Quants 26 - Interests

 Practice Questions:

Question 1:

What is the difference between the simple interest on a principal of ` 500 being calculated at 5% per annum for 3 years and 4% per annum for 4 years?

5% for 3 years (SI) = 15% of the amount; At the same time 4% SI for 4 years means 16% of the amount. The difference between the two is 1% of the amount. 1% of 500 = ` 5

(or)

SI = PNR/100 = (500*5*3)/100 = 75

SI = PNR/100 = (500*4*4)/100 = 80

Difference = 80 - 75 = 5.

Question 2:

In what time will ` 3300 become ` 3399 at 6% per annum interest compounded half-yearly?

Since compounding is half yearly, it is clear that the rate of interest charged for 6 months would be 3% 

3300 -3%- 3399.

Question 3:

What is the simple interest for 9 years on a sum of ` 800 if the rate of interest for the first 4 years is 8% per annum and for the last 4 years is 6% per annum?

8% of 800 for 4 years + 6% of 800 for 4 years = 64 × 4 + 48 × 4 = 256 + 192 = 448. 

However, we do not know the rate of interest applicable in the 5th year and hence cannot determine the exact simple interest for 9 years.

Question 4:

What is the difference between compound interest and simple interest for the sum of ` 20,000 over a 2 year period if the compound interest is calculated at 20% and simple interest is calculated at 23%?

Simple interest @ 23% = 4600 × 2 = 9200
Compound interest @ 20%

20000 -20%- 24000 -20%- 28800 Æ ` 8800 compound interest.
Difference = 9200 – 8800 = ` 400.

Question 5:

Find the compound interest on ` 1000 at the rate of 20% per annum for 18 months when interest is compounded half-yearly.

1000 -10%- 1100 -10%- 1210 -10%- 1331.
Compound interest = 1331 – 1000 = ` 331

Time to Think:

What is the simple interest on a sum of `700 if the rate of interest for the first 3 years is 8% per annum and for the last 2 years is 7.5% per annum?

Quants 25 - Interest

Practice Questions:

Question 1:

` 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

The annual interest would be ` 60 (PNR/100 = (1200*1*5)/100). 

After 3 years the total value would be 1200 + 60 × 3 = 1380.

Question 2:

Interest obtained on a sum of ` 5000 for 3 years is ` 1500. Find the rate percent.

The interest earned per year would be 1500/3=500. This represents a 10% rate of interest.

SI = PNR/100 => 1500 = (5000 x 3 x R)/100 => R = 1500 x 100 / ( 5000 x 3 ) = 10%.

Question 3:

` 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.

2100 + 5% of 2100 = 2100 + 105 = 2205 (after 1 year). 

Next year it would become:
2205 + 5% of 2205 = 2205 +110.25 = 2315.25

Question 4:

Find the difference between the simple and the compound interest at 5% per annum for 2 years on a principal of ` 2000.

Simple Interest for 2 years = 100 + 100 = 200.
Compound interest for 2 years: 

Year 1 = 5% of 2000 = 100.
Year 2: 5% of 2100 = 105 Æ Total compound interest = ` 205.
Difference between the Simple and Compound interest = 205 – 200 = ` 5

Question 5:

After how many years will a sum of ` 12,500 become ` 17,500 at the rate of 10% per annum?

12500 @ 10% simple interest would give an interest of ` 1250 per annum. For a total interest of ` 5000, it would take 4 years.

Time to Think:

Find the rate of interest if the amount after 2 years of simple interest on a capital of ` 1200 is ` 1440.

This week and the following week is a special week for Quantitative aptitude. Full fledged to concepts in 10 days!! Happy learning!!

Answers Week 12

Quants 23 - Stocks and Shares

A 9% stock yields 8%. The market value of the stock is:

To obtain Rs.8, investment, Rs. 100

To obtain Rs.9, investment, Rs. (100/8) * 9 = Rs.112.50

Market value of Rs. 100 stock = Rs.112.50

 

Logical 23 - Classification

a) Pistol

b) Sword

c) Gun

d) Rifle

e) Cannon

Here, all except Sword are fire arms, and can be used from a distance.

a) Cathedral

b) Mosque

c) Church

d) Monastery

e) Temple

All except Monastery are places of worship, while monastery is a place where monks stay.


Verbal 12 - Sentence Completion

We have to bear our own burdens.

Napoleon was killed in the battle of Waterloo.

The world is on the verge of a third world war.

Barter is an economic transaction involving exchange of articles.

The premises is for sale.

