
Mr. Thomas invested an amount of Rs. 13,900 divided into two different schemes A and B at the simple interest rate of 14 % p.a. and 11 % respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
(a). Rs. 6400
(b). Rs. 6500
(c). Rs. 7200
(d). Rs. 7500
Answer
604.2k+ views
Hint: Recall the formula for the simple interest which is \[SI = \dfrac{{PNR}}{{100}}\]. Determine the amount invested in scheme A and subtract it from the total amount to get the amount invested in scheme B.
Complete step-by-step answer:
Mr. Thomas invested an amount of Rs. 13,900 into two schemes A and B.
Let the amount invested into scheme A be x.
Then the amount invested in scheme B is 13,900 – x.
We know the formula for the simple interest for a principal amount P invested for N years at the rate of R % p.a. is given as follows:
\[SI = \dfrac{{PNR}}{{100}}\]
It is given that the total amount of simple interest earned at a rate of 14 % p.a. is Rs. 3508. Then, we have:
\[3508 = \dfrac{{x \times 2 \times 14}}{{100}} + \dfrac{{(13,900 - x) \times 2 \times 11}}{{100}}\]
\[350800 = 28x + 22(13,900 - x)\]
Simplifying, we have:
\[350800 = 28x + 305800 - 22x\]
\[6x = 350800 - 305800\]
\[6x = 45000\]
Solving for x, we have:
\[x = \dfrac{{45000}}{6}\]
\[x = 7500\]
Hence, the amount invested in scheme A is Rs. 7500.
The amount invested in scheme B is given by:
\[13900 - x = 13900 - 7500\]
\[13900 - x = 6400\]
Hence, the amount invested in scheme B is Rs. 6400.
Note: The total interest amount is Rs. 3508 and it is not just the interest amount gained in scheme A. Hence, you must use the simple interest formula for both scheme A and scheme B and add them to get Rs. 3508.
Complete step-by-step answer:
Mr. Thomas invested an amount of Rs. 13,900 into two schemes A and B.
Let the amount invested into scheme A be x.
Then the amount invested in scheme B is 13,900 – x.
We know the formula for the simple interest for a principal amount P invested for N years at the rate of R % p.a. is given as follows:
\[SI = \dfrac{{PNR}}{{100}}\]
It is given that the total amount of simple interest earned at a rate of 14 % p.a. is Rs. 3508. Then, we have:
\[3508 = \dfrac{{x \times 2 \times 14}}{{100}} + \dfrac{{(13,900 - x) \times 2 \times 11}}{{100}}\]
\[350800 = 28x + 22(13,900 - x)\]
Simplifying, we have:
\[350800 = 28x + 305800 - 22x\]
\[6x = 350800 - 305800\]
\[6x = 45000\]
Solving for x, we have:
\[x = \dfrac{{45000}}{6}\]
\[x = 7500\]
Hence, the amount invested in scheme A is Rs. 7500.
The amount invested in scheme B is given by:
\[13900 - x = 13900 - 7500\]
\[13900 - x = 6400\]
Hence, the amount invested in scheme B is Rs. 6400.
Note: The total interest amount is Rs. 3508 and it is not just the interest amount gained in scheme A. Hence, you must use the simple interest formula for both scheme A and scheme B and add them to get Rs. 3508.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

