Simple and Compound Interest: CAT Previous Year Questions

Ans: A

Explanation:Let the principal sum be P and the rate of simple interest per annum be R%.

According to the first condition, the sum amounts to Rs 13920 in 3 years.
Amount = Principal + Simple Interest
SI for 3 years = P * R * 3 / 100
13920 = P + (P * R * 3 / 100) — (1)

According to the second condition, the sum amounts to Rs 18960 in 6 years and 6 months, which is 6.5 years.
SI for 6.5 years = P * R * 6.5 / 100
18960 = P + (P * R * 6.5 / 100) — (2)

Subtract equation (1) from equation (2):
18960 – 13920 = [P + (P * R * 6.5 / 100)] – [P + (P * R * 3 / 100)]
5040 = (P * R * 6.5 / 100) – (P * R * 3 / 100)
5040 = P * R / 100 * (6.5 – 3)
5040 = P * R / 100 * 3.5

This means the simple interest earned in 3.5 years is Rs 5040.
SI for 1 year = 5040 / 3.5 = 5040 / (7/2) = 5040 * 2 / 7 = 720 * 2 = 1440.
So, the simple interest earned per year is Rs 1440.

Now we can find the principal sum P using the first condition:
SI for 3 years = 1440 * 3 = 4320
13920 = P + 4320
P = 13920 – 4320 = 9600.
The principal sum is Rs 9600.

Now we can find the rate of simple interest R:
SI for 1 year = P * R * 1 / 100
1440 = 9600 * R / 100
1440 = 96 * R
R = 1440 / 96 = 15.
The rate of simple interest is 15% per annum.

Now, we need to find the total interest earned if the same sum (Rs 9600) had been invested for 2 years at the same rate (15% per annum) but with interest compounded every 6 months.
The rate of interest per compounding period is R/2 = 15/2 = 7.5% per half-year.
The number of compounding periods for 2 years is 2 * 2 = 4 periods.
The formula for the amount with compound interest is A = P(1 + r/n)^(nt), where r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the number of years.
Alternatively, A = P(1 + i)^N, where i is the rate per period and N is the number of periods.
Here, P = 9600, i = 7.5% = 0.075, N = 4.

Amount after 2 years = 9600 * (1 + 0.075)^4
Amount = 9600 * (1.075)^4
Calculate (1.075)^4:
1.075^2 = 1.155625
1.075^4 = (1.155625)^2 = 1.335469140625

Amount = 9600 * 1.335469140625
Amount = 12820.50375

Total interest earned = Amount – Principal
Total interest earned = 12820.50375 – 9600
Total interest earned = 3220.50375

The total interest earned is approximately Rs 3220.50.
Comparing this with the given options:
Option A: 3221
Option B: 3180
Option C: 3150
Option D: 3096

The nearest value to 3220.50 is 3221.

Correct_Option:A

Ans: D

Explanation:Let P be the principal amount of the loan, which is Rs 1000.
Let $I_1$ be the first installment paid at the end of the first year, which is Rs 530.
Let $I_2$ be the second installment paid at the end of the second year, which is Rs 594.
Let r be the annual rate of interest in percentage.

The present value of the installments must be equal to the principal amount.
The present value of the first installment is $I_1$ discounted back by one year. So, it is $\frac{I_1}{(1+r)}$.
The present value of the second installment is $I_2$ discounted back by two years. So, it is $\frac{I_2}{(1+r)^2}$.

Therefore, we have the equation:
$P = \frac{I_1}{(1+r)} + \frac{I_2}{(1+r)^2}$

Substitute the given values:
$1000 = \frac{530}{(1+r)} + \frac{594}{(1+r)^2}$

To solve for r, we can let $x = \frac{1}{(1+r)}$. The equation becomes:
$1000 = 530x + 594x^2$

Rearrange the equation into a quadratic form:
$594x^2 + 530x – 1000 = 0$

We can test the given options for r to find the correct value.

If r = 10% = 0.10, then $1+r = 1.10$.
$x = \frac{1}{1.10} \approx 0.90909$
Check if $594(0.90909)^2 + 530(0.90909) – 1000 = 0$
$594(0.82644) + 530(0.90909) – 1000$
$491.11 + 481.81 – 1000 = 972.92 – 1000 = -27.08 \neq 0$

If r = 11% = 0.11, then $1+r = 1.11$.
$x = \frac{1}{1.11} \approx 0.90090$
Check if $594(0.90090)^2 + 530(0.90090) – 1000 = 0$
$594(0.81162) + 530(0.90090) – 1000$
$482.12 + 477.47 – 1000 = 959.59 – 1000 = -40.41 \neq 0$

If r = 9% = 0.09, then $1+r = 1.09$.
$x = \frac{1}{1.09} \approx 0.91743$
Check if $594(0.91743)^2 + 530(0.91743) – 1000 = 0$
$594(0.84167) + 530(0.91743) – 1000$
$500.23 + 486.24 – 1000 = 986.47 – 1000 = -13.53 \neq 0$

If r = 8% = 0.08, then $1+r = 1.08$.
$x = \frac{1}{1.08} \approx 0.92593$
Check if $594(0.92593)^2 + 530(0.92593) – 1000 = 0$
$594(0.85735) + 530(0.92593) – 1000$
$509.44 + 490.74 – 1000 = 1000.18 – 1000 = 0.18 \approx 0$

Let’s re-check the calculation for r=8% more precisely.
$1+r = 1.08$
$x = \frac{1}{1.08}$
$1000 = 530 \times \frac{1}{1.08} + 594 \times \frac{1}{(1.08)^2}$
$1000 = \frac{530}{1.08} + \frac{594}{1.1664}$
$1000 = 490.7407… + 509.2592…$
$1000 = 1000$

The rate of interest is 8%.

