Compound Interest: Concepts & Shortcuts
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
Scenario-Based Learning for Competitive Exams
By Wisdom Helps Mahesh Sir
Q1. Rahul invested ₹2,000 in a fixed deposit. If the bank offers 10% per annum compounded annually, after how many years will he receive a total maturity amount of ₹2,420?
Detailed Solution:
Given: Principal ($P$) = ₹2,000, Amount ($A$) = ₹2,420, Rate ($R$) = 10% p.a.
Using the formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 2420 = 2000\left(1 + \frac{10}{100}\right)^n$$ $$\Rightarrow \frac{2420}{2000} = (1.1)^n \Rightarrow \frac{121}{100} = (1.1)^n$$ $$\Rightarrow 1.21 = (1.1)^n$$ Since $(1.1)^2 = 1.21$, comparing both sides gives n = 2 years.
✅ Correct Option: (C)
Given: Principal ($P$) = ₹2,000, Amount ($A$) = ₹2,420, Rate ($R$) = 10% p.a.
Using the formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 2420 = 2000\left(1 + \frac{10}{100}\right)^n$$ $$\Rightarrow \frac{2420}{2000} = (1.1)^n \Rightarrow \frac{121}{100} = (1.1)^n$$ $$\Rightarrow 1.21 = (1.1)^n$$ Since $(1.1)^2 = 1.21$, comparing both sides gives n = 2 years.
✅ Correct Option: (C)
Q2. Anjali's savings account balance grew from ₹1,00,00 to ₹1,331. If the bank pays 10% p.a. compound interest annually, find the duration of her investment?
Detailed Solution:
Given: Principal ($P$) = ₹1,000, Amount ($A$) = ₹1,331, Rate ($R$) = 10% p.a.
Using the formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 1331 = 1000\left(1 + \frac{10}{100}\right)^n$$ $$\Rightarrow \frac{1331}{1000} = (1.1)^n \Rightarrow 1.331 = (1.1)^n$$ Since $(1.1)^3 = 1.331$, comparing the exponents gives n = 3 years.
✅ Correct Option: (B)
Given: Principal ($P$) = ₹1,000, Amount ($A$) = ₹1,331, Rate ($R$) = 10% p.a.
Using the formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 1331 = 1000\left(1 + \frac{10}{100}\right)^n$$ $$\Rightarrow \frac{1331}{1000} = (1.1)^n \Rightarrow 1.331 = (1.1)^n$$ Since $(1.1)^3 = 1.331$, comparing the exponents gives n = 3 years.
✅ Correct Option: (B)
Q3. A farmer borrowed a certain sum from a cooperative bank at 4% p.a. CI. If he cleared his debt by paying ₹270.40 after 2 years, what was the original sum borrowed?
Detailed Solution:
Given: Amount ($A$) = ₹270.40, Rate ($R$) = 4% p.a., Time ($n$) = 2 years.
Using the standard formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 270.40 = P\left(1 + \frac{4}{100}\right)^2$$ $$\Rightarrow 270.40 = P(1.04)^2 \Rightarrow 270.40 = P \times 1.0816$$ $$\Rightarrow P = \frac{270.40}{1.0816} = \mathbf{₹250}$$ ✅ Correct Option: (C)
Given: Amount ($A$) = ₹270.40, Rate ($R$) = 4% p.a., Time ($n$) = 2 years.
Using the standard formula: $A = P\left(1 + \frac{R}{100}\right)^n$
$$\Rightarrow 270.40 = P\left(1 + \frac{4}{100}\right)^2$$ $$\Rightarrow 270.40 = P(1.04)^2 \Rightarrow 270.40 = P \times 1.0816$$ $$\Rightarrow P = \frac{270.40}{1.0816} = \mathbf{₹250}$$ ✅ Correct Option: (C)
Q4. A business investment grows to ₹9,680 at the end of 2 years and ₹10,648 at the end of 3 years. What is the annual rate of compound interest being applied?
