When we have to find the average of a given set of values, we just add those values and divide by the number of values in consideration. But what do we do when we have to find the average of 2 given sets of values, each containing different number of elements? We use alligation method, which is a simplified technique to solve the complex average problems.

Generally, in competitive exams like CAT or XAT, the problems are not asked directly from these concepts, rather the techniques involved are applied for high difficulty level problems of CAT. So, consider the concepts of allegation & mixtures as a means and not as an end.

Let’s start with the types of mixtures.

- Simple mixtures: When 2 or more different ingredients are mixed together , a simple mixture is formed. Ex: mixing of milk and water, mixing rice of 2 different prices.
- Compound mixtures: When 2 or more different mixtures are mixed together, a compound mixture is formed. Ex: Mixing 2 alloys containing 2 or more types of metals in different proportion.

Concept of alligation.

Let’s try to understand it by examples.

Ex-1. Average weight of 60 girls is 15 kg and average weight of 40 boys is 30 kg . Find the average weight of the group in which boys and girls are taken together.

Weighted average method.

Average weight = ( 60×15 + 40×30 )/( 60+40 ) = 21 kg

Alligation Method

In the alligation chart above, x is the required average of the whole group. The individual average values have been written on the top, required average in the middle and the number of elements in both the groups at the base, correspondingly.

Take the difference of the top and middle values and equate it with the base value as the line indicates. So, equations obtained would be, x-15 = 40 and 30-x = 60.

Now, take the ratio of both equations.

( x-15/30-x ) = 40/60

Solving it would give x = 21 kg.

Suppose the question was framed a bit differently, as below.

Average weight of boys is 30kg and average weight of girls is 15kg. The average weight of whole group, boys and girls taken together, is 21 kg. Find the ratio of number of girls and boys.

We would again apply the alligation method.

Here a is the number of girls and b is the number of boys.

(21-15)/(30-21) = b/a => a/b = 2/3.

Now, we will see different formulae in mixture and alligations and their application.

**Application in normal mixtures.**

(Quantity of expensive item)/(Quantity of cheaper item) =(Mean Price- Price of cheaper item)/ (Price of expensive item – Mean price)

Average price of goods :

Ex-2. A shopkeeper mixes 30 kgs of sugar at Rs 20/kg and 45 kgs of sugar at 10/kg. What is the average price of the mixture?

Sol: Let average price be x.

x-10/20-x = 30/45 => x= Rs. 14/kg

Average Speed :

Ex-3. A car travels at 20kmph for 30 minutes and at 10 kmph for 45 minutes. Find the average speed of the car for entire journey. (Note that we are using the same values as in the above example )

Sol: x-10/ 20-x = 30/45 => x= 14 km/hr .

NOTE : In the questions involving speed, we never use distance directly while applying alligation method. Only speed and time are used. Speeds are always written at the top, average speed in the middle and time at the base.

Ex-4. Amit covers 200 km in 10 hours. The first part of journey is travelled by bus at the speed of 15 km/hr and second part by a car at the speed of 25 km/hr. What is the ratio of distances covered by bus and car?

Sol: Average speed= 250/10 = 10 km/hr. Let time travelled by bus be ‘a’ and by car be ‘b’. Applying alligation method

(20-15)/(25-20) = b/a = 1:1

This means that both vehicles have travelled equal amount of time out of 10 hours i.e. 5 hours each.

Distance travelled by bus = 15×5 = 75 and distance travelled by car = 25×5 = 125 . So, the ratio is

75/125 = 3/5

Profit/Loss :

Ex-5. A milkman has 30 litres of milk. He mixes 10 litres of water in that milk. What is his profit percentage, when he sells all the mixture at cost price. Assume the water is free of cost.

Sol: Water, which is free of cost, is being sold at the cost of milk, then the water gives profit on the cost of 30 litres of milk. Profit% = (10/30)x 100 = 33.33% .

Understand that all the milk in the mixture is being sold at its cost price, so profit from milk component is 0%, while water in the mixture is being sold at the cost of milk which is the source of profit.

In questions like these, where a cost-free item is being added to a non-free item in the ratio of a/b, and the mixture is being sold at the cost of non-free item, the profit% = (a/b) x 100

Ex-6. In what ratio should water be mixed with the wine worth Rs. 60 per litre so that after selling the mixture at Rs.50 per litre, the profit will be 25%?

Sol: Selling price= cost price x 1.25 => 50= cost price x 1.25 => C.P. of mixture = Rs. 40.

Let amount of water be ‘a’ and wine be ‘b’

40-0)/(60-40) = a/b = 1/2

**Application in compound mixture.**

2 mixtures A and B contain components x and y in the ratio p/q and r/s respectively. In what ratios A and B should be mixed so that in the new mixture ratio of x and y is t/u.

In questions involving such mixtures , select any particular component, say x, and find its proportion in each mixture, i.e. p/(p+q) and r/(r+s) . The required proportion is t / (t+u). Let the quantity of each mixture needed to be mixed be a and b. Then apply the alligation method by writing proportions on top, required proportion in the middle and corresponding quantities needed at the base.

Solving it would give the required ratio a/b .

Ex-7. Vessel A contains milk and water in the ratio 4:5. Vessel B contains milk and water in the proportion 5:1.In what proportion should quantities be taken from A & B to form a mixture in which milk and water are in the ratio 5:4?

