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Maths-7 Simple Interest : Questions and Step-by-Step Solutions for all Competitive Exams

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Maths-7 Simple Interest : Questions and Step-by-Step Solutions for all Competitive Exams

Simple Interest — a beginner’s guide (clear formulas + step-by-step examples) -I’ll explain the terms, give the core formulas, then walk through several competition-style examples with every arithmetic step shown so there’s no confusion.

1) What is Simple Interest (SI)?

Simple interest is the interest calculated only on the original amount (the principal), not on previously earned interest.

2) Key terms

  • Principal (P) — the initial amount of money (₹, $, etc.).
  • Rate (R) — interest rate per year (percent per annum).
  • Time (T) — time the money is lent/borrowed, expressed in years (can be fractional).
  • Simple Interest (SI) — amount of interest earned or paid.
  • Amount (A) — total after interest = Principal + Interest.

3) Core formulas

SI = P X R X T/100
Amount ( A) = P + SI = P(1 + R T /100)

If time is given in months, use T = months /12
If time is given in days, use (T = days/365 (or 366 where specified).

You can rearrange the SI formula to find any unknown:

  • (P = 100 x SI/ R xT
  • (R = 100 x SI / P x T
  • (T = 100 x SI/ P x R

4) Basic example (compute SI)

Q: Find SI on ₹5,000 at 6% p.a. for 2 years.

Step-by-step:

  1. Formula: PxRxT/100
  2. Substitute: (P=5000,\ R=6,\ T=2).
  3. Multiply numerator: (5000 x 6 = 30000).
  4. Then (30000 x 2 = 60000).
  5. Divide by 100 =  (60000 / 100 = 600).

Answer: SI = ₹600.
Amount (A = P + SI = 5000 + 600 = ₹5600.)

5) Find principal (P) — example

Q: Simple interest is ₹600 for 2 years at 5% per annum. What was the principal?

Step-by-step:

  1. Use (P = 100x SI / R x T
  2. Substitute: SI = 600, R = 5, T = 2.
  3. Compute denominator: (R x T = 5x 2 = 10).
  4. Compute numerator: (100 x 600 = 60000).
  5. Divide: (60000 / 10 = 6000).

Answer: Principal (P = ₹6000)

6) Find rate (R) — example

Q: Principal ₹1,500 yields SI ₹135 in 3 years. Find R.

Step-by-step:

  1. Use (R = 100 x SI / P x T ).
  2. Substitute: SI = 135, P = 1500, T = 3.
  3. Compute denominator: (1500 x 3 = 4500).
  4. Compute numerator: (100 x 135 = 13500).
  5. Divide: (13500 / 4500 = 3).

Answer: (R = 3%) per annum.

7) Find time (T) — example with months

Q: SI = ₹90 on ₹1,200 at 10% p.a. How long (in months)?

Step-by-step:

  1. (T = 100 x SI
  2. Substitute: SI = 90, P = 1200, R = 10.
  3. Compute denominator: (1200 x10 = 12000).
  4. Compute numerator: (100 x 90 = 9000).
  5. Divide: (T = 9000 / 12000 = 0.75) years.
  6. Convert to months: (0.75 x 12 = 9) months.

Answer: 9 months.

8) Time with years example (exam style)

Q: A sum becomes ₹1,920 in 2 years and ₹2,016 in 3 years under simple interest. Find the principal and rate.

Step-by-step:

  1. Let principal = (P).
    After 2 years: (A2 = P + 2  (P R /100} = 1920.)
    After 3 years: (A3 = P + 3 ( P R/ 100} = 2016.)
  2. Subtract: A3 – A2 = (P + 3 x PR/100} – (P + 2 x PR /100)  =PR / 100).
  3. Compute (A3 – A2 = 2016 – 1920 = 96).

So , PR / 100  = 96. → (PR = 9600).

  1. Use (A2 = 1920 = P + 2

 PR/100 = P + 2 x 96 = P + 192).

  1. So (P = 1920 – 192 = 1728).
  2. Now (PR = 9600) and (P = 1728) ⇒ (R = 9600 / 1728).
    Compute (1728x 5 = 8640). Remainder (9600-8640=960).

(1728 x 0.5=864) remainder 96 .

So approx (5.55…). But better compute exact: (9600 /1728). Simplify dividing numerator & denominator by 48

 (9600//48=200), (1728/48=36) ⇒ (200/36 = 50/9 approx  5.555 …%.

  1. So (R = 50/9}%  per annum ≈ 5.555…%.

Answer: Principal (= ₹1728)

           Rate (= 50 /9 % =  5.555% ) p.a.

9) Classic exam puzzle — relate two SIs

Q: Simple interest on a sum for 3 years at rate R% is equal to the simple interest on the same sum for 2 years at rate (R + 3)%. Find R.

