How to Prepare for Quantitative Aptitude for CAT (2014)

Block III: Arithmetic and Word-based Problems

Chapter 6. PROFIT & LOSS

INTRODUCTION

Traditionally, Profit & Loss has always been an important chapter for CAT. Besides, all other Management entrance exams like SNAP, CMAT, MAT, ATMA as well as Bank P.O. exams extensively use questions from this chapter. From the point of view of CAT, the relevance of this chapter has been gradually reducing. However, CAT being a highly unpredictable exam, my advice to students and readers would be to go through this chapter and solve it at least up to LOD II, so that they are ready for any changes in patterns.

Further, the Level of Difficulties at which questions are set in the various exams can be set as under:

LOD I: CAT, XLRI, IRMA, IIFT, CMAT, Bank PO aspirants, MAT, NIFT, NMAT, and SNAP and all other management exams.

LOD II: CAT, XLRI, IRMA (partially), etc.

LOD III: CAT, XLRI (students aiming for 60% plus in Maths in CAT).

THEORY

Profit & Loss are part and parcel of every commercial transaction. In fact, the entire economy and the concept of capitalism is based on the so called “Profit Motive”.

Profit & Loss in Case of Individual Transactions

We will first investigate the concept of Profit & Loss in the case of individual transactions. Certain concepts are important in such transactions. They are:

The price at which a person buys a product is the cost price of the product for that person. In other words, the amount paid or expended in either purchasing or pro- ducing an object is known as its Cost Price (also written as CP).

The price at which a person sells a product is the sales price of the product for that person. In other words, the amount got when an object is sold is called as the Selling Price (SP) of the object from the seller’s point of view.

When a person is able to sell a product at a price higher than its cost price, we say that he has earned a profit. That is,

If SP > CP, the difference, SP – CP is known as the profit or gain.

Similarly, if a person sells an item for a price lower than its cost price, we say that a loss has been incurred.

The basic concept of profit and loss is as simple as this.

If, however, SP < CP, then the difference, CP – SP is called the loss.

It must be noted here that the Selling Price of the seller is the Cost Price of the buyer.

Thus we can say that in the case of profit the following formulae hold true:

Notice that

SP = CP + Gain

 = CP + (Gain on Re 1) × CP

 = CP + (Gain%/100) × CP

Example: A man purchases an item for ` 120. If he sells it at a 20 per cent profit find his selling price.

Solution: The selling price is given by 120 + 120 × 0.2 = 144

= CP + (Gain%/100) × CP = CP

For the above problem, the selling price is given by this method as: Selling Price = 1.2 × 120 = 144.

Hence, we also have the following:

The above situation (although it is the basic building block of Profit and Loss) is not the normal situation where we face Profit and Loss problems. In fact, there is a wide application of profit and loss in day-to-day business and economic transactions. It is in these situations that we normally have to work out profit and loss problems.

Having investigated the basic concept of profit and loss for an individual transaction of selling and buying one unit of a product, let us now look at the concept of profit and loss applied to day-to-day business and commercial transactions.

Profit & Loss as Applied to Business and Commercial Transactions

Profit & Loss when Multiple Units of a Product are Being Bought and Sold The basic concept of profit and loss remains unchanged for this situation. However, a common mistake in this type of problem can be avoided if the following basic principle is adopted:

Profit or Loss in terms of money can only be calculated when the number of items bought and sold are equal.

That is, Profit or Loss in money terms cannot be calculated unless we equate the number of products bought and sold.

This is normally achieved by equating the number of items bought and sold at 1 or 100 or some other convenient figure as per the problem asked.

Overlooking of this basic fact is one of the most common mistakes that students are prone to making in the solving of profit and loss problems.

Types of Costs In any business dealing, there is a situation of selling and buying of products and services. From the sellers point of view, his principle interest, apart from maximising the sales price of a product/service, is to minimise the costs associated with the selling of that product/service. The costs that a businessman/trader faces in the process of day-to-day business transaction can be subdivided into three basic categories:

1.Direct Costs or Variable Costs This is the cost associated with direct selling of product/service. In other words, this is the cost that varies with every unit of the product sold. Hence, if the variable cost in selling a pen for `20 is ` 5, then the variable cost for selling 10 units of the same pen is 10 × 5 = ` 50.

As is clear from the above example, that part of the cost that varies directly for every additional unit of the product sold is called as direct or variable cost.

Typical examples of direct costs are: Raw material used in producing one unit of the product, wages to labour in producing one unit of the product when the wages are given on a piece rate basis, and so on. In the case of traders, the cost price per unit bought is also a direct cost (i.e. every such expense that can be tied down to every additional unit of the product sold is a direct cost).

2.Indirect Costs (Overhead Costs) or Fixed Costs There are some types of costs that have to be incurred irrespective of the number of items sold and are called as fixed or indirect costs. For example, irrespective of the number of units of a product sold, the rent of the corporate office is fixed. Now, whether the company sells 10 units or 100 units, this rent is fixed and is hence a fixed cost.

Other examples of indirect or fixed costs: Salary to executives and managers, rent for office, office telephone charges, office electricity charges.

Apportionment of indirect (or fixed) costs: Fixed Costs are apportioned equally among each unit of the product sold. Thus, if n units of a product is sold, then the fixed cost to be apportioned to each unit sold is given by

3.Semi-Variable Costs Some costs are such that they behave as fixed costs under normal circumstances but have to be increased when a certain level of sales figure is reached. For instance, if the sales increase to such an extent that the company needs to take up additional office space to accommodate the increase in work due to the increase in sales then the rent for the office space becomes a part of the semi-variable cost.

The Concept of Margin or Contribution per unit The difference between the value of the selling price and the variable cost for a product is known as the margin or the contribution of the product. This margin goes towards the recovery of the fixed costs incurred in selling the product/service.

The Concept of the Break-even Point The break-even point is defined as the volume of sale at which there is no profit or no loss. In other words, the sales value in terms of the number of units sold at which the company breaks even is called the break-even point. This point is also called the break-even sales.

Since for every unit of the product the contribution goes towards recovering the fixed costs, as soon as a company sells more than the break-even sales, the company starts earning a profit. Conversely, when the sales value in terms of the number of units is below the break-even sales, the company makes losses.

The entire scenario is best described through the following example.

Let us suppose that a paan shop has to pay a rent of ` 1000 per month and salaries of ` 4000 to the assistants.

Also suppose that this paan shop sells only one variety of paan for ` 5 each. Further, the direct cost (variable cost) in making one paan is ` 2.50 per paan, then the margin is ` (5 – 2.50) = ` 2.50 per paan.

Now, break-even sales will be given by:

Break-even-sales = Fixed costs/Margin per unit = 5000/2.5 = 2000 paans.

Hence, the paan shop breaks-even on a monthly basis by selling 2000 paans.

Selling every additional paan after the 2000th paan goes towards increasing the profit of the shop. Also, in the case of the shop incurring a loss, the number of paans that are left to be sold to break-even will determine the quantum of the loss.

Note the following formulae:

Also in the case of a loss:

Also, if the break-even sales equals the actual sales, then we reach the point of no profit no loss, which is also the technical definition of the break-even point.