Detailed Solution:
In compound interest, the amount at the end of the 2nd year acts as the principal for the 3rd year.
Interest earned in the 3rd year = $\text{Amount after 3 years} - \text{Amount after 2 years}$
$$\Rightarrow \text{Interest} = 10648 - 9680 = ₹968$$ Now, calculate the rate of interest on the 2nd year's amount:
$$\text{Rate } (R) = \frac{\text{Interest}}{\text{Principal for that year}} \times 100$$ $$R = \frac{968}{9680} \times 100 = \mathbf{10\%}$$ ✅ Correct Option: (B)
In compound interest, the amount at the end of the 2nd year acts as the principal for the 3rd year.
Interest earned in the 3rd year = $\text{Amount after 3 years} - \text{Amount after 2 years}$
$$\Rightarrow \text{Interest} = 10648 - 9680 = ₹968$$ Now, calculate the rate of interest on the 2nd year's amount:
$$\text{Rate } (R) = \frac{\text{Interest}}{\text{Principal for that year}} \times 100$$ $$R = \frac{968}{9680} \times 100 = \mathbf{10\%}$$ ✅ Correct Option: (B)
Q5. A jewelry shop owner finds that a gold deposit worth ₹2,304 appreciates to ₹2,500 in exactly 2 years under compound growth. Find the rate percent per annum?
Detailed Solution:
Given: Initial value ($P$) = ₹2,304, Final value ($A$) = ₹2,500, Time ($n$) = 2 years.
$$A = P\left(1 + \frac{R}{100}\right)^2 \Rightarrow \frac{2500}{2304} = \left(1 + \frac{R}{100}\right)^2$$ Taking square root on both sides:
$$\sqrt{\frac{2500}{2304}} = 1 + \frac{R}{100} \Rightarrow \frac{50}{48} = 1 + \frac{R}{100}$$ $$\Rightarrow \frac{R}{100} = \frac{50}{48} - 1 = \frac{2}{48} = \frac{1}{24}$$ $$\Rightarrow R = \frac{100}{24} = \frac{25}{6} = \mathbf{4 \frac{1}{6}\%}$$ ✅ Correct Option: (C)
Given: Initial value ($P$) = ₹2,304, Final value ($A$) = ₹2,500, Time ($n$) = 2 years.
$$A = P\left(1 + \frac{R}{100}\right)^2 \Rightarrow \frac{2500}{2304} = \left(1 + \frac{R}{100}\right)^2$$ Taking square root on both sides:
$$\sqrt{\frac{2500}{2304}} = 1 + \frac{R}{100} \Rightarrow \frac{50}{48} = 1 + \frac{R}{100}$$ $$\Rightarrow \frac{R}{100} = \frac{50}{48} - 1 = \frac{2}{48} = \frac{1}{24}$$ $$\Rightarrow R = \frac{100}{24} = \frac{25}{6} = \mathbf{4 \frac{1}{6}\%}$$ ✅ Correct Option: (C)
Q6. Vikram needs ₹1,352 to buy a gadget after 2 years. If his bank offers 4% p.a. compound interest, how much should he deposit today?
Detailed Solution:
Given: Target Amount ($A$) = ₹1,352, Rate ($R$) = 4% p.a., Time ($n$) = 2 years.
$$A = P\left(1 + \frac{R}{100}\right)^n \Rightarrow 1352 = P(1.04)^2$$ $$\Rightarrow 1352 = P \times 1.0816$$ $$\Rightarrow P = \frac{1352}{1.0816} = \mathbf{₹1250}$$ ✅ Correct Option: (D)
Given: Target Amount ($A$) = ₹1,352, Rate ($R$) = 4% p.a., Time ($n$) = 2 years.
$$A = P\left(1 + \frac{R}{100}\right)^n \Rightarrow 1352 = P(1.04)^2$$ $$\Rightarrow 1352 = P \times 1.0816$$ $$\Rightarrow P = \frac{1352}{1.0816} = \mathbf{₹1250}$$ ✅ Correct Option: (D)
Q7. The compound interest on a sum for 2 years at 4% is ₹102. What would be the Simple Interest on the same sum at the same rate and for the same period?
