Aptitude - Simple Interest - Discussion

I see. the key issue here is they did not say that they are substituting (total money in - "? scheme-A money in") for "? scheme-B money in".

A part that may be confusing (was for me at first) is the use of x: x being the variable and the multiplicative mechanism. the variable should be noted differently than the multiplication - capitalizing X at least or using * to denote multiplication. Let me draw this out in simple terms.

A = money put into scheme A.
Simple interest for scheme A is 14%

B = money put into scheme B.
Simple interest for scheme B is 11%

3508 is the total interest returned from scheme A and scheme B.

13900 is the total money invested into A and B.
(A + B) = 13900

Because the interest accumulates over 2 years, the interest gained is doubled. Therefore the interests in both schemes are doubled.

(A * 14 * 2)/100 +(B * 11 *2)/100 = total interest gained.

note the 100 in the denominator is because the 14 and 11 are percentages. To simplify things the .14 is represented as 14/100 and .11 to 11/100. The numbers workout in the end without any converting or anything.

since we cannot do anything with the 2 variables A and B, we substitute B for ("total money invested" - A). With this we can solve for just A.

(A * 14 * 2)/100 +((13900- A)* 11 * 2)/100 = 3508

Getting rid of the common denominators yields:

(A * 14 * 2)+((13900- A) * 11 * 2) = 350800

Simplify:
(28 * A)+((13900- A) * 22) = 350800

distribute the 22 into the parenthesis

(28 * A) + (13900 *22) -(22 * A) = 350800
Simplify:
28A - 22A + (305800) = 350800

Add like terms:
6A + (305800) = 350800

6A = 350800 - 305800

A = 7500.

Since A + B = 13900:

7500 + B = 13900

B = 6400

$7500 was initially invested into Scheme A and $6400 was initially invested into Scheme B. At the interest 14% and 10%, Scheme A and Scheme B yielded $3508.

Sorry if it is too simple, i just wanted to make sure there was a little confusion as possible.

Invested amount in Two Scheme 13900.
In scheme A invested ratio 14% and in Scheme B invested ratio 11%.
Simple Interest for 2 years 3508.
Question- B scheme's invested amount .

Answer-

Method-.
PTR/100 + PTR/100( A and B scheme)=3508(SI).
Let's P for A scheme- x
Then P of B scheme - 13900-x.
(You have total 10 mangoes. You give 4 mangoes to A and then rest mangoes which will give to B = Total mangoes(10)- A's mangoes(4)= 6 mangoes for B).


x*2*14/100+(13900-x)*2*11)/100 = 3508.
= 28x/100+(13900-x)*22/100 = 3508.
= 28x/100+{(22*13900)-(22*x)} = 3508.
= 28x/100+(305800)-22x = 3508.
= 28x-22x=350800-305800.
= 6x = 45000.
= x= 4500/6 = 7500(A's investment in 13900).

Then B's investment= 13900 - A's investment.
= 13900-7500= 6400.

Amount invested is nothing but the principal.

Now principal is divided into two parts.

Therefore P1+P2 = 13900 Rs-------equation no.1.

Now,

S.I with rate 14%=S.I1=(P1*2*14)/100 -----as for two years N=2.

S.I with rate 11%=S.I2=(P2*2*11)/100 -----as for two years N=2.

Therefore total S.I=S.I1+S.I2 ---------equation 2.

But S.I = 3508 Rs.

Therefore Equation 2 becomes,

3508 = [ (P1*2*14)/100 ] + [(P2*2*11)/100 ].

3508 = [ P1*14 +P2*11 ] [2/100]---TAKE 2& 100 common.

(3508*100)/2=[ P1*14 +P2*11 ] -----equation 3.

Solve equation 1 & 3 simultaneously,

Hence P1 = 7500 ----for A scheme at rate 14%.

P2 = 6400 ----for B scheme at rate 11%.

@Priya @Swathi @Kasamsetty Deepshika.


Many people asked that what if we take A as 13900-x,

Here the solution is;

Let us take A=(13900-x) , B=x;
S.I of A + S.I of B = 3508.

(13900-x)*14*2 / 100 + x*11*2/100 = 3508,
(13900-x)*28/100 + 22x/100 = 3508,
389200-28x/100 + 22x/100 3508,
389200-28x + 22x = 3508*100,
389200-28x + 22x = 350800,
389200- 350800 = 28x-22x.
38400 = 6x.

x = 38400/6 =>6400.
Therefore we found x which is A.

