SHARE AND MUTUAL FUND - Q&A

SHARES AND MUTUAL FUNDS

Example 1                                                                

      Mr. Prashant invested Rs. 75,375/- to purchase equity shares of a company at market price of Rs. 250 /- through a brokerage firm, charging 0.5% brokerage. The face value of a share is Rs. 10/-. How many shares did Mr. Prashant purchase?

Example 2

            Mr. Sandeep received Rs. 4,30,272 /- after selling shares of a company at market price of Rs. 720 /- through Sharekhan Ltd., with brokerage 0.4%. Find the number of shares he sold.

Example 3

            Ashus Beauty World ' has issued 60,000 shares of par value of Rs. 10/- each. The company declared a total dividend of Rs. 72,000 /- . Find the rate of dividend paid by the company.

Example 4

            The capital of ABC Company consists of Rs. 15 lakhs in 6 % cumulative preference shares of Rs. 100 each and Rs. 30 lakhs in equity shares of Rs.10/- each. The dividends on cumulative preference shares for earlier year was not paid . This year , the company has to distribute profit of Rs . 3 lakh after keeping 20 % as reserve fund. Find the percentage rate of dividend paid to the equity shareholders.

Example 5

            Mr. Dinesh bought some shares of a company which had a face value of Rs.100 /-. The company declared a dividend of 15 % but Mr. Dinesh's  rate of return on investment was only 12% . At what market price did he purchase the shares ? There was no brokerage involved .

Example 6: Comparison of two stocks

            Mr. Subu invested Rs. 20,000 /- in Rs. 100/- shares of company A at the rate of Rs. 125/- per share . He received 10 % dividend on these shares. Mr. Subu also invested Rs. 24,000/- in Rs. 10/- shares of company B at Rs.12/- per share and he received 15 % dividend. Which investment is more beneficial?

Example 7

            Ms. Ashma Mehta bought   300 shares of a company of face value Rs. 100/- each at a market price of Rs. 240 /- each . After receiving a dividend at 8% , she sold the shares at Rs . 256 /- each . Find her rate of return on investment. There was no brokerage involved .

Example 8

            Mr. Joshi purchased 30 shares of Rs. 10/- each of Medi computers Ltd. on 20th Jan. 2007, at Rs. 36/- per share . On 3rd April 2007 , the company decided t split their shares from the face value of Rs. 10/- per share to Rs. 2/- per share . On 4th April 2007 , the market value of each share was Rs. 8/- per share . Find Mr. Joshi's gain or loss , if he was to sell the shares on 4th April 2007? ( No brokerage was involved in the transaction ).

Example 9

            Rahul purchased 500 shares of Rs. 100 of company A at Rs. 700 /-. After 2 months , he received a dividend of 25 % . After 6 months, he also got one bonus share for every 4 shares held . After 5 months , he sold all his shares at Rs. 610/- each. The brokerage was 2% on both, purchases & sales . Find his percentage return on the investment.

Example 11

            Mr. Deore invested Rs. 25,000/- to purchase 2,500 uits of ICICI MF - B plan on 4th April 2007 . He decided to sell the  units on  14th Nov. 2007  at NAV of Rs. 16.4 /-. The exit load was 2.5 % . Find his profit (Calculations are upto 2 decimal points)

Example 12

            Ragini invested Rs. 94,070/- in mutual Fund when NAV was Rs. 460 /- with entry load of 2.25 % . She received a dividend of Rs. 5/- per unit . She, later sold all units of fund with an exit load of 0.5 % . If her gain was Rs. 1654/-, find NAV at which she sold the units . (Calculations are upto 2 decimal points)

Example 13

            If a mutual fund had NAV of Rs. 28 /- at the beginning of the year and  Rs. 38/- at the end of the year , find the absolute change and the percentage change in NAV during the year .

Example 14

            If NAV was Rs. 72/- at the end of the year , with 12.5 % increase during the year , find NAV at the beginning of the year .