Logical 24 - Verification of Truth of the Statement

A factory always has

a) Electricity

b) Chimney

c) Workers

d) Files

e) Sellers

A clock always has

a) Battery

b) Numbers

c) Alarm

d) Needles

e) Frame


Quants 23 - Stocks and Shares

A can do a work in 15 days and B in 20 days. If they work together for 4 days, then the fraction of the work left is:

A's 1 day work = 1/15

B's 1 day work = 1/20
A+B's 1 day work = 1/15 + 1/20 = 7/60
A+B's 4 days work = 7/60 * 4 = 7/15
Thus, remaining work = 1 - 7/15 = 8/15

Quants 24 - Time and Work

Practice Questions:

Question 1:

Sakshi can do a piece of work in 20 days. Ziva is 25% more efficient than Sakshi. The number of days taken by Ziva to do the same piece of work is:

Ratio of times taken by Sakshi and Ziva is 125:100 = 5:4
Suppose Ziva takes x days to do the work,
5 : 4 :: 20 : x => x = (4 * 20) / 5 = 80/5 = 16 days.
Hence, Ziva takes 16 days to complete the work.

Question 2:

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could do in 23 days?

Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work,
Then 10 : 13 :: 23 : x => x = (23 * 13) / 10 = 299/10
A's 1 day work = 1/23; B's 1 day work = 10/299
A+B 's 1 day work = 1/23 + 10/299 = 23/299 = 1/13
A and B together can complete the work in 13 days.

Question3:

A does half as much as B in three-fourth of the time. If together, they take 18 days to complete the work, how much time shall B take to do it?

Suppose B takes x days to complete the work,
A takes 2*(3/4)x = 3/2 x days
A+B 's 1 day work = 1/18
1/x + 2/3x = 1/18
x = 36/5 days.

Question 4:

A is 50% as efficient as B. C does half of the work done by A and B together. If C alone does the work in 40 days, then A, B and C together can do the work in:

A's 1 day work : B's 1 days work = 150 : 100 = 3 : 2
Let A's and B's 1 day work be 3x and 2x respectively.
Then C's 1 day work = (3x+2x)/2 = 5x/2
5x/2 = 1/40 => 5x = 1/20 => 100x = 1
x= 1/100
A's 1 day work = 3/100
B's 1 day work = 2/100 = 1/50
C's 1 day work = 5/100 / 2 = 5/200 = 1/40
A+B+C 's 1 day work = 3/100 + 2/100 + 5/200 = 15/200 = 3/40
So, A, B and C can complete the work in 40/3 = 13 1/3 days.

Question 5:

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. A alone could complete the work in:

Let A's 1 day work = x
Let B's 1 day work = y
Then x+y = 1/5 and 2x + 1/3y = 1/3
Solving the above 2 equations,
x = 4/25 and y = 1/25
A's 1 day work = 4/25
So, A alone could complete the work in 25/4 = 6 1/4 days.

Time to Think:

A can do a work in 15 days and B in 20 days. If they work together for 4 days, then the fraction of the work left is:

Logical 24 - Verification of Truth of the Statement

Key Points to Remember:

In this type of questions, we are required to stress only on the truth of the facts that always hold. Questions are asked in context of a particular thing or factor that is always characterized by a specific part or feature. The alternatives other than the correct answer also seem to bear a strong relationship with the thing mentioned. So, the absolute truth is to be followed. 

Examples:

1. Atmosphere always has

a) Oxygen

b) Air

c) Germs

d) Moisture

e) Dust

Clearly, though all the alternatives may form a part of the atmosphere, the air is the most vital part, without which there can be no atmosphere. 

2. A train always has

a) Engine

b) Rails

c) Driver

d) Guard

e) Passengers

Clearly, rails are necessary for the train to move on. Driver alone can move the train. A guard is also necessary for safety. A train is moved for the passengers. But all these do not constitute a train. A train cannot be called so without the engine. 

3. Which one of the following is always found in Bravery?

a) Courage

b) Experience

c) Power

d) Knowledge

Clearly, bravery is a quality exhibited only by a person who possesses courage. 

Practice Questions:

4. Which of the following an animal always has?

a) Lungs

b) Skin

c) Mind

d) Heart

e) Life

5. A race always has

a) Referee

b) Spectators

c) Rivals

d) Prize

e) Victory

6. Which of the following a Drama must have?

a) Actors

b) Story

c) Sets

d) Director

e) Spectators

7. A book always has

a) Chapters

b) Pages

c) Contents

d) Pictures

e) Illustrations

8. A mirror always

a) Reflects

b) Retracts

c) Distorts

d) Refracts

e) Reveals the truth

Time to Think

9. A factory always has

a) Electricity

b) Chimney

c) Workers

d) Files

e) Sellers

10. A clock always has

a) Battery

b) Numbers

c) Alarm

d) Needles

e) Frame

Verbal 12 - Sentence Completion

Fill in the blanks with the appropriate words from the options given below. Do not use the same word twice. 