Sol: For this question, we will consider the proportion of milk in each mixture. In Vessel A, the proportion of milk is 4/(4+5) =4/9. In vessel B, the proportion of milk is 5/(5+1) = 5/6.

The amount of milk in the final mixture= 5/(5+4) = 5/9

a = (5/6)-(5/9) = 5/18

b = (5/9)-(4/9) = 1/9

The ratio is (5/18):(1/9) = a/b = 5/2.

Ex-8. Product P is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product P with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product P?

Sol: First, we need to find the proportion of B in Product P. It can be easily found as its proportion in chemical X and Y is given and proportion of X and Y in P is also given.

P is produced by mixing chemical X and chemical Y in the ratio 5 : 4. Hence, 5/9th of product P is chemical X and 4/9th of product P is chemical Y.

Chemical X has A and B in the ratio 1 : 3.

So, 3/4th of X is B.

Therefore, proportion of B in product P from chemical X = 5/9×3/4

Chemical Y has B and C in the ratio 2 : 1.

So, 2/3rd of Y is B.

Therefore, proportion of B in product P from chemical Y = 4/9×2/3

Adding the two, the proportion of B in Product P

= 5/9×3/4 + 4/9×2/3

= 77/108

The final mixture is obtained by mixing 864 units of product P with water.

Proportion of B in final mixture= 1/2 = 54/108. Proportion of B in water is 0.

Let units of water to be added be x. Applying alligation,

x = 368 units.

**Replacing of part.**

In cases where a certain amount of an item A is being drawn out and being replaced by another item B and this process is repeated n times then,

Final or reduced amount of A = Initial amount of A x {1 – ( amount being drawn out in each operation/total amount)}^n

Let’s see some examples.

Ex-9. A container has 50 litres of milk in it. 5 litres of milk is taken out and is replaced by 5 litres of water. This process is repeated 4 more times. What is the amount of milk in the container after final replacement?

Sol: Total amount = 50 litres

Amount being replaced each time = 5 litres

n = 5

Initial amount of milk= 50 litres. Therefore,

Final amount of milk = 50 x (1-(5/50))^5 = 29.5 litres

NOTE: Above formula is valid for pure items as well as mixtures. But in case of mixtures, the component into consideration must be that which is getting reduced after replacement.

Ex-10. In a mixture of milk and water, there is only 75% milk. After replacing the mixture with 5 litres of water, percentage of milk in the mixture becomes 60%.. What is the total quantity of the mixture?

Sol: Let total quantity of mixture be T. We will consider milk as it is getting reduced.

Initial amount = 75% (if absolute values are not given, proportion/percentage can also be used)

Final amount = 60%

n = 1. Then,

60 = 75 x (1 – 5/T) => T = 25 litres.

Questions similar to above questions can also be solved using alligation. Let’s take an example.

Ex-11. Some part of a sugar solution which contains 40% sugar is replaced with another sugar solution which contains 19% sugar. Concentration of sugar in the new mixture became 26%. What fraction of the first sugar solution was replaced with second sugar solution?

Sol: Applying alligation method,

So, the second and first solutions are in the ratio 2:1 in the final mixture. Since volume of mixture is same as that of initial volume of first sugar solution, we can see that only 1/3 of first sugar solution is left in final mixture. This means 2/3 of it was replaced by second sugar solution.

To sum up, this topic is one of the scoring areas though the weight age is not that big as compared to other topics but, we should focus on scoring easy marks first. I hope you have got a gist of the topic and you can start practicing questions on Alligations and Mixtures.

Best of luck for your preparation.

Besides, there is tremendous scope for clinical research. Doctors with management background can do a phenomenal job in terms of product development and management.

It is not necessary that you have to move away from your medical profession once you have a management degree. You can very well continue to practice. In fact, a doctor with managerial skills can provide the best treatment, which is cheapest for the patients. Now, more and more students are showing interest in health-care management.

FAQ’s:

**1. Where should one do an MBA from?**

There are around are various good MBA institutes in India. Some of these include:

a. Indian Institutes of Management (IIMs)

b. Faculty of Management Studies (FMS) at the University of Delhi

c. Xavier Labor Relations Institute (XLRI) Jamshedpur

d. SP Jain Institute of Management Mumbai

e. Management Development Institute (MDI) Gurgaon.

f. Indian School of Business (ISB) Hyderabad

FMS & XLRI conduct independent entrance tests in January while the rest admit students based on the performance in CAT exam. GMAT scores are also accepted by most of the MBA institutes.

**2. Should I Really go in for an MBA?**

Physicians, in general, tend to be respected & relatively well-paid professionals, and changing career to MBA will largely make you forfeit contact with direct patient care. So, you should be doubly sure that you are ready to leave it for good. Consult your family & friends and take some time to make this decision. Consider all the pros and cons as this decision is very individual and once you steer towards management, it might become difficult to get back into only medicine practice (though, of course, you could still stay in the management aspect of medicine).