Note that the break-even point can be calculated on the basis of any time period (but is normally done annually or monthly).

Profit Calculation on the Basis of Equating the Amount Spent and the Amount Earned

We have already seen that profit can only be calculated in the case of the number of items being bought and sold being equal. In such a case, we take the difference of the money got and the money given to get the calculation of the profit or the loss in the transaction.

There is another possibility, however, of calculating the profit. This is done by equating the money got and the money spent. In such a case, the profit can be represented by the amount of goods left. This is so because in terms of money the person going through the transaction has got back all the money that he has spent, but has ended up with some amount of goods left over after the transaction. These left over items can then be viewed as the profit or gain for the individual in consideration.

Hence, profit when money is equated is given by Goods left. Also, cost in this case is represented by Goods sold and hence percentage profit = × 100.

Example: A fruit vendor recovers the cost of 25 mangoes by selling 20 mangoes. Find his percentage profit.

Solution: Since the money spent is equal to the money earned the percentage profit is given by:

% Profit = × 100 = 5 × 100/20 = 25%

Concept of Mark Up

Traders/businessmen, while selling goods, add a certain percentage on the cost price. This addition is called percentage mark up (if it is in money terms), and the price thus obtained is called as the marked price (this is also the price printed on the product in the shop).

The operative relationship is

The product is normally sold at the marked price in which case the marked price = the selling price

If the trader/shopkeeper gives a discount, he does so on the marked price and after the discount the product is sold at its discounted price.

Hence, the following relationship operates:

Use of PCG in Profit and Loss

1.The relationship between CP and SP is typically defined through a percentage relationship. As we have seen earlier, this percentage value is called as the percentage mark up. (And is also equal to the percentage profit if there is no discount).

Consider the following situation ––

Suppose the SP is 25% greater than the CP. This relationship can be seen in the following diagram.

In such a case the reverse relationship will be got by the AÆBÆA application of PCG and will be seen as follows:

If the profit is 25% :

Example:

Suppose you know that by selling an item at 25%, profit the Sales price of a bottle of wine is ` 1600. With this information, you can easily calculate the cost price by reducing the sales price by 20%. Thus, the CP is

WORKED-OUT PROBLEMS

Before we go into problems based on profit and loss, the reader should realize that there are essentially four phases of a profit and loss problem. These are connected together to get higher degrees of difficulty.

These are clues for (a) Cost calculations (b) Marked price calculations (c) Selling price calculations (d) Over-heads/fixed costs calculations.

It is left to the reader to understand the interrelationships between a, b, c and d above. (These have already been stated in the earlier part of this chapter.)

Problem 6.1 A shopkeeper sold goods for ` 2000 at a profit of 50%. Find the cost price for the shopkeeper.

Solution The shopkeeper sells his items at a profit of 50%. This means that the selling price is 150% of cost price (Since CP + % Profit = SP)

For short you should view this as SP = 1.5 CP.

The problem with this calculation is that we know what 150% of the cost price is but we do not know what the cost price itself is. Hence, we have difficulty in directly working out this problem. The calculation will become easier if we know the percentage calculation to be done on the basis of the selling price of the goods.

Hence look at the equation from the angle Æ CP = SP/1.5.

Considering the SP as SP/1, we have to find CP as SP/1.5. This means that the denominator is increasing by 50%. But from the table of denominator change to ratio change of the chapter of percentages, we can see that when the denominator increases by 50% the ratio decreases by 33.33%.

Interpret this as the CP can be got from the SP by reducing the SP by 33.33%. Hence, the answer is 2000 – (1/3) × 2000 = ` 1333.33

Also, this question can also be solved through options by going from CP (assumed from the value of the option) to the SP by increasing the assumed CP by 50% to check whether the SP comes out to 2000. If a 50% increase in the assumed CP does not make the SP equal 2000 it means that the assumed CP is incorrect. Hence, you should move to the next option. Use logic to understand whether you go for the higher options or the lower options based on your rejection of the assumed option.

Problem 6.2 A man buys a shirt and a trousers for ` 371. If the trouser costs 12% more than the shirt, find the cost of the shirt.

Solution Here, we can write the equation:

s + 1.12s = 371 Æ s = 371/2.12 (however, this calculation is not very easily done)

An alternate approach will be to go through options. Suppose the options are

(a)` 125    (b) ` 150

(c)` 175    (d) ` 200

Checking for, say, ` 150, the thought process should go like:

Let s = cost of a shirt

If s = 150, 1.12s will be got by increasing s by 12% i.e. 12% of 150 = 18. Hence the value of 1.12s = 150 + 18 =168 and s + 1.12s = 318 is not equal to 371. Hence check the next higher option.

If s = 175, 1.12s = s + 12% of s = 175 + 21 = 196. i.e. 2.12 s = 371.

Hence, Option (c) is correct.

Problem 6.3 A shopkeeper sells two items at the same price. If he sells one of them at a profit of 10% and the other at a loss of 10%, find the percentage profit/loss.

Generic question: A shopkeeper sells two items at the same price. If he sells one of them at a profit of x% and the other at a loss of x%, find the percentage profit/loss.

Solution The result will always be a loss of [x/10]²%. Hence, the answer here is [10/10]²% = 1% loss.

Problem 6.4 For Problem 6.3, find the value of the loss incurred by the shopkeeper if the price of selling each item is ` 160.

Solution When there is a loss of 10% Æ 160 = 90% of CP₁. \ CP₁ = 177.77

When there is a profit of 10% Æ 160 = 110% of CP₂ \ CP₂ = 145.45

Hence total cost price = 177.77 + 145.45 = 323.23 while the net realisation is ` 320.

Hence loss is ` 3.23.

Short cut for calculation: Since by selling the two items for ` 320 the shopkeeper gets a loss of 1% (from the previous problem), we can say that ` 320 is 99% of the value of the cost price of the two items. Hence, the total cost is given by 320/0.99 (solution of this calculation can be approximately done on the percentage change graphic).

Problem 6.5 If by selling 2 items for ` 180 each the shopkeeper gains 20% on one and loses 20% on the other, find the value of the loss.

Solution The percentage loss in this case will always be (20/10)² = 4% loss.

We can see this directly as 360 Æ 96% of the CP Æ CP = 360/0.96. Hence, by percentage change graphic 360 has to be increased by 4.166 per cent = 360 + 4.166% of 360 = 360 + 14.4 + 0.6 = ` 375.

Hence, the loss is ` 15.

Problem 6.6 By selling 15 mangoes, a fruit vendor recovers the cost price of 20 mangoes. Find the profit percentage.

Solution Here since the expenditure and the revenue are equated, we can use percentage profit = (goods left × 100)/goods sold = 5 × 100/15 = 33.33%.

Problem 6.7 A dishonest shopkeeper uses a 900 gram weight instead of 1 kilogram weight. Find his profit percent if he sells per kilogram at the same price as he buys a kilogram.

Solution Here again the money spent and the money got are equal. Hence, the percentage profit is got by goods left × 100/goods sold.

This gives us 11.11%.

Problem 6.8 A manufacturer makes a profit of 15% by selling a colour TV for ` 6900. If the cost of manufacturing increases by 30% and the price paid by the retailer is increased by 20%, find the profit percent made by the manufacturer.

