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Successive Borrowing / Lending

Introduction

Successive borrowing/lending problems में एक rate पर पैसा borrow करना और दूसरे rate पर उसे lend या दोबारा borrow करना शामिल होता है, और ये time periods के overlap को भी test करते हैं। ऐसे questions में principal, rate और सही time tracking बहुत महत्वपूर्ण होती है।

Pattern: Successive Borrowing / Lending

Pattern

Key concept: हर transaction को अलग-अलग treat करें, simple interest formula से उसका interest निकालें और अंत में interests को जोड़कर या घटाकर net result निकालें।

Formula:
SI = (P × R × T) / 100
जहाँ P = principal, R = rate (%) और T = time (years) है।

Step-by-Step Example

Question

Ramesh ₹12,000 को 10% p.a. simple interest पर 1 year के लिए borrow करता है। 3 months बाद वह ₹5,000 को 14% p.a. पर 9 months के लिए Suresh को lend करता है। Year के end में Ramesh द्वारा दिया गया net interest ज्ञात करें।

Options:

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

Solution

  1. Step 1: Borrowed amount पर interest

    Borrowed: ₹12,000 at 10% for 1 year
    SIpaid = (12000 × 10 × 1) / 100 = ₹1200

  2. Step 2: Lent amount पर interest

    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: Net interest (paid - received)

    Net interest = 1200 - 525 = ₹675

  4. Final Answer:

    ₹675 → Option B

  5. Quick Check:

    Paid = 1200, Received = 525 → Difference = 675 ✅

Quick Variations

1. अलग-अलग समय पर multiple lend या borrow transactions।

2. Borrow और lend की durations अलग-अलग हो सकती हैं।

3. Net gain या loss percentage निकालने की requirement हो सकती है।

4. Time months या days में हो तो उन्हें सही तरीके से years में convert करें।

Trick to Always Use

  • Step 1 → Timeline को blocks में बांटकर transactions अलग-अलग देखें।
  • Step 2 → हर block के लिए SI formula लगाएँ।
  • Step 3 → Interest paid और received values को जोड़-घटाकर net result निकालें।

Summary

Summary

  • Borrowing और lending transactions का SI अलग-अलग निकालें।
  • Months/days को हमेशा years में convert करें।
  • Paid और received SI को combine करके net result खोजें।
  • Block-wise SI calculation से timing mistakes से बचा जा सकता है।

Example to remember:
₹12,000 borrow @10% → ₹5,000 lend @14% → Net SI 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.

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