**3. What is the future for medicos in the corporate world?**

a. Healthcare Consulting:

Most top management consulting firms today have large practices dedicated to consulting healthcare organizations. Healthcare consultants, though, do not just work with hospitals (or other clinical organizations). They also work with large pharma companies, insurers and payers, and even public organizations that deal with healthcare (governments and international bodies such as the World Health Organization). A career in Healthcare Consulting can be a rewarding alternative to leveraging one’s prior expertise in healthcare.

b. Hospital Administration:

Hospital Administrators look after the operations of large hospitals (sometimes even a network of hospitals) and ensure coordination between different departments, and overall patient care and process implementation. Not all hospital administrators have an MBA (those who do not are usually very senior doctors), but having one is a definite advantage. This position is closely aligned with everyday clinical operations. So, having an undergraduate education in medicine or healthcare is most often a requirement.

c. Private Equity specializing in Healthcare:

There are a number of big as well as niche players that invest in Healthcare organizations to help such organizations meet their growth goals while earning the investor a good return at the same time. Healthcare professionals bring the intimate knowledge of the healthcare ecosystem, factors which contribute to growth, and strategies to eliminate cost without reducing quality.

As an MBBS holding an MBA degree, you belong to a very unique category. There is a huge gap between demand and supply for doctor MBA's and this especially pertains to the healthcare sector (hospitals, pharmaceuticals, healthcare organizations, etc).

Doctors aspiring for an MBA degree are required to start CAT preparation early to be able to complete all the important topics before the exam. MBA aspirants can also subscribe to BYJU’S YouTube channel and learn various CAT topics more effectively from the video lessons.

]]>Interest charged on the loan can be of any type either Simple Interest or Compound Interest. Though we have discussed regarding it but for revision’s sake.

Simple interest is a the one where interest once credited does not earn interest on it.

SI = (P * R * T)/ 100

Compound Interest is where interest earns itself interest. It is the most common form of interest that is being charged nowadays.

CI = P(1 + r/100)^n

**INSTALLMENTS UNDER SIMPLE INTEREST**

Suppose Ravi bought a T.V. worth ₹20000 on EMI’s and every month a fix installment has to be for next n months where interest is charged @ r% per annum on simple interest.

Now, if the loan is for n months then Ravi will pay end the of 1st month interest for (n-1) months, at the end of second month he’ll pay interest for (n-2) months, at the end of 3rd month he’ll pay interest for (n-3) months and similarly, at the end of nth month he’ll pay no interest i.e.

Therefore, total amount paid by Ravi = [x+ (x* (n-1) * r)/ 12* 100] + [x+ (x* (n-2) * r)/ 12* 100] + [x+ (x* (n-3) * r)/ 12* 100] … [x+ (x* 1* r)/ 12* 100] + x

This will be equal to the total interest charged for n months i.e. [P+ (P* n* r)/ 12* 100].

Thus, [P+ (P* n* r)/ 12* 100] = [x+ (x* (n-1) * r)/ 12* 100] + [x+ (x* (n-2) * r)/ 12* 100] + [x+ (x* (n-3) * r)/ 12* 100] … [x+ (x* 1* r)/ 12* 100] + x

Simplifying and generalizing the above equation we get the following formula, x = P (1 + nr/100)/ (n + n(n-1)/2 * r/100))

And instead of principal sum total amount (Principal + Interest) to be repaid is given then, x = 100A/ 100n + n(n-1) r/2

**INSTALLMENTS UNDER COMPOUND INTEREST**

Let a person takes a loan from bank at r% and agrees to pay loan in equal installments for n years. Then, the value of each installment is given by

P (1 + r/100)^n = X (1 + r/100)^(n-1) + X (1 + r/100)^(n-2) + X (1 + r/100)^(n-3) + …. + X (1 + r/100)

Using the Present Value Method,

P = X/ (1 + r/100)^n … X/ (1 + r/100)^2 + X/ (1 + r/100)

**MISCELLANEOUS CASES OF INSTALLMENTS ON SIMPLE INTEREST AND COMPOUND INTEREST**

**CASE 1**: To calculate the installment when interest is charged on SI

A mobile phone is available for ₹2500 or ₹520 down payment followed by 4 monthly equal installments. If the rate of interest is 24%p.a. SI, calculate the installment.

Sol: This is one basic question. You have to just use the above formula and calculate the amount of installment.

Therefore, x = P (1 + nr/100)/ (n + n(n-1)/2 * r/100))

Here P = 2500 – 520 = 1980

R = 25% p.a.

T = 4 months

Hence, x = 1980(1 + 15 * 12/ 1200)/ (4 + 4* 3* 12/ 2 * 12 * 100)

= ₹520

**CASE 2:** To calculate the installment when interest is charged on CI

What annual payment will discharge a debt of ₹7620 due in 3 years at 50/3% p.a. compounded interest?

Sol: Again, we will use the following formula,

P (1 + r/100)^n = X (1 + r/100)^(n-1) + X (1 + r/100)^(n-2) + X (1 + r/100)^(n-3) + …. + X (1 + r/100)

Here P = ₹7620

n = 3 years

r = 50/3% p.a.

7620(1 + 50/300)^3 = x (1 + 50/300)^2 + x (1 + 50/300) + x

12100.2778 = x (1.36111 + 1.1667 + 1)

x = ₹3430

**CASE 3:** To calculate loan amount when interest charged is Compound Interest

Ram borrowed money and returned it in 3 equal quarterly installments of ₹17576 each. What sum he had borrowed if the rate of interest was 16 p.a. compounded quarterly?

Sol: In this case, we will use present value method as we need to find the original sum borrowed by Ram.