Solution For this problem, the first line gives us that the cost price of the TV for the manufacturer is ` 6000.

(By question stem analysis you should be able to solve this part of the problem in the first reading and reach at the figure of 6000 as cost, before you read further. This can be achieved advantageously if your percentage rule calculations are strong. Hence, work on it. The better you can get at it the more it will benefit you. In fact, one of the principal reasons I get through the CAT every year is the strength in percentage calculation. Besides, percentage calculation will also go a long way in improving your scores in data Interpretation.)

Further, if you have got to the 6000 figure by the end of the first line, reading further you can increase this advantage by calculating while reading as follows:

Manufacturing cost increase by 30% Æ New manufacturing cost = 7800 and new selling price is 6900 + 20% of 6900 = 6900 + 1380 = 8280.

Hence, profit = 8280 – 7800 = 480 and profit percent = 480 × 100/7800 = 6.15%.

Problem 6.9 Find a single discount to equal three consecutive discounts of 10%, 12% and 5%.

Solution Using percentage change graphic starting from 100: we get 100 Æ 88 Æ 83.6 Æ 75.24 (Note we can change percentages in any order).

Hence, the single discount is 24.76%.

Problem 6.10 A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol.

Solution 10% reduction in price Æ 11.11% increase in consumption.

But 11.11% increase in consumption is equal to 5 gallons. Hence, original consumption is equal to 45 gallons for $180. Hence, original price = 4$ per gallon.

Problem 6.11 Ashok bought an article and spent ` 110 on its repairs. He then sold it to Bhushan at a profit of 20%. Bhushan sold it to Charan at a loss of 10%. Charan finally sold it for ` 1188 at a profit of 10%. How much did Ashok pay for the article.

(a)` 890    (b) ` 1000

(c)` 780    (d) ` 840

Solution Solve through options using percentage rule and keep checking options as you read. Try to finish the first option-check before you finish reading the question for the first time. Also, as a thumb rule always start with the middle most convenient option. This way you are likely to be required lesser number of options, on an average.

Also note that LOD II and LOD III questions will always essentially use the same sentences as used in LOD I questions. The only requirement that you need to have to handle LOD II and III questions is the ability to string together a set of statements and interconnect them.

Problem 6.12 A dishonest businessman professes to sell his articles at cost price but he uses false weights with which he cheats by 10% while buying and by 10% while selling. Find his percentage profit.

Solution Assume that the businessman buys and sells 1 kg of items. While buying he cheats by 10%, which means that when he buys 1 kg he actually takes 1100 grams. Similarly, he cheats by 10% while selling, that is, he gives only 900 grams when he sells a kilogram. Also, it must be understood that since he purportedly buys and sells the same amount of goods and he is trading at the same price while buying and selling, money is already equated in this case. Hence, we can directly use: % Profit = (Goods left × 100/Goods sold) = 200 × 100/900 = 22.22% (Note that you should not need to do this calculation since this value comes from the fraction to percentage conversion table).

If you are looking at 70% plus net score in quantitative ability you should be able to come to the solution in about 25 seconds inclusive of problem reading time. And the calculation should go like this:

Money is equated Æ % profit = 2/9 = 22.22%

The longer process of calculation in this case would be involving the use of equating the amount of goods bought and sold and the money value of the profit. However, if you try to do this you will easily see that it requires a much higher degree of calculations and the process will tend to get messy.

The options for doing this problem by equating goods would point to comparing the price per gram bought or sold. Alternatively, we could use the price per kilogram bought and sold (which would be preferable to equating on a per gram basis for this problem).

Here the thought process would be:

Assume price per kilogram = ` 1000. Therefore, he buys 1100 grams while purchasing and sells 900 grams while selling.

To equate the two, use the following process:

Problem 6.13 RFO Tripathi bought some oranges in Nagpur for ` 32. He has to sell it off in Yeotmal. He is able to sell off all the oranges in Yeotmal and on reflection finds that he has made a profit equal to the cost price of 40 oranges. How many oranges did RFO Tripathi buy?

Solution Suppose we take the number of oranges bought as x. Then, the cost price per orange would be ` 32/x, and his profit would by 40 × 32/x = 1280/x.

To solve for x, we need to equate this value with some value on the other side of the equation. But, we have no information provided here to find out the value of the variable x. Hence, we cannot solve this equation.

Problem 6.14 By selling 5 articles for ` 15, a man makes a profit of 20%. Find his gain or loss percentage if he sells 8 articles for ` 18.4?

Questions of this type normally appear as part of a more complex problem in an exam like the CAT.

Remember, such a question should be solved by you as soon as you finish reading the question by solving-while-reading process, as follows.

By selling 5 articles for ` 15, a man makes a profit of 20% Æ SP = 3. Hence, CP = 2.5, if he sells 8 articles for ` 18.4 Æ SP = 2.3. Hence percentage loss = 8%. For solving this question through this method with speed you need to develop the skill and ability to calculate percentage changes through the percentage change graphic. For this purpose, you should not be required to use a pencil and a paper.

Problem 6.15 Oranges are bought at 12 for a rupee and are sold at 10 for a rupee. Find the percentage profit or loss.

Solution Since money spent and got are equated, use the formula for profit calculation in terms of goods left/goods sold.

This will give you percentage profit = 2/10 = 20%.

Alternatively, you can also equate the goods and calculate the percentage profit on the basis of money as

CP of 1 orange = 8.33 paise

SP of 1 orange = 10 paise

8.33 paise Æ 10 paise (corresponds to a percentage increase of 20% on CP)

Problem 6.16 In order to maximise its profits, AMS Corporate defines a function. Its unit sales price is ` 700 and the function representing the cost of production = 300 + 2p², where p is the total units produced or sold. Find the most profitable production level. Assume that everything produced is necessarily sold.

Solution The function for profit is a combination of revenue and costs. It is given by Profit = Revenue – Costs = 700 p – (300 + 2p²) = –2p² + 700p – 300.

In order to find the maxima or minima of any quadratic function, we differentiate it and equate the differentiated equation to zero.

Thus, the differentiated profit function is –4p + 700 = 0 Æ p = 175. This value of production will yield the maximum profits in this case.

Problem 6.17 For Problem 6.16, what is the value of the maximum profits for AMS Corporate?

Solution For this, continuing from the previous question’s solution, we just put the value of p = 175 in the equation for profit. Thus, substitute p = 175 in the equation. Profit = –2p² + 700p – 300 and get the answer.

Problem 6.18 A shopkeeper allows a rebate of 25% to the buyer. He sells only smuggled goods and as a bribe, he pays 10% of the cost of the article. If his cost price is ` 2500, then find what should be the marked price if he desires to make a profit of 9.09%.

Solution Use solving-while-reading as follows: Cost price (= 2500) + Bribe (= 10% of cost of article = 250) = Total cost to the shopkeeper (2500 + 250 = 2750).

He wants a profit of 9.09 percent on this value Æ Using fraction to percentage change table we get 2750 + 9.09% of 2750 = 2750 + 250 = ` 3000.

