on compound interest comparison to simple interest

In how much time will Rs 10000 amounts to Rs. 10,600:00 with simple interest .rate of interest is 6%. With the same time period  find the compound interest rate.

amount in Rs.	year	interest rate= 6/100=0.06	interest	total
10000	after 1st year	0.06	600	10600
10000	after 2 year	0.06	600	10600
10000	after 3 year	0.06	600	10600
			1800
After three years Rs 10000.00 will become Rs.11800 .00. Total interest received is Rs. 1800.00
While calculating with compound interest
amount in Rs.	year	interest rate= 6/100=0.06	interest	total
10000	after 1st year	0.06	600	10600
10600	after 2 year	0.06	636	11236
11236	after 2 year	0.06	674.16	11910.16
With compound interest rate , interest received  Rs.			1910.16

On mathematics, it is real fun. Mathematics on compound interest

Mathematics is real fun.Generally teachers failed to extract the fun from the mathematics and percolate this to their students.  Objectives and usefulness of the particular math. is not known to the students and they failed to grasp where they will apply those stuffs in real life. So they felt the burden of mathematics and the fear of burden is felt life long.   
For some mathematics is a hobby and they can take this as their profession. 


A practical aspect on mathematics. If students know the power of compounding they will be much matured at early stage of life on financial freedom.

In how much time will Rs. 50000 amount to Rs. 560000 at 4 % compound interest

amount	interest rate	interest	year	total amount
50000	4/100= 0.04	2000	after 1 st year	52000
52000	0.04	2080	2 year end	54080
54080	0.04	2163.2	3 year end	56243.2
56243.2	0.04	2249.728	4 year end	58492.928
58492.928	0.04	2339.71712	5 year end	60832.64512
60832.6451	0.04	2433.3058	6 year end	63265.95092
63265.9509	0.04	2530.63804	7 year end	65796.58896
65796.589	0.04	2631.86356	8 year end	68428.45252
68428.4525	0.04	2737.1381	9 year end	71165.59062
71165.5906	0.04	2846.62362	10 year end	74012.21425
74012.2142	0.04	2960.48857	11 year end	76972.70282
76972.7028	0.04	3078.90811	12 year end	80051.61093
80051.6109	0.04	3202.06444	13 year end	83253.67537
83253.6754	0.04	3330.14701	14 year end	86583.82238
86583.8224	0.04	3463.3529	15 year end	90047.17528
90047.1753	0.04	3601.88701	16 year end	93649.06229
93649.0623	0.04	3745.96249	17 year end	97395.02478
97395.0248	0.04	3895.80099	18 year end	101290.8258
101290.826	0.04	4051.63303	19 year end	105342.4588
105342.459	0.04	4213.69835	20 year end	109556.1572
109556.157	0.04	4382.24629	21 year end	113938.4034
113938.403	0.04	4557.53614	22 year end	118495.9396
118495.94	0.04	4739.83758	23 year end	123235.7772
123235.777	0.04	4929.43109	24 year end	128165.2082
128165.208	0.04	5126.60833	25 year end	133291.8166
133291.817	0.04	5331.67266	26 year end	138623.4892
138623.489	0.04	5544.93957	27 year end	144168.4288
144168.429	0.04	5766.73715	28 year end	149935.166
149935.166	0.04	5997.40664	29 year end	155932.5726
155932.573	0.04	6237.3029	30 year end	162169.8755
162169.876	0.04	6486.79502	31 year end	168656.6705
168656.671	0.04	6746.26682	32 year end	175402.9373
175402.937	0.04	7016.11749	33 year end	182419.0548
182419.055	0.04	7296.76219	34 year end	189715.817
189715.817	0.04	7588.63268	35 year end	197304.4497
197304.45	0.04	7892.17799	36 year end	205196.6277
205196.628	0.04	8207.86511	37 year end	213404.4928
213404.493	0.04	8536.17971	38 year end	221940.6725
221940.673	0.04	8877.6269	39 year end	230818.2994
230818.299	0.04	9232.73198	40 year end	240051.0314
240051.031	0.04	9602.04126	41 year end	249653.0727
249653.073	0.04	9986.12291	42 year end	259639.1956
259639.196	0.04	10385.5678	43 year end	270024.7634
270024.763	0.04	10800.9905	44 year end	280825.7539
280825.754	0.04	11233.0302	45 year end	292058.7841
292058.784	0.04	11682.3514	46 year end	303741.1354
303741.135	0.04	12149.6454	47 year end	315890.7809
315890.781	0.04	12635.6312	48 year end	328526.4121
328526.412	0.04	13141.0565	49 year end	341667.4686
341667.469	0.04	13666.6987	50 year end	355334.1673
355334.167	0.04	14213.3667	51 year end	369547.534
369547.534	0.04	14781.9014	52 year end	384329.4354
384329.435	0.04	15373.1774	53 year end	399702.6128
399702.613	0.04	15988.1045	54 year end	415690.7173
415690.717	0.04	16627.6287	55 year end	432318.346
432318.346	0.04	17292.7338	56 year end	449611.0798
449611.08	0.04	17984.4432	57 year end	467595.523
467595.523	0.04	18703.8209	58 year end	486299.3439
486299.344	0.04	19451.9738	59 year end	505751.3177
505751.318	0.04	20230.0527	60 year end	525981.3704
525981.37	0.04	21039.2548	61 year end	547020.6252
547020.625	0.04	21880.825	62 year end	568901.4502
568901.45	0.04	22756.058	63 year end	591657.5082