Since, P = X/ (1 + r/100)^n … X/ (1 + r/100)^2 + X/ (1 + r/100)

Therefore, P = 17576/ (1 + 4/100)^3 + 17576/ (1 + 4/100)^2 + 17576/ (1 + 4/100)

= 17576 (0.8889 + 0.92455 + 0.96153)

= 17576 * 2774988

= 48773.1972

**CASE 4:**

Gopal borrows ₹1,00,000 from a bank at 10% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are ₹10,000, ₹20,000, ₹30,000 and ₹40,000 respectively, what amount should be paid at the end of the fifth year to clear the debt?

Sol: Total principal amount left after 5th year = 100000 – (10000 + 20000 + 30000 + 40000)

= 100000 – 100000 = 0

Therefore, only interest component has to be paid in the last installment.

Hence, Interest for the first year = (100000 * 10 * 1) /100 = ₹10000

Interest for the second year = (100000 – 10000) * 10/ 100 = ₹9000

Interest for the third year = (100000 – 10000 – 20000) * 10/ 100 = ₹7000

Interest for the fourth year = (100000 – 10000 – 20000 – 30000) * 10/ 100 = ₹4000

Thus, Amount that need to paid in the fifth installment = (10000 + 9000 + 7000 + 4000) = ₹30000

**CASE 5:**

An amount of ₹12820 due in 3 years, hence is fully repaid in three annual installments starting after 1 year. The first installment is 1/2 the second installment and the second installment is 2/3 of the third installment. If rate of interest is 10% p.a. Find the first installment.

Sol: Let the third installment be x.

Since, second installment is 2/3 of the third, it will be 2/3 x. And finally, 1st installment will be 1/2 * 2/3 * x

Now proceeding in the similar fashion as we did earlier and using the compound interest formula to calculate the installment amount.

P (1 + r/100)^n = X (1 + r/100)^(n-1) + X (1 + r/100)^(n-2) + X (1 + r/100)^(n-3) + …. + X (1 + r/100)

12820 (1 + 10/100)^3 = 1/3 X (1 + 10/100)^2 + 1/2 X (1 + 10/100)^1 + X

12820(1.1)^3 = x (1/3 * (1.1)^2 + 1/2 * (1.1) + 1)

17063.42 = x * (0.40333 + 0.55 + 1)

17063.42 = x * 1.953333

x = ₹8735.53

Therefore, the amount of first installment will be 2/3 * 1/2 * x, i.e. 1/3 * 8735.53 = ₹2911.84

**CASE 6:**

Ravi lent out ₹9 to Sam on the condition that the amount is payable in 10 months by 10 equal installments of ₹1 each payable at the start of every month. What is the rate of interest per annum if the first installment has to paid one month from the date the loan is availed.

Sol: The value of money coming in should equal the value of the money going out for the loan to be completely paid off. Therefore,

₹9 + Interest on ₹9 for 10 months = (₹1 + interest on ₹1 for 9 months) + (₹1 + interest on ₹1 for 8 months) + (₹1 + interest on ₹1 for 7 months) + (₹1 + interest on ₹1 for 6 months) + …. + (₹1 + interest on ₹1 for 2 months) + (₹1 + interest on ₹1 for 1 month) + ₹1

₹9 + interest on ₹1 for 90 months = ₹10 + interest on ₹1 for 45 months

interest on ₹1 for 90 months – interest on ₹1 for 45 months = ₹10 – ₹9

interest on ₹1 for 45 months = ₹1 (i.e. money would double in 45 months)

Hence, the rate of interest = 100%/ 45 = 2.2222%

I hope you are now clear with the installment concept and can now numerous other these and varied type of questions!

]]>Before determining the reason of this why? Let’s first know what is interest and these interest rates?

Interest is the amount charged by the lender from the borrower on the principal loan sum. It is basically the cost of renting money. And, the rate at which interest is charged on the principal sum is known as the interest rate. The rate at which interest is charged depends on two factors

The value of money doesn’t remain same over time. It changes with time. The net worth of ₹ 100 today will not be same tomorrow i.e. If 5 pens could be bought presently with a INR 100 note then in future, maybe only 4 pens can be bought with the same ₹ 100 note. The reason behind this the inflation or price rise. So, the interest rate includes this factor of inflation

The credibility of the borrower, if there is more risk and chance of default on borrower’s part then more interest will be charged. And, if there is less chance of payment failure on the part of borrower then the rate of interest would be lower.

The above two reason becomes the basis of why interest rates are so important and have a great effect on markets and economy. Since a minor rise in interest rates increases the cost of borrowing for the borrower and as a result, he has to pay more interest on his loan amount and thus, a decline in his money income that he could spend on other products which create a ripple effect of decreased spending throughout the economy and vice versa. Since change in interest rate has a chain effect in the market, it has a great deal of importance in the study of market, finance, and economy. And that’s why, forms an integral part of the curriculum in the MBA programs. But, a relatively simpler level of questions is asked in the CAT based on the concepts learned at the time of high school.

These concepts are categorized into type of interests

- Simple Interest
- Compound Interest

Let’s first start and understand Simple Interest because as the name suggests it is simple and comparatively easy to comprehend.

**Simple Interest**

Simple interest is that type of interest which once credited does not earn interest on itself. It remains fixed over time.