But this ` 3000 is got after a rebate of 25%. Since we do not have the value of the marked price on which 25% rebate is to be calculated, it would be a good idea to work reverse through the percentage change graphic:

Going from the marked price to ` 3000 requires a 25% rebate. Hence the reverse process will be got by increasing ` 3000 by 33.33% and getting ` 4000.

Problem 6.19 A man sells three articles, one at a loss of 10%, another at a profit of 20% and the third one at a loss of 25%. If the selling price of all the three is the same, find by how much percent is their average CP lower than or higher than their SP.

Solution

We have to calculate: (average CP – average SP)/average SP.

Here, the selling price is equal in all three cases. Since the maximum number of calculations are associated with the SP, we assume it to be 100. This gives us an average SP of 100 for the three articles. Then, the first article will be sold at 111.11, the second at 83.33 and the third at 133.33. (The student is advised to be fluent at these calculations) Further, the CP of the three articles is 111.11 + 83.33 + 133.33 = 327.77.

The average CP of the three articles is 327.77/3 = 109.2566.

Hence, (average CP – average SP)/average SP = 9.2566%. higher

Any other process adopted for this problem is likely to require much more effort and time.

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

Directions for Questions 14 and 15: Read the following and answer the questions that follow.

Doctors have advised Renu, a chocolate freak, not to take more than 20 chocolates in one day. When she went to the market to buy her daily quota, she found that if she buys chocolates from the market complex she would have to pay ` 3 more for the same number of chocolates than she would have spent had she bought them from her uncle Scrooge’s shop, getting two sweets less per rupee. She finally decided to get them from Uncle Scrooge’s shop paying only in one rupee coins.

Directions for Questions 18 and 19: Read the following and answer the questions that follow.

Ramu and Shyamu decided to sell their cars each at ` 36,000. While Ramu decided to give a discount of 8% on the first ` 8000, 5% on next ` 12,000 and 3% on the rest to buyer Sashi, Shyamu decided to give a discount of 7% on the first 12,000, 6% on the next 8,000 and 5% on the rest to buyer Rajesh. These discounts were, however, subject to the buyers making the payment on time failing which the discount gets reduced by 1% for every delay of a week. In each case, the selling price of 36,000 was arrived at by increasing the cost price by 25%.

For questions 33 to 36 use the following data:

LEVEL OF DIFFICULTY (III)

The charges of a taxi journey are decided on the basis of the distance covered and the amount of the waiting time during a journey. Distance wise, for the first 2 kilometers (or any part thereof) of a journey, the metre reading is fixed at ` 10 (if there is no waiting). Also, if a taxi is boarded and it does not move, then the meter reading is again fixed at ` 10 for the first ten minutes of waiting. For every additional kilometre the meter reading changes by ` 5 (with changes in the meter reading being in multiples of Re. 1 for every 200 meters travelled). For every additional minute of waiting, the meter reading changes by ` 1. (no account is taken of a fraction of a minute waited for or of a distance less than 200 meters travelled). The net meter reading is a function of the amount of time waited for and the distance travelled.

The cost of running a taxi depends on the fuel efficiency (in terms of mileage/liter), depreciation (straight line over 10 years) and the driver’s salary (not taken into account if the taxi is self owned).

Depreciation is ` 100 per day everyday of the first 10 years. This depreciation has to be added equally to the cost for every customer while calculating the profit for a particular trip. Similarly, the driver’s daily salary is also apportioned equally across the customers of the particular day. Assume, for simplicity, that there are 50 customers every day (unless otherwise mentioned). The cost of fuel is ` 15 per liter (unless otherwise stated).

The customer has to pay 20% over the meter reading while settling his bill. Also assume that there is no fuel cost for waiting time (unless otherwise stated).

Based on the above facts, answer the following:

Directions for Questions 5 to 10: Answer questions based on this additional information:

Mr. Vikas Verma owns a fleet of 3 taxis, where he pays his driver ` 3000 per month. He also insists on keeping an attendant for ` 1500 per month in each of his taxis. Idling requires 50 ml of fuel for every minute of idling. For a moving taxi, the fuel consumption is given by 12 km/per litre. On a particular day, he received the following reports about the three taxis.

Directions for Questions 11 to 15: Read the following and answer the questions that follow.

The Coca-Cola Company is setting up a plant for manufacture and sale of the soft drink.

The investment for the plant is ` 10 crore (to be invested in plant, machinery, advertising, infrastructure, etc.).

The following information is available about the different bottle sizes planned:

Based on this information answer the questions given below:

Directions for Questions 23 to 28: Read the following and answer the questions that follow.

Prabhat Ranjan inaugurates his internet cafe on the 1st of January 2003. He invests in 10 computers @ ` 30,000 per computer. Besides, he also invests in the other infrastructure of the centre, a sum of ` 1 lakh only. He charges his customers on the time spent on the internet a flat rate of ` 50 per hour. His initial investment on computers has to be written off equally in 3 years (1 lakh per year) and the infrastructure has to be written off in 5 years (@ ` 20,000 per year).

He has to pay a fixed rental of ` 8000 per month for the space and also hires an assistant at ` 2000 per month.

For every hour that he is connected to the internet, he has to bear a telephone charge of ` 20 irrespective of the number of machines operational on the internet at that time. On top of this, he also has to pay an electricity charge of ` 5 per computer per hour. Assume that there are no other costs involved unless otherwise mentioned. The internet cafe is open 12 hours a day and is open on all 7 days of the week. (Assume that if a machine is not occupied, it is put off and hence consumes no electricity).

Directions for Questions 29 to 33: Read the following and answer the questions that follow.

A train journey from Patna to Delhi by the Magadh Express has 4 classes:

The fares of the 4 classes are as follows:

Patna to Delhi distance: 1100 kilometres. Assume the train does not stop at any station unless otherwise indicated. Running cost per kilometre: AC bogey Æ ` 25, non AC bogey Æ ` 10.

Level of Difficulty (I)

Hints

Level of Difficulty (II)

1.Cost of 750 articles = ` 450 Æ outlay.

Hence, SP of 600 articles = ` 630. (@40% profit on his outlay)

\ SP of each article = ` 1.05.

2.When the manufacturer makes 100 articles, he sells only 88.

Revenues = 88 × 7.5 = ` 660

But profit being 20% of outlay, total outlay is ` 550.

3.Total cost = Type + Running cost of printing machine + paper, ink

= 1000 + 120 × 9 + 540 = 2620

\ Net sum to be recovered = ` 2882.

4.Assume CP of goods = 100.

Then marked price = 130.

Revenues = × 130 + × 0.85 × 130 + × 0.7 × 130

5.If CP = x, then marked price = x + 205 and

Selling price = 0.9x + 184.5.

Profit = 184.5 – 0.1x

7.Assume CP as 1 Re/gram.

Then, he sells 900 grams for ` 1080,

While the CP of 900 grams is ` 900.

8.He buys 1100 grams for `1000 and sells 900 grams for ` 1080.

To calculate profit percentage, either equate the money or the goods.

9.He buys 1100 grams for ` 900 and sells 900 grams for ` 1080.

To calculate percentage profit, either equate the money or equate the goods.

10.Take the LCM of 4 and 5 and assume 20 apples bought at each price.

Ensure that you equate either the goods or the money.