Detailed Solution:
Method 1: Effective Interest Rate
Effective Simple Interest (SI) rate for 2 years = $4\% + 4\% = 8\%$
Effective Compound Interest (CI) rate for 2 years = $4 + 4 + \frac{4 \times 4}{100} = 8.16\%$
Given that the CI value is ₹102:
$$\Rightarrow 8.16\% \text{ of Principal} = 102$$ $$\Rightarrow \text{Principal} = \frac{102}{8.16} \times 100 = ₹1250$$ Now, calculate SI on ₹1,250 for 2 years at 4%:
$$\text{SI} = \frac{1250 \times 4 \times 2}{100} = \mathbf{₹100}$$ ✅ Correct Option: (D)
Method 1: Effective Interest Rate
Effective Simple Interest (SI) rate for 2 years = $4\% + 4\% = 8\%$
Effective Compound Interest (CI) rate for 2 years = $4 + 4 + \frac{4 \times 4}{100} = 8.16\%$
Given that the CI value is ₹102:
$$\Rightarrow 8.16\% \text{ of Principal} = 102$$ $$\Rightarrow \text{Principal} = \frac{102}{8.16} \times 100 = ₹1250$$ Now, calculate SI on ₹1,250 for 2 years at 4%:
$$\text{SI} = \frac{1250 \times 4 \times 2}{100} = \mathbf{₹100}$$ ✅ Correct Option: (D)
Q8. A financial consultant tells you that the difference between SI and CI on your principal for 2 years at 4% is ₹4. What is the value of your principal?
Detailed Solution:
Using the direct shortcut formula for the difference between CI and SI for 2 years:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹4$, $R = 4\%$
$$\Rightarrow 4 = P\left(\frac{4}{100}\right)^2$$ $$\Rightarrow 4 = P \times \frac{16}{10000}$$ $$\Rightarrow P = \frac{4 \times 10000}{16} = \frac{40000}{16} = \mathbf{₹2500}$$ ✅ Correct Option: (A)
Using the direct shortcut formula for the difference between CI and SI for 2 years:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹4$, $R = 4\%$
$$\Rightarrow 4 = P\left(\frac{4}{100}\right)^2$$ $$\Rightarrow 4 = P \times \frac{16}{10000}$$ $$\Rightarrow P = \frac{4 \times 10000}{16} = \frac{40000}{16} = \mathbf{₹2500}$$ ✅ Correct Option: (A)
Q9. In a loan scheme, the difference between CI and SI for 2 years at 8% per annum is ₹768. Find the total sum involved in this scheme.
Detailed Solution:
Using the standard 2-year difference formula:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹768$, $R = 8\%$
$$\Rightarrow 768 = P\left(\frac{8}{100}\right)^2$$ $$\Rightarrow 768 = P \times \frac{64}{10000}$$ $$\Rightarrow P = \frac{768 \times 10000}{64}$$ Since $\frac{768}{64} = 12$, we get:
$$\Rightarrow P = 12 \times 10000 = \mathbf{₹1,20,000}$$ ✅ Correct Option: (C)
Using the standard 2-year difference formula:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹768$, $R = 8\%$
$$\Rightarrow 768 = P\left(\frac{8}{100}\right)^2$$ $$\Rightarrow 768 = P \times \frac{64}{10000}$$ $$\Rightarrow P = \frac{768 \times 10000}{64}$$ Since $\frac{768}{64} = 12$, we get:
$$\Rightarrow P = 12 \times 10000 = \mathbf{₹1,20,000}$$ ✅ Correct Option: (C)
Q10. For a long-term bond, the difference between CI and SI for 3 years at 5% p.a. is calculated to be ₹15.25. What is the face value (principal) of the bond?