[ abolished, abated, abdicated, abandoned, able, rebates, capable, abridged, banished, capacious, abbreviation, about, bear, exile, bared, sale, war, barter, battle, above ]

Practice Questions:

1. Lord Buddha abdicated his kingship and became a hermit.

2. The Indian Government abolished slavery.

3. The residents abandoned the haunted house.

4. The rain abated after some time.

5. The shops offer rebates in the off season.

6. Our government is able to confront any type of situation.

7. He is capable of solving this problem.

8. Confucius possessed a capacious mind.  

9. B.A. is an abbeviation  of "Bachelor of Arts".

10. Abridged versions of classics are easier to read than the original versions.
11.  It was about ten in the night when we reached home.

12. The painting was hung above the window.

13. The Shah of Iraq was banished from his country.

14. Prospero, a character of Shakespeare's play "The Tempest" was living in exile.

15. The patient bared his chest when the Doctor examined him.

Time to Think:

 

16. We have to _______ our own burdens.

17. Napoleon was killed in the _______ of Waterloo.

18. The world is on the verge of a third world _______.

19. _______ is an economic transaction involving exchange of articles.

20. The premises is for _______.

Logical 23 - Classification

Classification means to assort the items of a given group on the basis of a certain common quality they possess and then spot the stranger or odd one out.

Here, we are given a group of certain items, out of which all except one are similar to one another in some manner. We are required to choose this one item which does not fit into the given group.

Choosing the Odd Word

Practice Questions: 

Question 1: 

a) Zebra

b) Lion

c) Tiger

d) Horse

e) Giraffe

Here, all except Horse, are wild animals. Horse can be domesticated.

Question 2: 

a) Parrot

b) Bat

c) Crow

d) Sparrow 

e) Pigeon

Here, all except Bat, belong to the class Aves - birds, while Bat is a mammal.

Question 3: 

a) Copper

b) Zinc

c) Brass

d) Aluminium

e) Iron

Here, all except Brass, are metals, while Brass is an alloy.

Question 4: 

a) Apple

b) Marigold

c) Rose

d) Lily

e) Lotus

Here, all except Apple are flowers, while Apple is a fruit. 

Question 5: 

a) January

b) May

c) July

d) August

e) November

Here, all except November are months having 31 days, while November has 30 days.

Question 6: 

a) Amethyst

b) Ruby

c) Marble

d) Sapphire

e) Diamond

 Here, all except Marble are precious stones.

Question 7: 

a) Ginger

b) Onion

c) Beetroot

d) Coriander

e) Potato

Here, all except Coriander are modified stems.

Question 8: 

a) Bake

b) Peel

c) Fry

d) Boil

e) Roast

Here, all except Peel are different forms of cooking.

Time to Think:

Question 9: 

a) Pistol

b) Sword

c) Gun

d) Rifle

e) Cannon

Question 10: 

a) Cathedral

b) Mosque

c) Church

d) Monastery

e) Temple

Quants 23 - Stocks and Shares

Practice Questions:

Question 1:

A 6% stock yields 8%. The market value of the stock is:

For an income of Rs.8, investment, Rs. 100

For an income of Rs.6, investment, Rs. (100/8) * 6 = Rs.75

Market value of Rs. 100 stock = Rs. 75

Question 2:

By investing Rs. 1620 in 8% stock, Ragu earns Rs. 135. The stock is then quoted at:

To earn Rs.135, investment = Rs. 1620

To earn Rs.8, investment = Rs. (1620/135) * 8 = Rs. 96

Market value of Rs. 100 stock = Rs.96

Question 3:

A 12% stock yielding 10% is quoted at:

To earn Rs.10, money invested = Rs. 100

To earn Rs.12, money invested = (100/10) * 12 = Rs.120

Market value of Rs. 100 stock = Rs.120

Question 4:

To produce an annual income of Rs. 1200 from a 12% stock at 90, the amount of stock needed is:

For an income of Rs.12, stock needed = Rs. 100

For an income of Rs.1200, stock needed = Rs. (100/12) * 1200 = Rs.10,000

Question 5:

In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of

To obtain Rs.10, investment = Rs.96

To obtain Rs.650, investment = Rs. (96/10) * 650 = Rs.6240

Time to Think:

A 9% stock yields 8%. The market value of the stock is:

Concept 7 - Alligations

Alligation is nothing but a faster technique of solving problems based on the weighted average situation as applied to the case of two groups being mixed together.

THEORY

To recollect, the weighted average is used when a number of smaller groups are mixed together to form one larger group.
If the average of the measured quantity was
A1 for group 1 containing n1 elements
A2 for group 2 containing n2 elements
A3 for group 3 containing n3 elements
Ak for group k containing nk elements
We say that the weighted average, Aw is given by:
Aw = (n1A1 + n2 A2 + n3 A3 + ……. + nkAk)/ (n1 + n2 + n3 … + nk)
That is, the weighted average

In the case of the situation where just two groups are being mixed, we can write this as:
Aw = (n1A1 + n2 A2 )/(n1 + n2)
Rewriting this equation we get: (n1 + n2) Aw = n1A1 + n2A2
n1(Aw – A1) = n2 (A2 – Aw)
or n1/n2 = (A2 – Aw)/(Aw – A1) Æ The alligation equation.

