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 5 of 12.
Renu said:
1 decade ago
Formula is simple interest(S.I)=P*T*R/100;
P=Principal amount or sum
T=Time
R=Interest rate
By using this formula we can solve this.
P=Principal amount or sum
T=Time
R=Interest rate
By using this formula we can solve this.
Swati Alone said:
1 decade ago
Can anyone tell me how to do this answer easily and fast without using pen ?
Pranali said:
1 decade ago
Can someone please explain me the 2nd step 28X-22X = 350800 - (13900x22) how it came?
Rajesh said:
1 decade ago
Assume the amount invested in scheme B be 6400.
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.
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.
Chaitanya said:
1 decade ago
Is we use the formula ptr/100?
Bikash kumar said:
1 decade ago
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
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.
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.
Sameth said:
1 decade ago
What is Rs. Stand for?
MANI SMART said:
1 decade ago
The formula is (pnr/100).
p=inicial amount,
n=no of years,
r=rate of interest
p=inicial amount,
n=no of years,
r=rate of interest
Teja said:
1 decade ago
See, it is given that both a&b together combinedly having principal amount of 13,900 for two years having si of 3508.
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.
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.
R.srinivasulu said:
1 decade ago
Shortcut method tell me sir.
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