Correct_Option:D

Ans: D

Explanation:Let the principal amount deposited in bank A be P_A = Rs 10000.
Let the investment period in bank A be T_A years.
The simple interest rate in bank A is R_A = 5% per annum.
The simple interest received from bank A is SI_A = (P_A * R_A * T_A) / 100.
SI_A = (10000 * 5 * T_A) / 100 = 500 * T_A.

On maturity, the total amount received from bank A is A_A = P_A + SI_A.
A_A = 10000 + 500 * T_A.

This amount A_A is deposited in bank B. So, the principal amount in bank B is P_B = A_A = 10000 + 500 * T_A.
The investment period in bank B is T_B = 5 years.
The simple interest rate in bank B is R_B = 6% per annum.
The simple interest received from bank B is SI_B = (P_B * R_B * T_B) / 100.
SI_B = ((10000 + 500 * T_A) * 6 * 5) / 100.
SI_B = ((10000 + 500 * T_A) * 30) / 100.
SI_B = (10000 + 500 * T_A) * 0.3.
SI_B = 3000 + 150 * T_A.

We are given that the interests received from bank A and bank B are in the ratio 10 : 13.
So, SI_A / SI_B = 10 / 13.
(500 * T_A) / (3000 + 150 * T_A) = 10 / 13.

Cross-multiply:
13 * (500 * T_A) = 10 * (3000 + 150 * T_A).
6500 * T_A = 30000 + 1500 * T_A.

Subtract 1500 * T_A from both sides:
6500 * T_A – 1500 * T_A = 30000.
5000 * T_A = 30000.

Divide by 5000:
T_A = 30000 / 5000.
T_A = 6 years.

Let’s check the options:
If T_A = 4 (Option A):
SI_A = 500 * 4 = 2000.
P_B = 10000 + 2000 = 12000.
SI_B = (12000 * 6 * 5) / 100 = 120 * 30 = 3600.
Ratio SI_A : SI_B = 2000 : 3600 = 20 : 36 = 5 : 9 (Not 10:13)

If T_A = 5 (Option B):
SI_A = 500 * 5 = 2500.
P_B = 10000 + 2500 = 12500.
SI_B = (12500 * 6 * 5) / 100 = 125 * 30 = 3750.
Ratio SI_A : SI_B = 2500 : 3750 = 250 : 375 = 10 : 15 = 2 : 3 (Not 10:13)

If T_A = 3 (Option C):
SI_A = 500 * 3 = 1500.
P_B = 10000 + 1500 = 11500.
SI_B = (11500 * 6 * 5) / 100 = 115 * 30 = 3450.
Ratio SI_A : SI_B = 1500 : 3450 = 150 : 345 = 30 : 69 = 10 : 23 (Not 10:13)

If T_A = 6 (Option D):
SI_A = 500 * 6 = 3000.
P_B = 10000 + 3000 = 13000.
SI_B = (13000 * 6 * 5) / 100 = 130 * 30 = 3900.
Ratio SI_A : SI_B = 3000 : 3900 = 30 : 39 = 10 : 13.

The calculation for T_A = 6 matches the given ratio.

Correct_Option:D

Ans: D

Let the sum contributed by Sunil be denoted by ‘S’.
The sum obtained by Anil after 6 years, assuming a compounded interest rate of 4% per annum, compounded semi-annually, can be calculated as:
$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$
The sum obtained by Sunil after 5 years, assuming the same compounded interest rate, would be:
$S\left(1.02\right)^{10}$
In the sixth year, Sunil then applies a simple interest rate of 10% to the sum he possesses at the end of 5 years. The interest earned during this year is:
$S\left(1.02\right)^{10}\times\ 0.1$
Thus, the total sum held by Sunil at the conclusion of 6 years is:
$S\left(1.02\right)^{10}\times\ \left(1+0.1\right)$
By setting the final sums held by Sunil and Anil equal, we establish the following equation:
$S\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$
Solving for S:
$S=\frac{22000\left(1.02\right)^2}{1.1}=20808$
Consequently, the correct option is D.

Ans: 9

Explanation:Let P be the principal amount invested, which is Rs 4000.
Let r be the annual rate of interest.
The value of the investment after t years, compounded annually, is given by $A(t) = P(1 + r)^t$.

We are given that the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36.
So, $\frac{A(3)}{A(5)} = \frac{P(1 + r)^3}{P(1 + r)^5} = \frac{25}{36}$.

