Home
Class 14
MATHS
A sum of 725 is lent in the beginning o...

A sum of 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of 362.50 more is lent but at the rate twice the former. At the end of the year, 33.50 is earned as interest from both the loans. What was the original rate of interest?

A

`3.6%`

B

`4.5%`

C

`5%`

D

`3.46%`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to find the original rate of interest (R) based on the information provided. ### Step 1: Understand the Problem We have two loans: 1. The first loan of ₹725 is lent for 12 months at an interest rate of R%. 2. The second loan of ₹362.50 is lent after 8 months, which means it is lent for 4 months at an interest rate of 2R%. ### Step 2: Calculate the Interest from the First Loan The formula for calculating simple interest is: \[ \text{Interest} = \frac{P \times R \times T}{100} \] Where: - P = Principal amount - R = Rate of interest - T = Time in years For the first loan: - P = 725 - R = R (unknown) - T = 1 year (12 months) So, the interest from the first loan (I1) is: \[ I_1 = \frac{725 \times R \times 1}{100} = \frac{725R}{100} \] ### Step 3: Calculate the Interest from the Second Loan For the second loan: - P = 362.50 - R = 2R (twice the original rate) - T = 4 months = \(\frac{4}{12} = \frac{1}{3}\) years So, the interest from the second loan (I2) is: \[ I_2 = \frac{362.50 \times 2R \times \frac{1}{3}}{100} = \frac{725R}{300} \] ### Step 4: Set Up the Equation for Total Interest The total interest earned from both loans is given as ₹33.50. Therefore, we can set up the equation: \[ I_1 + I_2 = 33.50 \] Substituting the values we calculated: \[ \frac{725R}{100} + \frac{725R}{300} = 33.50 \] ### Step 5: Find a Common Denominator To solve the equation, we need a common denominator. The least common multiple of 100 and 300 is 300. We can rewrite the equation: \[ \frac{725R \times 3}{300} + \frac{725R}{300} = 33.50 \] This simplifies to: \[ \frac{2175R}{300} = 33.50 \] ### Step 6: Solve for R Now, we can multiply both sides by 300 to eliminate the fraction: \[ 2175R = 33.50 \times 300 \] Calculating the right side: \[ 2175R = 10050 \] Now, divide both sides by 2175: \[ R = \frac{10050}{2175} \] Calculating this gives: \[ R = 4.62 \] ### Step 7: Conclusion The original rate of interest is approximately **4.62%**. ---
Promotional Banner

Topper's Solved these Questions

  • INTEREST

    DISHA PUBLICATION|Exercise Practice Exercises (Expert Level)|17 Videos
  • INTEREST

    DISHA PUBLICATION|Exercise Test Yourself|15 Videos
  • INTEREST

    DISHA PUBLICATION|Exercise Practice Exercises (Foundation Level)|27 Videos
  • INEQUALITIES

    DISHA PUBLICATION|Exercise Test Yourself|15 Videos
  • LINEAR EQUATIONS

    DISHA PUBLICATION|Exercise TEST YOUSELF|15 Videos

Similar Questions

Explore conceptually related problems

A sum of Rs 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs 362.50 more is lent but at the rate twice the former. At the end of the year, Rs 33.50 is earned as interest from both the loans. What was the original rate of interest? (a) 3.6% (b) 4.5% (c) 5% (d) 6% (e) None of these

A sum of Rs. 825 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 412.50 more is lent but at the rate twice the former. At the ned of the year, Rs. 35.75 is earned as interest from the loans. What was the original rate of interest ?

A sum of Rs.400 is lent at the rate of 5% per annum.Find the interest at the end of 2 years.

A sum of Rs.400 is lent for 3 years at the rate of 6% per annum.Find the interest.

A sum of Rs4000 is lent for 5 years at the rate of 15% per annum.Find the interest.

A sum of Rs 800is lent for one year at the rate of 18% per annum.Find the interest.

A sum of Rs4000 is lent for 5 years at the rate of 15% per annum.Find the interest.

DISHA PUBLICATION-INTEREST-Practice Exercises (Standard Level)
  1. The population of towns A and B is the ratio of 1 : 4. For the next 2 ...

    Text Solution

    |

  2. If the population of a town at the beginning of a year was 1530000, an...

    Text Solution

    |

  3. Arun borrowed a sum of money from Jayant at the rate of 8% per annum s...

    Text Solution

    |

  4. What will be the difference between simple interest and compound inter...

    Text Solution

    |

  5. Aniket deposited two parts of a sum of 25000 in different banks at the...

    Text Solution

    |

  6. Mr. X invested an amount for 2 years at 15 percent per annum at simple...

    Text Solution

    |

  7. Mr Sridharan invested money in two schemes A and B, offering compound ...

    Text Solution

    |

  8. Parmeshwarn invested an amount of Rs 12,000 at the rate of 10% per ann...

    Text Solution

    |

  9. Simplify:- √400 + (5+2)^2 = ?

    Text Solution

    |

  10. Two equal sums of money were invested, one at 4% and the other at 4.5%...

    Text Solution

    |

  11. A sum of 725 is lent in the beginning of a year at a certain rate of ...

    Text Solution

    |

  12. David invested certain amount in three different schemes A, B and C wi...

    Text Solution

    |

  13. Simplify:- 192 * 4 = ?^2 - √256

    Text Solution

    |

  14. A owes B 1,573, payable 1(1)/(2) years hence. Also B owes A 1,444.50,...

    Text Solution

    |

  15. Seema invested an amount of 16,000 for two years on compound interest ...

    Text Solution

    |

  16. A finance company declares that, at a certain compound interest rate, ...

    Text Solution

    |

  17. Two friends A and B jointly lent out 81,600 at 4% per annum compound ...

    Text Solution

    |

  18. A money-lender, lends a part of his money at 10% per annum and the res...

    Text Solution

    |

  19. The simple interest on a sum of money is (1)/(9) th of the principal, ...

    Text Solution

    |

  20. Amin borrowed some money from Vishwas. The rate of interest for first ...

    Text Solution

    |