IT is called UNIFICATION method on Mathematics

Problem with Chintu and Mintu

Chintu can do a task in 24 days while Mintu can do it in 30 days. In how many days can they complete the task when work to gether?

Now let us take the complete work as 1 

Chintu can complete the work in 24 days. 
So in one day Chintu can complete 1/24 part of the work

Similarly Mintu is capable of doing the work in 30 days.
So in one day Mintu completes 1/30 parts of work.

Now when they both works  in one day total work completion  is  1/24 + 1/30 parts of work
 1/24+ 1/30 
= 54/720= 3/40 parts of the total work


Now  we can say that 

3/40 parts of work is being completed in 1 day

so 1 work(complete) will be completed in 40/3 days= 13.33 days

Mathematics on Time and Distance

Problem given 

One car has a speed  of  204.2  Km/hr and other car has the speed  200 km/hr. Both the car started simultaneously , but the second car reached the same distance after  7 seconds of the first car. What is the distance and what will be time required for both the cars .


Solution:

Let , the distance be X   km and let the time taken for the  first car be T hr. to cover the entire distance.
Time taken for the first car T =  X/ 204.2 Hr.
Naturally other car has taken T Hr + 7 secs to cover that distance

Now , for the first car , at the speed of  204.2 km/hr at T hr it will traverse 204.2 T Km.

and for the second car at the speed of 200 Km/hr. at (T+ 7/3600) Hrs. it will cover the same distance of  200 *( T + 7/3600 ) Km , which we have already assumed as X km. 


204*T Km = 200 *(T+ 7/3600 ) KM

204 T = 200 T + 200 x 7/3600
4 T = 7/ 18
T= 28/18 Hrs. = 1.5 Hrs. 

Distance covered by the first car in 1.5 Hrs. at the speed of 204.2 Km/Hr  204.2 x 1.5= 306.30 Km

Another problem
Two train starts , one from station A towards station B at morning 8:00 AM and reached at station B at 10:30 AM. 
Another train starts from station B at 9:00 AM and reached station B at 12:00 noon. When both the trains will meet? 

Solution

Let the distance between the stations A and B be X km. and let after a time T    both the trains meet at a same point. 
Speed of train A is ( X /2.5 ) Km/hr and speed of train B is ( X/3) Km/hr
So in time T train A travels (X/2.5)*T Km. 
and in time T train B travels (X/3)* T Km.

Now , we know that to total distance is  X Km.   

(X/2.5)*T +(X/3)*T = X
or, XT/2.5 + XT/3= X
or,X(T/2.5+ T/3)= X
or, T/2.5+ T/3= 1
or,( 3T +2.5T)/7.5= 1
or,5.5 T= 7.5
or,T = 7.5/5.5 Hr= 15/11 Hr= 

One simple mathematics

Factorise
  4 X2 - 12 X + 9 = 0

OR, (2X)2 - 2. 2X. 3 + (3)2 = 0
OR, (2X -3)(2X-3)
OR, X = 3/2

PROBLEM

3 ^(X-Y) = 27 = 3 ^{3  ______________________}1

3^(X+Y)= 243= 3 ^{5________________________}2

OR,    X- Y = 3

           X+ Y= 5   OR , 2 X      =  8   OR  X = 4    When  X = 4 , Y = 1

Other  problem

 Find value of    3 * square root of 5 *  5 * square root of 3  

Let,  3 * square root 5 * 5 square root 3 =  X

So , X ² =  9 *5 *5*3

Or, X ² =  3*3 *5*5*3

Or, X = 3*5* square root 3

Or, 15 square root 3

Now, if the  integers are   3 square root 5 *  5 Square root 3

Let it is X ,

So X =  3 square root 5 * 5 square root 3

Squaring both sides,   X ² = 45 * 75

Or , X = 15 square root 15

some mathematics on percentage

problem

In an examination total students appeared are 40% girls and rest boys. Now 80% of the boys and 60% of the girls has succeeded in examination. How many % of the total have failed in the examination.


