Shortcuts & Speed Methods

Multipliers, the Fraction Table and CP = 100

Profit and loss is percentage work with prices attached. In Tier 1 you get about 36 seconds a question, and nearly every method below comes down to writing each change as a multiplier. Each one says when to use it, shows it on a short sum and lists where it goes wrong. Here that is usually a percentage taken on the wrong price: the SP in place of the CP, or the CP in place of the MP.

The multiplier

When to use it: CP and a profit or loss % are known and SP is asked, or SP is known and CP is asked.

Fast way: a p% profit multiplies CP by $\frac{100 + p}{100}$, and an l% loss multiplies it by $\frac{100 - l}{100}$. To go back from SP to CP, divide by the same multiplier.

$$\text{SP} = \text{CP} \times \frac{100 + p}{100} \qquad \text{CP} = \text{SP} \times \frac{100}{100 + p}$$

Write the multiplier as a fraction and the sums shrink. A 25% profit is × 1.25 = × $\frac{5}{4}$, and a 20% loss is × 0.8 = × $\frac{4}{5}$.

For example: a mixer sells for ₹1,197 at a 12.5% profit. 12.5% is $\frac{1}{8}$, so SP is $\frac{9}{8}$ of CP, and CP = 1,197 × $\frac{8}{9}$ = ₹1,064.

The fraction table

Multiples come from the same rows: 37.5% = $\frac{3}{8}$, 62.5% = $\frac{5}{8}$, $66\frac{2}{3}\%$ = $\frac{2}{3}$ and $28\frac{4}{7}\%$ = $\frac{2}{7}$. So a $16\frac{2}{3}\%$ loss on a heater bought for ₹1,470 means SP = 1,470 × $\frac{5}{6}$ = ₹1,225.

Assume CP = 100

When to use it: the question gives only percentages, or one rupee figure right at the end.

Fast way: take CP = 100 (or any handy number that clears the fractions), do all the work in those units, and change units into rupees only at the end.

For example: a shopkeeper's profit is $16\frac{2}{3}\%$ of his selling price. What is his profit % on cost? $16\frac{2}{3}\%$ is $\frac{1}{6}$, so the handy number is SP = 6. Profit = 1, so CP = 5, and profit on cost = $\frac{1}{5}$ = 20%. In general, a profit of $\frac{1}{n}$ of SP is $\frac{1}{n - 1}$ of CP.

Where it fails:

• Going back by taking the % off the SP. In the mixer example, 1,197 minus 12.5% of 1,197 is about ₹1,047.38, not ₹1,064. The 12.5% was on the CP, which you don't know yet.
• Profit or loss % is on CP unless the question says "on the selling price". $16\frac{2}{3}\%$ of SP is 20% of CP, and both numbers are often among the choices.
• A loss works the other way: a loss of $\frac{1}{n}$ of SP is $\frac{1}{n + 1}$ of CP. A loss of a quarter of SP is a fifth of CP, which is 20%.

[SSC Tip] Learn the table both ways. When you see a $9\frac{1}{11}\%$ profit, think "× $\frac{12}{11}$" straight away. Questions with odd-looking percentages are nearly always built on one of these fractions.

CP from Two Selling Prices

Rupee gap over percentage gap

When to use it: an article is sold (or could be sold) at two prices, with a profit or loss % at each, and CP is asked.

Fast way: both SPs are percentages of the same CP, so the gap between them in rupees is the gap between the percentages.

• A gain of x% at one price and a loss of y% at the other: the gap is (x + y)% of CP.
• Two gains, or two losses: the gap is (x − y)% of CP.

$$\text{CP} = \frac{\text{difference of the SPs}}{\text{difference of the percentages}} \times 100$$

For example: a cooler sold for ₹3,510 gives an 8% gain, and sold for ₹3,120 it gives a 4% loss. The gap is ₹390, which is 8 + 4 = 12% of CP. So CP = 390 × $\frac{100}{12}$ = ₹3,250. Check: 3,250 × 1.08 = 3,510.

