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Question: Susan invested a certain amount of money in two schemes A and B, which offers interest at the rate o...

Susan invested a certain amount of money in two schemes A and B, which offers interest at the rate of 8% per annum and 9% per annum respectively. She received Rs.1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs.20 more as annual interest. How much did she invest in each scheme?

a.Rs.13000 in scheme A, Rs.12000 in scheme B
b.Rs.12000 in scheme A, Rs.10000 in scheme B
c.Rs.11000 in scheme A, Rs.10000 in scheme B
d.Rs.10000 in scheme A, Rs.13000 in scheme B

Explanation

Solution

Hint: Here, take the amount invested in A as xx and B as yy.Now, calculate the simple interests of A and B. For that apply the formula: SI=P×R×T100SI=\dfrac{P\times R\times T}{100}. For total interest Rs.1860 we will get the equation, 0.08x+0.09y=18600.08x+0.09y=1860.When the total interest is 20 more we will get the equation, 0.09x+0.08y=18800.09x+0.08y=1880. By solving these two equations we will get the values of xx and yy.

Complete step-by-step answer:

Here, given two schemes A and B.
Hence, the interest is not specified. We can take the interest as simple interest.
Here we have to calculate the money invested in schemes A and B.
Now, let the amount invested in A, Principal P = xx
The rate of interest in A is R=8R=8%
Here, the time period is taken for one year. Therefore,
Time, T = 1

Now, let us calculate the Simple interest of A.
SI1=P×R×T100 SI1=x×8×1100 SI1=8x100 \begin{aligned} & S{{I}_{1}}=\dfrac{P\times R\times T}{100} \\\ & S{{I}_{1}}=\dfrac{x\times 8\times 1}{100} \\\ & S{{I}_{1}}=\dfrac{8x}{100} \\\ \end{aligned}
By dividing 8 by 100 we obtain:
SI1=0.08xS{{I}_{1}}=0.08x ..... (1)

Let the amount invested in B, Principal P = yy
Rate of interest of B, R=9R=9%
Here, also the time period is taken as 1 year. Therefore, we have:
Time, T = 1
Now, let us calculate the simple interest of B.
SI2=P×R×T100 SI2=y×9×1100 SI2=9x100 \begin{aligned} & S{{I}_{2}}=\dfrac{P\times R\times T}{100} \\\ & S{{I}_{2}}=\dfrac{y\times 9\times 1}{100} \\\ & S{{I}_{2}}=\dfrac{9x}{100} \\\ \end{aligned}
By dividing 9 by 100 we get:
SI2=0.09xS{{I}_{2}}=0.09x …… (2)

We are given that the total interest is 1860. That is:
SI1+SI2=1860S{{I}_{1}}+S{{I}_{2}}=1860
By substituting equation (1) and equation (2) in the above equation we get:
0.08x+0.09y=18600.08x+0.09y=1860
It is also given that if Susan interchanges the amount of investments she will receive 20 more as annual interest. Then we will get:
SI1=0.08y ...... (3) SI2=0.09x ...... (4) \begin{aligned} & S{{I}_{1}}=0.08y\text{ }......\text{ (3)} \\\ & S{{I}_{2}}=0.09x\text{ }......\text{ (4)} \\\ \end{aligned}
Now, the total interest becomes:
SI1+SI2=1860+20 SI1+SI2=1880 \begin{aligned} & S{{I}_{1}}+S{{I}_{2}}=1860+20 \\\ & S{{I}_{1}}+S{{I}_{2}}=1880 \\\ \end{aligned}
By substituting equation (3) and equation (4) in the above equation we get:
0.08y+0.09x=18800.08y+0.09x=1880
By rearranging the above equation we get:
0.09x+0.08y=18800.09x+0.08y=1880
Hence, let us write the two equations:
0.08x+0.09y=1860 .... (5) 0.09x+0.08y=1880 .... (6) \begin{aligned} & 0.08x+0.09y=1860\text{ }....\text{ (5)} \\\ & 0.09x+0.08y=1880\text{ }....\text{ (6)} \\\ \end{aligned}

In the next step, we have to eliminate xx by multiplying 9 in equation (1) and multiplying 8 in equation (6). Hence, we obtain:
0.72x+0.81y=16740 0.72x+0.64y=15040 \begin{aligned} & 0.72x+0.81y=16740 \\\ & 0.72x+0.64y=15040 \\\ \end{aligned}
Next, by subtracting the above two equations we obtain:
0.17y=17000.17y=1700
Now, by cross multiplication we get:
y=17000.17y=\dfrac{1700}{0.17}
Next, by multiplying 100 in the numerator and denominator we obtain:
y=17000017y=\dfrac{170000}{17}
By cancellation we get:
y=10000y=10000
Now, we have to calculate xx by substituting the value of y=10000y=10000 in equation (5). Therefore, we get:
0.08x+0.09×10000=1860 0.08x+900=1860 \begin{aligned} & 0.08x+0.09\times 10000=1860 \\\ & 0.08x+900=1860 \\\ \end{aligned}
Next, by taking 900 to the right side, 900 becomes -900, we get:
0.08x=1860900 0.08x=960 \begin{aligned} & 0.08x=1860-900 \\\ & 0.08x=960 \\\ \end{aligned}
By cross multiplication we obtain:
x=9600.08x=\dfrac{960}{0.08}
Next, by multiplying 100 in the numerator and denominator we will get:
x=960008x=\dfrac{96000}{8}
By cancellation we get:
x=12000x=12000
Therefore, the values of xx and yy are x=12000x=12000 and y=10000y=10000.
Hence we can say that the money invested in scheme A is Rs.12000 and the money invested in B is Rs.10000.
Therefore, the correct answer for this question is option (b).

Note: If in the question it is not specified which interest should be calculated then you have to calculate the simple interest. The interest is calculated annually, therefore, the time period should be taken as one year.