Example 15

            Rohit purchased some units in open end equity fund at Rs. 16/- . The fund distributed interim dividend of Rs. 5/- per unit , and the NAV of the fund at the end of the year was Rs. 25/- . Find the total percentage return . (Calculations are upto 2 decimal points)

Example 16

            Mr. Hosur purchased some units in open- end fund at Rs. 30/- and its NAV after 18 months was Rs. 45/- . Find the annualised change in NAV as a percentage Mr. Kamble purchased 586.909 units of Kotak cash plus retail Growth on 1st June 2007 when the NAV was RS. 20.4461. Its NAV as on 3^rd December, 2007 was Rs. 21.1960/- . The fund has neither entry load nor an exit load. Find the amount invested on 1st June 2007 and the value of Mr. Kamble's investment on 3^rd December 2007 .

Ans .  12,000 , 12440.12 .

EXTRA QUESTION’S

1.     Mr. Amar invested Rs 1,20,480/- to buy equity shares of a company at market price of Rs . 480 /- at 0.4 % brokerage . Find the No. of shares he purchased .      Ans: 250

2.     Aditi invested Rs. 19,890 /- to purchase shares of a company with face value of Rs.10/- each , at market price of Rs. 130/- . She received dividend of 20 % as well Afterwards , she sold these shares at market price of Rs. 180/- . She had to pay brokerage of 2 % for both purchase and sales of shares. Find her net profit.                        Ans: No. of shares =150 , sales = 26460 , Dividend = 300 , purchase = 19,890, profit= 6870

3.     Amol wants to invest some amount in company A or company B , by purchasing equity shares of face value of Rs. 10 /- each , with market price of R. 360/- and Rs. 470/- respectively . The companies are expected to declare dividends at 20 % and 45% respectively . Advise him on the choice of shares of company.                                                         Ans: company B is a better choice .

4.     Find the percentage gain or loss if 200 shares of face value Rs. 10/- were purchased at Rs . 350/- each and sold later at Rs. 352 /- , the brokerage being 0.5 % on each of the transaction .                                                                                                                        Ans:  -0.43 % i.e. a loss of 43 %

5.     Find the number of shares if the total dividend at 8% on the shares with face value Rs.10/- was Rs. 240.                                                                                                                   Ans :- 300

6.     Ms . Kannan purchased 113.151 units of 'FT India Prima Plus' on 9th April 2007 and redeemed all the units on 7th Aug 2007 when the NAV was Rs. 35.5573 . The entry load was 2.25 % and the exit load was 1 % . If she gained Rs. 483.11 , find the NAV on 9th April 2007 . (Calculations are upto 2 decimal points)                   Ans . 30.2514

7.     Mr. Pandit invested Rs. 10,000/- in Birla Sunlife Equity Fund- Divjdend plan ' on 10/07/2007 , when the NAV was Rs. 78.04 ,and redeemed all the units on 12/11/2007 when the NAV was Rs. 84.54 . In the meanwhile , on 31/08/2007 , she had received a dividend @ Rs. 10  per unit . Find her total gain and the rate of return considering loads as follows: The entry load was 2.25 % and the exit load was 0.5 % The number of units were calculated correct upto 3 decimal places.

                                                Ans . Total gain = Rs . 1794.46, Rate of return = 17.94%

8.     Given the following information , calculate NAV of the mutual fund :- No. of units =15000 Market value of investments in Govt . securities = Rs. 20 lakhs Market value of investments in corporate Bonds = Rs. 25 lakhs Other Assets of the fund = Rs. 15 lakhs Liabilities of the fund = 6 lakhs

                                                Ans . Rs. 360/- .