The Alligation Situation
Two groups of elements are mixed together to form a third group containing the elements of both the groups. If the average of the first group is A1 and the number of elements is n1 and the average of the second group is A2 and the number of elements is n2, then to find the average of the new group formed, we can use either the weighted average equation or the alligation equation.
As a convenient convention, we take A1 < A2. Then, by the principal of averages, we get A1 < Aw < A2. 

Illustration 1

Two varieties of rice at ` 10 per kg and ` 12 per kg are mixed together in the ratio 1 : 2. Find the average price of the resulting mixture.
Solution 

1/2 = (12 – Aw)/(Aw – 10) Æ Aw – 10 = 24 – 2Aw
fi 3Aw = 34 fi Aw = 11.33 `/kg.

Illustration 2

On combining two groups of students having 30 and 40 marks respectively in an exam, the resultant group has an average score of 34. Find the ratio of the number of students in the first group to the number of students in the second group.
Solution 

n1/n2 = (40 – 34)/(34 – 30) = 6/4 = 3/2

Graphical Representation of Alligation
The formula illustrated above can be represented by the following cross diagram:

[Note that the cross method yields nothing but the alligation equation. Hence, the cross method is nothing but a graphical representation of the alligation equation.]
As we have seen, there are five variables embedded inside the alligation equation. These being: the three averages Æ A1, A2 and Aw and the two weights Æ n1 and n2
Based on the problem situation, one of the following cases may occur with respect to the knowns and the unknown, in the problem.

Now, let us try to evaluate the effectiveness of the cross method for each of the three cases illustrated
above:
Case 1: A1, A2, Aw are known; may be one of n1 or n2 is known.
To find: n1 : n2 and n2 if n1 is known OR n1 if n2 is known.
Let us illustrate through an example:

Illustration 3 

On mixing two classes of students having average marks 25 and 40 respectively, the overall average
obtained is 30 marks. Find
(a) The ratio of students in the classes
(b) The number of students in the first class if the second class had 30 students. 

Solution

(a) Hence, solution is 2 : 1.
(b) If the ratio is 2 : 1 and the second class has 30 students, then the first class has 60 students.

Note: The cross method becomes pretty effective in this situation when all the three averages are known and the ratio is to be found out.
Case 2: A1, A2, n1 and n2 are known, Aw is unknown.

Illustration 4

4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.

Solution

= (6 – Aw) : (Aw – 5)
= (6 – Aw)/(Aw – 5) = 4/8 Æ12 – 2 Aw = Aw – 5
3Aw = 17
Aw = 5.66 `/kg.

Note: The cross method becomes quite cumbersome in this case, as this method results in the formula being written. Hence, there seems to be no logic in using the cross method in this case.
Case 3: A1, Aw, n1 and n2 are known; A2 is unknown.

Illustration 5

5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice.
Solution 

Using the cross method:

= (x – 7) : 1
(x – 7)/1 = 5/4 Æ 4x – 28 = 5
x = ` 8.25.

Note: The cross method becomes quite cumbersome in this case since this method results in the formula being written. Hence, there seems to be no logic in using the cross method in this case.
The above problems can be dealt quite effectively by using the straight line approach, which is explained below.

The Straight Line ApproachAs we have seen, the cross method becomes quite cumbersome in Case 2 and Case 3. We will now proceed to modify the cross method so that the question can be solved graphically in all the three cases.
Consider the following diagram, which results from closing the cross like a pair of scissors. Then the
positions of A1, A2, Aw, n1 and n2 are as shown.

Visualise this as a fragment of the number line with points A1, Aw and A2 in that order from left to right.
Then,
(a) n2 is responsible for the distance between A1 and Aw or n2 corresponds to Aw – A1
(b) n1 is responsible for the distance between Aw and A2. or n1 corresponds to A2 – Aw
(c) (n1 + n2) is responsible for the distance between A1 and A2. or (n1 + n2) corresponds to A2 – A1.
The processes for the 3 cases illustrated above can then be illustrated below:

Illustration 6

On mixing two classes of students having average marks 25 and 40 respectively, the overall average
obtained is 30 marks. Find
(a) the ratio in which the classes were mixed.
(b) the number of students in the first class if the second class had 30 students.

Solution

Hence, ratio is 2 : 1, and the second class has 60 students.

Case 2 A1, A2, n1 and n2 are known; Aw is unknown.

Illustration 7

4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.

Solution

Then, by unitary method:
n1 + n2 corresponds to A2 – A1
Æ 1 + 2 corresponds to 6 – 5
That is, 3 corresponds to 1
\ n2 will correspond to

In this case (1/3) × 2 = 0.66.
Hence, the required answer is 5.66.