Simplifying the equation:
$\frac{(1 + r)^3}{(1 + r)^5} = \frac{1}{(1 + r)^2} = \frac{25}{36}$.

Taking the reciprocal of both sides:
$(1 + r)^2 = \frac{36}{25}$.

Taking the square root of both sides:
$1 + r = \sqrt{\frac{36}{25}} = \frac{6}{5}$.
Since the interest rate must be positive, we take the positive square root.

Now, we can find the annual rate of interest:
$r = \frac{6}{5} – 1 = \frac{6 – 5}{5} = \frac{1}{5}$.
So, the rate of interest is $\frac{1}{5}$ or 20%.

The value of the investment after t years is given by $A(t) = 4000 \left(\frac{6}{5}\right)^t$.

We need to find the minimum number of years required for the value of the investment to exceed Rs 20000.
So, we need to find the smallest integer t such that $A(t) > 20000$.
$4000 \left(\frac{6}{5}\right)^t > 20000$.

Divide both sides by 4000:
$\left(\frac{6}{5}\right)^t > \frac{20000}{4000}$.
$\left(\frac{6}{5}\right)^t > 5$.

Now, we can test values of t:
For t = 1: $\left(\frac{6}{5}\right)^1 = 1.2$ (not > 5)
For t = 2: $\left(\frac{6}{5}\right)^2 = \left(\frac{36}{25}\right) = 1.44$ (not > 5)
For t = 3: $\left(\frac{6}{5}\right)^3 = \frac{216}{125} = 1.728$ (not > 5)
For t = 4: $\left(\frac{6}{5}\right)^4 = \frac{1296}{625} = 2.0736$ (not > 5)
For t = 5: $\left(\frac{6}{5}\right)^5 = \frac{7776}{3125} = 2.48832$ (not > 5)
For t = 6: $\left(\frac{6}{5}\right)^6 = \left(\frac{6}{5}\right)^5 \times \frac{6}{5} = 2.48832 \times 1.2 = 2.985984$ (not > 5)
For t = 7: $\left(\frac{6}{5}\right)^7 = \left(\frac{6}{5}\right)^6 \times \frac{6}{5} = 2.985984 \times 1.2 = 3.5831808$ (not > 5)
For t = 8: $\left(\frac{6}{5}\right)^8 = \left(\frac{6}{5}\right)^7 \times \frac{6}{5} = 3.5831808 \times 1.2 = 4.29981696$ (not > 5)
For t = 9: $\left(\frac{6}{5}\right)^9 = \left(\frac{6}{5}\right)^8 \times \frac{6}{5} = 4.29981696 \times 1.2 = 5.159780352$ ( > 5)

Alternatively, we can use logarithms:
$t \log\left(\frac{6}{5}\right) > \log(5)$
$t (\log(6) – \log(5)) > \log(5)$
$t (0.7781 – 0.6990) > 0.6990$
$t (0.0791) > 0.6990$
$t > \frac{0.6990}{0.0791} \approx 8.836$.

Since t must be an integer, the minimum number of years required is 9.

Final_Answer:9

Ans: 20808

Explanation:
Anil’s investment:
Principal (P_Anil) = Rs. 22000
Time (t_Anil) = 6 years
Rate of interest (r) = 4% per annum = 2% per half-year
Number of compounding periods (n_Anil) = 6 years * 2 = 12

The formula for compound interest is A = P(1 + r/100)^n.
Amount received by Anil (A_Anil) = 22000 * (1 + 2/100)^12
A_Anil = 22000 * (1.02)^12

Sunil’s investment:
Let Sunil’s initial investment be P_Sunil.
For the first 5 years:
Principal = P_Sunil
Time = 5 years
Rate of interest = 4% per annum compounded half-yearly
Number of compounding periods (n_Sunil1) = 5 years * 2 = 10
Amount received by Sunil at the end of 5 years (A_Sunil1) = P_Sunil * (1 + 2/100)^10
A_Sunil1 = P_Sunil * (1.02)^10

For the next 1 year (reinvestment):
Principal = A_Sunil1
Time = 1 year
Rate of simple interest = 10% per annum
Simple Interest (SI) = (Principal * Rate * Time) / 100
SI = (A_Sunil1 * 10 * 1) / 100
SI = A_Sunil1 * 0.10

Amount received by Sunil at the end of 6 years (A_Sunil2) = A_Sunil1 + SI
A_Sunil2 = A_Sunil1 + A_Sunil1 * 0.10
A_Sunil2 = A_Sunil1 * (1 + 0.10)
A_Sunil2 = A_Sunil1 * 1.10

Substitute the expression for A_Sunil1:
A_Sunil2 = (P_Sunil * (1.02)^10) * 1.10

Given that the amounts received by both at the end of 6 years are the same:
A_Anil = A_Sunil2
22000 * (1.02)^12 = P_Sunil * (1.02)^10 * 1.10

Now, we need to solve for P_Sunil:
P_Sunil = (22000 * (1.02)^12) / ((1.02)^10 * 1.10)
P_Sunil = (22000 * (1.02)^(12-10)) / 1.10
P_Sunil = (22000 * (1.02)^2) / 1.10
P_Sunil = (22000 * 1.0404) / 1.10
P_Sunil = 22888.8 / 1.10
P_Sunil = 20808

Final Answer: The initial investment made by Sunil is Rs. 20808.