Detailed Solution:
Using the shortcut formula for the difference between CI and SI for exactly 3 years:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2 \times \left(\frac{300 + R}{100}\right)$$ Given: $D = ₹15.25$, $R = 5\%$
$$\Rightarrow 15.25 = P\left(\frac{5}{100}\right)^2 \times \left(\frac{300 + 5}{100}\right)$$ $$\Rightarrow 15.25 = P\left(\frac{1}{400}\right) \times \left(\frac{305}{100}\right)$$ $$\Rightarrow 15.25 = P \times \frac{305}{40000}$$ $$\Rightarrow P = \frac{15.25 \times 40000}{305} = \frac{610000}{305} = \mathbf{₹2,000}$$ ✅ Correct Option: (A)
Using the shortcut formula for the difference between CI and SI for exactly 3 years:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2 \times \left(\frac{300 + R}{100}\right)$$ Given: $D = ₹15.25$, $R = 5\%$
$$\Rightarrow 15.25 = P\left(\frac{5}{100}\right)^2 \times \left(\frac{300 + 5}{100}\right)$$ $$\Rightarrow 15.25 = P\left(\frac{1}{400}\right) \times \left(\frac{305}{100}\right)$$ $$\Rightarrow 15.25 = P \times \frac{305}{40000}$$ $$\Rightarrow P = \frac{15.25 \times 40000}{305} = \frac{610000}{305} = \mathbf{₹2,000}$$ ✅ Correct Option: (A)
Q11. A real estate property doubles its value every 5 years under compound growth. In how many years will the value of the property become 8 times the original price?
Detailed Solution:
Let the initial value be $1$.
It becomes 2 times ($2^1$) in 5 years.
We need it to become 8 times, which can be written in powers of 2 as $8 = 2^3$.
Shortcut Formula: If a sum becomes $x^1$ times in $T$ years, it becomes $x^m$ times in $T \times m$ years.
Here, $x = 2$, $T = 5$ years, and $m = 3$.
$$\text{Total Time} = 5 \times 3 = \mathbf{15\text{ years}}$$ ✅ Correct Option: (B)
Let the initial value be $1$.
It becomes 2 times ($2^1$) in 5 years.
We need it to become 8 times, which can be written in powers of 2 as $8 = 2^3$.
Shortcut Formula: If a sum becomes $x^1$ times in $T$ years, it becomes $x^m$ times in $T \times m$ years.
Here, $x = 2$, $T = 5$ years, and $m = 3$.
$$\text{Total Time} = 5 \times 3 = \mathbf{15\text{ years}}$$ ✅ Correct Option: (B)
Q12. A stock market index doubles itself in 4 years. If it continues to grow at the same compound rate, how many years will it take to quadruple (4 times)?
Detailed Solution:
The index becomes 2 times ($2^1$) its value in 4 years.
We want to find out when it quadruples, which means becoming 4 times ($4 = 2^2$) its value.
Using the shortcut rule: Multiply the initial time cycle by the power exponent.
$$\text{Required Time} = \text{Initial Time } (T) \times \text{Power component } (m)$$ $$\text{Required Time} = 4 \times 2 = \mathbf{8\text{ years}}$$ ✅ Correct Option: (C)
The index becomes 2 times ($2^1$) its value in 4 years.
We want to find out when it quadruples, which means becoming 4 times ($4 = 2^2$) its value.
Using the shortcut rule: Multiply the initial time cycle by the power exponent.
$$\text{Required Time} = \text{Initial Time } (T) \times \text{Power component } (m)$$ $$\text{Required Time} = 4 \times 2 = \mathbf{8\text{ years}}$$ ✅ Correct Option: (C)
Q13. Mrs. Sharma borrowed ₹2,550 at 4% p.a. CI to be paid back in 2 equal yearly instalments. What is the value of each instalment?