9.     Mumtaz purchased 1200 units of TATA BIG Bond- G Rs. 12,000 /-  on 14th April 2007 . She sold her units on 9th Dec 2007 at NAV of Rs. 15.36/- . The short term gain tax (STGT) was 10% of the profit . Find her net profit . (Calculations upto 2 decimal points)

                                                Ans . profit = 6432 , STGT = 643.2 , Net profit = 5788.8

MUTUAL FUND

Consider the following example :-

            Mr. Shaikh keeps Rs. 5000/- on 3rd of every month for 4 months as follows :-(Calculations are correct to 2 points of decimal)

Month	Amount (in Rs.)	NAV	No. of units he gets
1	5000	109.48	5000/109.48=45.67
2	5000	112.36	5000/112.36=44.50
3	5000	108.14	5000/108.14 =46.24
4	5000	105.62	5000/105.62=47.34
Total	20,000		183.75

Avg price of units = 20,000 / 183.75 = 108.84

            If Mr. Shaikh would have invested the entire amount of Rs. 20,000/- 0n 3rd of first month only , with NAV Rs. 109.48/- , the no. of units purchased would have been 20,000/ 109.48 = 182.68

Thus he gained more units and average price of units also was Rs.108.84 instead of Rs.109.48 which was NAV on 3rd of the first month

If SIP is followed for a long period of time , it can create wealth to  meet a person's future needs like housing , higher education etc .

Now , we will study the following examples to understand SIP .

Example 17

            Mr. Patil invested in a SIP of a M.F. , a fixed sum of Rs. 10,000/- on 5th of every month , for 4 months . The NAV on these dates were Rs. 34.26 , 46.12 , 39.34 and 41.85 . The entry load was 2.25 % through out the period  .  Find  the  average  price  ,  including  the  entry  load  ,  using the Rupee-cost-Averaging method .How does it compare with the Arithmetic mean of the prices ? (Calculations are to 4 digits decimal)

Solution :

Month	NAV	Entry load = 2.25%	Total price	No.ofunits=1000/Total price
1	34.2600	0.7708	35.0308	285.4627
2	46.1200	1.0377	47.1577	212.0544
3	39.3400	0.8851	40.2252	248.6006
4	41.8500	0.9141	42.7916	233.6906
TOTAL			165.2053	979.8083

By using Rupee-cost-Averaging method :- Avg Price = Total amount

                                                                        Total No. of units

=  40,000                = 40 .8243

                                                            979.8083

A.M. of price = Total price= 165.2053= 41.3013

                                    4                                 4

Avg. price using Rupee-cost- Averaging method is less than A.M. of prices .

Example 18

            Mr. Desai invested Rs. 5000/- on 1st of every month for 5 months in a  SIP  of  a  M.F.  with  NAV's  as  48.15  ,  52.83  ,41.28, 35.44  &  32.65 respectively .There was no entry load charged . Find the average price , Mr. Desai paid using the Rupee-cost-Averaging method . After 6 months ,he sold all his units , when NAV was Rs. 51.64 with contingent deferred sales charge (CDSC) as 2.25 % . Find his net gain. (Calculations are correct to 2 digits decimal)

Example 19

            Mr. Thomas started a SIP in 'HDFC long term advantage Fund ' . On the 10th July , Aug and Sept 2007 he invested Rs. 1000/- each at the NAVs Rs. 44.100, Rs. 43.761/-, s. 45.455 respectively . The entry load was 2.25% . Find his average acquisition cost per unit upto 3 decimal places . (Calculations are up to 3 decimal points).

                                    Ans. Rs. 45.427/- .

Example 20

Maneeshad Rs. 20,000/- on 2nd of every month for 5 months  in a SIP  of a mutual fund , with NAVs as Rs. 53.12 , Rs. 56.26 , Rs. 48.86 ,Rs.50.44 and Rs. 54.62 respectively . The entry load was 2.25 % throughout this period .Find average price , including the entry load , using the Rupee-cost -Averaging method and compare it with Arithmetic mean of prices . (Calculate up to 2 decimal points)

                                    Ans . 53.70 , 53.84 .