Turning a loss into a gain

"Had it been sold for ₹322 more, there would have been an 8% gain instead of a 6% loss." You don't need either SP. The ₹322 covers 6% + 8% = 14% of CP, so CP = 322 × $\frac{100}{14}$ = ₹2,300.

Equal gain and loss, or one gain twice another

• If the gain at one price equals the loss at another, CP sits exactly halfway. Gain at ₹1,286 = loss at ₹958 gives CP = (1,286 + 958) ÷ 2 = ₹1,122.
• If the gain at one price is double the gain at another, CP = 2 × lower price − higher price. The gain at ₹1,404 is twice the gain at ₹1,272, so CP = 2,544 − 1,404 = ₹1,140 (gains of ₹264 and ₹132).

Where it fails:

• Adding when you should subtract. Gains of 18% and 10% at two prices are 8% apart, not 28%.
• Both percentages must be on CP. If one is on the selling price, change it to CP first.
• With a gain at one price and a loss at the other, CP must lie between the two prices. If your answer doesn't, you swapped which price gave the gain.

Articles: Bought at One Rate, Sold at Another

x for ₹a, y for ₹b

When to use it: "buys x articles for ₹a and sells y articles for ₹b".

Fast way: bring both to the same number of articles (the LCM of x and y) and compare rupees. Turn dozens (12) and scores (20) into articles before you take the LCM. In one line, the profit factor is $\frac{b \times x}{a \times y}$.

For example: a fruit seller buys oranges at 8 for ₹30 and sells them at 6 for ₹27. For 24 oranges, CP = ₹90 and SP = ₹108, so the profit is 18 on 90 = 20%. One line: $\frac{27 \times 8}{30 \times 6} = \frac{216}{180} = \frac{6}{5}$.

Same money, how many to sell: a man buys 15 notebooks for ₹40. How many must he sell for ₹40 to gain 25%? For the same money he hands over fewer: 15 × $\frac{100}{125}$ = 12.

SP of x articles = CP of y articles

Fast way: profit or loss % = $\frac{y - x}{x} \times 100$. Always divide by x, the number sold.

For example: the SP of 16 pens equals the CP of 20 pens. Gain = $\frac{4}{16}$ = 25%. If the SP of 25 equals the CP of 20, it is a loss of $\frac{5}{25}$ = 20%.

Profit equal to the SP or CP of some articles

• Selling n articles gains the SP of m articles: profit % = $\frac{m}{n - m} \times 100$. Selling 33 and gaining the SP of 11 gives $\frac{11}{22}$ = 50%.
• Gaining the CP of m articles: profit % = $\frac{m}{n} \times 100$. Selling 40 and gaining the CP of 6 gives 15%.
• Losing the SP of m articles: loss % = $\frac{m}{n + m} \times 100$.

Where it fails:

• Dividing by the wrong count. In the pens example, $\frac{4}{20}$ = 20% is the trap. The base is the 16 sold.
• "Profit equal to the SP of 11" and "profit equal to the CP of 11" are different. For 33 articles they give 50% and $33\frac{1}{3}\%$.
• If a few articles break or are given away, the per-article rates change. That is wastage, which comes later in this section.

Two Items Sold Together

Same SP, x% gain on one and x% loss on the other

When to use it: two articles are sold at the same price, one at a gain and the other at a loss of the same %.

Fast way: the result is always a loss of $\frac{x^2}{100}\%$.

For example: two phones are sold for ₹9,100 each, one at a 30% gain and the other at a 30% loss. Loss = $\frac{30^2}{100}$ = 9%. In rupees, the CPs are ₹7,000 and ₹13,000, so ₹20,000 went in and ₹18,200 came back, a loss of ₹1,800.

For the loss in rupees straight away: total SP × $\frac{x^2}{100^2 - x^2}$ = 18,200 × $\frac{900}{9100}$ = ₹1,800.

