Deepak invests rs 600 at 5% simple interest per annum in how much time will his money get doubled

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

Answer: Option A

Explanation:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then,		x x 14 x 2		+		(13900 - x) x 11 x 2		= 3508
100	100

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Video Explanation: https://youtu.be/Xi4kU9y6ppk

Can you please tell me that method.

See, 28x/100 + (13900 - x)* 22 / 100 = 3508
= 28x/100 + 305800 - 22x /100 = 3508
take 100 lcm
28x + 305800 - 22x/100 = 3508
or 28x + 305800 - 22x = 350800
or 28x + 305800 - 22x - 350800 = 0
0r 28x - 45000 - 22x = 0
or 6x - 45000 = 0
or 6x = 45000
x = 7500

You cannot take 25% directly from Principal since money is divided into two parts, so that diff is there in your calculation

Now see, 7500 x 14% + 6400 x 11% in 2 years will give you 3508.

P=Principal amount or sum
T=Time
R=Interest rate

By using this formula we can solve this.

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%.

Then the amount invested in scheme A is 13900-6400 = 7500.

14% in 7500 = 1050.

For two years (1050*2) = 2100.

11% in 6400 = 704.

For two years (704*2) = 1408.

Adding these two gives (2100 + 1408) = 3508.

Therefore we assumed correctly and the answer is 6400.

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.

Then, (X*14*2/100)+{(13900-X)*11*2/100} = 3508.
28x/100 + (13900*22)-(x*22)/100 = 3508.
28x/100 + 305800-22x/100 = 3508.
28x+305800-22x/100 = 3508.

Now,
28x+305800-22x=3508*100.
28x-22x=350800-305800.
6x=45000.
x=45000/6.
x=7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

p=inicial amount,
n=no of years,
r=rate of interest

Now by adding both schemes si's with the total si given for two years by using formula
si = p*t*r/100.

By considering scheme b = 13,900-a,
We can find investment on b scheme.

P1 = ((100*I) - (P*T*R2))/(difference of R1 and R2). And then find P2 by:

P2 = P-P1.

Here,

P = 13900.
P1 = Part1;
P2 = Part2;
I = 3508;
T = 2;
R1 = 14;
R2 = 11;

Say 3500. 3500 is approximately 25% of 14000 (13900). So for one year it slightly more than 12.5%. So invest in A (14%) must be more.

So the answer must be less than 6950(13900/2). Two opinion A are B. Now pair up. (6400/7500) or (6500/7400).

Take any one pair find out their simple interest. And then add up both SI which result in 3508 is the answer.

Could you please explain it in your way? Your previous answer seems good.

Formula : RT = (n-1)100.

Here: N = 2 (amount is double).

T = 5yr.
R?

Then 5R = 100.
R = 100/5.
R = 20.

Here p = Total amount shared between A & B as P = P1+P2.

I = Total interest; R1 = P1's rate of interest, R2 = P2's rate of interest.

And wish formula use.

Please, can anyone explain it in detail?

Total amount can be given in 2 types of interests as one A is @14% & B is @11%.

He gets average in totally 2 years is given 3508/2 = 1754.
Its mean % is 1754/13900 = 12.6%.

Apply Allegation principle on interests.
Then, we get;
-> A : B.
-> 8 : 7.

Totally we have 15 parts = 13900

We require the amount invested in Scheme:

B = 7/15 * 13900.

=> 6486 (approx).

Because I tried and could not get the answer.

Total interest for 2 years - 3508.
Then for a year - 3508/2 = 1754.
Assume investment in A scheme as x and B as Y.

14/100 * x + 11/100 * y = 1754 --------> 1st eqn
x + y = 13900 -------> 2nd eqn.

Solve this. You will get the answer.

28% of 13900 = 3892.
3892 -> given amount, then3892 - 3508 = 384.
Two schemes difference is 6%(28 - 22).
6% = 384.
Sum 100% =?

Then cross multiplication 384 * 100/6 = 6400 is the answer.

Now you'll get 28, & 22.
=> 28 + 22 = 50 * 3508 * 2 = 350800.
Now you need to find the amount of B = 22 * 13900 = 305800.
So. 350800 - 305800 = 45000 .
28 - 22 difference is = 6.
Now 45000/6 = 7500.
=> 13900 - 7500 = 6400.

