = 2, and a : b = 2 : 1. Check: $\frac{14 + 4}{14 - 4} = \frac{18}{10} = \frac{9}{5}$.

Time check: about 15 seconds. Cross-multiplying (35a + 20b = 63a − 36b) also works but takes longer.

Question 7

The price of a round rug varies as the square of its radius. A rug of radius 1.2 m costs ₹2,160. What does a rug of radius 1.8 m cost?

Answer. The radius grows by $\frac{1.8}{1.2} = \frac{3}{2}$, so the price grows by $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$. New price = 2,160 × $\frac{9}{4}$ = ₹4,860. If you don't square, you get ₹3,240.

Time check: about 15 seconds.

Question 8

The incomes of Kavya and Meher are in the ratio 7 : 5 and their expenditures in the ratio 3 : 2. Each of them saves ₹5,000. Find Kavya's income.

Answer. Cross-multiplying: ad − bc = 14 − 15 = −1, and d − c = 2 − 3 = −1. One unit = 5,000 × $\frac{-1}{-1}$ = 5,000. Kavya's income = 7 × 5,000 = ₹35,000. Check with the gap method: the incomes have a gap of 2 and the expenditures a gap of 1, so the expenditures become 6 : 4. Kavya saves 7 − 6 = 1 unit, which is ₹5,000.

Time check: about 20 seconds.

Question 9

A box has ₹2, ₹1 and 50-paise coins in the ratio 2 : 5 : 6 by number. Their total value is ₹480. How many 50-paise coins are there?

Answer. Value ratio = 2 × 2 : 5 × 1 : 6 × 0.5 = 4 : 5 : 3. That is 12 units, so 1 unit = ₹40. The 50-paise coins are worth 3 × 40 = ₹120, which is 240 coins.

Time check: about 20 seconds. Splitting ₹480 in 2 : 5 : 6 doesn't even come out exact, a sure sign it is the wrong ratio.

Question 10

Six years ago, the ages of Ravi and Sunil were in the ratio 5 : 8. Six years from now, they will be in the ratio 7 : 10. Find Ravi's present age.

Answer. Both ratios have a gap of 3, so the units already match. Ravi goes from 5 units to 7 units over 12 years, so 2 units = 12 years and 1 unit = 6. Six years ago Ravi was 30, so now he is 36. Check: Sunil was 48 and is now 54, and in six years 42 : 60 = 7 : 10.

Time check: about 15 seconds.

Question 11

Boys and girls in a school are in the ratio 8 : 7. The number of boys goes up by 25% and the number of girls by 20%, and the school now has 920 students. How many girls were there at first?

Answer. New ratio = 8 × 1.25 : 7 × 1.2 = 10 : 8.4 = 25 : 21. That is 46 units for 920 students, so 1 unit = 20 and there are now 21 × 20 = 420 girls. Before the 20% rise there were 420 ÷ 1.2 = 350 girls. Check: the school had 400 boys and 350 girls, and now has 500 + 420 = 920.

Time check: about 30 seconds.

Question 12

P starts a business with ₹81,000. After 4 months Q joins with ₹1,08,000. After 2 more months P takes out ₹27,000, and Q leaves 2 months before the year ends. The profit for the year is ₹99,000. Find Q's share.

Answer. In thousands, P has 81 for 6 months and 54 for 6 months: 486 + 324 = 810. Q is in from month 4 to month 10, which is 6 months: 108 × 6 = 648. So P : Q = 810 : 648 = 5 : 4. One unit is 99,000 ÷ 9 = 11,000, so Q gets ₹44,000.

Time check: about 35 seconds. Drawing the year from 0 to 12 and marking months 4, 6 and 10 makes sure no stretch is missed.

Question 13

A and B invest ₹84,000 and ₹56,000. A runs the business and draws a salary of ₹3,000 a month. The year's profit is ₹1,06,000. How much does A get in all?

Answer. Salary first: 3,000 × 12 = 36,000. That leaves 70,000 to split in 84 : 56 = 3 : 2, so A gets 42,000 of it. A's total = 36,000 + 42,000 = ₹78,000. B gets ₹28,000.

Time check: about 20 seconds.

Question 14

X and Y invest ₹1,40,000 and ₹84,000 for a year. From the profit of ₹62,400, each first gets 10% interest on capital. Then Y takes 20% of what is left for running the business, and the rest is shared in the ratio of capitals. What is Y's total?

Answer. Interest: X gets 14,000 and Y gets 8,400, so 22,400 goes out and 40,000 is left. Y's 20% of that is 8,000, which leaves 32,000. The capitals are 140 : 84 = 5 : 3, so Y's part is $\frac{3}{8}$ × 32,000 = 12,000. Y's total = 8,400 + 8,000 + 12,000 = ₹28,400. X gets ₹34,000, and the two add up to ₹62,400.

Time check: about 45 seconds. That is long, but every step is a single multiplication. Tier 2 sometimes stacks two fixed parts like this.

Question 15

A, B and C rent a field for ₹10,400. A grazes 18 sheep for 4 months, B grazes 6 cows for 5 months and C grazes 24 sheep for 3 months. If 1 cow eats as much as 3 sheep, how much does B pay?

Answer. I change everything into sheep. B's 6 cows count as 18 sheep. Sheep-months: A = 72, B = 90 and C = 72, which is 4 : 5 : 4. That is 13 units, and 10,400 ÷ 13 = 800. B pays 5 × 800 = ₹4,000.

Time check: about 25 seconds. Without converting, B has only 30 animal-months against 72 and 72, and a much smaller share.