The formula to calculate Simple Interest is

SI = {(P x R x T)/ 100}

Where, P = Principal Sum (the original loan/ deposited amount)

R = rate of interest (at which the loan is charged)

T = time period (the duration for which money is borrowed/ deposited)

So, if P amount is borrowed at the rate of interest R for T years then the amount to be repaid to the lender will be

A = P + SI

Consider a basic example of SI to understand the application of above formula such as Find the simple interest on ₹ 68000 at 50/3 % p.a. for 9 months.

Here, P = ₹ 68000

R = 50/3% p.a.

T = 9 months = 9/12 years = 3/4 years

SI = (68000 x 50/3 x 3/4 x 1/100) = ₹ 8500

**Some useful results based on Simple Interest**

Result : If rate to interest is r1% for T1 years, r2% for T2 years …. rn for Tn years for an investment. And if the Simple Interest obtained is ₹a on the investment. Then the principal amount is given by a x 100/ (r1T1 + r2T2 + … + rnTn)

Example : Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond 5 years. If he pays a total interest of ₹ 11400 at the end of nine years, how much money did he borrow?

In this case, r1 = 6%, T1 = 2 years

r2 = 9%, T2 = 3 years

r3 = 14%, T3 = 4 years (since, beyond 5 years rate is 14%)

and Simple interest = ₹11400

Therefore, P = (11400 x 100)/ (6 * 2 + 9 * 3 + 14 * 4)

= 1140000/ (12 + 27 + 56)

= 1140000/ 95

= ₹12000

Result : If a person deposits sum of ₹A at r1% p.a. and sum of ₹B at r2% p.a. then the rate of interest for whole sum is R = {(Ar1 + Br2)/ (A + B)}

Example : A man invested 1/3 of his capital at 7%; ¼ at 8% and the remainder at 10%. If his annual income is ₹561, What is his capital?

Let x be his capital or principal.

Therefore, R = (1/3 x * 0.07 + 1/4 x * 0.08 + 5/12 x * 0.10)/x

R = (1/3 * 0.07 + 1/4 * 0.08 + 5/12 * 0.10)

R = 0.08496

Total SI = ₹561

₹561 = 0.08496x

x = ₹6602

Result : If a sum of money becomes “n” times in “T years” at Simple Interest, then the rate of interest p.a. is R = 100(n – 1) / T

Example : The rate at which a sum becomes 4 times of itself in 15 years at S.I will be?

It’s a very easy question you just need to use this formula and you will directly reach to an answer.

Therefore, R = (100 x 3)/15

= 20%

Result : If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R1, R2, … , Rn respectively and time periods are T1, T2, … , Tn respectively, then the ratio in which the sum will be divided in n parts can be given by 1/R1T1: 1/R2T2 : ... : 1/RnTn

Example : A person invests money in three different schemes for 6 years, 10 years and 12 years at 10%, 12% and 15% Simple Interest respectively. At the completion of each scheme, he gets the same interest. What is the ratio of his investment?

Here, T1 = 6, T2 = 10 and T3 = 12 years resp.

And, R1 = 10%, R2 = 12%, and R3 = 15% resp.

Hence, the ratio of his investment will be

100/60 : 100/120 : 100/180

1/6 : 1/12 : 1/18

1 : 1/2 : 1/3

6 : 3 : 2

**Compound Interest**

This the most usual type of interest that is used in the banking system and economics. In this kind of interest along with one principal further earns interest on it after the completion of 1-time period. Suppose an amount P is deposited in an account or lent to the borrower that pays compound interest at the rate of R% p.a. Then after n years the deposit or loan will accumulate to: P(1 + R/100)^n

**Important Formulas**

When the interest is compounded Annually: Amount= P (1 + R/100)^n

When the interest is compounded Half-yearly: Amount = P (1 + (R/2)/100)^(2n)

When the interest is compounded Quarterly: Amount = P (1 + (R/4)/100)^(4n)

When the rates are different for different years, say R1%, R2% and R3% for 1 year, 2 years and 3-year resp. Then, Amount = P (1 + R1/100) (1 + R2/100) (1 + R3/100)

Present worth of ₹ x due n years hence is given by: Present worth = x/ (1 + R/100)^n

If a certain sum becomes “x” times in n years, then the rate of compound interest will be R = 100(x^(1/n) – 1)

If a sum of money P amounts to A1 after T years at CI and the same sum of money amounts to A2 after (T + 1) years at CI, then R = (A2 – A1)/ A1 x 100

**Miscellaneous Examples of application of Compound Interest**

Question 1: A man invests ₹ 5000 for 3 years at 5% p.a. compounded interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of third year.

Sol: Here, P = ₹5000, T = 3 years, r = 5%

Therefore, Interest at the end of 1st year = 5000 (1 + 0.05) – 5000 = ₹250

Now Income tax is 20% on the interest income so the leftover interest income after deducing income tax = (1 – 0.2) * 250 = ₹200

Total Amount at the end of 1st year = ₹5000 + 200 = ₹5200

Interest at the end of 2nd year = 5200 (1 + 0.05) – 5200 = ₹260

Interest income after Income tax = 0.8 * ₹260 = ₹208

Total Amount at the end of 2nd year = ₹5200 + 208 = ₹5408

Interest at the end of 3rd year = ₹5408 (1.05) – 5408 = ₹270.4

Interest income after Income tax = 0.8 * ₹270.4 = ₹216.32

Total Amount at the end of 2rd year = ₹5408 + 216.32 = ₹5624.32

Question 2: A sum of ₹12000 deposited at compound interest becomes double after 5 years. After 20 years, it will become?