Note: In this case, the problem associated with the cross method is overcome and the solution becomes graphical.
Case 3: A1, Aw, n1 and n2 are known; A2 is unknown.

Illustration 8

5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice.
Using straight line method:

4 corresponds to 7 – 6 and 5 corresponds to x – 7.
The thought process should go like:

4 Æ1
\ 5 Æ1.25
Hence, x – 7 = 1.25
and x = 8.25

SOME TYPICAL SITUATIONS WHERE ALLIGATIONS CAN BE USED
Given below are typical alligation situations, which we should be able to recognize. This will help us improve upon the time required in solving questions.
The following situations should help us identify alligation problems better as well as spot the way A1, A2, n1 and n2 and Aw are mentioned in a problem.
In each of the following problems the following magnitudes represent these variables:
A1 = 20, A2 = 30, n1 = 40, n2 = 60
Each of these problems will yield an answer of 26 as the value of Aw.
1. A man buys 40 kg of rice at ` 20/kg and 60 kg of rice at ` 30/kg. Find his average price. (26/kg)
2. Pradeep mixes two mixtures of milk and water. He mixes 40 litres of the first containing 20% water and 60 litres of the second containing 30% water. Find the percentage of water in the final mixture. (26%)
3. Two classes are combined to form a larger class. The first class having 40 students scored an average of 20 marks on a test while the second having 60 students scored an average of 30 marks on the same test. What was the average score of the combined class on the test. (26 marks)
4. A trader earns a profit of 20% on 40% of his goods sold, while he earns a profit of 30% on 60% of his goods sold. Find his percentage profit on the whole. (26%)
5. A car travels at 20 km/h for 40 minutes and at 30 km/h for 60 minutes. Find the average speed of the car for the journey. (26 km/hr)
6. 40% of the revenues of a school came from the junior classes while 60% of the revenues of the school came from the senior classes. If the school raises its fees by 20% for the junior classes and by 30% for the senior classes, find the percentage increase in the revenues of the school. (26%)

Some Keys to spot A1, A2 and Aw and differentiate these from n1 and n2
1. Normally, there are 3 averages mentioned in the problem, while there are only 2 quantities. This isn’t foolproof though, since at times the question might confuse the student by giving 3 values for quantities representing n1, n2 and n1 + n2 respectively.
2. A1, A2 and Aw are always rate units, while n1 and n2 are quantity units.
3. The denominator of the average unit corresponds to the quantity unit (i.e. unit for n1 and n2).
4. All percentage values represent the average values.

A Typical Problem
A typical problem related to the topic of alligation goes as follows:
4 litres of wine are drawn from a cask containing 40 litres of wine. It is replaced by water. The process is repeated 3 times
(a) What is the final quantity of wine left in the cask.
(b) What is the ratio of wine to water finally.
If we try to chart out the process, we get: Out of 40 litres of wine, 4 are drawn out.
This leaves 36 litres wine and 4 litres water. (Ratio of 9 : 1)
Now, when 4 litres are drawn out of this mixture, we will get 3.6 litres of wine and 0.4 litres of water (as the ratio is 9 : 1). Thus at the end of the second step we get: 32.4 litres of wine and 7.6 litres of water. Further, the process is repeated, drawing out 3.24 litres wine and 0.76 litres water leaving 29.16 litres of wine and 10.84 litres of water.
This gives the final values and the ratio required.
A closer look at the process will yield that we can get the amount of wine left by:
40 × 36/40 × 36/40 × 36/40 = 40 × (36/40)3
fi 40 × (1 – 4/40)3
This yields the formula:
Wine left : Capacity × (1 – fraction of wine withdrawn)n for n operations.
Thus, you could have multiplied:
40 × (0.9)3 to get the answer
That is, reduce 40 by 10% successively thrice to get the required answer.
Thus, the thought process could be:
40 – 10% Æ 36 – 10% Æ 32.4 – 10% Æ 29.16

Concept 6 - Average

The average of a number is a measure of the central tendency of a set of numbers. In other words, it is an estimate of where the center point of a set of numbers lies.
The basic formula for the average of n numbers x1, x2, x3, … xn is

This also means A_n × n = total of the set of numbers.

The average is always calculated for a set of numbers.

Concept of weighted average: 

When we have two or more groups whose individual averages are known, then to find the combined average of all the elements of all the groups we use weighted average. 

Thus, if we have k groups with averages A1, A2 ... Ak and having n1, n2 ... nk elements then the weighted average is given by the formula:

Another meaning of average 

The average [also known as arithmetic mean (AM)] of a set of numbers can also be defined as the number by which we can replace each and every number of the set without changing the total of the set of numbers.