Final_Answer:20808

Ans: B

Anil obtained a loan of Rs 2 lakhs with an annual interest rate of 8%, calculated semi-annually. He made a repayment of Rs 10320 after the first year and settled the entire loan at the conclusion of the third year.

The loan balance after one year is calculated as follows:
$200000\times\ \left(1 + \frac{8\%}{2}\right)^2 = 200000 \times (1.04)^2 = 216320$

After the first year’s repayment of Rs 10320, the remaining outstanding principal is Rs 206000 (216320 – 10320).

This remaining balance will accumulate interest for the subsequent two years. The total amount due at the end of the third year is:
$206000\times\ \left(1 + \frac{8\%}{2}\right)^4 = 206000 \times (1.04)^4 = 240990.86$

The interest accumulated during these two years amounts to Rs 34990.86 (240990.86 – 206000).

The total interest paid over the entire three-year period is the sum of the interest from the first year (Rs 16320, which is 216320 – 200000) and the interest from the subsequent two years (Rs 34990.86), totaling Rs 51311.

The correct option is B

Ans: A

Explanation:Let the first part of Alex’s savings be P1 and the second part be P2.
Let the interest rate for the first part be R1 = 15% per annum and the time period be T1 = 4 years.
Let the interest rate for the second part be R2 = 12% per annum and the time period be T2 = 3 years.

The simple interest (SI) earned on the first part is given by:
SI1 = (P1 * R1 * T1) / 100
SI1 = (P1 * 15 * 4) / 100
SI1 = (P1 * 60) / 100

The simple interest (SI) earned on the second part is given by:
SI2 = (P2 * R2 * T2) / 100
SI2 = (P2 * 12 * 3) / 100
SI2 = (P2 * 36) / 100

According to the problem, the simple interest earned on the first part is the same as the simple interest earned on the second part.
So, SI1 = SI2
(P1 * 60) / 100 = (P2 * 36) / 100

Multiplying both sides by 100, we get:
P1 * 60 = P2 * 36

Now, we need to find the ratio of P1 to P2.
P1 / P2 = 36 / 60

Simplify the fraction:
P1 / P2 = (12 * 3) / (12 * 5)
P1 / P2 = 3 / 5

This means that for every 3 units invested in the first part, 5 units are invested in the second part.
The total savings is P1 + P2.
The total number of units is 3 + 5 = 8 units.

The percentage of his savings invested in the first part is given by:
(P1 / (P1 + P2)) * 100

Since P1/P2 = 3/5, we can assume P1 = 3x and P2 = 5x for some value x.
Then P1 + P2 = 3x + 5x = 8x.

Percentage of savings in the first part = (3x / 8x) * 100
= (3 / 8) * 100
= 300 / 8
= 75 / 2
= 37.5%

Therefore, the percentage of his savings invested in the first part is 37.5%.

Let’s check the options:
Option A: 37.5%

Correct_Option:A

Ans: 20

Explanation:Let the initial capital of Mr. Pinto be C.
The capital invested at 6% is $\frac{1}{5}C$.
The capital invested at 10% is $\frac{1}{3}C$.
The remaining capital is $C – \frac{1}{5}C – \frac{1}{3}C = C \left(1 – \frac{1}{5} – \frac{1}{3}\right) = C \left(\frac{15-3-5}{15}\right) = \frac{7}{15}C$.
This remaining capital is invested at 1%.

Let the number of years be $n$.
The simple interest from the investment at 6% is $I_1 = \frac{\frac{1}{5}C \times 6 \times n}{100} = \frac{6Cn}{500}$.
The simple interest from the investment at 10% is $I_2 = \frac{\frac{1}{3}C \times 10 \times n}{100} = \frac{10Cn}{300}$.
The simple interest from the investment at 1% is $I_3 = \frac{\frac{7}{15}C \times 1 \times n}{100} = \frac{7Cn}{1500}$.

The cumulative interest income is $I_{total} = I_1 + I_2 + I_3$.
$I_{total} = \frac{6Cn}{500} + \frac{10Cn}{300} + \frac{7Cn}{1500}$
To add these fractions, we find a common denominator, which is 1500.
$I_{total} = \frac{6Cn \times 3}{500 \times 3} + \frac{10Cn \times 5}{300 \times 5} + \frac{7Cn}{1500}$
$I_{total} = \frac{18Cn}{1500} + \frac{50Cn}{1500} + \frac{7Cn}{1500}$
$I_{total} = \frac{18Cn + 50Cn + 7Cn}{1500}$
$I_{total} = \frac{75Cn}{1500}$
$I_{total} = \frac{Cn}{20}$

We want the cumulative interest income to equal or exceed his initial capital, so $I_{total} \ge C$.
$\frac{Cn}{20} \ge C$
Since C is the initial capital, $C > 0$. We can divide both sides by C.
$\frac{n}{20} \ge 1$
$n \ge 20$

The minimum number of years required is 20.