Same SP, different rates

Fast way: turn each SP back into its CP with its own multiplier, then compare totals. Take the SP as a number that both multipliers divide.

For example: two lamps are sold at the same price, one at a 25% gain and the other at a $16\frac{2}{3}\%$ loss. Take SP = 30 each. The CPs are 30 × $\frac{4}{5}$ = 24 and 30 × $\frac{6}{5}$ = 36. That's 60 in and 60 out: no profit, no loss.

Same CP, different rates

Then the overall % is just the average of the two rates. One at an 18% gain and one at a 6% loss gives (18 − 6) ÷ 2 = 6% gain.

Where it fails:

• The $\frac{x^2}{100}$ rule needs the same SP and the same %. With the same CP, an x% gain and an x% loss cancel out exactly.
• Averaging the rates when the SPs (not the CPs) are equal. In the lamp example, the average of 25 and $-16\frac{2}{3}$ is about 4.2%, but the real answer is 0.
• The loss % is on the total CP, not the total SP. ₹1,800 is 9% of ₹20,000 but about 9.89% of ₹18,200.

Parts of a Stock at Different Rates

When to use it: a stock is sold in parts at different profit or loss rates, and the overall %, or the rate on one part, is asked.

Fast way: the overall % is the weighted average of the rates, weighted by cost (or by quantity, when every unit cost the same). Write a loss as a negative rate.

$$\text{overall rate} = \frac{w_1 r_1 + w_2 r_2}{w_1 + w_2}$$

For example: a trader sells $\frac{2}{5}$ of his stock at a 15% gain and the rest at a 5% loss. Overall = $\frac{2 \times 15 + 3 \times (-5)}{5}$ = 3% gain.

Another one (finding one rate): a grocer buys 80 kg of sugar and sells 30 kg of it at a 10% loss. At what gain must he sell the other 50 kg to gain 15% on the whole? Balance the two sides: 30 × (−10) + 50 × r = 80 × 15. So 50r = 1,200 + 300 = 1,500, and r = 30%.

Finding the parts: when the rates and the overall % are known and the parts are asked, it's the alligation cross (Chapter 6) in one line. Loss part : gain part = (gain rate − overall) : (overall + loss rate). With a 5% loss, a 15% gain and 3% overall: (15 − 3) : (3 + 5) = 12 : 8 = 3 : 2, which matches the first example.

Where it fails:

• The weights are costs, not selling prices. Two parts that cost ₹6,000 and ₹4,000 weigh 3 : 2, whatever they sold for.
• A plain average works only for equal parts. In the sugar example it gives (−10 + r) ÷ 2 = 15, so r = 40%, which is the trap. The 50 kg part carries more weight.
• When the cost per kg differs between the parts, weight by the rupees spent, not by kg.

[SSC Tip] The overall rate always lies between the part rates, and closer to the bigger part. An option outside that range is out without any working.

Changed CP or SP: Two Equations in Units

When to use it: "had he bought it x% cheaper and sold it for ₹k less (or more), he would have gained y%". Or the CP rises while the SP stays the same.

Fast way: take the original CP = 100 units. Write the actual SP and the "what if" SP in units. Their gap in units equals the rupee gap, which gives one unit.

For example: a man sells a fan at a 20% gain. Had he bought it 10% cheaper and sold it for ₹42 less, he would have gained 25%. Find the CP.

The SPs differ by 7.5 units, which is ₹42. So one unit is ₹5.60 and CP = ₹560. Check: SP = ₹672, and in the "what if" case CP = ₹504 and SP = ₹630, which is 672 − 42.

CP rises, SP stays

For example: the profit on a shirt is 45%. Its CP goes up by 16% but the SP doesn't change. Take CP = 100 and SP = 145. The new CP is 116, so the profit is 29 on 116 = 25%.

Both change by a %

Profit 20%, then the CP falls by 10% and the SP by 4%. New SP = 120 × 0.96 = 115.2 on a new CP of 90, which is a factor of 1.28, so the profit becomes 28%.

