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Successive Borrowing / Lending

Introduction

Successive borrowing/lending problems involve borrowing money at one rate and lending or re-borrowing it at another rate for overlapping time periods. These problems test how well you can track principal, rate, and exact time periods for each transaction.

Pattern: Successive Borrowing / Lending

Pattern

Key concept: Treat each transaction separately using the simple interest formula and then add or subtract the interest amounts to find the net result.

Formula:
SI = (P × R × T) / 100
where P = principal, R = rate (%), and T = time (years).

Step-by-Step Example

Question

Ramesh borrows ₹12,000 at 10% p.a. simple interest for 1 year. After 3 months, he lends ₹5,000 to Suresh at 14% p.a. for 9 months. Find the net interest Ramesh pays at the end of the year.

Options:

  • A: ₹650
  • B: ₹675
  • C: ₹700
  • D: ₹725

Solution

  1. Step 1: Calculate SI on the amount borrowed

    Borrowed: ₹12,000 at 10% for 1 year → SIpaid = (12000 × 10 × 1) / 100 = ₹1200.
  2. Step 2: Calculate SI on the amount lent to the other party

    Lent: ₹5,000 at 14% for 9 months → 9 months = 9/12 = 0.75 year → SIreceived = (5000 × 14 × 0.75) / 100 = ₹525.
  3. Step 3: Compute net interest (paid - received)

    Net interest = 1200 - 525 = ₹675.
  4. Final Answer:

    ₹675 → Option B
  5. Quick Check:

    Interest paid = 1200, interest received = 525, difference = 675 ✅

Quick Variations

1. Lending or borrowing in multiple parts at different times.

2. Different durations for borrow and lend.

3. Questions may ask for net gain or loss percentage.

4. Time may be in months or days, convert to years correctly.

Trick to Always Use

  • Step 1 → Break the timeline into blocks for each transaction.
  • Step 2 → Apply SI formula separately for each block.
  • Step 3 → Add/subtract interests to find the net result.

Summary

Summary

  • Calculate SI separately for each borrowing or lending transaction.
  • Convert all months or days into years correctly.
  • Combine interest paid and interest received to get the net result.
  • Use block-wise SI computation to avoid timing mistakes.

Example to remember:
Borrow ₹12,000 at 10% → Lend ₹5,000 at 14% → Net interest paid = ₹675.

Practice

(1/5)
1. Aman borrows ₹8000 at 10% simple interest for 1 year. He lends ₹4000 to Raj at 12% for the same period. Find his net interest paid.
easy
A. ₹320
B. ₹400
C. ₹440
D. ₹480

Solution

  1. Step 1: Compute interest paid

    Interest paid = (8000 × 10 × 1)/100 = ₹800.
  2. Step 2: Compute interest received

    Interest received = (4000 × 12 × 1)/100 = ₹480.
  3. Step 3: Compute net interest

    Net interest = 800 - 480 = ₹320.
  4. Final Answer:

    ₹320 → Option A
  5. Quick Check:

    800 - 480 = 320 ✅
Hint: Net interest = SI on borrowed - SI on lent.
Common Mistakes: Forgetting to subtract interest received.
2. Ravi borrows ₹5000 at 8% simple interest for 2 years. He lends ₹3000 at 10% for 2 years. Find his net interest paid.
easy
A. ₹200
B. ₹220
C. ₹240
D. ₹260

Solution

  1. Step 1: Calculate interest paid

    Interest paid = (5000 × 8 × 2)/100 = ₹800.
  2. Step 2: Calculate interest received

    Interest received = (3000 × 10 × 2)/100 = ₹600.
  3. Step 3: Subtract to find net interest

    Net interest = 800 - 600 = ₹200.
  4. Final Answer:

    ₹200 → Option A
  5. Quick Check:

    800 - 600 = 200 ✅
Hint: Apply SI formula on both sides, subtract.
Common Mistakes: Confusing borrowed and lent principal.
3. Kiran borrows ₹10,000 at 12% for 1 year. After 6 months, he lends ₹5000 at 15% for 6 months. Find the net interest paid.
easy
A. ₹750
B. ₹825
C. ₹900
D. ₹1000

Solution

  1. Step 1: Compute interest paid

    Interest paid = (10000 × 12 × 1)/100 = ₹1200.
  2. Step 2: Compute interest received

    Interest received = (5000 × 15 × 0.5)/100 = ₹375.
  3. Step 3: Compute net interest

    Net interest = 1200 - 375 = ₹825.
  4. Final Answer:

    ₹825 → Option B
  5. Quick Check:

    1200 - 375 = 825 ✅
Hint: Convert 6 months → 0.5 years in SI formula.
Common Mistakes: Forgetting to adjust time in years.
4. Suresh borrows ₹20,000 at 10% for 2 years. He lends ₹12,000 at 12% for 2 years. Find his net interest paid.
medium
A. ₹1100
B. ₹1120
C. ₹1200
D. ₹1220

Solution

  1. Step 1: Interest paid

    Interest paid = (20000 × 10 × 2)/100 = ₹4000.
  2. Step 2: Interest received

    Interest received = (12000 × 12 × 2)/100 = ₹2880.
  3. Step 3: Net interest

    Net interest = 4000 - 2880 = ₹1120.
  4. Final Answer:

    ₹1120 → Option B
  5. Quick Check:

    4000 - 2880 = 1120 ✅
Hint: When time is equal, compare paid vs received directly.
Common Mistakes: Misplacing lending rate/principal.
5. Anil borrows ₹15,000 at 9% for 3 years. He lends ₹10,000 at 12% for 3 years. Find his net interest paid.
medium
A. ₹450
B. ₹500
C. ₹540
D. ₹600

Solution

  1. Step 1: Interest paid

    Interest paid = (15000 × 9 × 3)/100 = ₹4050.
  2. Step 2: Interest received

    Interest received = (10000 × 12 × 3)/100 = ₹3600.
  3. Step 3: Net interest

    Net interest = 4050 - 3600 = ₹450.
  4. Final Answer:

    ₹450 → Option A
  5. Quick Check:

    4050 - 3600 = 450 ✅
Hint: If time is same, use Net = (P1×R1 - P2×R2)×T / 100.
Common Mistakes: Ignoring equal time simplification.

Mock Test

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