NOTE:
if we substitute B = (13900-x) we have to subtract x from 13900 to get an answer
Here we took B as x so we find the answer in the first step. (SI of A + SI of B = 3508).

Assume that the principal invested in scheme A be rupees x.

Then according to question the principal invested in scheme B is (13900-x) rupees.
Number of years: 2 years.

Rate of interest in scheme A:14%.
Rate of interest in scheme B: 11%.
Total interest earned : rs 3508.
SI: PNR/100 formula.
(x*2*14/100)+((13900-x)*2*11/100)=3508.
(28x/100)+((305800-22x)/100)=3508.
(28x-22x)+305800=3508*100.
6x+305800=350800,
6x=350800-305800,
6x=45000,
x=7500.
Now we got x ie.. principal invested in scheme A.

Now we can find the principal invested in scheme B.
13900-7500= 6400.
Principal invested in scheme B is rupees 6400.

Let us assume that the scheme A be as X & scheme B be as Y.

The total amount invested is 13900.
So, X+Y=13900
X=13900-Y--------(1).

Now, we use simple interest formula SI=PRT/100
Time T=2years
Per annum R=14% & 11%
Now, substitute the values in SI formula.

Amount invested scheme B=
[Xx14x2/100]+[Yx11x2]=3508
[28X/100] +[22Y/100]=3508
28X+22Y=3508x100
28X+22Y=350800--------------> (2)

Now substitute eq(1) in eq(2).

28(13900-Y) + 22Y = 350800
389200-28Y + 22Y = 350800
- 28Y + 22Y = 350800 - 389200
- 6Y = - 38400.
6Y = 38400.
Y = 38400/6.
Y = 6400.
Amount invested scheme B= 6400.

@Renu.

It has Some correction @Phani Kumar .

Given -----> P = P1+P2 (Principle is divided between two)
Sum of interests on P1 and P2 is given as I = SI1+ SI2 = 3508 S11---> Simple interest on P1
SI2 ---> Simple interest on P2
I = ( P1*T1*R1 /100) + (P2*T2*R2/100)
100*I = (P1*T1*R1) + (P2*T2*R2)
But Given T1= T2=T
so,
100*I /T = P1*R1 + P2*R2
= P1*R1 + (P-P1)*R2
= P1*R1 + P*R2 - P1*R2

(100*I/T) - P*R2 = P1*(R1-R2)
((100*I)-( P*T*R2))/ T = P1*(R1-R2)
P1 = (100*I - P*T*R2) / T*(R1-R2) (R1-R2 = 14 -11 =3)
Find P2 = P - P1

From the question we know 13900 is the amount he invested.
Let assume, 'x' be the amount invested on A.
so 13900-x be the amount B have

SI=PNR/100.
Case of A:
for 2 years, as per the question.
SI1=x*2*14/100
Similarly
Case of B:
for 2 years
SI2=(13900-x)*2*11/100

its given that the total amount of simple interest is 3508
ie,SI1+SI2=3508

substituting the above two equations
Then, x x 14 x 2 /100 + (13900 - x) x 11 x 2/100 = 3508
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Given,

Time = 2yrs
Total interest earn 2 yrs = 3508.
Interest for A and B is 14% and 11q respectively,
Let Scheme A be X and Scheme B be 13900 - X,
Then; A will be = (X*14*2/100),
B will be = ((13900-X)*11*2/100).

(X*14*2/100)+((13900-X)*11*2/100) = 3508,
(28X/100)+(13900*22-22X/100)=3508,
28X-22X + 305800/100 = 3508,
28X-22X + 305800 = 3508*100,
6X + 305800 = 350800,
6X = 350800 - 305800,
6x = 45000,
X = 45000/6,
X = Rs. 7500.

Substitute x with 7500 in (13900-X).
= 13900-7500.
= Rs. 6400.

Let A amount be "x" rupees so B must be "13900-x "(because 13900 divided two members we don't know the exact amount so we consider "x" rs by A remaining amount 13900-x by B).

Apply formula.
PTR/100.
P = PRINCIPAL
T = TIME
R = RATE OF INTEREST

X * 14 * 2/100 + (13,900-X) * 11 * 2/100 = 3508.
(x is the principal amount we don't know),
(14 is rate of interest A),
(2 is the year),
After calculation we get 6400.