Where it fails:

• The new profit % is on the new CP. In the shirt example, 29 on 100 gives 29%, a trap.
• Putting the ₹42 on the CP instead of the SP. Read which price the rupees belong to.
• Cutting the SP by 10% too just because the CP fell 10%. The SP changes only in the way the question says, here by ₹42.

[SSC Tip] Choose the starting units to avoid decimals. CP = 100 suits most questions, but if $16\frac{2}{3}\%$ or $33\frac{1}{3}\%$ shows up, start from 60 or 300 instead.

Markup and Discount: The MP : CP Route

When to use it: an article is marked m% above CP and sold at a d% discount, and the profit or loss % is asked. This type comes up often in Tier 1.

Fast way: multiply the two multipliers. The profit % is

$$\frac{(100 + m)(100 - d)}{100} - 100$$

and a negative answer is a loss. Another way to see it: MP : CP = (100 + m) : 100, and SP = MP × $\frac{100 - d}{100}$.

For example: marked 36% above CP and sold at 15% off: $\frac{136 \times 85}{100}$ − 100 = 115.6 − 100 = 15.6% profit.

Ratio form: MP : CP = 15 : 11 and the discount is 12%. SP = 15 × 0.88 = 13.2 against a CP of 11, so the profit is 2.2 on 11 = 20%.

Going backwards: MP : CP = (100 + p) : (100 − d). If a 16% discount still leaves a 5% profit, MP : CP = 105 : 84 = 5 : 4, which is a 25% markup.

With fractions: when the markup and the discount are in the fraction table, use fractions. $33\frac{1}{3}\%$ up and 25% off is $\frac{4}{3} \times \frac{3}{4}$ = 1, so there is no profit and no loss.

Where it fails:

• Subtracting the rates. 36 − 15 = 21% is the classic wrong option. The discount is on the bigger MP, so it takes away more than 15% of CP.
• The discount is on MP and the profit is on CP. Never take the discount % of the CP.
• Equal markup and discount always lose. x% up and x% off is a loss of $\frac{x^2}{100}\%$, the same rule as two items sold at one price.

MP Needed for a Target Profit

When to use it: the shopkeeper wants p% profit even after a d% discount, and the MP or the markup % is asked.

Fast way:

$$\text{MP} = \text{CP} \times \frac{100 + p}{100 - d}$$

and the markup % = $\frac{100 + p}{100 - d} \times 100 - 100$.

For example: CP = ₹1,800, and he wants 20% profit after a 25% discount. MP = 1,800 × $\frac{120}{75}$ = ₹2,880, a 60% markup. Check: 25% off ₹2,880 is ₹2,160, which is 1,800 × 1.2.

Another one: to gain 8% after a 28% discount, the factor is $\frac{108}{72}$ = 1.5, so mark the goods 50% above CP.

Discount equal to profit

• Equal in rupees: SP is halfway between CP and MP. With CP = ₹1,600 and MP = ₹2,400, SP = ₹2,000. That is a $16\frac{2}{3}\%$ discount and a 25% profit, equal as amounts (₹400 each) but not as percentages.
• Equal as percentages: call the common rate r. Then MP : CP = (100 + r) : (100 − r). If MP : CP = 9 : 7, then 7(100 + r) = 9(100 − r), so 16r = 200 and r = 12.5%.

Where it fails:

• Adding the rates. A 20 + 25 = 45% markup is too small: 1.45 × 0.75 = 1.0875, only an 8.75% profit.
• Dividing by (100 + d) instead of (100 − d). 1,800 × $\frac{120}{125}$ = ₹1,728 is below the CP, which can't be right for a profit.

[SSC Tip] "By what % above CP must he mark it" wants the markup %, which is MP ÷ CP − 1. "What price must he mark" wants MP in rupees. Both are usually among the options, so read the last line twice.

Successive Discounts and Comparing Schemes

When to use it: two or three discounts one after another, a single equivalent discount, or two offers to compare.

