Rs. $x$ is invested in scheme A at compound interest rate of $r\%$ per annum compounded annually for $t$ years. The same amount is invested in scheme B at compound interest rate of $p\%$ per annum compounded half-yearly for $t$ years. If the interest earned from both schemes is equal, which of the following pairs of $(r, p)$ (both integers) satisfy this condition?

(i) $(20, 24)$

(ii) $(24, 30)$

(iii) $(44, 40)$

(iv) $(21, 20)$

(v) $(12, 16)$

Question

Rs. $x$ is invested in scheme A at compound interest rate of $r\%$ per annum compounded annually for $t$ years. The same amount is invested in scheme B at compound interest rate of $p\%$ per annum compounded half-yearly for $t$ years. If the interest earned from both schemes is equal, which of the following pairs of $(r, p)$ (both integers) satisfy this condition?

(i) $(20, 24)$

(ii) $(24, 30)$

(iii) $(44, 40)$

(iv) $(21, 20)$

(v) $(12, 16)$

Accepted Answer

**Solution:**

1. Let the principal be $x$ and time be $t$ years.

2. For scheme A (compounded annually at rate $r\%$), the amount after $t$ years is:
   $$A_A = x 	imes \left(1 + \frac{r}{100}ight)^t$$

3. For scheme B (compounded half-yearly at rate $p\%$ per annum), the half-yearly rate is $\frac{p}{2}\%$, and the number of half-year periods is $2t$. So the amount after $t$ years is:
   $$A_B = x 	imes \left(1 + \frac{p}{200}ight)^{2t}$$

4. Given the interest earned from both schemes is equal, so the amounts are equal:
   $$A_A = A_B$$
   $$x 	imes \left(1 + \frac{r}{100}ight)^t = x 	imes \left(1 + \frac{p}{200}ight)^{2t}$$

5. Cancel $x$ and take the $t$th root:
   $$\left(1 + \frac{r}{100}ight) = \left(1 + \frac{p}{200}ight)^2$$

6. Expanding the right side:
   $$\left(1 + \frac{p}{200}ight)^2 = 1 + 2 	imes \frac{p}{200} + \left(\frac{p}{200}ight)^2 = 1 + \frac{p}{100} + \frac{p^2}{40000}$$

7. Equate and simplify:
   $$1 + \frac{r}{100} = 1 + \frac{p}{100} + \frac{p^2}{40000}$$
   $$\Rightarrow \frac{r}{100} = \frac{p}{100} + \frac{p^2}{40000}$$
   $$\Rightarrow r = p + \frac{p^2}{400}$$

8. Now check each pair $(r, p)$:

- (i) $r=20$, $p=24$:
     $$p + \frac{p^2}{400} = 24 + \frac{24^2}{400} = 24 + \frac{576}{400} = 24 + 1.44 = 25.44 
eq 20$$

- (ii) $r=24$, $p=30$:
     $$30 + \frac{900}{400} = 30 + 2.25 = 32.25 
eq 24$$

- (iii) $r=44$, $p=40$:
     $$40 + \frac{1600}{400} = 40 + 4 = 44$$ (Matches)

- (iv) $r=21$, $p=20$:
     $$20 + \frac{400}{400} = 20 + 1 = 21$$ (Matches)

- (v) $r=12$, $p=16$:
     $$16 + \frac{256}{400} = 16 + 0.64 = 16.64 
eq 12$$

9. Hence, only pairs (iii) and (iv) satisfy the condition.

**Answer:** Only (iii) and (iv)

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