Solution:
When 40 % are the girls , so 60 % are boys. { In totality it is 100%}

When the answer is asking in % number, let us assume that total no os girls and boys are 100.

Now what is happening that out of 60 boys, 80% of the boys have passed.

No of boys have passed is  =  60* 80/100 = 48 No of boys or 48 % of boys have passed.
No of girls passed is =  40* 60/100 = 24 % of the girls have passed.

so  no of boys and girls failed = 100 - (48 + 24) = 28%

Other problem

A person makes a profit of 12 % on selling an article at 10 % discount on the printed price. Find out the ratio of the cost price and printed price 

Cost price is the price in which the seller has purchased the article on which he has to make the profit. 
Printed price is the price which is printed on the packaging of the article. like MRP ( maximum retail price) 

Let us assume the printed price is  Rs. x 

the person sold the article at 10 % discount 

Selling price  x - (10/100)*x

=  Rs. (  x- 0.1x)= Rs.0.9x  

Now the other statement is showing that on selling at Rs. 0.9 he makes a 12 % profit. 
So his purchasing price is 12 % less of Rs. 0.9 x

so cost price or the sellers purchasing price is  =Rs. 0.9 x- (12/100)0.9x = Rs. 0.792 x 

so in terms of ratio  of cost price/ printed price = 0.792 x/ x= 0.792:1


other problem 

By selling an umbrella for Rs. 500 a shopkeeper gains 20 % . During a clearance sale the shopkeeper allows a discount of 10 % on the marked price. What is his gain percentage.


Here marked price is Rs. 500

Cost price of the umbrella is  Rs. 500 - (20/100)*500 = Rs. 400 

Now on clearance sale it is discounted to  10 % , so the new selling price is Rs. 500- (10/100)*500= Rs 450


Now  the item is sold at Rs. 450 against cost price of Rs. 400 

percentage(%) gain  (450-400)/400*100 = 50/400*100 =12.5 %

some mathematics on percentage

problem

In an examination total students appeared are 40% girls and rest boys. Now 80% of the boys and 60% of the girls has succeeded in examination. How many % of the total have failed in the examination.


Solution:
When 40 % are the girls , so 60 % are boys. { In totality it is 100%}

When the answer is asking in % number, let us assume that total no os girls and boys are 100.

Now what is happening that out of 60 boys, 80% of the boys have passed.

No of boys have passed is  =  60* 80/100 = 48 No of boys or 48 % of boys have passed.
No of girls passed is =  40* 60/100 = 24 % of the girls have passed.

so  no of boys and girls failed = 100 - (48 + 24) = 28%

Other problem

In a school 83 % of students passed in mathematics, 79% students passed in English, 75% students passed in both the subjects and if total of students failed is 26. Find the total no of students.

Solution
83% or 83 no of students passed in mathematics.

So students passed in mathematics but failed in english  = 83- 75= 8%

Similarly students passed in english but failed in mathematics = 79-75= 4%

So the students who have passed in one subject and both subjects are 75 +4+8= 87%

and students failed 100- 87= 13%

Now 13% of the total students are 26 


Total no of students are  26* 100/13 = 200 nos. 



Aklbert buys 4 horses and 9 cows for Rs. 56,00000.00 . If he sells the horses at 10 % profit and cows with 20 % profit , then he earns a total profit of Rs.60000.00 . The cost of the horse and the cows.


Solution:

4x  + 9y =  5600000.________ (1) 

4* (x+ 10/100*x) + 9 * (y+ 20/100 *y) = 56000000 + 600000_________(2)



Another problem in percentage

A person gives 12 % additional discount on the discounted price after giving an initial discount on 20 % .If the final sale price is Rs. 5400.00 what it is labelled  price?


Let us start the math backwardly.

Final sale price is Rs. 5400.00 
He gave additional discount of 12 %

Before giving the discount price was Rs. 5400.00 + 12 % of 5400= Rs 6048.00
This price is already a discounted one by 20%

The price will be adding this discount =  6048 + 20% of 6048

= 6048 + 20/100 * 6048
=7257.60

the actual price before the discounting was  Rs. 7257.60   


another problem  SBI PO 2000

a shopkeeper sells 25 articles at Rs. 45 per article after giving 10 % discount and earns 50 % profit. If the discount is not given the profit gained ?