Properties of average (AM) 

The properties of averages [arithmetic mean] are:

Property 1: The average of 4 numbers 12, 13, 17 and 18 is:

Solution: Required average = (12 + 13 + 17 + 18)/4 = 60/4 = 15

This means that if each of the 4 numbers of the set were replaced by 15 each, there would be no change in the total. This is an important way to look at averages. In fact, whenever you come across any situation where the average of a group of ‘n’ numbers is given, we should visualise that there are ‘n’ numbers, each of whose value is the average of the group. This view is a very important way to visualise averages. This can be visualised as

Property 2: In Example 1, visualise addition of a fifth number, which increases the average by 1.

15 + 1 = 16

15 + 1 = 16

15 + 1 = 16

15 + 1 = 16

The +1 appearing 4 times is due to the fifth number, which is able to maintain the average of 16 first and then ‘give one’ to each of the first 4. Hence, the fifth number in this case is 20

Property 3: The average always lies above the lowest number of the set and below the highest number of the set.

Property 4: The net deficit due to the numbers below the average always equals the net surplus due to the numbers above the average.

Property 5: Ages and averages: If the average age of a group of persons is x years today then after n

years their average age will be (x + n). Also, n years ago their average age would have been (x – n). This happens due to the fact that for a group of people, 1 year is added to each person’s age every year.

Property 6: A man travels at 60 kmph on the journey from A to B and returns at 100 kmph. Find his

average speed for the journey.

Solution: Average speed = (total distance) / (total time)

If we assume distance between 2 points to be d. Then,

Average speed = 2d / [(d/60) + (d/100)] = (2 × 60 × 100)/ (60 + 100) = (2 × 60 × 100)/160 = 75

Average speed = (2S1 ◊ S2)/(S1 + S2) [S1 and S2 are speeds] of going and coming back respectively.

Short Cut The average speed will always come out by the following process:

The ratio of speeds is 60 : 100 = 3 : 5 (say r1 : r2)

Then, divide the difference of speeds (40 in this case) by r1 + r2 (3 + 5 = 8, in this case) to get one part. (40/8 = 5, in this case)

The required answer will be three parts away (i.e. r1 parts away) from the lower speed.

Check out how this works with the following speeds:

S1 = 20 and S2 = 40

Step 1: Ratio of speeds = 20 : 40 = 1 : 2

Step 2: Divide difference of 20 into 3 parts (r1 + r2) Æ = 20/3 = 6.66

Required average speed = 20 + 1 × 6.66

This process is essentially based on alligations.

Concept 5 - Profit & Loss

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:

1. Profit = SP – CP
2. SP = Profit + CP
3. CP = SP – Profit
4. Percentage Profit = (Profit x 100) / CP
Percentage Profit is always calculated on CP unless otherwise stated.

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[1 + (%Gain/100)]
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:

 

In case of loss

1. Loss = CP – SP
2. SP = CP – Loss
3. CP = SP + Loss
4. Loss% = Loss on ` 100 = (Loss x 100) / CP
Percentage Loss is always calculated on CP unless otherwise stated.

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-today 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.

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 maximizing the sales price of a product/service, is to minimize 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 Fixed Costs/n.

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:
Profit = (Actual sales – Break-even sales) × Contribution per unit
Also in the case of a loss:
Loss = (Break-even sales – Actual sales) × Contribution per unit
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 = (Goods Left / Goods Sold) × 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 = (Goods Left / Goods Sold) × 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
CP + Mark up = Marked price
or CP + % Mark up on CP = Marked Price
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:
CP + % Mark up (Calculated on CP) = Marked Price
Marked price – % Discount = Selling price
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

Concept 4 - Interest

CONCEPT OF TIME VALUE OF MONEY

The value of money is not constant. This is one of the principal facts on which the entire economic world is based. A rupee today will not be equal to a rupee tomorrow. Hence, a rupee borrowed today cannot be repaid by a rupee tomorrow. This is the basic need for the concept of interest. The rate of interest is used to determine the difference between what is borrowed and what is repaid.

There are two basis on which interests are calculated:

Simple Interest

It is calculated on the basis of a basic amount borrowed for the entire period at a particular rate of interest. The amount borrowed is the principal for the entire period of borrowing.

Compound Interest 

The interest of the previous year/s is/are added to the principal for the calculation of the compound interest.

This difference will be clear from the following illustration:

A sum of ` 1000 at 10% per annum will have

Simple interest                                                             Compound interest

      ` 100                                  First year                                  ` 100

      ` 100                                  Second year                              ` 110

      ` 100                                  Third year                                 ` 121

      ` 100                                  Fourth year                               ` 133.1

Note that the previous years’ interests are added to the original sum of ` 1000 to calculate the interest to be paid in the case of compound interest.

Terminology Pertaining to Interest

The man who lends money is the Creditor and the man who borrows money is the Debtor.

The amount of money that is initially borrowed is called the Capital or Principal money.