Final_Answer:20

Ans: B

Explanation:
Nitu’s initial capital is ₹20,000.

Investment in bank A: ₹8,000 at 5.5% simple interest per annum.
Annual interest from bank A = 8000 * (5.5/100) = 8000 * 0.055 = ₹440.

Investment in bank B: ₹5,000 at 5.6% simple interest per annum.
Annual interest from bank B = 5000 * (5.6/100) = 5000 * 0.056 = ₹280.

Remaining amount invested in bank C = Initial capital – Investment in A – Investment in B
Remaining amount = 20000 – 8000 – 5000 = ₹7,000.
This amount is invested at x% in bank C.
Annual interest from bank C = 7000 * (x/100) = 70x.

Her combined annual interest income is equal to 5% of the initial capital.
Combined annual interest income = 5% of 20000 = (5/100) * 20000 = ₹1,000.

The combined annual interest income is also the sum of interests from the three banks:
Interest from A + Interest from B + Interest from C = Combined interest
440 + 280 + 70x = 1000
720 + 70x = 1000
70x = 1000 – 720
70x = 280
x = 280 / 70
x = 4

So, the interest rate in bank C is 4% per annum.

Now, if she had invested her entire initial capital (₹20,000) in bank C alone at 4% per annum, her annual interest income would be:
Annual interest income from bank C alone = 20000 * (4/100) = 20000 * 0.04 = ₹800.

Correct_Option: B

Ans: C

Explanation:Let P be the principal amount and r be the annual rate of interest.
The interest accrued during the second year is the interest on the amount at the end of the first year.
Amount at the end of the first year = P(1 + r)
Interest accrued during the second year = P(1 + r) * r = ₹ 806.25 (Equation 1)

The interest accrued during the third year is the interest on the amount at the end of the second year.
Amount at the end of the second year = P(1 + r)^2
Interest accrued during the third year = P(1 + r)^2 * r = ₹ 866.72 (Equation 2)

We can find the rate of interest by dividing Equation 2 by Equation 1:
(P(1 + r)^2 * r) / (P(1 + r) * r) = 866.72 / 806.25
(1 + r) = 866.72 / 806.25
1 + r = 1.075
r = 1.075 – 1
r = 0.075 or 7.5%

Now we need to find the interest accrued during the fourth year. This is the interest on the amount at the end of the third year.
Amount at the end of the third year = P(1 + r)^3
Interest accrued during the fourth year = P(1 + r)^3 * r

We can also find the interest accrued during the fourth year by multiplying the interest accrued during the third year by (1 + r):
Interest accrued during the fourth year = Interest accrued during the third year * (1 + r)
Interest accrued during the fourth year = 866.72 * (1 + 0.075)
Interest accrued during the fourth year = 866.72 * 1.075
Interest accrued during the fourth year = 931.724

The interest accrued during the fourth year is approximately ₹ 931.72.

Comparing this with the given options:
Option A: 929.48
Option B: 934.65
Option C: 931.72
Option D: 926.84

The calculated value is nearest to Option C.

Correct_Option:C

Ans: B

Explanation:Let the amount invested by Raju in Bank B be P.
Let the period for which Raju invests in Bank B be T years.
Let the annual interest rate offered by Bank B be R% per annum.
Then, the annual interest rate offered by Bank C is 2R% per annum.

Raju invests P in Bank B for T years at R% simple interest.
Interest accrued to Raju from Bank B = (P * R * T) / 100

Raju invests the same amount P in Bank A for one year at 6% interest per annum compounded half-yearly.
For half-yearly compounding, the interest rate per period is 6%/2 = 3% = 0.03.
The number of periods in one year is 2.
Amount in Bank A after 1 year = P * (1 + 0.03)^2 = P * (1.03)^2 = P * 1.0609
Interest accrued to Raju from Bank A in 1 year = P * 1.0609 – P = 0.0609 * P

According to the problem, the interest accrued to Raju from Bank B is equal to the interest accrued from Bank A for one year.
(P * R * T) / 100 = 0.0609 * P
R * T = 6.09 (Equation 1)

Rupa invests ₹ 10,000 in Bank C for twice the period Raju invested in Bank B, which is 2T years.
The annual interest rate offered by Bank C is 2R% per annum.
Interest accrued to Rupa from Bank C = (10000 * 2R * 2T) / 100
Interest accrued to Rupa = (10000 * 4 * R * T) / 100
Interest accrued to Rupa = 400 * R * T

Substitute the value of R*T from Equation 1 into the expression for Rupa’s interest:
Interest accrued to Rupa = 400 * 6.09
Interest accrued to Rupa = 2436

Correct_Option:B

Ans: 12

Explanation:Let $V_P$ be Veeru’s principal and $V_R$ be Veeru’s interest rate.
Let $J_P$ be Joy’s principal and $J_R$ be Joy’s interest rate.

