A sum of money invested for 7years in Scheme 1 which offers SI at a ra...
SI ⇒ Amount = x*8*7/100 + x = 56x+100x/100 = 156x/100 = 39x/25
CI⇒ 39x/25[(1+10/100)2 – 1] 1638 = 39x/25[121/100 – 1] = 39x/100[21/100] X
= 1638*100*25/21*39 = 5000
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A sum of money invested for 7years in Scheme 1 which offers SI at a ra...
Given information:
- Scheme 1 offers simple interest (SI) at a rate of 8% pa.
- The sum of money is invested for 7 years in Scheme 1.
- The amount received from Scheme 1 is invested for 2 years in Scheme 2.
- Scheme 2 offers compound interest (CI) at a rate of 10% pa.
- The interest received from Scheme 2 is Rs. 1638.
To find:
- The sum invested in Scheme 1.
Solution:
Let the sum of money invested in Scheme 1 be Rs. x.
Calculation of interest from Scheme 1:
- SI formula: SI = (P * R * T)/100, where P = principal, R = rate of interest, and T = time period.
- In this case, P = x, R = 8%, and T = 7 years.
- Therefore, interest from Scheme 1 = (x * 8 * 7)/100 = 0.56x.
Calculation of amount received from Scheme 1:
- Amount = Principal + Interest
- Amount received from Scheme 1 after 7 years = x + 0.56x = 1.56x.
Calculation of interest from Scheme 2:
- CI formula: A = P * (1 + R/100)^T, where A = amount, P = principal, R = rate of interest, and T = time period.
- In this case, A = 1.56x, R = 10%, and T = 2 years.
- Let the interest received from Scheme 2 be Rs. y.
- Therefore, 1.56x * (1 + 10/100)^2 - 1.56x = y
Simplifying the above equation:
- 1.56x * 1.1^2 - 1.56x = y
- 1.56x * 1.21 - 1.56x = y
- 0.87x = y
Given that y = Rs. 1638, we can find x:
- 0.87x = 1638
- x = 1885.06
- Rounding off to the nearest thousand, x = Rs. 5000.
Therefore, the sum invested in Scheme 1 is Rs. 5000.
Answer: Option (b) Rs. 5000.
A sum of money invested for 7years in Scheme 1 which offers SI at a ra...
SI ⇒ Amount = x*8*7/100 + x = 56x+100x/100 = 156x/100 = 39x/25
CI⇒ 39x/25[(1+10/100)2 – 1] 1638 = 39x/25[121/100 – 1] = 39x/100[21/100] X
= 1638*100*25/21*39 = 5000