Then the value of A is 7500 and B is 6400.

A farmer has a rectangular garden plot surrounded by 200ft of the fence. Find the length and width of the garden if its area is 2400ft2.

Area of rectangle = l * w
Perimeter of rectangle = 2 * (l + w)
--------------------------------------------

2 * (l + w) = 200
l + w = 100 ---> equa(1)

l * w = 2400
so, w = 2400/l ---> equa(2)
Putting equa(2) in equa(1).

We get,
l + (2400/l) = 100
l^2 + 2400 = 100 * l
l^2 - 100 * l -2400 = 0
l^2 - 60 * l - 40 * l -2400 = 0
l (l - 60) - 40 (l - 60) = 0
(l - 60) (l - 40) = 0

Hence, if l = 60 w = 40 (from equa(1))
Else, l = 40 then w = 60.

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.

Thanks for giving the explanation of answer.

Can you please solve using the formula you stated?

i.e P1= {(100*1)-(PTR2)} / (R1-R2),
P2=P-P1.

Please anyone help me with this solution.

Where did the other denomination (ie 100) gone?

14x+11y=175400 ---> equation 2.
To solve the equation we are multiplying the equation 1 by 14.
14x+14y=194600 ---> equation 1 after multiply by 14.
14x+11y=175400.

After subtracting the equation we will get.
3y = 19200,
Y = 6400.

Nice explanation.

they asked for scheme B so we shld take x=13900-y instead of y=13900-x.

It is not necessary that A will invest X and B will invest the difference.

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.

Can you plz tell me how to take a difference between r1 and r2 (14% - 11%) or (14*2-11*2).

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

SI=P*R*T/100.

The difference is 3%(14%-11%) in one year.
If we got 14% on both then 14% of 13900 = 1946.
But we got interest 3504 which is in two years so in one year 3504/2 =1754.
The difference is 1946-1754=192.
So we got 3% difference on 192.
3% =192.
1%=192/3=64.
100%=6400.
So, one part is 6400 and other (13900-6400) = 7500.

& b be y.

The total amt invested is 13900,
so x+y=13900 or y=13900-x,
by using formula i=ptr/100,
x*14*2/100 + y*11*2/100,
= 28x/100+22y/100=3508,
28x+22y=350800,

Substitution of y frm 1
28x+22(13900-x) = 350800 now solve.

Hence we need to fine out B scheme.
Use option given 1st option is 6400,
6400*2 year*11%÷100=1408,
And remaining intrest is 3508-1408=2100,
Now A scheme,
13900-6400=7500,
7500*2year*14%÷100=2100.

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.

B = 11%.
A = 14%.

Calculate Rate for 1 year on 100*1754/13900*1 = 12.61%.
1754 is one-year int.
mean is 12.61%.
then 14%-12.61% = 1.38%.
12.61-11 = 1.61.
1.38/3*13900 = 6400.

Given intrerest for 2 year so 3508/2 = 1754 p.a.
By using alligation:

1956-1754 = 192
1754 - 1529 = 225.

Tatio;-B:A = 192 : 225.

Sum of B = 13900 * 192/417 = 6400.

13900/30*14 = 6500.

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.

2nd case
SI = (13900-x).2.11/100.
SI = 3508,
3508 = 28x/100+ (13900-x).2.11/100,
350800= 28x +305800-22x,
-45000 = -6x.
X = 45000/6.
X = 7500.
B = 13900 - 7500.
B = 6400.

13900 X 11% = 1529 (They asked for B whose interest is 11%).
2 YR = 3508
1 YR = 1754,

EXTRA INTERST = 1754-1529 = 225.

That extra interest nothing but 14%-11%=3%,
3% = 225,
100% = 7500,
13900-7500 = 6400,

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).

A:14% for 2 years = 28% (In si rate of interest is same for every year i.e. for 2yrs 14 is directly multiplied).

B:11%for 2years = 22%.

28% of principal13900=3892 (interest) but the interest of A+B scheme is given=3508.

i.e 384 extra (3892-3508) =384.

28%~22%=6%.
6%---> 384.
100%---> 6400.

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.