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 |
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
Video Explanation: https://youtu.be/Xi4kU9y6ppk
Discussion:
115 comments Page 8 of 12.
Loke said:
9 years ago
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.
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.
Krishna Kant Sharma said:
9 years ago
Thanks for giving the solution. I understand it now.
Sri said:
8 years ago
@Depu.
Thanks for giving the explanation of answer.
Thanks for giving the explanation of answer.
Madhu said:
8 years ago
@Phani Kumar
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.
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.
Manish said:
8 years ago
Let x be the amount of B and 13900-x be the amount of A then without subtract we can find the value o B.
Tiama said:
8 years ago
I am confused with the following stem.
28x + 305800 - 22x = 3808 * 100.
Where did the other denomination (ie 100) gone?
28x + 305800 - 22x = 3808 * 100.
Where did the other denomination (ie 100) gone?
Karthik said:
8 years ago
Why the amount invested in the scheme a should be taken as x and in scheme b should be taken as 13900-x, as both amounts are part of 13900?
Jitender said:
8 years ago
Simple interest for two year 3508, therefore simple interest for one year will be1754
x+y=13900 ---> Equation (1).
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.
x+y=13900 ---> Equation (1).
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.
Ganesh naik said:
8 years ago
@ Aditi.
Nice explanation.
they asked for scheme B so we shld take x=13900-y instead of y=13900-x.
Nice explanation.
they asked for scheme B so we shld take x=13900-y instead of y=13900-x.
Tushar waghchaure said:
8 years ago
Let's assume any answer between option, suppose 6400,
and total amount is given 13900 so, 13900-6400=7500 so we got two values of principle(p),
we know formula i,e SI= P*R*T/100.
so put value...SI=7500*14*2/100=2100 AND SI=6400*11*2/100=1408 and addition of this is =3508. So, A invest 7500 AND B invest 6400.
and total amount is given 13900 so, 13900-6400=7500 so we got two values of principle(p),
we know formula i,e SI= P*R*T/100.
so put value...SI=7500*14*2/100=2100 AND SI=6400*11*2/100=1408 and addition of this is =3508. So, A invest 7500 AND B invest 6400.
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