12.The loss is covered by the sale of 20 extra kgs of rice.

i.e.CP of 100 kg = SP of 120 kg

13.Profit maximisation is totally dependent on revenue maximisation, since the cost is constant.

14-15.Solve using options.

16.Difference between original CP and SP is Q%, while the reverse difference is 62.5% of Q.

The only values satisfying these are 60% and 37.5%. Hence Q = 60%.

17.If CP = 100, original marked price = 115 and new marked price = 126.5. But he now sells on 80% of his marked price.

21.Assume the SP to be 125 for both the salesmen.

24.If initial cost price is 100, then initial selling price will be 120. Then, the new cost price is 90, while the new selling price is 132.

25.The successive percentage drops in marked price to get the selling price are 10% each.

27.Primary cost = 0.35 × 12600

Gross profit = 12600 – (1400 + 650 + 252) – 0.35 × 12600.

\ Trading cost = 25% of Gross profit.

\ Net profit = Gross profit – 25% of Gross profit = 75% of Gross profit.

and percent profit = .

28.Solve using options.

29.Assume David’s cost price as 100. Then, Goliath buys at 80, sells at 100 and buys back from Hercules at 90.

30.Buys 1200 when he pays for 1000 and sells 900 when he receives money for 1000.

33-36.Investments are ` 55000 for men and ` 45000 for women.

The rate of return being given, we can find out the percentage profit overall.

37.Solve through options.

39.A’s share is only dependent on the ratio of capital investment.

40.Assume cost = 100 and SP = 120.

Then, when price of petrol is reduced Æ

Cost = 80 and SP = 160.

41.The number of washing machines plus the number of microwaves = 25.

Unsold items = 2 washing machines + 3 microwaves.

Goods sold = 8 washing machines + 12 microwaves = 80% of total value.

44.If L represents the loss, then 20 L represents the profit. Hence, L = = 25.

Hence, CP = 300.

45.A sells at 50% profit, while B sells at 30% profit.

46.The required profit will be given by:

Maximum possible selling price – Minimum possible purchase price.

50.Sales of 3000 copies for `160 and 1500 copies for ` 200 each.

Level of Difficulty (III)

1-10.Concentrate on creation of the revenue equation and the cost equations separately. Revenue from a journey will depend on

(a)length of journey (over 2 kilometres)

(b)time of waiting.

Besides, the fixed metre reading of ` 10 at the start is used up at the rate of 1 Re per 200 metres and/or 1 Re/minute of waiting.

11.Coca-Cola earns (10 – 3) – (2 + 0.6 + 0.1) = ` 4.3 per 300 ml bottle.

Similarly, for 500 ml bottle, the profit is

12 – 6.15 = ` 5.85.

and for 1500 ml bottle, 28 – 13.2 = 14.8 for 1500 ml bottle.

The profit per ml sold has to be maximised.

12. = 2,32,55,814.

13-14.The earning for one set of 5 bottles = 4.3 × 4 + 5.85 = ` 23.05.

15.Maximum profitability = .

16-19.Profit = Revenue – Expenses.

20-22.Observe the profitability rates for each type of item.

23.Revenues = 8 computers × ` 50/hr × 12 hours × 30 days

Costs = monthly cost + + Hourly cost × 12 hours × 30 days.

24-28.Will be solved on the same principle as question 23.

29-33.Revenues = occupancy × cost/ticket.

× no. of kilometres.

Solutions and Shortcuts

Level of Difficulty (I)

1.0.9 × Price = 495 Æ Price 550.

2.The SP = 107.5% of the CP. Thus, CP = 34.4/1.075 = ` 32.

3.1.15 × Price = 4600 Æ Price = 4000.

4.A loss of 20% means a cost price of 100 corresponding to a selling price of 80. CP as a percentage of the SP would then be 125%

5.2400 = 1.25 × cost price Æ Cost price = 1920

Profit at 2040 = ` 120

Percentage profit = (120/2040) × 100 = 6.25%

6.CP = 935/1.1 = 850. Selling this at 810 would mean a loss of `40 on a CP of ` 850.

7.The CP will be ` 5400. Hence at an S.P. of 5703.75 the percentage profit will be 5.625%

8.CP = 63/1.05 = 60. Thus, the required SP for 10% profit = 1.1 × 60 = 66.

9.The buying price is ` 9 per dozen, while the sales price is ` 12 per dozen – a profit of 33.33%

10.Sales tax = 120/5 =24. Thus, the SP contains ` 24 component of sales tax. Of the remainder (120 – 24 = 96) 1/3^rd is the profit. Thus, the profit = 96/3 = 32. Cost price = 96 – 32 = 64.

11.C.P × 1.2 = 25 Æ CP = 20.833

At a selling price of ` 22.5, the profit percent 1.666/20.833 = 8%

12.Solve using options. Option (c) gives you ` 175 as the cost of the trouser. Hence, the shirt will cost 12% more i.e. 175 + 17.5 + 3.5 = 196.

This satisfies the total cost requirement of ` 371.

13.The formula that satisfies this condition is:

Loss of a²/100% (Where a is the common profit and loss percentage). Hence, in this case 400/100 = 4% loss.

14.Use PCG as follows:

Hence, 8% is the correct answer.

15.The cost per toffee = 75/125= ` 0.6 = 60 paise. Cost of 1 million toffees = 600000. But there is a discount of 40% offered on this quantity. Thus, the total cost for 1 million toffees is 60% of 600000 = 360000.

16.On a marked price of ` 80, a discount of 10% would mean a selling price of ` 72. Since this represents a 20% profit we get:

1.2 × CP = 72 Æ CP = 60.

17.The thought process in this question would go as follows:

80 – 5% of 80 = 76 (after the first discount). 76 – 5% of 76 = 76 – 3.8 = 72.2 (after the second discount)

18.For ` 72, we can buy a dozen pair of gloves. Hence, for ` 24 we can buy 4 pairs of gloves.

19.100 Æ 80 (after 20% discount) Æ 72 (after 10% discount) Æ 68.4 (after 5% discount).

Thus, the single discount which would be equivalent would be 31.6%.

20.180 × 0.9 × x = 137.7 Æ x = 0.85

Which means a 15% discount.

21.If you assume the cost price to be 100 and we check from the options, we will see that for Option (c) the marked price will be 120 and giving a discount of 12.5% would leave the shopkeeper with a 5% profit.

22.Solve by trial and error using the options. If he marks his goods 30% above the cost price he would be able to generate a 17% profit inspite of giving a 10% discount.

23.The customer pays ` 38 after a discount of 5%. Hence, the list price must be ` 40.

This also means that at a 20% discount, the retailer buys the item at ` 32.

Hence, the profit for the retailer will be ` 6 (38 – 32).

24.The profit would be given by the percentage value of the ratio 6/32 = 18.75%.

25.The labour price accounts for ` 400. Since the profit percentage gives a 20% profit on this component i.e. ` 80.

Hence, the marked price is ` 980.

26.The costs in 2006 were 300,400 and 200 respectively. An increase of 20% in material Æ increase of 60. An increase of 30% in labor Æ increase of 120. Increase of 10% in overheads Æ increase of 20.