Fast way: for two discounts,

$$d = d_1 + d_2 - \frac{d_1 d_2}{100}$$

For three, multiply the "pay" fractions: you pay $\frac{100 - d_1}{100} \times \frac{100 - d_2}{100} \times \frac{100 - d_3}{100}$ of MP. The order of the discounts doesn't matter.

For example: 30% and 12% give 30 + 12 − 3.6 = 38.4%.

Three discounts: 10%, 20% and 25%. You pay 0.9 × 0.8 × 0.75 = 0.54 of MP, so the single discount is 46%.

Three equal discounts: cube the pay fraction. Three 25% discounts leave $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ of MP, about 0.42, so the discount is about 57.8%, well short of 75%.

Comparing schemes: on a ₹6,500 cooler, is 34% off better than 30% followed by 5%? The pair gives 30 + 5 − 1.5 = 33.5%, so 34% wins by 0.5% of MP, which is ₹32.50. Check: ₹4,290 against ₹4,322.50.

Finding the second discount: a single 44% is the same as 30% followed by what? You pay 0.56 = 0.7 × (second pay fraction), so the second pay fraction is 0.8 and the second discount is 20%.

Where it fails:

• Adding the discounts. 30% + 12% = 42% is nearly always one of the wrong options. Successive discounts always come to less than their sum.
• Taking the second discount on MP. It is on the price left after the first.
• A rise and a fall are not zero. 10% up and then 10% off use the same formula with +10 and −10, which is a 1% fall overall.

Buy x Get y Free

When to use it: "buy 5, get 1 free" or "buy 6, get 2 free", alone or with a discount or a markup.

Fast way: the customer takes x + y articles and pays for x. The effective discount is $\frac{y}{x + y}$, as a %.

For example: buy 5 get 1 free is $\frac{1}{6}$ = $16\frac{2}{3}\%$ off. Buy 6 get 2 free is $\frac{2}{8}$ = 25% off.

With a discount too: multiply the pay fractions. Buy 6 get 2 free (pay $\frac{3}{4}$) and then 4% off the bill (pay 0.96) means paying 0.72, a 28% discount in all.

With a markup, the profit: goods are marked 35% above CP with buy 8 get 1 free. The SP of 9 articles is the MP of 8, so each sells for 1.35 × $\frac{8}{9}$ = 1.2 times CP, a 20% profit.

Comparing: buy 6 get 1 free is only $\frac{1}{7}$ = $14\frac{2}{7}\%$ off, so a plain 15% discount beats it, even though "1 free" sounds like a lot.

Backwards: run the formula the other way to find the free articles. A 20% discount on "buy 8" needs 2 free, because $\frac{2}{8 + 2}$ = 20%.

Where it fails:

• Dividing by x instead of x + y. Buy 5 get 1 free is not 20% off, because the free article is 1 out of the 6 taken.
• "The discount to the customer" and "the shopkeeper's profit" are different numbers. Check which one the question asks.
• "Buy 2, get the third at half price" is a different deal. You pay for 2.5 out of 3, a $16\frac{2}{3}\%$ discount.

False Weights and Measures

When to use it: a dealer gives less than he says (a light weight or a short metre), sometimes with a profit, a loss or a discount on top.

Fast way (selling at CP):

$$\text{gain} = \frac{\text{error}}{\text{true value} - \text{error}} \times 100$$

For example: 980 g given for 1 kg is a gain of $\frac{20}{980}$ = $2\frac{2}{49}\%$. A 96 cm "metre" gives $\frac{4}{96}$ = $4\frac{1}{6}\%$.

With a profit or a discount: multiply the factors

Fast way: gain factor = price factor × $\frac{\text{true weight}}{\text{weight given}}$.

For example: a grocer charges 4% above CP and uses 960 g for a kg. 1.04 × $\frac{1000}{960}$ = 1.04 × $\frac{25}{24}$ = $\frac{13}{12}$, a gain of $8\frac{1}{3}\%$.