No of articles does not bear any significance in percentage computation as all are computed in 100 

At Rs. 45  he earns 50 % profit.

So his cost price is = Rs 45 - (50/100)45 =Rs. 22.5  

Now at Rs. 45 he gave 10 % discount , if he did not give the discount price will be  45 + (10/100)*45 = Rs.49.50

Now at cost price of Rs 22.50 , selling price is Rs. 49.50

Profit=  49.50 -22.50 =Rs.27.00 

profit percentage= 27/22.50 *100

A shopkeeper sold sarees  at RS 266 each after giving 5 % discount on labelled price . Had he not given the discount , he would have earned a profit of 12 % on the cost price . What was the cost price. (sbi po 1999)

If labelled price is X then  5% discount of Rs X is 266

x- (5/100)*x = 266
or 0.95 x = 266
or x = 280

Now the condition is the at Rs 280 he will earn 12 % profit above cost price

let cost price be Rs. y 
Then,  y + 12 % y =  280
or y + 0.12 Y = 280
or 1.12 Y =280
or y= Rs. 250 

magic of compounding , compound interest

When we deposit some amount of money to any bank, or any financial institution they give some percentage of their profit. Otherwise who is going to deposit his money if return is less when no interest is assumed. Why return is always less (discounted) that is another subject of discussion.

Now, interest payment are of various types. We will discuss simple interest and compound interest.

In simple interest ,  no interest is paid on the interest earned

Suppose if I deposit Rs. 100 .00 for an interest rate of 2 % per year rate . At the end of the year  I will get Rs. 102.00 .
Again for the  next year I will again get Rs. 102.00 , as Rs. 100 is principal .Interest rate is 2% per year.

Another system is (compounding) , at the second year , interest will be calculated on Rs. 102 .00

It is said that  compounding is the eighth. wonder. Power of compounding is enormous.


Any one can  calculate like below

Rs. 100.00 will become after 1 year  Rs. 102.00
Rs. 102.00 will become after 1 year  Rs. 104.00
Rs. 104.00 will become after 1 year  Rs. 106.00
next year                                                    Rs. 108.00

like on  it will increase. But in comparison to simple interest   interest gain is huge if year after after interest is added to principal

amount	year	interest rate	interest	total amount
100	start
100	1	2%	2	102
102	2	2%	2.04	104.04
104.04	3	2%	2.0808	106.1208
106.1208	4	2%	2.122416	108.243216
108.24322	5	2%	2.164864	110.4080803
110.40808	6	2%	2.208162	112.6162419
112.61624	7	2%	2.252325	114.8685668
114.86857	8	2%	2.297371	117.1659381
117.16594	9	2%	2.343319	119.5092569
119.50926	10	2%	2.390185	121.899442
121.89944	11	2%	2.437989	124.3374308
124.33743	12	2%	2.486749	126.8241795

compare this with simple rate of interest and with long time  frame

Linear Equation of Two Variables. remember the points ( Important)

let us start with a small story but have some wisdom thoughts

A person was walking in the morning at the sea shore. He saw so many fishes has swept away by the wave to the shore and water had returned back.The fishes were thriving for water. Seeing this the man started throwing away as much fish he could to the water. But the no of fishes are more. But without losing hope the man was engaged to his work. Being tired also he could not left the work.

Seeing this another man who was also morning walking told the man , When You would not be able to  save all the fishes , why the man is wasting his time.

The man replied , If i could save one fish only, it will be the change  to the life of that fish. I tried to save as much lives as I could.

Similarly If one student is benefited to my effort , that will be  my big achievement.

Any one can  communicate through mail  at  kaacconsultant@gmail.com. to get mathematics solution. Solved problem will be shown in this web site for the benefit of many.    

Point no 1

Where a,b,c all are real numbers and an  equation in the form of       ax +by +c =0  can be formed , also a and b both are not zero ,

then    ax +by + c = 0  is called a linear equation of two variables.

Point no 2

A linear equation in two variables has infinite no of solutions.

Point no 3 

Linear equation of two variables will create a straight line when plotted in  graph form.

Point no 4

x=0 is the equation of the y axis and y=0 is the equation of the x axis

Point no 5

the graph of x = a is a straight line parallel to y axis and y=a is a straight line parallel to x axis.

Point no 6

the equation y= mx is a straight line passing through the origin.

Mathematics on LCM

 problem is  what is the least number which when doubled will be exactly divisible by

12,18, 21 and 30



LCM  is  lowest common multiplier  

in LOWEST  is the minimum number which can be divided  not only by  12 but also 18, 21, and 30,

COMMON  means the number is common multiplier of 12, 18, 21,and 30 .

