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Equal Installments in SI

Introduction

Equal installments वाले problems exams में काफी देखे जाते हैं: borrower loan को बराबर किस्तों में चुकाता है, जबकि simple interest outstanding amount पर लगाया जाता है। यह pattern important है क्योंकि इसमें time conversion, outstanding principal को track करना और basic algebra - तीनों का use होता है, ताकि installment, principal, rate या time निकाला जा सके।

Pattern: Equal Installments in SI

Pattern

Key concept: हर installment से पहले outstanding principal पर उतने समय का interest लगता है; installment के बाद outstanding update करते जाएँ और final outstanding = 0 रखकर linear equation बनती है।

Core steps:
1. Installment timings को years में convert करें (months ÷ 12).
2. हर period के लिए: Interest = (Outstanding_before × R × T)/100.
3. Payment के बाद: Outstanding_after = Outstanding_before + Interest - Installment.
4. Final outstanding = 0 रखें और solve करें।

Step-by-Step Example

Question

एक व्यक्ति ₹6,000 उधार लेता है और 3 equal annual installments में इसे चुकाने का वादा करता है। Rate of simple interest 10% per annum है। हर installment की राशि निकालें।

Options:

  • A. ₹2000
  • B. ₹2200
  • C. ₹2412
  • D. ₹2600

Solution

  1. Step 1: Unknown define करें

    हर installment = X.

  2. Step 2: Given values

    Outstanding = 6000, Rate = 10% p.a., Time per installment = 1 year.

  3. Step 3: हर साल interest निकालें और outstanding update करें

    • End of Year 1:
      Interest = (6000 × 10)/100 = 600
      Outstanding after payment = 6000 + 600 - X = 6600 - X
    • End of Year 2:
      Interest = (6600 - X) × 10% = 660 - 0.1X
      Outstanding = (6600 - X) + (660 - 0.1X) - X = 7260 - 2.1X
    • End of Year 3:
      Interest = (7260 - 2.1X) × 10% = 726 - 0.21X
      Outstanding = (7260 - 2.1X) + (726 - 0.21X) - X = 7986 - 3.31X

  4. Step 4: Final condition लगाएँ

    Final outstanding = 0 → 7986 - 3.31X = 0

  5. Step 5: Solve

    X = 7986 ÷ 3.31 = 2412.08

  6. Final Answer:

    ₹2412 → Option C

  7. Quick Check:

    X = 2412 लगाने पर तीसरी किस्त के बाद outstanding लगभग 0 आ जाता है (हल्का rounding difference)। ✅

Quick Variations

1. Installments half-yearly या monthly हों → time को years में convert करें।

2. Installments period की शुरुआत में हों → interest period adjust करें।

3. Principal या rate unknown हों → outstanding की equation बनाकर solve करें।

4. Installments बहुत ज्यादा हों → average outstanding method use करें।

Trick to Always Use

  • सबसे पहले सभी installment intervals को years में बदलें।
  • Interest = (Outstanding × R × T)/100 से निकालें।
  • हर payment के बाद outstanding update करें।
  • Final outstanding = 0 रखकर equation solve करें।

Summary

Summary

  • हर outstanding period पर SI formula apply करें।
  • Installments के बीच outstanding balance को track करें।
  • Final outstanding = 0 से required equation मिलती है।
  • Installment, rate, time या principal - सब निकाले जा सकते हैं।

याद रखने लायक example: ₹6000 को 3 equal installments में 10% पर चुकाने के लिए हर installment ≈ ₹2412 होती है।

Practice

(1/5)
1. A loan of ₹1200 is to be repaid in 2 equal annual installments at 10% simple interest. Find each installment.
easy
A. ₹691.43
B. ₹700.38
C. ₹660.27
D. ₹720.65

Solution

  1. Step 1: Define the unknown

    Let each installment = X. Given P = 1200, R = 10% p.a., payments at end of year 1 and year 2.

  2. Step 2: Compute interest for Year 1

    Interest₁ = (1200 × 10 × 1)/100 = 120. Outstanding after payment₁ = 1200 + 120 - X = 1320 - X.

  3. Step 3: Compute interest for Year 2

    Interest₂ = (1320 - X) × 10/100 = 132 - 0.10X. Outstanding after payment₂ = (1320 - X) + (132 - 0.10X) - X = 1452 - 2.10X.

  4. Step 4: Apply final outstanding condition

    Set final outstanding = 0 → 1452 - 2.10X = 0 → X = 1452 ÷ 2.10 = 691.428571... ≈ ₹691.43.

  5. Final Answer:

    ₹691.43 → Option A

  6. Quick Check:

    Year1 outstanding before payment = 628.57; Year2 interest ≈ 62.86; outstanding before final payment ≈ 691.43; paying X clears the loan → confirms ✅
Hint: Form outstanding after each payment: Outstanding_after = Outstanding_before + Interest - Installment, then set final outstanding = 0.
Common Mistakes: Using only principal ÷ number of installments (ignores interest) or forgetting to add interest before subtracting the installment.
2. A sum of ₹1500 is to be paid back in 3 equal yearly installments at 8% simple interest. Find each installment.
easy
A. ₹520.10
B. ₹582.05
C. ₹560.5
D. ₹600.25

Solution

  1. Step 1: Define the unknown

    Let each installment = X. Given P = 1500, R = 8% p.a.

  2. Step 2: Compute Year 1 interest

    Interest₁ = (1500 × 8 × 1)/100 = 120. Outstanding after payment₁ = 1500 + 120 - X = 1620 - X.