With a markup and a discount: goods are marked 50% above CP, sold at a 20% discount, and weighed with a 750 g weight. 1.5 × 0.8 × $\frac{1000}{750}$ = 1.2 × $\frac{4}{3}$ = 1.6, a 60% profit.

x% less weight and y% more price: the factor is $\frac{100 + y}{100 - x}$. Giving 15% less and charging 2% more is $\frac{102}{85}$ = 1.2, a 20% gain.

Cheating both ways: a trader takes 25% extra when he buys and gives 25% less when he sells. For each kg he pays for he gets 1.25 kg, and each 0.75 kg he hands over is paid as a kg, so the factor is $\frac{1.25}{0.75} = \frac{5}{3}$, a $66\frac{2}{3}\%$ gain even at CP.

A loss on the price: a dealer who sells at a loss can still gain through the weight. A 6% loss with an 800 g weight gives 0.94 × $\frac{1000}{800}$ = 0.94 × 1.25 = 1.175, a 17.5% gain.

Where it fails:

• Dividing by the true weight. $\frac{20}{1000}$ = 2% is the trap, because the dealer's cost is for the 980 g he actually gives.
• "15% less weight" is not a 15% gain. It is $\frac{15}{85}$ = $17\frac{11}{17}\%$.
• Adding the parts. A 4% markup and a 960 g weight is not 4% + $4\frac{1}{6}\%$ = $8\frac{1}{6}\%$. Multiply the factors to get $8\frac{1}{3}\%$.

Chains, Commission and Wastage

Successive transactions

When to use it: A sells to B at a gain or loss, B sells to C, and so on.

Fast way: the multipliers multiply. To go forward, multiply; to get back to the first CP, divide.

For example: A sells a bike to B at a 25% gain, B sells it to C at a 12% gain, and C pays ₹35,000. A's CP = 35,000 ÷ (1.25 × 1.12) = 35,000 ÷ 1.4 = ₹25,000.

With a loss in the chain: A to B at a 25% gain and B to C at a 20% loss is 1.25 × 0.8 = 1, so C pays exactly what A paid. Equal steps up and down, though, always lose: a 10% gain and then a 10% loss leave 0.99.

Commission

An agent takes c% of the sale price, so the owner keeps SP × $\frac{100 - c}{100}$. To make p% profit after paying the commission, SP = CP × $\frac{100 + p}{100 - c}$, the same form as a discount. For CP = ₹8,000, a 14% profit and a 5% commission: SP = 8,000 × $\frac{114}{95}$ = ₹9,600.

Wastage

When to use it: part of the goods rot, break or go missing, and the rest must still give the target profit.

Fast way: the target is on the whole cost. Find the total SP needed, then divide by what's left. Per unit, SP = CP per unit × $\frac{100 + p}{100 - w}$, where w is the % wasted.

For example: 160 kg of potatoes are bought at ₹25 a kg and 20% rot. For a 20% profit on the whole: 4,000 × 1.2 = ₹4,800 from 128 kg, which is ₹37.50 per kg. In one line: 25 × $\frac{120}{80}$ = 37.5.

Where it fails:

• Adding the rates in a chain. 25% + 12% = 37% is wrong; 1.25 × 1.12 = 1.4 gives 40%.
• Going back by subtracting. ₹35,000 less 40% is ₹21,000, the trap. Divide by 1.4 instead.
• In wastage, spreading the ₹4,800 over all 160 kg gives ₹30 a kg, which forgets the rotten potatoes.

Option Elimination

When to use it: as a first look at the options before you start the sum. Sometimes it settles the answer on its own.

Fast way: test the options against these four checks.

• Direction. A profit means SP above CP, and a discount means SP below MP. An option that breaks this is out.
• Limits. With an m% markup, any discount leaves a profit below m%. Successive discounts total less than their sum. The same SP with an x% gain and an x% loss is always a loss.
• Between. CP from a gain price and a loss price lies between the two prices. The overall rate on parts of a stock lies between the part rates.
• False weight. 900 g for 1 kg gains more than 10%, never exactly 10%.