Step-by-step:

  1. SI for 3 years at R%:  P xR x 3 /100
  2. SI for 2 years at (R+3)%:  SI}2 = P x (R+3) x 2 /100
  3. Given: (SI 1 = SI 2).
  4. Cancel (P/100) (nonzero): (3R = 2(R+3)).
  5. Expand RHS: (3R = 2R + 6) ⇒ (R = 6).

Answer: (R = 6%) p.a.

10) When SI equals principal (find rate)

Q: A sum yields simple interest equal to itself in 5 years. What is the rate?

Step-by-step:

  1. SI = P. Using formula: P x R x 5 /100 = P
  2. Cancel P (nonzero): R x 5/100 = 1
  3. So (R = 100 / 5 = 20).

Answer: (20%) p.a.

11) Multiple parts / mixed durations (typical competitive problem)

Q: ₹10,000 is invested for 1½ years at 8% p.a. simple interest. What is the SI and the amount?

Step-by-step:

  1. Convert time: (1and 1/2}) years = (1.5) years.
  2. SI = (P R T )/ 100 = (10000 x 8 x  1.5)/100
  3. Compute (10000 x 8 = 80000).
  4. Multiply (80000 x 1.5 = 120000).
  5. Divide by 100: (120000/100 = 1200).
  6. Amount = (P + SI) = 10000 + 1200 = 11200).

Answer: SI = ₹1,200; Amount = ₹11,200.

12) Interest for months (shortcut)

If rate is R% p.a., the interest for m months on principal P is:

SI = (P x R x m)/100 x 12

Example: ₹2,400 at 9% p.a. for 8 months.

  1. Multiply: (2400 x 9=21600).
  2. Multiply by months: 21600 x 8 = 172800
  3. Divide by (100 x 2 = 1200): (172800/ 1200 = 144).

Answer: SI = ₹144.

13) Quick tips/shortcuts for exams

  • 1% of P = (P/100). So SI for 1 year at 1% is (P/100).
  • For quick mental math: SI for (R%) in 1 year ≈ , multiply P by R, then move two decimals left.
  • If time is in months, divide by 12; if in days, divide by 365 (or 366 if specified).
  • For multiple consecutive rates/time segments, compute SI for each segment and add (since interest is always on principal).
    • e.g., SI for 1 year at 5% + SI for next year at 7% = (P(5 + 7)/100 = P(12) /100).
  • If two sums are invested at the same rate for same time, SI ratio = ratio of sums.

14) Common mistakes to avoid

  • Mixing units (years vs months) — always convert to years first.
  • Using the compound interest formula for SI problems. (They are different.)
  • Forgetting to divide by 100 in the formula.
  • Forgetting to add SI to principal to get the Amount.

15) Practice problems (try these; solutions below)

  1. Find SI on ₹7,500 at 4% p.a. for 3 years.
  2. SI on a sum is ₹450 for 2 years at 5% p.a. Find the principal.
  3. ₹6,000 at 12% p.a. for 7 months — find SI.
  4. A sum becomes ₹6,720 in 2 years and ₹7,056 in 3 years under SI. Find P and R.
  5. SI on ₹10,000 for 4 years equals SI on ₹15,000 for how many years at the same rate?
  6. A man borrows ₹8,000 and repays ₹9,200 after 2 years. What rate was charged (simple interest)?
  7. If SI on a sum for 3 years at 6% is ₹180, find the sum.
  8. Two sums ₹2,000 and ₹3,000 earn SI and the difference of interest for 2 years is ₹50. Find R.

Quick answers (verify by applying formulas)

  1. SI = ₹900.
  2. P = ₹4,500.
  3. SI = ₹420.
  4. P = ₹6,000; R = 12% (you can show work like example 8 earlier).
  5. Let years = x: (10000\times R \times 4/100 = 15000\times R \times x/100) → (4\times10000 = x\times15000) → (x = \dfrac{40000}{15000} = \dfrac{8}{3} = 2\frac{2}{3}) years.
  6. Interest = 9200 – 8000 = 1200. Rate (= \dfrac{100\times 1200}{8000\times2} = \dfrac{120000}{16000} = 7.5%).
  7. Sum = (100\times180/(6\times3) = 100). Wait compute: denominator (6\times3=18). Numerator (100\times180=18000). (18000/18 = 1000). So sum = ₹1,000.
  8. Difference in interest = (\dfrac{(3000-2000)\times R\times 2}{100} = 50). So (1000\times R \times 2 /100 = 50). Simplify (20R = 50) ⇒ (R = 2.5%).

16) Final quick comparison: Simple vs Compound Interest

  • Simple interest: interest only on the original principal. Linear growth. Use SI formula.
  • Compound interest: interest on principal + accumulated interest. Growth is exponential; use CI formulas (not covered here unless you ask).
BSSC CGL

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