The period for which money is deposited or borrowed is called Time.

The extra money, that will be paid or received for the use of the principal after a certain period is called the Total interest on the capital.

The sum of the principal and the interest at the end of any time is called the Amount.

So, Amount = Principal + Total Interest.

Rate of Interest is the rate at which the interest is calculated and is always specified in percentage terms.

SIMPLE INTEREST

The interest of 1 year for every ` 100 is called the Interest rate per annum. If we say “the rate of interest per annum is r%”, we mean that ` r is the interest on a principal of ` 100 for 1 year.

Relation Among Principal, Time, Rate Percent of Interest Per Annum and Total Interest

Suppose, Principal = ` P, Time = t years, Rate of interest per annum = r% and Total interest = ` I,

Then I = ( P x t x r ) / 100

i.e. Total interest = ( Principal x Time x Rate of Interest per annum ) / 100

Since the Amount = Principal + Total interest, we can write,

Amount A = P + ((P x t x r) / 100)

Time = (Total Interest / Interest on the Principal for 1 year) years

Thus, if we have the total interest as ` 300 and the interest per year is ` 50, then we can say that the number of years is 300/50 = 6 years.

Note: The rate of interest is normally specified in terms of annual rate of interest. In such a case we take the time t in years. However, if the rate of interest is specified in terms of 6-monthly rate, we take time in terms of 6 months. Also, the half-yearly rate of interest is half the annual rate of interest. That is if the interest is 10% per annum to be charged six-monthly, we have to add interest every six months @ 5%.

COMPOUND INTEREST

In monetary transactions, often, the borrower and the lender, in order to settle an account, agree on a certain amount of interest to be paid to the lender on the basis of specified unit of time. This may be yearly or half-yearly or quarterly, with the condition that the interest accrued to the principal at a certain interval of time be added to the principal so that the total amount at the end of an interval becomes the principal for the next interval. Thus, it is different from simple interest. In such cases, the interest for the first interval is added to the principal and this amount becomes the principal for the second interval, and so on. The difference between the amount and the money borrowed is called the compound interest for the given interval.

Formula

Case 1: Let principal = P, time = n years and rate = r% per annum and let A be the total amount at the

end of n years, then

Case 2: When compound interest is reckoned half-yearly. If the annual rate is r% per annum and is to be calculated for n years. Then in this case, rate = (r/2)% half-yearly and time = (2n) half-years.

Case 3: When compound interest is reckoned quarterly. In this case, rate = (r/4)% quarterly and time = (4n) quarter years.

Note: The difference between the compound interest and the simple interest over two years is given by 

DEPRECIATION OF VALUE

The value of a machine or any other article subject to wear and tear, decreases with time. This decrease is called its depreciation.

Thus if V0 is the value at a certain time and r% per annum is the rate of depreciation per year, then the value V1 at the end of t years is

POPULATION

The problems on Population change are similar to the problems on Compound Interest. The formulae applicable to the problems on compound interest also apply to those on population. The only difference is that in the application of formulae, the annual rate of change of population replaces the rate of compound interest.

However, unlike in compound interest where the rate is always positive, the population can decrease. In such a case, we have to treat population change as we treated depreciation of value illustrated above. In the case of compound interests, the percentage rule for calculation of percentage values will be highly beneficial. 

Thus, a calculation: 4 years increase at 6% pa CI on ` 120 would yield an expression: 120 × 1.064. It

would be tedious. So, we should look at it from the following percentage change graphic perspective:

If you try to check the answer on a calculator, you will discover that you have a very close approximation. Besides, given the fact that you would be working with options and given sufficiently comfortable options, you need not calculate so closely; instead, save time through the use of approximations.

APPLICATIONS OF INTEREST IN D.I.

The difference between Simple Annual Growth Rate and Compound Annual Growth Rate:

The Measurement of Growth Rates is a prime concern in business and Economics. While a manager might be interested in calculating the growth rates in the sales of his product, an economist might be interested in finding out the rate of growth of the GDP of an economy.

In mathematical terms, there are basically two ways in which growth rates are calculated. To familiarize yourself with this, consider the following example.

The sales of a brand of scooters increase from 100 to 120 units in a particular city. What does this mean to you? Simply that there is a percentage increase of 20% in the sales of the scooters. Now read further:

What if the sales moves from 120 to 140 in the next year and 140 to 160 in the third year? Obviously, there is a constant and uniform growth from 100 to 120 to 160 – i.e. a growth of exactly 20 units per year.

In terms of the overall growth in the value of the sales over there years, it can be easily seen that the sale has grown by 60 on 100 i.e. 60% growth.

In this case, what does 20% represent? If you look at this situation as a plain problem of interests 20% represents the simple interest that will make 100 grow to 160.