Veeru’s principal, $V_P = 10000$.
Veeru’s interest rate, $V_R = 5\% = 0.05$ per annum.
Let $t$ be the number of years after Veeru’s investment.

The simple interest earned by Veeru after $t$ years is $V_I = V_P \times V_R \times t = 10000 \times 0.05 \times t = 500t$.
The balance of Veeru’s investment after $t$ years is $V_B = V_P + V_I = 10000 + 500t$.

Joy invested Rs 8000 exactly after two years of Veeru’s investment.
Joy’s principal, $J_P = 8000$.
Joy’s interest rate, $J_R = 10\% = 0.10$ per annum.

Joy’s investment starts 2 years after Veeru’s investment. So, if Veeru’s investment has been for $t$ years, Joy’s investment has been for $t-2$ years, assuming $t \ge 2$.

The simple interest earned by Joy after $t-2$ years is $J_I = J_P \times J_R \times (t-2) = 8000 \times 0.10 \times (t-2) = 800(t-2)$.
The balance of Joy’s investment after $t$ years from Veeru’s investment is $J_B = J_P + J_I = 8000 + 800(t-2)$.

We need to find the time $t$ when their balances are equal, i.e., $V_B = J_B$.
$10000 + 500t = 8000 + 800(t-2)$
$10000 + 500t = 8000 + 800t – 1600$
$10000 + 500t = 6400 + 800t$

Now, we need to solve for $t$. Subtract $500t$ from both sides:
$10000 = 6400 + 800t – 500t$
$10000 = 6400 + 300t$

Subtract $6400$ from both sides:
$10000 – 6400 = 300t$
$3600 = 300t$

Divide by $300$:
$t = \frac{3600}{300}$
$t = 12$

So, after 12 years from Veeru’s investment, their balances will be equal. Let’s check the balances at $t=12$.
Veeru’s balance: $10000 + 500 \times 12 = 10000 + 6000 = 16000$.
Joy’s investment duration = $12 – 2 = 10$ years.
Joy’s balance: $8000 + 800 \times 10 = 8000 + 8000 = 16000$.
The balances are equal.

Final_Answer:12

Ans: 90000

Explanation:Let the principal amount be P.
Compound interest (CI) for two years at 5% per annum is given by:
CI = P * [(1 + R/100)^n – 1]
Here, R = 5% and n = 2.
CI = P * [(1 + 5/100)^2 – 1]
CI = P * [(1.05)^2 – 1]
CI = P * [1.1025 – 1]
CI = P * 0.1025

Simple interest (SI) for three years at 3% per annum is given by:
SI = (P * R * n) / 100
Here, R = 3% and n = 3.
SI = (P * 3 * 3) / 100
SI = (9 * P) / 100
SI = P * 0.09

According to the question, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125.
So, CI – SI = 1125
P * 0.1025 – P * 0.09 = 1125
P * (0.1025 – 0.09) = 1125
P * 0.0125 = 1125
P = 1125 / 0.0125
To simplify the division, we can multiply the numerator and denominator by 10000:
P = (1125 * 10000) / (0.0125 * 10000)
P = 11250000 / 125
Now, we can perform the division:
P = 90000

So, the principal amount is Rs 90000.

Let’s verify:
CI = 90000 * 0.1025 = 9225
SI = 90000 * 0.09 = 8100
CI – SI = 9225 – 8100 = 1125. This matches the given condition.

Final_Answer:90000

Ans: 16000

Explanation:Let P be the principal amount invested.
The annual interest rate is 10%.
The interest is compounded half-yearly, so the rate per compounding period is 10%/2 = 5% or 0.05.
The investment period is one and a half years, which is 1.5 years.
Since the compounding is half-yearly, the number of compounding periods is 1.5 years * 2 = 3 periods.

The formula for the amount A after n compounding periods is:
A = P * (1 + r)^n
where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
r = the interest rate per period
n = the number of periods

In this problem:
A = Rs. 18522
r = 5% = 0.05
n = 3

So, we have:
18522 = P * (1 + 0.05)^3
18522 = P * (1.05)^3

Now, we need to calculate (1.05)^3:
1.05 * 1.05 = 1.1025
1.1025 * 1.05 = 1.157625

So, the equation becomes:
18522 = P * 1.157625

To find P, we divide 18522 by 1.157625:
P = 18522 / 1.157625

P = 16000

Therefore, the amount the person had invested is Rs. 16000.

Final_Answer:16000

Ans: D

Explanation:Let the amounts invested by Amala, Bina, and Gouri be 3x, 4x, and 5x respectively.
Let the annual interest rates for Amala, Bina, and Gouri be 6y, 5y, and 4y respectively.