Detailed Solution:
Rate ($R$) = 4% = $\frac{4}{100} = \frac{1}{25}$. This means a principal unit of 25 grows to a maturity unit of 26 ($25+1$) in a year.
Let the equal annual installment be $x$.
The formula for the present value of 2 installments under CI is:
$$\text{Principal } (P) = \frac{x}{\left(1 + \frac{R}{100}\right)^1} + \frac{x}{\left(1 + \frac{R}{100}\right)^2}$$ $$\Rightarrow 2550 = \frac{x}{\frac{26}{25}} + \frac{x}{\left(\frac{26}{25}\right)^2}$$ $$\Rightarrow 2550 = \frac{25x}{26} + \frac{625x}{676}$$ Take the LCM of 26 and 676, which is 676:
$$\Rightarrow 2550 = \frac{25 \times 26x + 625x}{676}$$ $$\Rightarrow 2550 = \frac{650x + 625x}{676} \Rightarrow 2550 = \frac{1275x}{676}$$ $$\Rightarrow x = \frac{2550 \times 676}{1275} = 2 \times 676 = \mathbf{₹1352}$$ ✅ Correct Option: (A)
Rate ($R$) = 4% = $\frac{4}{100} = \frac{1}{25}$. This means a principal unit of 25 grows to a maturity unit of 26 ($25+1$) in a year.
Let the equal annual installment be $x$.
The formula for the present value of 2 installments under CI is:
$$\text{Principal } (P) = \frac{x}{\left(1 + \frac{R}{100}\right)^1} + \frac{x}{\left(1 + \frac{R}{100}\right)^2}$$ $$\Rightarrow 2550 = \frac{x}{\frac{26}{25}} + \frac{x}{\left(\frac{26}{25}\right)^2}$$ $$\Rightarrow 2550 = \frac{25x}{26} + \frac{625x}{676}$$ Take the LCM of 26 and 676, which is 676:
$$\Rightarrow 2550 = \frac{25 \times 26x + 625x}{676}$$ $$\Rightarrow 2550 = \frac{650x + 625x}{676} \Rightarrow 2550 = \frac{1275x}{676}$$ $$\Rightarrow x = \frac{2550 \times 676}{1275} = 2 \times 676 = \mathbf{₹1352}$$ ✅ Correct Option: (A)
Q14. In a private lending agreement, the difference between CI and SI at 10% for 2 years is ₹464. Find the total sum lent?
Detailed Solution:
Using the 2-year difference formula for simple and compound interest:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹464$, $R = 10\%$
$$\Rightarrow 464 = P\left(\frac{10}{100}\right)^2$$ $$\Rightarrow 464 = P\left(\frac{1}{10}\right)^2 \Rightarrow 464 = P \times \frac{1}{100}$$ $$\Rightarrow P = 464 \times 100 = \mathbf{₹46,400}$$ ✅ Correct Option: (A)
Using the 2-year difference formula for simple and compound interest:
$$\text{Difference } (D) = P\left(\frac{R}{100}\right)^2$$ Given: $D = ₹464$, $R = 10\%$
$$\Rightarrow 464 = P\left(\frac{10}{100}\right)^2$$ $$\Rightarrow 464 = P\left(\frac{1}{10}\right)^2 \Rightarrow 464 = P \times \frac{1}{100}$$ $$\Rightarrow P = 464 \times 100 = \mathbf{₹46,400}$$ ✅ Correct Option: (A)
Q15. Find the compound interest on a sum of ₹16,000 for 9 months at 20% p.a., if the interest is payable quarterly?
