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 3 of 12.
Umesh gandla said:
7 years ago
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.
100 100
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
(2)
Swati said:
7 years ago
Why it has not taken X for B's principal? Please explai me.
(2)
Mangesh said:
7 years ago
P=13900
A=14%
B=11%
So,total SI= 3508.
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.
A=14%
B=11%
So,total SI= 3508.
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.
(1)
Priya said:
7 years ago
Thank you so much @Jitendar.
(1)
Sree said:
7 years ago
Good explanation. Thank you all.
(1)
Sreelatha said:
7 years ago
C let us assume that principle invested in a is x.
& 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.
& 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.
(2)
Manish Kumar said:
7 years ago
In this question money is divided into two parts. On one part he got 14% interest and on other part he got 11%.
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.
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.
Dhilip said:
8 years ago
It is a PTR/100 formula total SI is given we find amount invested in B so subtract x from p here x is an amount invested and p total amount. Two different scheme so we add two amount and the SI is placed other side by the formula
SI=P*R*T/100.
SI=P*R*T/100.
Punna vandana said:
8 years ago
Please, can anyone explain this problem in percentages?
Pragna said:
8 years ago
@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
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
(1)
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