The interest income for each person after one year is calculated as:
Interest Income = Principal * Rate of Interest

Amala’s interest income = (3x) * (6y) = 18xy
Bina’s interest income = (4x) * (5y) = 20xy
Gouri’s interest income = (5x) * (4y) = 20xy

We are given that Bina’s interest income exceeds Amala’s by Rs 250.
Bina’s interest income – Amala’s interest income = 250
20xy – 18xy = 250
2xy = 250
xy = 125

The total interest income after a year is the sum of the interest incomes of Amala, Bina, and Gouri.
Total Interest Income = Amala’s interest income + Bina’s interest income + Gouri’s interest income
Total Interest Income = 18xy + 20xy + 20xy
Total Interest Income = 58xy

Substitute the value of xy = 125 into the total interest income equation:
Total Interest Income = 58 * 125

To calculate 58 * 125:
58 * 100 = 5800
58 * 25 = 58 * (100 / 4) = 5800 / 4 = 1450
Total Interest Income = 5800 + 1450 = 7250

Therefore, their total interest income after a year is Rs 7250.

Comparing this with the given options:
Option A: 6350
Option B: 6000
Option C: 7000
Option D: 7250

The calculated total interest income matches Option D.

Correct_Option:D

Ans: 9

Explanation:Let the total amount invested be $T = 15$ lakh.
Let the amount invested in fixed deposit be $x$ lakh.
The interest rate for the fixed deposit is $r_1 = 6\%$.

The remaining amount is $T – x = 15 – x$ lakh.
This remaining amount is invested in two other deposits in the ratio $2:1$.
Let the amounts invested in these two deposits be $2y$ and $y$ lakh.
So, $2y + y = 15 – x$
$3y = 15 – x$
$y = \frac{15-x}{3}$ lakh.

The first of these two deposits is $2y = 2 \times \frac{15-x}{3} = \frac{30-2x}{3}$ lakh, earning an interest rate of $r_2 = 4\%$.
The second of these two deposits is $y = \frac{15-x}{3}$ lakh, earning an interest rate of $r_3 = 3\%$.

The total annual interest income is Rs 76000.
We need to express the interest income in lakh rupees to match the invested amounts.
Total annual interest income = $\frac{76000}{100000}$ lakh = 0.76 lakh.

The interest earned from the fixed deposit is $I_1 = x \times \frac{6}{100}$.
The interest earned from the first of the other two deposits is $I_2 = \frac{30-2x}{3} \times \frac{4}{100}$.
The interest earned from the second of the other two deposits is $I_3 = \frac{15-x}{3} \times \frac{3}{100}$.

The total annual interest income is $I_1 + I_2 + I_3 = 0.76$ lakh.

$x \times \frac{6}{100} + \frac{30-2x}{3} \times \frac{4}{100} + \frac{15-x}{3} \times \frac{3}{100} = 0.76$

Multiply by 300 to clear the denominators:
$x \times 6 \times 3 + (30-2x) \times 4 + (15-x) \times 3 = 0.76 \times 300$
$18x + 120 – 8x + 45 – 3x = 228$
$7x + 165 = 228$
$7x = 228 – 165$
$7x = 63$
$x = \frac{63}{7}$
$x = 9$

The amount invested in the fixed deposit was Rs 9 lakh.

Let’s check the answer:
Amount in fixed deposit = 9 lakh (6% interest) -> Interest = $9 \times 0.06 = 0.54$ lakh.
Remaining amount = 15 – 9 = 6 lakh.
This 6 lakh is invested in ratio 2:1, so amounts are $\frac{2}{3} \times 6 = 4$ lakh and $\frac{1}{3} \times 6 = 2$ lakh.
First part = 4 lakh (4% interest) -> Interest = $4 \times 0.04 = 0.16$ lakh.
Second part = 2 lakh (3% interest) -> Interest = $2 \times 0.03 = 0.06$ lakh.
Total interest = $0.54 + 0.16 + 0.06 = 0.76$ lakh.
This matches the given total annual interest income of Rs 76000.

Final Answer: The amount (in Rs lakh) invested in the fixed deposit was 9.

Final_Answer:9

Ans: 20920

Explanation:
Let’s break down the calculations for Amal and Bimal.

**Amal’s Investments:**

Amal has two investments for one year.

* **Investment 1:** Rs 12000 at 8% compounded annually.
The formula for compound interest is A = P(1 + r/n)^(nt), where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
r = the annual interest rate (as a decimal)
n = the number of times that interest is compounded per year
t = the number of years the money is invested or borrowed for

Here, P = 12000, r = 0.08, n = 1, t = 1.
Amount after 1 year = 12000 * (1 + 0.08/1)^(1*1) = 12000 * (1.08) = 12960
Interest from Investment 1 = Amount – Principal = 12960 – 12000 = 960

* **Investment 2:** Rs 10000 at 6% compounded semi-annually.
Here, P = 10000, r = 0.06, n = 2 (semi-annually means twice a year), t = 1.
Amount after 1 year = 10000 * (1 + 0.06/2)^(2*1) = 10000 * (1 + 0.03)^2 = 10000 * (1.03)^2 = 10000 * 1.0609 = 10609
Interest from Investment 2 = Amount – Principal = 10609 – 10000 = 609

* **Total Interest for Amal:**
Total interest = Interest from Investment 1 + Interest from Investment 2
Total interest for Amal = 960 + 609 = 1569