Sol: The rate of interest at which ₹12000 doubles after 5 years is given by

R = 100(x^(1/n) – 1)

= 100(2^(1/5) – 1)

=100 x (1.1486 – 1)

= 100 x 0.1486 = 14.86%

Therefore, after 20 years it becomes,

A = ₹12000(1 + 14.86/100)^20

= ₹12000 (1.1486)^20

= ₹12000 x 15.97

= ₹ 191671.474

**Compound Interest Instalments**

Let a person takes a loan from bank at r% and agrees to pay loan in equal instalments for n years. Then, the value of each instalment is given by

P = X/ (1 + r/100)^n … X/ (1 + r/100)^2 + X/ (1 + r/100)

For better understanding, let’s understand with the help of example.

One can purchase a flat from a house building society for ₹ 55000 or on the terms that he should pay ₹ 4275 as cash down payment and the rest in three equal installments. The society charges interest at the rate 16% p.a. compounded half-yearly. If the flat is purchased under installment plan, find the value of each installment.

Sol: The cost of the flat is ₹ 55000. Now, if the person could either buy flat by paying ₹55000 or through installment plan. Since the flat was purchased through installment plan then the loan amount = ₹55000 – 4275 (down payment) = ₹50725.

Here r = 16% compounded Half-yearly in 3 equal instalments. Let x be the amount of installment. Then,

₹50725 = x/ (1 + 16/200)^3 + x/ (1 + 16/200)^2 + x/ (1 + 16/200)

₹50725 = x (1/1.2591 + 1/1.1664 + 1/1.08)

₹50725 = x (0.79421 + 0.85722 + 0.9259)

₹50725 = x (2.577)

₹50725/2.5777 = x

x = ₹19683

The above examples are just few types based on compound interest, there could be numerous others complex and difficult questions that could come in the CAT exam. Compound Interest is very significant topic in today’s world. It has vast and diverse application. In exam you could also find problems that involve both simple and compound interest. You will get thorough and fluent in this topic with time and practice.

]]>To understand the concept let’s first open a shop say a local retail stationary shop. Now suppose a student shows up and wants to buy 5 gel pens worth ₹ 5 each. So, what is the profit earned by the shopkeeper in net terms and its profit percentage?

Now this ₹ 5 is the selling price, the price at which the commodity is sold to its buyer. And let ₹ 4 is the price at which the shopkeeper has bought these pens from his supplier and this price is known as Cost Price. Thus, this difference between the price at which shopkeeper buys his pen and at which it is sold is known as profit/ loss earned. Now if Selling Price > Cost Price then he will earn profit and if Selling Price < Cost Price, then he will earn loss i.e.

Therefore, Profit = Selling Price – Cost Price

= ₹ 5 – ₹4 = ₹1 per pen

And consequently, Loss = Cost Price – Selling Price.

Now profit % = (profit/ Cost price) * 100 or profit % = {(selling price – cost price)/ Cost price} *100

= 1/5 * 100 = 20%

Similarly, Loss % = (loss/ Cost price) * 100 or Loss % = {(Cost price – selling price)/ Cost price} * 100

Now suppose another customer comes to the shop and bought 2 registers worth ₹ 50 each and a pencil box from him. And this time the shopkeeper has earned 40% profit on the registers. He earned profit of ₹10 on pencil box and the profit % on pencil box is 20%. Then what is the cost price and profit on register and selling price and cost price of pencil box?

In this case, we are given S.P. = ₹ 50/ each and profit% as 40%. Let C.P. be x. Now,

Profit % = (50 – x / x) * 100

40 = (50 – x / x) * 100

4x= 500 – 10x

14x = 500

x = 35.71

Profit = 50 – 35.71

= 14.83

What if the profit of 40% is on selling price instead?

Then, Profit = profit% * S.P

= 0.4 * 50

= ₹20

C.P. = S.P. – Profit

= ₹50 – ₹20

= ₹30

Let’s now move on to calculate S.P. and C.P. of pencil box. Let C.P. of pencil box be y

Since, profit % = (profit/ Cost price) *100

20 = (10/y) *100

2y = 100

y = ₹ 50

Therefore, S.P. = ₹50 + ₹10

= ₹60

Now since not many customers showed up on the first day of shop. Therefore, to popularize the shop the shopkeeper puts up discount of 20% on all the products. The first customer shows up and bought a packet of pencil and 3 erasers and still making up the profit of 30% on both the items. Then what is the actual cost price of both the items when the pencil is marked as ₹30 and eraser ₹ 5/ each.

Now what does the underlined marked mean? Here marked means Marked Price is the price that is offered to customer before discount basically, discount is just difference between marked price and Selling price i.e. Discount = M.P. – S.P.