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PERMUTATIONS AND COMBINATIONS

PERMUTATIONS  AND  COMBINATIONS

The other day, I wanted to travel from Bangalore to Allahabad by train. There is no direct train from Bangalore to Allahabad, but there are trains from Bangalore to Itarsi and from Itarsi to Allahabad. From the railway timetable I found that there are two trains from Bangalore to Itarsi and three trains from Itarsi to Allahabad. Now, in how many ways can I travel from Bangalore to Allahabad?

            There are counting problems which come under the branch of Mathematics called combinatorics.

            Suppose you have five jars of spices that you want to arrange on a shelf in your kitchen. You would like to arrange the jars, say three of them, that you will be using often in a more accessible position and the remaining two jars in a less accessible position. In how many ways can you do it?

            In another situation suppose you are painting your house. If a particular shade or colour is not available, you may be able to create it by mixing different colours and shades. While creating new colours this way, the order of mixing is not important. It is the combination or choice of colours that determine the new colours; but not the order of mixing.

                To give another similar example, when you go for a journey, you may not take all your dresses with you. You may have 4 sets of shirts and trousers, but you may take only 2 sets. In such a case you are choosing 2 out of 4 sets and the order of choosing the sets doesn’t matter. In these examples, we need to find out the number of choices in which it can be done.

In this lesson we shall consider simple counting methods and use them in solving such simple  counting problems.

After studying this lesson, you will be able to :

·        find out the number of ways in which a given number of objects can be arranged;

·        state the Fundamental Principle of Counting;

·        define n! and evaluate it for defferent values of n;

·        state that permutation is an arrangement and write the meaning of n P

COUNTING PRINCIPLE :

            Let us now solve the problem mentioned in the introduction. We will writet1, t2 to denote trains from Bangalore to Itarsi and T1, T2, T3, for the trains from Itarsi to Allahabad. Suppose I take t1 to travel from Bangalore to Itarsi. Then from Itarsi Ican take T1 or T2 or T3. So the possibilities are t1T1, t2T2 and t3T3 where t1T1 denotes travel from Bangalore to Itarsi by t1 and travel from Itarsi to Allahabad by T1. Similarly, if I take t2 to travel from Bangalore to Itarsi, then the possibilities are t2T1, t2T2 and t2T3. Thus, in all there are 6 (2 x 3) possible ways of travelling from Bangalore to Allahabad.

            Here we had a small number of trains and thus could list all possibilities. Had there been 10 trains from Bangalore to Itarsi and 15 trains from Itarsi to Allahabad, the task would have been verytedious. Here the Fundamental Principle of Counting or simply the Counting  Principle comes in use :

            If any event can occur in m ways and after it happens in any one of these ways, a second event can occur in n ways, then both the events together can occur in m x n ways.

EX. 1) How many multiples of 5 are there from 10 to 95 ?

Solution :As you know, multiples of 5 are integers having 0 or 5 in the digit to the extreme right

(i.e. the unit’s place).

The first digit from the right can be chosen in 2 ways. The second digit can be any one of 1,2,3,4,5,6,7,8,9.

i.e. There are 9 choices for the second digit.

Thus, there are 2 9  18 multiples of 5 from 10 to 95.

EX. 2)  In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A,B,C,D,E and F. How many different bus routes are possible?

Solution : The number can be any one of the natural numbers from 1 to 99. There are 99 choices for the number.

The letter can be chosen in 6 ways.

 Number of possible bus routes are 99 6  594.

Let us now state the General Counting Principle

            If there are n events and if the first event can occur in m1 ways, the second event can occur in m2  ways after the first event has occured, the third event can occur in m3  ways after the second event has ocurred, and so on, then all the n events can occur in m1 x  m2   mn 1  mn ways.

EX. 6) Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?

Solution : The bus from A to B can be selected in 3 ways. The bus from B to C can be selected in 4 ways. The bus from C to D can be selected in 2 ways. The bus from D to E can be selected in 3 ways. So, by the General Counting Principle, one can travel from A to E in 3 4 23 ways = 72 ways.