In the context of D.I., this value of 20% interest is also called the Simple Annual Growth Rate. (SAGR)

The process for calculating SAGR is simply the same as that for calculating Simple Interest.

Suppose a value grows from 100 to 200 in 10 years – the SAGR is got by the simple calculation 100%/10 = 10%

What is Compound Annual Growth Rate (CAGR)?

Let us consider a simple situation. Let us go back to the scooter company.

Suppose, the company increases it’s sales by 20% in the first year and then again increases its’ sales by 20% in the second year and also the third year. In such a situation, the sales (taking 100 as a starting value) trend can be easily tracked as below:

As you must have realised, this calculation is pretty similar to the calculation of Compound interests. In the above case, 20% is the rate of compound interest which will change 100 to 172.8 in three years. This 20% is also called as the Compound Annual Growth Rate (CAGR) in the context of Data interpretation. Obviously, the calculation of the CAGR is much more difficult than the calculation of the SAGR and the Compound Interest formula is essentially a waste of time for anything more than 3 years. (upto three years, if you know your squares and the methods for the cubes you can still feasibly work things out – but beyond three years it becomes pretty much infeasible to calculate the compound interest). 

So is there an alternative? Yes there is and the alternative largely depends on your ability to add well. Hence, before trying out what I am about to tell you, I would recommend you should strengthen yourself at addition. Suppose you have to calculate the C.I. on ` 100 at the rate of 10% per annum for a period of 10 years. You can combine a mixture of PCG used for successive changes with guesstimation to get a pretty accurate value.

In this case, since the percentage increase is exactly 10% (Which is perhaps the easiest percentage to

calculate), we can use PCG all the way as follows:

Thus, the percentage increase after 10 years @ 10% will be 159.2 (approx).

However, this was the easy part. What would you do if you had to calculate 12% CI for 10 years. The percentage calculations would obviously become much more difficult and infeasible. How can we tackle this situation?

In order to understand how to tackle the second percentage increase in the above PCG, let’s try to evaluate where we are in the question. We have to calculate 12% of 112, which is the same as 12% of 100 + 12% of 12. But we have already calculated 12% of 100 as 12 for the first arrow of the PCG. Hence, we now have to calculate 12% of 12 and add it to 12% of 100. 

Hence the addition has to be: 12 + 1.44 = 13.44

Take note of the addition of 1.44 in this step. It will be significant later. The PCG will now look like:

We are now faced with a situation of calculating 12% of 125.44. Obviously, if you try to do this directly, you will have great difficulty in calculations. We can sidestep this as follows:

12% of 125.44 = 12% of 112 + 12% of 13.44.

But we have already calculated 12% of 112 as 13.44 in the previous step.

Hence, our calculation changes to:

12% of 112 + 12% of 13.44 = 13.44 + 12% of 13.44

But 12% of 13.44 = 12% of 12 + 12% of 1.44. We have already calculated 12% of 12 as 1.44 in the

previous step.

Hence 12% of 13.44 = 1.44 + 12% of 1.44 = 1.44 + 0.17 = 1.61 (approx)

Hence, the overall addition is 13.44 + 1.61 = 15.05

Now, your PCG looks like:

You are again at the same point—faced with calculating the rather intimidating looking 12% of 140.49

Compare this to the previous calculation:

The only calculation that has changed is that you have to calculate 12% of 15.05 instead of 12% of 13.44. (which was approx 1.61). In this case it will be approximately 1.8. Hence you shall now add 16.85 and the PCG will look as:

If you evaluate the change in the value added at every arrow in the PCG above, you will see a trend—

The additions were: +12, +13.44 (change in addition = 1.44), +15.05(change in addition = 1.61), +16.85 (change in addition = 1.8)

If you now evaluate the change in the change in addition, you will realize that the values are 0.17, 0.19. This will be a slightly increasing series (And can be easily approximated).

Thus, the following table shows the approximate calculation of 12% CI for 10 years with an initial value of 100. Thus, 100 becomes 309.78 (a percentage increase of 209.78%)

Similarly, in the case of every other compound interest calculation, you can simply find the trend that the first 2 – 3 years interest is going to follow and continue that trend to get a close approximate value of the overall percentage increase. Thus for instance 7% growth for 7 years at C.I. would mean:

An approximate growth of 60.24%

The actual value (on a calculation) is around 60.57% – Hence as you can see we have a pretty decent approximation for the answer.

Note: The increase in the addition will need to be increased at a greater rate than as an A.P. Thus, in this case if we had considered the increase to be an A.P. the respective addition would have been:

+7, +7.49, +8.01, +8.55, +9.11, +9.69, +10.29.

However +7, +7.49, +8.01, +8.55, +9.11, +9.75, +10.35 are the actual addition used. Notice that using 9.75 instead of 9.69 is a deliberate adjustment, since while using C.I. the impact on the addition due to the interest on the interest shows an ever increasing behavior.