For example: an article is marked 30% above CP and sold at a 12% discount. The profit is (a) 18% (b) 14.4% (c) 42% (d) 26.4%.

The profit must be below 30%, so (c) is out. (a) is 30 − 12, the subtract-the-rates trap, and (d) takes the discount off the markup only (30 × 0.88). 1.3 × 0.88 = 1.144, so the answer is (b) 14.4%.

Another one: discounts of 18% and 10% one after the other equal a single discount of (a) 28% (b) 26.2% (c) 29.8% (d) 25.2%. The sum is 28%, so (a) and (c) are out at once. 18 + 10 − 1.8 = 26.2, so (b).

Where it fails:

• When all four options pass the checks. Then you have to work the sum out in full.
• The markup limit is for plain markup and discount. A false weight on top can push the profit above the markup.

[SSC Tip] Trap options usually come from one slip: adding or subtracting rates instead of multiplying, dividing by the wrong base (SP for CP, the true weight for the false one), or forgetting the free or wasted articles. If your answer is what one of those slips would give, redo that step before you mark it.

Solved Examples

Every answer below uses the quick method, and the time check at the end says roughly how long it should take.

Question 1

A cycle is sold for ₹4,760 at a gain of $16\frac{2}{3}\%$. What did it cost?

Answer. $16\frac{2}{3}\%$ is $\frac{1}{6}$, so SP is $\frac{7}{6}$ of CP. CP = 4,760 × $\frac{6}{7}$ = ₹4,080. Taking $\frac{1}{6}$ off ₹4,760 gives about ₹3,966.67, which is the trap.

Time check: about 10 seconds.

Question 2

A trader makes a profit of 12.5% on the selling price of a bag that costs him ₹1,400. Find the selling price and his profit % on cost.

Answer. 12.5% is $\frac{1}{8}$, so I take SP = 8 units. Profit is 1 unit and CP is 7 units = ₹1,400, so one unit is ₹200. SP = ₹1,600, and the profit on cost is $\frac{1}{7}$ = $14\frac{2}{7}\%$.

Time check: about 15 seconds.

Question 3

A table is sold at an 8% loss. Had it been sold for ₹399 more, there would have been a 13% gain. Find its cost price.

Answer. The ₹399 covers 8% + 13% = 21% of CP, so 1% is ₹19 and CP = ₹1,900. Check: SP = ₹1,748, and 1,748 + 399 = 2,147 = 1,900 × 1.13.

Time check: about 12 seconds.

Question 4

A vendor buys bananas at 12 for ₹40 and sells them at 10 for ₹45. Find his gain %.

Answer. The LCM of 12 and 10 is 60. 60 bananas cost 5 × 40 = ₹200 and sell for 6 × 45 = ₹270, so the gain is 70 on 200 = 35%. In one line: $\frac{45 \times 12}{40 \times 10} = \frac{540}{400}$ = 1.35.

Time check: about 15 seconds.

Question 5

The cost price of 21 notebooks equals the selling price of 28 notebooks. Find the gain or loss %.

Answer. He sells 28 and gets back the cost of only 21, so he loses 7 notebooks' worth. I divide by the 28 sold: $\frac{7}{28}$ = 25% loss.

Time check: about 8 seconds.

Question 6

Two fans are sold for ₹3,150 each, one at a 12.5% gain and the other at a 12.5% loss. Find the overall gain or loss in rupees.

Answer. Same SP and the same %, so it's a loss of $\frac{12.5^2}{100}$ = 1.5625% of the total CP. For the rupees I use the multipliers: the CPs are 3,150 × $\frac{8}{9}$ = ₹2,800 and 3,150 × $\frac{8}{7}$ = ₹3,600. Total CP ₹6,400 against total SP ₹6,300, so a loss of ₹100.

Time check: about 20 seconds.

Question 7

A trader buys 150 kg of rice and sells 60 kg of it at a 6% loss. At what gain % must he sell the rest to gain 12% on the whole?