PERMUTATION :

            Suppose you want to arrange your books on a shelf. If you have only one book, there is only one way of arranging it. Suppose you have two books, one of History and one of Geography.

                You can arrange the Geography and History books in two ways. Geographybook first and the History book next, GH or History book first and Geography book next; HG. In other words, there are two arrangements of the two books.

            Now, suppose you want to add a Mathematics book also to the shelf. After arranging History and Geography books in one of the two ways, say GH, you can put Mathematics book in one of the following ways: MGH, GMH or GHM. Similarly, corresponding to HG, you have three other ways ofarrangingthe books. So, bythe Counting Principle, youcan arrange Mathematics, Geography and History books in 3 2 ways = 6 ways.

            By permutation we meanan arrangement of objects in a particular order. In the above example, we were discussing the number of permutations of one book or two books.

In general, if you want to find the number of permutations of n objects n 1, how can you do it? Let us see if we can find an answer to this.

            Similar to what we saw in the case of books, there is one permutation of 1 object, 2 1 permutations of two objects and 3 21 permutations of 3 objects. It may be that, there are n  (n 1)  (n  2) ... 2 1 permutations of n objects. In fact, it is so, as you will see when we prove the following result.

Evaluate (a) 3!    (b) 2! + 4!   (c) 2! 3!

                 

EX. 7) Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

Solution : We have to arrange 6 books.

The number of permutations of n objects is n! = n. (n – 1) . (n – 2)       2.1

Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720

PERMUTATION OF R OBJECT OUT OF N OBJECT

Suppose you have five story books and you want to distribute one each to Asha, Akhtar and Jasvinder. In how many ways can you do it? You can give any one of the five books to Asha and after that you can give anyone of the remaining four books to Akhtar. After that, you can give one of the remaining three books to Jasvinder. So, by the Counting Principle, you can distribute the books in 5 4  3 ie.60 ways.

More generally, suppose you have to arrange r objects out of n objects. In how manyways can you do it? Let us view this in the following way. Suppose you have n objects and you have to arrange r of these in r boxes, one object in each box.

EX. 8) Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

Solution : Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1. Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways. So, in this case the beds can be allotted in 5 5!ways  600 ways.

Case 2 : Anju is allotted bed number 7. Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For

example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin. Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways. Therefore, in each of these cases, the beds can be allotted in 4  5!  480 ways.

        The beds can be allotted in........

COMBINATIONS :

                  Let us consider the example of shirts and trousers as stated in the introduction. There you have 4 sets of shirts and trousers and you want to take 2 sets with you while going on a trip. In how many ways can you do it?

Let us denote the sets by S1, S2, S3, S4. Then you can choose 2 pairs in the following ways :

Now as you maywant to know the number of ways of wearing 2 out of 4 sets for two days, say Monday and Tuesday, and the order of wearing is also important to you. We know from section 7.3, that it can be done in 4P  12 ways. But note that each choice of 2 sets gives us

two ways of wearing 2 sets out of 4 sets as shown below :

1.     {S1,S2}  S1 on Monday and S2 on Tuesday or S2 on Monday and S1 on Tuesday

2.     {S1,S3}  S1 on Monday and S3 on Tuesday or S3 on Monday and S1 on Tuesday

3.     {S1,S4}  S1 on Monday and S4 on Tuesday or S4 on Monday and S1 on Tuesday

4.     {S2,S3}  S2 on Monday and S3 on Tuesday or S3 on Monday and S2 on Tuesday

5.     {S2,S4}  S2 on Monday and S4 on Tuesday or S4 on Monday and S2 on Tuesday

6.     {S3,S4}  S3 on Monday and S4 on Tuesday or S4 on Monday and S3 on Tuesday Thus, there are 12 ways of wearing 2 out of 4 pairs.

This argument holds good in general as we can see from the following theorem.

EX. 14 )Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

Solution : Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements. By relation (7.6), this can be done in ………..ways

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