Total increase = 60 + 120 + 20 = 200. New cost = 900 + 200 = 1100

27.For a 20% profit on labout cost, he should mark his goods at 1100 + 20% of 520 = 1204. Note, 520 is the new cost of labout after a 30% increase as described in Question 26.

28.SP = 960 = 0.96 × CP Æ CP = 1000. To gain a profit of 16%, the marked price should be 116% of 1000 = 1160.

29.The SP per article = ` 3. This represents a profit of 20%. Thus, CP = 3/1.2 = 2.5. 8 articles would cost ` 20 and hence selling at 18.40 would represent a loss of ` 1.6 which would mean an 8% loss on ` 20.

30.The percentage profit = × 100

= 10/40 × 100 = 25%

31.In the question, A’s investment has to be considered as ` 10,000 (the house he puts up for sale).

He sells at 11,500 and buys back at ` 9775. Hence his profit is ` 1725.

Required answer = × 100 = 17.25

32.For 12 locks, he would have paid ` 51, and sold them at ` 57. This would mean a profit percentage of 11.76%

33.230/200 = 1.15 Æ the profit percentage would be 15% if sold at 230. Thus, the increase in profit percent = 15 – 10 = 5%.

34.A’s selling price = 1.25 × 120 = 150. C’s Cost price = B’s selling price = 198/1.1 = 180. Thus, B’s profit = ` 30 and his profit percent = 30 × 100/150 = 20%.

35.A 10% reduction in price increases the consumption by 11.11% (Refer Table 4.1). But the increase in consumption is 6.2 kg.

Hence, the consumption (original) will be 6.2 × 9 = 55.8 kg.

Hence, original price = 279/55.8 = `5.

Hence, reduced price = ` 4.5

36.Total cost = 50 × 10 + 40 × 12 = 980. Total revenue = 90 × 11 = 990. Gain percent = (10 × 100)/980 = 100/98 %.

37.Percentage profit = × 100

= 10/20 × 100 = 50%

38.The profit percent would be equal to 50 × 100 /950 = 5000/950 = 100/19% = 5 (5/19)%

39.A gross means 144 eggs. Thus, the cost price per egg = 25 paise and the selling price after a 12.5% profit = 28 paise approx.

40.B sold the table at 25% profit at ` 75. Thus cost price would be given by: CP_B × 1.25 = 75

B’s Cost price = ` 60.

We also know that A sold it to B at 20% profit.

Thus,

A’s Cost price × 1.2 = 60

Æ A’s cost price = 50.

41.300 (A buys at this value) Æ 345 (sells it to B at a profit of 15%) Æ 404 (B sells it back to A at a profit of 20% gaining ` 69 in the process). Thus, A’s original cost = `300.

42.Net loss = (20/100)² = 4% of cost price. Thus, 4800 (total money realized) represents 96% of the value. Thus, the cost price would be ` 5000 and the loss would be ` 200.

43.CP × 0.9 = 40 Æ CP = ` 44.444. Loss per kg = ` 4.44. To incur a loss of `80, we need to sell 80/4.44 = 18 kgs of tea.

44.The CP of the TV Æ CP_TV × 0.8 =12000 Æ CP_TV = 15000

The CP of the VCP Æ CP_VCP × 1.2 = 12000 Æ CP_VCP = 10000.

Total sales value = 12000 × 2 = 24000.

Total cost price = 15000 + 10000 = 25000. Loss = 25000 – 24000= 1000.

45.The profit of 15% amounts to ` 450. This should also be the actual loss on the second TV.

Thus, the actual loss = ` 450 (10% of C.P.)

Hence, the CP of the second set = ` 4500.

46.Let the cost price be P. Then, P × 0.95 × 1.1 = P × 1.05 – 2 Æ P = 400. Alternately, you could have solved this using options.

47.Let the cost price be P. Then, P × 0.95 × 1.2 = P × 1.1 + 7 Æ P = 175. Alternately, you could have solved this using options.

48.4% of the cost price = ` 200.

Thus, cost price = ` 5000

and selling price @ 6% profit = ` 5300.

49.From the last statement we have: Charan’s cost price = 1188/1.1 = 1080 = Bhushan’s selling price. Then, Bhushan’s CP would be given by the equation: CP × 0.9 = 1080 Æ CP for Bhushan = 1200 = SP for Ashok.

Also, Ashok gains 20%. Hence, CP for Ashok Æ CP × 1.2 = 1200 Æ CP for Ashok = 1000.

This includes a ` 110 component of repairs. Thus, the purchase price for Ashok would be 1000 – 110 = 890.

50.Solve through the values given in the options. Option (c) is correct because at 4/5 × 300 + 5/4 × 600 we see that the profit earned = ` 90.

51.Solve this question using the options. The first thing you should realize is that the cost of the lower priced item should be less than 240. Thus, we can reject options a and d. Checking option (c) we can see that if the lower priced item is priced at 200, the higher priced item would be priced at ` 280. Then:

1.19 × 200 = 238 and 0.85 × 280 = 238. It can be seen that in this condition the values of the selling price of both the items would be equal (as required by the conditions given in the problem). Thus, option (c) is the correct answer.

52.Original Cost Price = ` 5000

New Cost Price = 1.3 × 5000 = ` 6500

Price paid by retailer = 1.2 × 5750 = ` 6900

Profit percentage = (400/6500) × 100 = 6 (2/13)%

53.The total manufacturing cost of the article = 60 + 45 + 30 = 135. SP = 180. Thus, profit = ` 45.

Profit Percent = 45 × 100/135 = 33.33%

54.Assume marked price for both to be 100.

X’s selling price = 100 × 0.75 × 0.95 = 71.25

Y’s selling price = 100 × 0.84 × 0.88 = 73.92.

Buying from ‘X’ is more profitable.

55.The total discount offered by A = 8% on 20000 + 5% on 16000 = 1600 + 800 = 2400.

If B wants to be as competitive, he should also offer a discount of ` 2400 on 3600. Discount percentage = 2400 × 100/36000 = 6.66% discount.

56.The trader pays 800 × 0.95 × 0.95 = ` 722

57.Manufacturer’s profit percentage

= (22/700) × 100 = (22/7)%

58.For a cost price of ` 400, he needs a selling price of 480 for a 20% profit. This selling price is arrived at after a discount of 4% on the marked price. Hence, the marked price MP = 480/0.96 = 500.

59.Solve using options. Option (b) fits the situation as a 12.5% discount on 288 would mean a discount of ` 36. This would leave us with a selling price of 252 which represents a profit percent of 20% on ` 210.

60.If he marks the camera at 800, a 10% discount would still allow him to sell at 720 – a profit of 20%.

61.Cost price to the watch dealer

= 250 + 10% of 250 = ` 275

Desired selling price for 20% profit

= 1.2 × 275 = 330

But 330 is the price after 25% discount on the marked price.

Thus,

Marked price × 0.75 = 330 Æ MP = 440

Hence, he should mark the item at ` 440.

62.If the cost price is 100, a mark up of 80% means a marked price of 180. Further a 15% discount on the marked price would be given by:

180 – 15% of 180 = 180 – 27 =153. Thus, the percentage profit is 53%.