In this case, Discount = 0.2 * Price of packet of pencil

= 0.2 *30

= ₹ 6

Therefore, S.P = 30 – 6

= ₹ 24

And profit is 30% on C.P., assume C.P. be x. Then,

0.3 = (24 – x/x)

0.3x = 24 – x

1.3x = 24

X = 18.46

Similarly, Discount = 0.2 * Price of packet of pencil

= 0.2* 3*5

= ₹ 3

Hence, S.P. = 12

And, 0.3 = 12 – y/y

1.3y = 12

Y (C.P. of 3 eraser) = 9.23

The above method of solving questions was a direct straightforward way of finding the solution. There’s also an alternative way of doing so. Let’s now learn that alternative method but this time instead of earning profit the seller was earning the loss. Consider this situation, that the stationer sold a parker pen at a loss of 20% for ₹ 100 and a pack of colored sketch pens at loss of 15% on S.P. What is the cost price and selling price of both the articles?

First let’s find out the C.P. and S.P. parker pen,

Since loss is 20% on C.P. Then,

S.P. = 0.8 of C.P.

100 = 0.8x

100/0.8 = C.P.

125 = C.P.

For sketch pens,

Loss is 15% on S.P. Let S.P. be y and C.P. be 100

Loss = 0.15y

C.P. = 1.15 y

100 = 1.15y

100/1.15 = y

Y = ₹87

The problems explained above are rather simple and easy ones. Now let’s do some problems that are bit complex and have come up in previous year CAT exam

Problem 1:

A watch dealer incurs an expense of Rs. 150 for producing every watch. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each watch for Rs. 100. If he is able to sell only 1,200 out of 1,500 watches he has made in the season, then he has made a profit of

a) ₹ 90000

b) ₹ 75000

c) ₹ 45000

d) ₹ 60000

This question is not basic and direct as the problems given above. This one came in CAT 2016 paper and you should expect to get this level of questions in the exam and not the ones explained above earlier. Now let’s see how to solve this problem

Here, first find out cost he has incurred to produce the watches.

Since, he made 1500 watches costing ₹ 150 each and an additional ₹ 30000 expense on them.

Thus, total Cost = (1500*150) + 30000

= ₹ 255000

Now we will calculate the revenue he earned from selling them. As, he’s able to sell 1200 watches in the season @ ₹250 each

So, the revenue earned by him during the season = ₹250 * 1200

= ₹ 300000

Also, the left over 300 pieces of clocks would have been sold by the watchmaker in off season @ 100 each.

Revenue earned through these 300 watches = 300*100

= ₹ 30000

Total Revenue = ₹ 300000 + ₹30000 = ₹330000

Profit = Revenue – Cost

= ₹ 330000 – 255000

= ₹ 75000

Problem 2:

Instead of a meter scale, a cloth merchant uses a 120cm scale while buying, but uses an 80cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?

This question above is a special one with the faulty dealer. Here, the dealer is earning profit by using false scale.

To solve this problem, first assume that price of cloth is ₹ 1/cm

Now he’s using 120 cm scale. Therefore his C.P. = (100/120)*₹1 = 0.8333/cm

Now this merchant again uses faulty scale to sell the cloth to his customers.

He uses a scale that measures 80cm as 100cm i.e. he sells 80cm for ₹100

Now he also gives discount of 20% on the cloth.

Thus, his mark up price is ₹100/80cm

So, after deducting discount @ 20%.

S.P. = ₹ 1/cm

Therefore, his profit % = (1 – 0.8333)/0.8333 * 100

= 20%

As you can see the above problem was a bit tricky, you will face similar problems in the exam. You will get more idea of the type of questions you need to practice through past year CAT papers. This is a very important topic. Every year around 3-4 questions come in the exam on profit and loss. So, practice this topic thoroughly.

]]>**Percentage as Fractions:**

One of the most useful tips, when it comes to percentage, is the use of fractions in calculating them. Given below is a table which can help you get started. There is no better way to remember these values than to start using them in your daily calculations.

Fraction | %age | Fraction | %age |
---|---|---|---|

1/2 | 50.00% | 1/9 | 11.11% |

1/3 | 33.33% | 1/10 | 10.00% |

1/4 | 25.00% | 1/11 | 9.09% |

1/5 | 20.00% | 1/12 | 8.33% |

1/6 | 16.66% | 1/13 | 7.69% |

1/7 | 14.28% | 1/14 | 7.14% |

1/8 | 12.50% | 1/15 | 6.66% |

Example: Find out 28.5% of 476

a) 135.66 b) 136 c) 136.28 d)136.43

As you can see here the option are really close and you might think that you would have to calculate the actual value. But if you are well versed with the table given above you can save some effort and time in doing so. These seconds you save might be vital in the exam.

To calculate 28.5%, I urge you to have a closer look at 1/7.

1/7 represents 14.28%, so 2/7 would represent 28.56% which is freakishly close to the percentage that we are looking for. Also, calculating 2/7 of 476 is not that difficult as 476/7 is divisible by 7 and it is 68. So, 2/7 of 476 will be 136. Now, we know that 2/7 is 28.56% and we also know that that it is 136. Our answer will be little less than 136 as we are trying to find out 28.5% which is a little less than 28.56%. So, our answer would be 135.66. Option A

**Percentage Change:**

To find out the percentage change, you can use the formula

% change = (Final value - Initial value) * 100 / Initial value

A lot of people make a mistakes in calculating values related to the given formula because they do not calculate % change based upon the initial value.

Example: My current salary is Rs. 1000 a day, which is 25% higher than what it was last year. What was my salary last year?

If you think that the answer to the above question is Rs. 750 per day then you are making the same mistake that lot of students make. The raise that I got (25%) was not on my current salary but the initial value, which was my salary previous year.