Answer. I balance the rates weighted by kg: 60 × (−6) + 90 × r = 150 × 12. So 90r = 1,800 + 360 = 2,160 and r = 24%. The plain average, (−6 + r) ÷ 2 = 12, gives 30%, which ignores that the second part is bigger.

Time check: about 20 seconds.

Question 8

A man sells a chair at a 30% gain. Had he bought it 10% cheaper and sold it for ₹62 less, he would have gained 40%. Find the cost price.

Answer. CP = 100 units and SP = 130. In the "what if" case CP = 90 and SP = 90 × 1.4 = 126. The SPs differ by 4 units = ₹62, so one unit is ₹15.50 and CP = ₹1,550. Check: SP = ₹2,015, and the "what if" SP is 1,395 × 1.4 = ₹1,953, which is 2,015 − 62.

Time check: about 25 seconds.

Question 9

A shopkeeper marks a saree 37.5% above cost and allows a 20% discount. Find his gain %.

Answer. 37.5% is $\frac{3}{8}$, so MP = $\frac{11}{8}$ of CP, and 20% off leaves $\frac{4}{5}$ of that: $\frac{11}{8} \times \frac{4}{5} = \frac{11}{10}$. So the gain is 10%.

Time check: about 8 seconds.

Question 10

A dealer buys a TV for ₹22,400. He wants a 25% profit even after a 12.5% discount. What price should he mark on it?

Answer. MP = 22,400 × $\frac{125}{87.5}$ = 22,400 × $\frac{10}{7}$ = ₹32,000. Check: 12.5% off ₹32,000 is ₹28,000, which is 22,400 × 1.25.

Time check: about 15 seconds.

Question 11

On an ₹8,000 sofa, which is better for the buyer: a single 40% discount, or successive discounts of 30%, 10% and 5%? By how much?

Answer. With three discounts I pay 0.7 × 0.9 × 0.95 = 0.5985 of ₹8,000, which is ₹4,788. The single 40% leaves ₹4,800. So the three discounts are better by ₹12, a single equivalent of 40.15%. The three add up to 45%, well above 40%, and still they only just win. Had they added up to exactly 40%, the single discount would have been better, because successive discounts always come to less than their sum.

Time check: about 25 seconds.

Question 12

A shop marks its goods 44% above cost, offers "buy 5, get 1 free", and then gives 6% off the bill. Find the profit or loss %.

Answer. I multiply the factors: 1.44 × $\frac{5}{6}$ = 1.2, and 1.2 × 0.94 = 1.128. So it's a profit of 12.8%.

Time check: about 12 seconds.

Question 13

A grocer marks his goods 30% above cost, gives a 3% discount, and uses a weight of 970 g for 1 kg. Find his gain %.

Answer. The factor is 1.3 × 0.97 × $\frac{1000}{970}$. The 0.97 and the $\frac{1000}{970}$ cancel out, so the factor is just 1.3 and the gain is 30%. The weight wins back exactly what the discount gave away.

Time check: about 12 seconds.

Question 14

A sells a scooter to B at a 10% loss, B sells it to C at a 20% gain, and C sells it to D at a 12.5% gain. D pays ₹72,900. What did A pay for it?

Answer. The chain multiplies to 0.9 × 1.2 × 1.125 = 1.215. So A paid 72,900 ÷ 1.215 = ₹60,000. Adding the rates (−10 + 20 + 12.5 = 22.5%) gives about ₹59,510, which is wrong.

Time check: about 20 seconds.

Question 15

A shopkeeper buys 200 eggs at ₹6 each, and 10% of them break. At what price per egg must he sell the rest to gain 35% on the whole?

Answer. Per egg: 6 × $\frac{135}{90}$ = ₹9. In full, the cost is ₹1,200 and the target is ₹1,620, which has to come from 180 eggs, so ₹9 each. Spreading the ₹1,620 over all 200 eggs gives ₹8.10, the trap.

Time check: about 12 seconds.