# Aptitude - Simple Interest - Discussion

Discussion Forum : Simple Interest - General Questions (Q.No. 2)

2.

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

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 |

28*x* - 22*x* = 350800 - (13900 x 22)

6*x* = 45000

*x* = 7500.

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

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

Discussion:

115 comments Page 1 of 12.
Swarnaprava said:
1 year ago

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.

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.

(25)

AreEn said:
2 years ago

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.

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.

(18)

Subham Prakash said:
2 years ago

@Priya @Swathi @Kasamsetty Deepshika.

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.

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.

(98)

Dharshini said:
2 years ago

@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).

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).

(3)

Kasamsetty Deepshika said:
3 years ago

What if I take A as 13900-x? Please explain.

(4)

Sourav Kalal said:
3 years ago

Let me solve it in a simple way.

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,

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,

(30)

Yuki said:
3 years ago

Thanks all for explaining the answer clearly.

(2)

Deepak said:
3 years ago

Thanks for explaining @Srini.

(2)

Yogeswara Rao said:
3 years ago

Invested money = 13900.

Divided into two schemes A and B.

Let the scheme A = x.

Scheme B = 13900-x.

SI = PTR/100.

SI = x.2.14/100,

SI = 28x/100.

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.

Divided into two schemes A and B.

Let the scheme A = x.

Scheme B = 13900-x.

SI = PTR/100.

SI = x.2.14/100,

SI = 28x/100.

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.

(4)

K. Uma said:
3 years ago

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.

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.

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