To calculate it correctly, let us assume my last year’s salary to be ‘x’ Rs. per day.

A 25% increase means an increase of 1/4

My current salary is 5/4 of my previous year’s salary.

This means that my salary last year was 4/5 of my current salary.

My last year’s salary = 4/5 * 1000 = 800

**Changing Quantities by Percents:**

The simplest way to change a number by a given percent is to simply multiply it by the ‘final’ percentage.

That is, to increase a number by x%, simply multiply it by (100 + x)% or (100 + x)/100 , and to decrease it by x%, simply multiply it by (100 – x)% or (100 - x)/100

Example: A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle?

a.16.665 b. 16.565 c. 15.465 d. 14.365

If the hypotenuse reduced by 35%, it became 100 – 35 = 65% of the original or in other words it became 0.65 times the original. The other two sides also become 0.65 times the original. Area will become (0.65)^2 = 0.4225 times the original.

New area = 0.4225*34 = 14.365. Option D

**Successive Percentage Change:**

I am sure you would have seen retail stores with offers like 50% + 40% off. If my memory serves me right, these type of offers were made extremely popular by Koutons. I always thought of it as a 90% off offer but clearly I was wrong. I guess some of you might have made the same mistake as well. A 50% + 40% off does not mean a 90% off. It actually means that you will be given a discount of 50% first and then on the reduced price you will given another 40% discount. Let us calculate to figure out how much this actually means.

Let us assume that the T-shirt you are trying to buy costs 100 Rs. A 50% discount would bring down the price of the T-shirt to 50 Rs. Another 40% discount on the reduced price of 50 Rs. would further bring down the price by 20 Rs., which is 40% of 50, to 30Rs. So, the price has effectively gone down from 100 Rs. to 30 Rs. which means that the effective discount has been 70% and not 90%.

Let me ask you another question. Try to answer it in your head before you actually scroll down and check the answer.

Example 1: Shop A is selling a T-shirt at a discount of 50% + 40% on the MRP whereas shop B is selling the same T-shirt at a discount of 40% + 50%. You should buy the T-shirt from Shop A or Shop B?

Now, there might be a few things going on in your head like ‘A’ is better because it is offering higher percentage first or ‘B’ is better because it is offering higher percentage later or this cannot be determined until and unless we know the MRP of the T-shirt. Well, stop your internal monologue.

The answer is that it would not make a difference whether you buy it from Shop A or from Shop B. Don’t believe me – do the math.

Let us assume that the MRP of the T-shirt is ‘x’

From Shop A, a 50% discount would bring down the price to 0.5x and another 40% discount would bring down the price to 0.3x

From Shop B, a 40% discount would bring down the price to 0.6x and another 50% discount would bring down the price to 0.3x

As you can see the sale price is coming out to be the same in both cases.

In case you are wondering as to why it is the same in both cases, it is because percentages are multiplicative in nature and p * q is always the same as q * p.

For future reference, you can use the formula for effective % change in case of successive % changes of a% and b% is (a + b + ab/100)

Note: Please keep in mind the +/- sign while using the formula.

If you put a = 50% and b = 40% (from the above example), you would get 110% which would be the incorrect answer. You should use -50% and -40% to get the correct answer of -70%. You will have to use –ive values for the above example because the % changes considered are discounts.

Example 2: A company’s revenue grew by 10% and 20% in the 2009 and 2010 but fell by 25% in 2011. What was the net % change?

Use the formula for the first two years = 10 + 20 + 10 * 20/100 = 32

Combine this with the last year = 32 – 25 + (32)(-25)/100 = 7 – 8 = -1%

**Compensating Percentage Change:**

I am sure you must have encountered questions like these:

a) The price of beer has gone up by 25%, by how much should you reduce your consumption so that the your expenses do not change.

b) Ravi got a salary hike of 10%. The new boss thought this shouldn’t have happened because Ravi doesn’t deserve it, so he slashes Ravi’s salary by 10%. Is Ravi back to the original? If no, then by what % should Ravi’s boss deduct his salary.

Now there are couple of ways of doing these questions. One is using the concepts of fractions, proportionality, etc. The other one is by using the formula. Let us discuss both of them

a) Price of beer has gone up by 25%

Price of beer has increased by 1/4th

Price of beer has become 5/4th

Consumption should become 4/5th

Consumption should reduce by 1/5th

Consumption should reduce by 20%

b) A hike of 10% and then a reduction of 10% won’t be fair as Ravi would end up getting less than the original. Using the successive % change formula {10 – 10 + 10* (-10)/100 = -1}, we can say that he would end up getting 1% lesser than the original. To get the correct deduction value:

Ravi’s salary was hiked by 10%

Ravi’s salary has increased by 1/10th

Ravi’s salary has become 11/10th

To go back to the original, it should become 10/11th of the current value

It should reduce by 1/11

It should reduce by 9.09%

We could have also used the formula for compensating a change of r% : **- 100r/(100 + r) %**

a) r = 25%. Compensating % change = – (100 * 25)/(100+25)

–20% = reduction of 20%

b) r = 10%. Compensating % change = – (100 * 10)/(100+10)

–9.09% = reduction of 9.09%

I hope you found this post useful and it would help you get a higher percentage and percentile in your exam.

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