63.A cost price of ` 650 would meet the conditions in the problem as it would give us a loss of 140 (if sold at 510) and a profit of 70 (when sold at 720)

64.Cost per 100 apples = 60 + 15% of 60 = ` 69.

Selling price @ 20% profit = 1.2 × 69 = ` 82.8

65.Profit percent = (100/900) × 100 = 11.11%

Level of Difficulty (II)

1.Total outlay (initial investment) = 750 × 0.6 = ` 450.

By selling 600, he should make a 40% profit on the outlay. This means that the selling price for 600 should be 1.4 × 450 Æ ` 630

Thus, selling price per article = 630/600 = 1.05. Since, he sells only 630 articles at this price, his total recovery = 1.05 × 630 = 661.5

Profit percent (actual) = (211.5/450) × 100 = 47%

2.In order to solve this problem, first assume that the cost of manufacturing 1 article is `1. Then 100 articles would get manufactured for `100. For a 20% profit on this cost, he should be able to sell the entire stock for `120. However since he would be able to sell only 88 articles (given that 12% of his manufactured articles would be rejected) he needs to recover `120 from selling 88 articles only. Thus, the profit he would need would be given by the ratio 32/88.

Now it is given to us that his selling price is `7.5. The same ratio of profitability i.e.32/88 is achieved if his cost per article is ` 5.5.

3.The total cost to print 900 copies would be given by:

Cost for setting up the type + cost of running the printing machine + cost of paper/ink etc

= 1000 + 120 × 9 + 900 × 0.6 = 1000 +1080 + 540 = 2620.

A 10% profit on this cost amounts to ` 262. Hence, the total amount to be recovered is ` 2882.

Out of this, 784 copies are sold for ` 2.75 each to recover ` 2156.

The remaining money has to be recovered through advertising.

Hence, The money to be recovered through advertising = 2882 – 2156 = ` 726. Option (c) is correct.

4.Total cost (assume) = 100.

Recovered amount = 65 + 0.85 × 32.5 + 0.7 × 32.5

= 65 + 27.625 + 22.75 = 115.375

Hence, profit percent = 15.375%

5.Cost price = x

Marked Price = x + 205

Selling Price = 0.9x + 184.5

Percentage Profit = [(–0.1x + 184.5)/x] × 100.

=

6.She should opt for a straight discount of 30% as that gives her the maximum benefit.

7.If you assume that his cost price is 1 Re per gram, his cost for 1000 grams would be ` 1000. For supposed 1 kg sale he would charge a price of 1080 (after an increase of 20% followed by a decrease of 10%).

But, since he gives away only 900 grams the cost for him would be ` 900.

Thus he is buying at 900 and selling at 1080 – a profit percentage of 20%

8.While buying

He buys 1100 gram instead of 1000 grams (due to his cheating).

Suppose he bought 1100 grams for ` 1000

While selling:

He sells only 900 grams when he takes the money for 1 kg.

Now according to the problems he sells at a 8% profit (20% mark up and 10% discount).

Hence his selling price is ` 1080 for 900 grams.

To calculate profit percentage, we either equate the goods or the money.

In this case, let us equate the money as follows:

Buying;

1100 grams for ` 1000

Hence 1188 grams for ` 1080

Selling: 900 grams for ` 1080

Hence, profit% = 288/900 = 32%

(using goods left by goods sold formula)

9.The new situation is

Buying:

1100 grams for ` 900

Hence, 1320 grams for ` 1080

Selling: 900 grams for ` 1080

Profit % = = 46.66%

10.Assume he bought 20 apples each. Net investment fi ` 5 + ` 4 = ` 9 for 40 apples. He would sell 40 apples @ (40 × 2)/9 = ` 8.888 Æ Loss of ` 0.111 on ` 9 investment

Loss percentage = 1.23%

11.600 – 10% of 600 = 540. 540 – 5% of 540 = 513. 513 + 5% of 513 = 538.65

12.The problem is structured in such a way that you should be able to interpret that if he had sold 120 kg of rice he would recover the investment on 100 kg of rice.

% Loss/Profit = × 100

(–20/120) × 100 = 16.66% loss.

Since, cost price for Deb is ` 11; selling price per kg would be ` 9.166.

13.Comparisons have to be made between:

192 × 34, 198 × 33, 204 × 32 and 210 × 31 for the highest product amongst them.

The highest value of revenue is seen at a price of ` 198.

14 & 15: Using options from question 15. Suppose she had spent ` 6 at the market complex, she would spend ` 3 at her uncle’s shop. The other condition (that she gets 2 sweets less per rupee at the market complex) gets satisfied in this scenario if she had bought 12 chocolates overall. In such a case, her buying would have been 2 per Rupee at the market and 4 per rupee at Uncle Scrooge’s shop.

Trial and error will show that this condition is not satisfied for any other option combination.

16.The given situation fits if we take Q as 60% profit and then the loss would be 37.5% (which is 62.5%) of Q). Thus, if ` 24 is the cost price, the selling price should be 24 × 1.6 = ` 38.4

17.Assume the price of 1 kg as 100. He initially sells the kg at 115. His original profit is 15%. When he is able to sell only 80% of his items: his new revenue would be given by 80 × 1.265 = 101.2 on a cost of 100. Profit percentage = 1.2%

Change in profit percent = –13.8 (It drops from 15 to 1.2)

18.Ramu’s total discount:

8% on 8000 = ` 640

5% on 12000 = ` 600

3% on 16000 =

Total = ` 1720 on ` 36000.

Hence, Realised value = 34280.

Shyamu’s Discounts:

7% on 12000 = 840

6% on 8000 = 480

5% on 16000 =

 ` 2120 on ` 36000

Hence, Realised value = 33880.

The higher profit is for Ramu.

Also, the CP has a mark up of 25% for the Marked price. Thus the CP must have been 28800 (This is got by 36000 – 20% of 36000 – PCG thinking)

Thus, the profit % for Ramu would be: (5480*100)/28800 Æ 19% approx.

19.In the case of the given defaults, the discount for Ramu would have gone down to:

4% on 12000 (the second payment) and the second discount would thus have been `480 meaning that the sale price would have risen by `120 (since there is a `120 drop in the discount)

1% on 16000 Æ A reduction of 2% of 16000 in the discount Æ a reduction of ` 320.

Hence, Ramu’s profit would have gone up by `440 in all & would yield his new profit as:

5480 + 440 = ` 5920

20.The following working would show the answer:

Ramu’s Discounts

7% on 8000 = ` 560

4% on 12000 = ` 480

2% on 16000 =

Total = ` 1360 on ` 36,000.

Shyamu’s discounts:

6% on 12000 = 720

5% on 8000 = 400

4% on 16000 =

` 1760 on ` 36000

Thus, their profits would vary by ` 400 (since their cost price is the same)

21.Solve using options. Option (a) fits as if we take SP as 2000, we get CP₁ as 1500 and CP₂ as 1600 which gives us the required difference of ` 100.

22.The first one would get a profit of ` 500 (because his cost would be 2500 for him to get a 20% profit on cost price by selling at 3000).

The second one would earn a profit of 600 (20% of 3000).

Difference in profits = `100

23.Find out the total revenue realization for both the cases:

Case 1: (Old) Total sales revenue = 2000 × 3.25 × 0.75.