**Bimal’s Investment:**

Bimal invests an unknown amount (let’s call it P_Bimal) at 7.5% simple interest for one year.
The formula for simple interest is I = P * r * t, where:
I = Simple Interest
P = Principal amount
r = Annual interest rate (as a decimal)
t = Time the money is invested or borrowed for, in years

Here, r = 0.075, t = 1.
Interest for Bimal = P_Bimal * 0.075 * 1 = 0.075 * P_Bimal

**Equating the Interests:**

The problem states that Amal and Bimal get the same amount of interest.
Interest for Amal = Interest for Bimal
1569 = 0.075 * P_Bimal

Now, we solve for P_Bimal:
P_Bimal = 1569 / 0.075
P_Bimal = 1569 / (75/1000)
P_Bimal = 1569 * (1000/75)
P_Bimal = 1569 * (40/3)
P_Bimal = (1569 / 3) * 40
P_Bimal = 523 * 40
P_Bimal = 20920

The amount invested by Bimal is Rs 20920.

Final_Answer:20920

Ans: 121000

Explanation:Let P be the principal amount borrowed, which is Rs. 2,10,000.
Let r be the annual interest rate, which is 10% or 0.10.
Let n be the number of equal installments, which is 2.
Let x be the amount of each equal installment.

After the first year, the interest accrued is P * r = 2,10,000 * 0.10 = Rs. 21,000.
The total amount due after the first year is P + Interest = 2,10,000 + 21,000 = Rs. 2,31,000.

The first installment of x is paid at the end of the first year. This installment covers the interest for the first year and a part of the principal.
Principal repaid in the first installment = x – Interest = x – 21,000.
The remaining principal amount after the first installment is P – (x – 21,000) = 2,10,000 – (x – 21,000) = 2,10,000 – x + 21,000 = 2,31,000 – x.

At the beginning of the second year, the outstanding principal is 2,31,000 – x.
The interest for the second year on this outstanding principal is (2,31,000 – x) * r = (2,31,000 – x) * 0.10 = 23,100 – 0.10x.

The second installment of x is paid at the end of the second year. This installment covers the remaining principal plus the interest for the second year.
So, the second installment is equal to the outstanding principal at the beginning of the second year plus the interest for the second year.
x = (2,31,000 – x) + (23,100 – 0.10x)
x = 2,31,000 – x + 23,100 – 0.10x
x = 2,54,100 – 1.10x
x + 1.10x = 2,54,100
2.10x = 2,54,100
x = 2,54,100 / 2.10
x = 25,41,000 / 21
x = 1,21,000

Let’s verify the installments:
First installment = Rs. 1,21,000.
Interest for the first year = Rs. 21,000.
Principal repaid in the first installment = 1,21,000 – 21,000 = Rs. 1,00,000.
Remaining principal after the first installment = 2,10,000 – 1,00,000 = Rs. 1,10,000.

Second installment = Rs. 1,21,000.
Interest for the second year on the remaining principal = 1,10,000 * 0.10 = Rs. 11,000.
Principal repaid in the second installment = Second installment – Interest for the second year = 1,21,000 – 11,000 = Rs. 1,10,000.
This matches the remaining principal.

Final_Answer:121000

Ans: 4000

Explanation:Let X be the amount Gopal borrows from Ankit.
Let Y be the amount Gopal adds from his own money.

Case 1:
Gopal borrows X from Ankit at 8% annual interest. Interest paid to Ankit = 0.08 * X.
Gopal lends X + Y to Ishan at 10% annual interest. Interest received from Ishan = 0.10 * (X + Y).
Net interest retained by Gopal = Interest received – Interest paid
Net interest = 0.10 * (X + Y) – 0.08 * X
Net interest = 0.10X + 0.10Y – 0.08X
Net interest = 0.02X + 0.10Y

The problem states that the net interest retained by Gopal is the same as that accrued to Ankit.
Interest accrued to Ankit = 0.08 * X.
So, 0.02X + 0.10Y = 0.08X
0.10Y = 0.08X – 0.02X
0.10Y = 0.06X
Multiplying by 100, we get:
10Y = 6X
Dividing by 2, we get:
5Y = 3X (Equation 1)

Case 2:
Gopal lends X + 2Y to Ishan at 10% annual interest. Interest received from Ishan = 0.10 * (X + 2Y).
Gopal still borrows X from Ankit at 8% annual interest. Interest paid to Ankit = 0.08 * X.
Net interest retained by Gopal = Interest received – Interest paid
Net interest = 0.10 * (X + 2Y) – 0.08 * X
Net interest = 0.10X + 0.20Y – 0.08X
Net interest = 0.02X + 0.20Y

The net interest retained by him would have increased by Rs. 150 compared to Case 1.
So, (0.02X + 0.20Y) – (0.02X + 0.10Y) = 150
0.02X + 0.20Y – 0.02X – 0.10Y = 150
0.10Y = 150
Multiplying by 10, we get:
Y = 1500

Now, substitute the value of Y into Equation 1:
5Y = 3X
5 * 1500 = 3X
7500 = 3X
X = 7500 / 3
X = 2500

We need to find the value of X + Y.
X + Y = 2500 + 1500
X + Y = 4000

Final Answer:4000