So if any set of numbers there is a common number can be achieved , which is divisible by every number of the set.

In this case the set of  numbers are  12,18,21, 30

Let us find this numbers LCM

3 || 12,18,21,30 |4
       2 || 4, 6, 7, 10
              2, 3, 7,  5

So LCM is   3x2x2x3x7x5 =  1260

1260 is lowest number which is divisible by 12.18. 21. 30

Now problem is asking  the lowest no which is doubled  Here doubled means twice or multiplied by two.

 Actual no is  1260 divided by 2 which is  630 is the required answer.

Problem 2
Now another problem is given like that

We have to find out the number which when diminished by 7 is exactly divisible by 12,16,18,21, and 28

Now we know that for each set or bunch of such nos there is a lowest number can be found out which is divisible by that numbers


So Lowest common multiplier (LCM) is the lowest no which we have to found out


12,16,18,21,28

on doing LCM  it is coming 1008

But the problem has still some twist

It is asking find a lowest number which is diminished (lowered) by 7

If 1008 is the lowest number the required number being asked to find out is 1008 +7= 1015

Mathematics on HCF

Problem is given that there are four digit highest number , which when divided by 4,7,13 leaves the reminder 3 in each case


So we are to find out the common number
Four digit highest number is 9999

suppose we were given only one no say 4

we will divide   9999 by 4


 4  |   9999 | 2499
         8
___________
        1 9
____16_____
            39
             36
____________
               39
                36
           _____
                   3

For 4  , highest no is 9999 giving a remainder of  3

But the problem is with a sets of numbers 4,7,13

let us do the LCM of  4,7,13  which is  364

364 | 9999| 27    
         728
_________
           2719
           2548
__________
              171     Now remainder is  171 , if this remainder is diminished from the 9999 it will be exactly divisible

So     9999 - 171 = 9828

9828 is the four digit highest no. which is exactly divisible by the sets of numbers 4,7,13

But the problem does not end here , It is asking a number in which there will be a remainder of 3 .

So the required no will be  9828 +3 = 9831



Now another issues related to HCF and LCM of some numbers.

HCF is always a factor of the LCM that is LCM is divisible by the HCF

 Now there is a practical examples

The maximum no of students among them 1001 pens and 910 pencils can be distributed in such a way that each students get the same no of pens and pencil


This problem can be solved by simply doing the HCF of the numbers


HCF of this nos are 91

so the max no of students are 91 .

Mensuration ,mathematics on right angled cone and on spheres

Students are  asking for the mathematics which they are not understanding in the email;    kaacconsultant@gmail.com . Solution is appearing in this page for helping others also. 

The problem is  The slant height of a right circular cone  is 10 m and its height is 8 m. find the area of the curved surface.


right circular cone means ,its base is a circle the meaning of the right is , its peak is straight  at 90 degrees. above the center of the base .

slant height means ,distance of the peak from the any point of base perimeter


height is   8 meter

slant height is 10 meter

from the above data we have to deduce another date , which is the radius of the base, otherwise we will not be able to find curved surface area.

radius of the base is 6 meter.  ( I have not deduced the same , students have to apply their skill) if not otherwise contact.

Now area of the curved surface is   pi * radius* slant height value of    pi has taken as = 3.142

which  will  become  187.52 sq meter     why  square meter ? because to units of meters has been multiplied.


Suppose there are two spheres  having radius R and r , is there any relation can be established   with this two radius. let us try

Surface area of the sphere with  radius  R  = 4 * pi* R2 ( R square)
similarly surface area of the sphere with radius r = 4*pi* r2 ( r square)

ratio is coming as R2 / r2 or  (R/r)2

Problem
A cylindrical shaped article has a height of 20 Cm.The radius of its base is 21 cm. and its volume is
2310  cm 3
It is open at the top
Find the cost of painting it from otside when rate is  Rs. 5 per Cm 2
Solution
Given data :
Height  H	20 Cm
Radius of the base R		21 Cm
Volume V		2310  cm 3
Now as the cylinder is open at the top , surface area we have to find out at the cylinder wall and the bottom
surface area of the cylinder			2 π R L
			2640	cm 2
surface area of the bottom			π R 2
			1386	cm 2
Total area to be painted			surface  + bottom
			4026	cm 2
And cost of painting is Rs. 5 per square Cm
That is one square Cm will cost  Rs. 5
But we have  4026		cm 2		Rs.	5*4026	cm 2
				Rs.	20,130	will be cost.