Profit_old = Total sales revenue – 4800

Case 2: (New) Total sales revenue = 3000 × 4.25 × 0.75

Profit_new = Total sales revenue – 4800

The ratio of profit will be given by Profit_new /Profit_old

24.Profit in original situation = 20%.

In new situation, the purchase price of 90 (buys at 10% less) would give a selling price of 132 (sells at 10% above 120).

The new profit percent = [(132 – 90) × 100]/90 = 46.66

Change in profit percent = [(46.66 – 20) × 100]/20 = 133.33%

25.The successive discounts must have been of 10% each. The required price will be got by reducing 25 by 10% twice consecutively. (use PCG application for successive change)

26.From the options you can work out that if the original price was ` 12 per dozen, the cost per apple would be `1.

If she is able to get a dozen apples at a reduced price (reduction of ` 1 per dozen), she would be able to purchase 1 extra apple for the 1 Rupee she saved. Thus, option (c) is correct.

27.The following calculations will show the respective costs:

Primary Cost: 35% of 12600 = 4410

Miscellaneous costs = 2% of 12600 = 252

Gross Profit = 12600 – 4410 – 1400 – 650 – 252 = 5888

Trading cost = 0.25 × 5888 = 1472

Hence, Net profit = 4416.

Percentage profit = 4416/14000 = 31.54%

28.If we assume the value of the first cycle as ` 900. Then 900 + 96 = 996 should be equal to twice the value of the second cycle. Hence, the value of the second cycle works out to be: 498.

Also 498 + 96 = 594 which is `306 less than 900.

Hence, Option (a) fits the situation perfectly and will be the correct answer.

Note here that if you had tried to solve this through equations, you would have got stuck for a very long time.

29.David (100) Æ Goliath (80) Æ Hercules (100) Æ Goliath (90)

Hercules loss corresponds to 10 when David buys the laptop for 100.

Hence, Hercules’s loss would be `17500 when David buys the laptop for 1,75,000.

30.While purchasing he would take 1200 grams for the price of 1000 grams.

While selling he would sell 900 grams for the price of 1000 grams. Since CP = SP, the profit earned is through the weight manipulations. It will be given by:

Goods left/goods sold = 300 × 100/900 = 33.33%

31.Assume that for 100 items the cost price is ` 100, then the selling price is ` 130. Since 24 is sold at half the price, he would recover 24 × 1/2 = `12 (since it is sold at half the cost price)

The remaining 70 would be sold at 70 × 1.3 = `91.

Total revenue = 91 + 12 = 103 Æ a profit of 3% (on a cost of 100).

32.An increase in the price by `12 will correspond to 50% of the CP.

Hence, The CP is `24 and initially the book was being sold at `19.2. Hence, if there is an increment of ` 4.8 in the selling price, there would be no profit or loss.

33.In the first year, the profit percentage would be:

Old Profit Percentage = = 6.35%

New Profit Percentage = = 6.65

34.Since the ratio remains unchanged the percentage profit of the village will remain unchanged too.

35.The profit would increase by 10% as there is no change in the percentage profit.

36.Since the answer to question 33 is 0.3, if we increase the percentage profit for both men and women by 1 % the overall percentage profit would also go up by 1% - thus 0.3 + 1 = 1.3%

37.x × 8 + 0.75x × 22 = 1.4 × 4725 Æ x = 270.

On an investment of ` 4725, a profit of 40% means a profit of 1890.

Hence, the targeted sales realization is ` 6615.

The required equation would be:

8p + 22 (3p)/4 = 6615

Æ 8p + 33p/2 = 6615

In this expression for LHS to be equal to RHS, we need 33p/2 to be an odd number. This can only happen when p is not a multiple of 4 (why?? Apply your mind). Hence, options a & c get eliminated automatically.

38.After 2 years, the flat would be worth ` 288000, while the land would be worth ` 266200. The profit percentage of the gainer would be given by:

(21800/266200) × 100 = 8.189%

Hence (d).

39.The total investment will be A + B + C.

C being 3000, B will be 2250 and A will be 1500.

The total investment is: 6750.

Returns to be given on their expectations:

A = 150, B = 337.5 and C = 0.

From this point calculate the total profit , subtract A’s and B’s expected returns and B’s share of the profits for managing the business before dividing the profits in the ratio of capital invested. However, most of this information is unknown. Hence option (d) is correct.

40.The cost of the trip would be proportional to the price of petrol. So, if initially the cost is 100, the new cost would be 80. Also, initially since his profit is 20%, his revenue would be 120. When he takes 4 passengers instead of 3 his revenue would go up to 160 – and his profit would become 100% (cost 80 and revenue 160).

41.Total number of microwave ovens = 15

Hence, washing machines = 10

Thus, He sells 80% of both at a profit of ` 40,000.

Cost of 80% of the goods = 0.8 × 2,05,000 = 1,64,000.

Total amount recovered = 1,64,000 + 40,000 = 2,04,000

Hence, loss = ` 1000

42.Since the actual initial loss was 10% and it is to be compared to a profit of 5%, it is 200% of the profit. Option (c) is correct.

43.He would be selling 800 grams for `12. Since a kg costs `10 800 grams would cost ` 8.

Hence, his profit percentage is 50%.

44.The interpretation of the first statement is that if the loss at 275 is L, the profit at 800 is 20L.

Thus, 21L = 800 – 275 = 525 Æ L = 25.

Thus, the cost price of the item is `300.

To get a profit of 25%, the selling price should be 1.25 × 300 = 375.

45.C’s purchase price = 2145 × 10/11 =1950

B’s rate of profit is 3 times C’s rate of profit. Hence, B sells to C at 30% profit.

B’s price + 30% profit = 1950 (C’s price).

Hence, B’s Price = 1500.

Further, since A’s profit rate is 5/3^rd the rate of profit of B, A’s profit percent would be 30 × 5/3 = 50%.

Thus, A’s Price + 50% profit = 1500 (B’s price)

Thus, A’s price = 1000

46.He would buy at 500 and sell at 888 to get a profit of 388

47.There were 5 printers (2 + 3) and 20 monitors. He sells 2 printers for a profit of ` 2000 each. Hence, profit from printer sales = ` 4000.

Then, profit from monitor sales = ` 45000

Thus, profit per monitor = = ` 3000

(Since, 15 monitors were sold in all.)

Hence, C.P. of monitor = ` 15000

And C.P. of Printer = ` 7500

Total cost = 15000 × 20 + 7500 × 5 = 3,37,500

Total Revenues = 18000 × 15 + 9500 × 2 = 28,900

Hence, loss of ` 48,500

48.Loss% = × 100 = 14.37%

49.By charging ` 1.2 more his profit should double to 40%. This means that his profit of 40% should be equal to ` 2.4. Thus, his cost price must be `6 and his original selling price should be 7.2. Hence, option (d) is correct.

50.Total cost = 5 lacs

Total revenue = 3000 × 160 + 1500 × 200 – vendors discount of 20% of revenues

= 7.8 lacs – 1.56 lacs = 6.24 lacs.

Profit percent = (1.24 × 100)/5 = 24.8%