Detailed Solution:
When interest is compounded quarterly:
* New Rate ($R'$) = $\frac{\text{Annual Rate}}{4} = \frac{20\%}{4} = 5\% \text{ per quarter}$
* Time periods ($n$) = $\frac{9 \text{ months}}{3 \text{ months}} = 3 \text{ quarters}$
Now apply the amount formula with these updated variables:
$$A = P\left(1 + \frac{R'}{100}\right)^n \Rightarrow A = 16000\left(1 + \frac{5}{100}\right)^3$$ $$\Rightarrow A = 16000 \times (1.05)^3 = 16000 \times 1.157625 = ₹18,522$$ Now, extract the Compound Interest (CI):
$$\text{CI} = \text{Amount } (A) - \text{Principal } (P)$$ $$\text{CI} = 18522 - 16000 = \mathbf{₹2522}$$ ✅ Correct Option: (C)
When interest is compounded quarterly:
* New Rate ($R'$) = $\frac{\text{Annual Rate}}{4} = \frac{20\%}{4} = 5\% \text{ per quarter}$
* Time periods ($n$) = $\frac{9 \text{ months}}{3 \text{ months}} = 3 \text{ quarters}$
Now apply the amount formula with these updated variables:
$$A = P\left(1 + \frac{R'}{100}\right)^n \Rightarrow A = 16000\left(1 + \frac{5}{100}\right)^3$$ $$\Rightarrow A = 16000 \times (1.05)^3 = 16000 \times 1.157625 = ₹18,522$$ Now, extract the Compound Interest (CI):
$$\text{CI} = \text{Amount } (A) - \text{Principal } (P)$$ $$\text{CI} = 18522 - 16000 = \mathbf{₹2522}$$ ✅ Correct Option: (C)
Q16. What sum of money, if invested today at 6% p.a. compound interest, will grow to exactly ₹5,618 in 2 years?
Detailed Solution:
Given: Amount ($A$) = ₹5,618, Rate ($R$) = 6% p.a., Time ($T$) = 2 years.
$$A = P \left(1 + \frac{R}{100}\right)^T \Rightarrow 5618 = P \left(1 + \frac{6}{100}\right)^2$$ $$\Rightarrow 5618 = P (1.06)^2 \Rightarrow 5618 = P \times 1.1236$$ $$\Rightarrow P = \frac{5618}{1.1236} = \mathbf{₹5000}$$ ✅ Correct Option: (B)
Given: Amount ($A$) = ₹5,618, Rate ($R$) = 6% p.a., Time ($T$) = 2 years.
$$A = P \left(1 + \frac{R}{100}\right)^T \Rightarrow 5618 = P \left(1 + \frac{6}{100}\right)^2$$ $$\Rightarrow 5618 = P (1.06)^2 \Rightarrow 5618 = P \times 1.1236$$ $$\Rightarrow P = \frac{5618}{1.1236} = \mathbf{₹5000}$$ ✅ Correct Option: (B)
Q17. Ashok has ₹1612 with him. He divided it amongst his sons Raj and Varun and asked them to invest it at 8% rate of interest compounded annually. It was seen that Raj and Varun got same amount after 14 and 15 years respectively. How much (in ₹) did Ashok give to Raj? (Most Important Problem) (SSC, Railway, Bank)
⚡ Shortcut Trick (Very Important)
When the rate is the same, final maturity amounts become equal, and the compounding time differs by exactly 1 year:
Step 1: Find the share ratio formula
$$\text{Ratio of investments (Lower Time : Higher Time)} = \left(1 + \frac{R}{100}\right) : 1$$ $$= \left(1 + \frac{8}{100}\right) : 1 = 1.08 : 1 = 108 : 100 = 27 : 25$$ Since Raj's money stays in the bank for less time (14 years) but ends up equal to Varun's (15 years), Raj must have been given the larger initial share.
Step 2: Calculate the Shares
Total parts = $27 + 25 = 52\text{ units}$.
$$\text{Value of 52 units} = ₹1612 \Rightarrow 1\text{ unit} = \frac{1612}{52} = ₹31$$ $$\text{Raj's Share (27 units)} = 27 \times 31 = \mathbf{₹837}$$ $$\text{Varun's Share (25 units)} = 25 \times 31 = ₹775$$ ✅ Correct Option: (B)
When the rate is the same, final maturity amounts become equal, and the compounding time differs by exactly 1 year:
Step 1: Find the share ratio formula
$$\text{Ratio of investments (Lower Time : Higher Time)} = \left(1 + \frac{R}{100}\right) : 1$$ $$= \left(1 + \frac{8}{100}\right) : 1 = 1.08 : 1 = 108 : 100 = 27 : 25$$ Since Raj's money stays in the bank for less time (14 years) but ends up equal to Varun's (15 years), Raj must have been given the larger initial share.