  3. Step 3: Compute Year 2 interest

    Interest₂ = (1620 - X) × 8/100 = 129.6 - 0.08X. Outstanding after payment₂ = (1620 - X) + (129.6 - 0.08X) - X = 1749.6 - 2.08X.

  4. Step 4: Compute Year 3 interest

    Interest₃ = (1749.6 - 2.08X) × 8/100 = 139.968 - 0.1664X. Outstanding after payment₃ = 1889.568 - 3.2464X.

  5. Step 5: Apply final outstanding condition

    Set final outstanding = 0 → 1889.568 - 3.2464X = 0 → X = 1889.568 ÷ 3.2464 ≈ ₹582.05.

  6. Final Answer:

    ₹582.05 → Option B

  7. Quick Check:

    3 × 582.05 ≈ 1746.15 ≈ principal 1500 + total interest ≈ 246.15 → matches stepwise interest sum → confirms ✅
Hint: Carry outstanding forward each year: Outstanding_after = Outstanding_before*(1+R/100) - X and set final outstanding = 0.
Common Mistakes: Dropping the X terms when computing interest in subsequent years or rounding too early.
3. A loan of ₹2000 is to be repaid in 4 equal annual installments at 5% simple interest. Find each installment.
easy
A. ₹525.08
B. ₹550.12
C. ₹564.02
D. ₹575.65

Solution

  1. Step 1: Define the unknown

    Let each installment = X. Given P = 2000, R = 5% p.a.

  2. Step 2: Compute Year 1 interest

    Interest₁ = (2000 × 5)/100 = 100. Outstanding after payment₁ = 2100 - X.

  3. Step 3: Compute Year 2 interest

    Interest₂ = (2100 - X) × 5/100 = 105 - 0.05X. Outstanding = 2205 - 2.05X.

  4. Step 4: Compute Year 3 interest

    Interest₃ = (2205 - 2.05X) × 5/100 = 110.25 - 0.1025X. Outstanding = 2315.25 - 3.1525X.

  5. Step 5: Compute Year 4 interest

    Interest₄ = (2315.25 - 3.1525X) × 5/100 = 115.7625 - 0.157625X. Outstanding = 2431.0125 - 4.310125X.

  6. Step 6: Apply final outstanding condition

    Set final outstanding = 0 → 2431.0125 - 4.310125X = 0 → X = 2431.0125 ÷ 4.310125 ≈ ₹564.02.

  7. Final Answer:

    ₹564.02 → Option C

  8. Quick Check:

    4 × 564.02 ≈ 2256.08 vs principal 2000 + total interest ≈ 256.08 → matches perfectly ✅
Hint: Keep precise coefficients during yearly computations; round only at the end.
Common Mistakes: Rounding intermediate steps too early leading to incorrect X.
4. A sum of ₹5000 is to be repaid in 2 equal yearly installments at 12% simple interest. Find each installment.
medium
A. ₹2800.57
B. ₹2900.43
C. ₹3000.17
D. ₹2958.49

Solution

  1. Step 1: Define the unknown

    Let each installment = X. Given P = 5000, R = 12% p.a.

  2. Step 2: Compute Year 1 interest

    Interest₁ = 600. Outstanding = 5600 - X.

  3. Step 3: Compute Year 2 interest

    Interest₂ = (5600 - X) × 12/100 = 672 - 0.12X. Outstanding = 6272 - 2.12X.

  4. Step 4: Apply final outstanding condition

    Set outstanding = 0 → 6272 - 2.12X = 0 → X = 6272 ÷ 2.12 ≈ ₹2958.49.

  5. Final Answer:

    ₹2958.49 → Option D

  6. Quick Check:

    Outstanding before final payment ≈ 2958.49; paying X clears loan → confirms accuracy ✅
Hint: Two-installment problems form a clean linear equation quickly.
Common Mistakes: Using (P + total interest)/2 without accounting for outstanding-based interest.
5. A loan of ₹4000 is to be cleared in 4 equal annual installments at 6% simple interest. Find each installment.
medium
A. ₹1154.37
B. ₹1080.89
C. ₹1100.32
D. ₹1120.15

Solution

  1. Step 1: Define the unknown

    Let installment = X. P = 4000, R = 6% p.a.

  2. Step 2: Compute Year 1 interest

    Interest₁ = 240. Outstanding = 4240 - X.

  3. Step 3: Compute Year 2 interest

    Interest₂ = (4240 - X) × 6/100 = 254.4 - 0.06X. Outstanding = 4494.4 - 2.06X.

  4. Step 4: Compute Year 3 interest

    Interest₃ = (4494.4 - 2.06X) × 6/100 = 269.664 - 0.1236X. Outstanding = 4764.064 - 3.1836X.

  5. Step 5: Compute Year 4 interest

    Interest₄ = (4764.064 - 3.1836X) × 6/100 = 285.84384 - 0.191016X. Outstanding = 5049.90784 - 4.374616X.

  6. Step 6: Apply final outstanding condition

    X = 5049.90784 ÷ 4.374616 ≈ ₹1154.37.

  7. Final Answer:

    ₹1154.37 → Option A

  8. Quick Check:

    Sum of installments ≈ 4617.48 vs principal + total interest ≈ 4617.48 → exact match ✅
Hint: Avoid rounding mid-calculation; coefficients accumulate across years.
Common Mistakes: Rounding mid-steps causes exponential error in final X.

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