Step 2: Calculate the Shares
Total parts = $27 + 25 = 52\text{ units}$.
$$\text{Value of 52 units} = ₹1612 \Rightarrow 1\text{ unit} = \frac{1612}{52} = ₹31$$ $$\text{Raj's Share (27 units)} = 27 \times 31 = \mathbf{₹837}$$ $$\text{Varun's Share (25 units)} = 25 \times 31 = ₹775$$ ✅ Correct Option: (B)
Q18. Gopal invests a sum of ₹5400 and Akshay invests a sum of ₹10200 at the same rate of simple interest per annum. If, at the end of 4 years, Akshay gets ₹720 more interest than Gopal, then find the rate of interest per annum.
⚡ Ultra Shortcut Method (Exam Trick)
Instead of writing separate long equations for both interests, look directly at the extra principal amount that creates the extra interest output.
Step 1: Find the difference in invested principals
$$\text{Difference in P } (\Delta P) = 10200 - 5400 = ₹4800$$
Step 2: Apply the simple interest difference formula
$$\text{Extra Interest} = \frac{\Delta P \times \text{Rate} \times \text{Time}}{100}$$
$$720 = \frac{4800 \times R \times 4}{100}$$
$$720 = 192 \times R \Rightarrow R = \frac{720}{192} = \mathbf{3.75\%}$$
✅ Correct Option: 1
Q19. Mahesh has two grandsons Kiran and Tushar. 11 year old Kiran gets some money from Mahesh’s wealth and 12 year old Tushar gets rest of the money. Both will get money only when they turn 22 years old. Till then the money is in a bank getting interest at 8% compounded annually. When both turn 22, they receive the same amount. How much had Mahesh given Tushar (in ₹) initially, if total money was ₹24700? (Very Important)
📌 Given Values:
* Target maturity age = 22 years old
* Kiran’s current age = 11 $\Rightarrow$ Time inside bank ($n_1$) = $22 - 11 = 11\text{ years}$
* Tushar’s current age = 12 $\Rightarrow$ Time inside bank ($n_2$) = $22 - 12 = 10\text{ years}$
* Interest Rate ($R$) = 8% p.a. | Combined principal pool = ₹24,700
📘 Standard Algebraic Method:
Let Kiran's initial assignment share be $x$. Tushar's share becomes $(24700 - x)$.
$$x\left(1 + \frac{8}{100}\right)^{11} = (24700 - x)\left(1 + \frac{8}{100}\right)^{10}$$
Divide both paths by $\left(1.08\right)^{10}$:
$$x(1.08)^1 = 24700 - x \Rightarrow 1.08x + x = 24700$$
$$2.08x = 24700 \Rightarrow x = \frac{24700}{2.08} = 11875\text{ (Kiran's Share)}$$
$$\text{Tushar's Share} = 24700 - 11875 = \mathbf{12825}$$
⚡ Fast Shortcut Trick:
The time gap between them is exactly 1 year ($11 - 10$). The grandson with less investment time (Tushar) must receive a larger base share to catch up to the same amount.
$$\text{Ratio (Kiran : Tushar)} = 1 : \left(1 + \frac{R}{100}\right) = 1 : 1.08 = 100 : 108$$
$$\text{Total parts} = 100 + 108 = 208\text{ units}$$
$$\text{Value of 1 part} = \frac{24700}{208} = 118.75$$
$$\text{Tushar's share} = 108 \times 118.75 = \mathbf{12825}$$
✅ Correct Option: B
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