Quick Concept Summary
True Discount is the interest concept applied to future payments.
Scenario: A person owes Rs X to be paid after T years. If he wants to settle the debt today, how much should he pay?
The answer is the Present Worth (PW) — less than Rs X.
The True Discount (TD) = X − PW = the reduction given for paying early.
Key distinction:
- Simple Interest (SI) = interest calculated on Principal
- True Discount (TD) = interest calculated on Present Worth (not on full amount due)
- SI is always greater than TD for the same principal and rate
Key Formulas
| Formula | Expression |
|---|
| Present Worth | PW = (Amount × 100) / (100 + R×T) |
| True Discount | TD = Amount − PW |
| True Discount (direct) | TD = (Amount × R × T) / (100 + R×T) |
| True Discount from PW | TD = (PW × R × T) / 100 |
| Amount from PW | Amount = PW + TD = PW(1 + RT/100) |
| Rate from TD | R = (TD × 100) / (PW × T) |
| Relation: TD and SI | SI on PW for T years = TD |
| Banker's Gain | BG = SI − TD = SI on TD |
Worked Examples
Example 1: Find Present Worth and True Discount
Problem:
Find the present worth and true discount of Rs 1,320 due in 2 years at 10% per annum.
Given:
Amount (A) = Rs 1,320
Rate (R) = 10%
Time (T) = 2 years
Step 1: Present Worth
PW = (A × 100) / (100 + R×T)
PW = (1320 × 100) / (100 + 10×2)
PW = 132000 / 120
PW = Rs 1,100
Step 2: True Discount
TD = A − PW = 1320 − 1100 = Rs 220
Verification: SI on Rs 1100 at 10% for 2 years
= (1100 × 10 × 2)/100 = Rs 220 = TD ✓
Answer: PW = Rs 1,100 | TD = Rs 220
Example 2: Find the Sum (Amount)
Problem:
The true discount on a certain sum due 3 years hence at 8% per annum is Rs 240.
Find the sum.
Given:
TD = Rs 240
R = 8%, T = 3 years
Step 1: Use TD = (A × R × T) / (100 + R×T)
240 = (A × 8 × 3) / (100 + 24)
240 = 24A / 124
A = (240 × 124) / 24
A = 29760 / 24
A = Rs 1,240
Alternatively:
Step 1: PW = (TD × 100) / (R × T) = (240 × 100)/(8×3) = 24000/24 = Rs 1,000
Step 2: A = PW + TD = 1000 + 240 = Rs 1,240
Answer: The sum = Rs 1,240
Example 3: True Discount vs Simple Interest
Problem:
The true discount on Rs 2,600 due after some time at 10% is Rs 260.
Find the simple interest on the same sum for the same time at the same rate.
Given:
Amount = Rs 2,600
TD = Rs 260, R = 10%
Step 1: Find PW
PW = A − TD = 2600 − 260 = Rs 2,340
Step 2: Find time T (from TD formula)
TD = (PW × R × T) / 100
260 = (2340 × 10 × T) / 100
260 = 234T
T = 260/234 ≈ 1.11 years
Step 3: Simple Interest on Amount
SI = (A × R × T) / 100
SI = (2600 × 10 × 260/234) / 100
Shortcut method:
SI on Amount A = TD × A / PW
SI = 260 × 2600 / 2340
SI = 676000 / 2340
SI ≈ Rs 288.89 ≈ Rs 289
Answer: SI ≈ Rs 289
Note: SI (Rs 289) > TD (Rs 260) as expected
Example 4: Find Rate
Problem:
The present worth of Rs 1,404 due in 2 years is Rs 1,200. Find the rate of interest.
Given:
A = Rs 1,404
PW = Rs 1,200
T = 2 years
Step 1: Find TD
TD = A − PW = 1404 − 1200 = Rs 204
Step 2: Find Rate
R = (TD × 100) / (PW × T)
R = (204 × 100) / (1200 × 2)
R = 20400 / 2400
R = 8.5% per annum
Answer: Rate = 8.5% per annum
Example 5: TD When Time is in Months
Problem:
Find the true discount on Rs 5,100 due 9 months hence at 8% per annum.
Given:
A = Rs 5,100
T = 9 months = 9/12 = 3/4 year
R = 8%
Step 1:
PW = (5100 × 100) / (100 + 8 × 3/4)
PW = 510000 / (100 + 6)
PW = 510000 / 106
PW = Rs 4,811.32
Step 2:
TD = 5100 − 4811.32 = Rs 288.68 ≈ Rs 289
Direct:
TD = (5100 × 8 × 3/4) / (100 + 6)
= (5100 × 6) / 106
= 30600 / 106
= Rs 288.68
Answer: True Discount ≈ Rs 289
Example 6: Two Parts of a Bill
Problem:
A sum of Rs 4,368 is due partly in 3 months and partly in 6 months.
At 8% per annum, the present worth of both parts are equal. Find each part.
Given:
R = 8%, T₁ = 3 months = 1/4 year, T₂ = 6 months = 1/2 year
PW₁ = PW₂
Step 1: Let the two parts be A₁ and A₂
PW₁ = A₁ × 100 / (100 + 8×1/4) = A₁ × 100/102
PW₂ = A₂ × 100 / (100 + 8×1/2) = A₂ × 100/104
Step 2: PW₁ = PW₂
A₁/102 = A₂/104
A₁/A₂ = 102/104 = 51/52
Step 3: A₁ + A₂ = 4368
51k + 52k = 4368
103k = 4368
k = 42.41...
A₁ = 51 × 42.41... = Rs 2,163.09 ≈ Rs 2,163
A₂ = 52 × 42.41... = Rs 2,205.33 ≈ Rs 2,205
Answer: The two parts are approximately Rs 2,163 and Rs 2,205.
Shortcuts and Tricks
Trick 1: Quick PW Formula
PW = A × 100 / (100 + RT) — just remember: "100 on top, (100 + RT) below"
Trick 2: SI vs TD Relationship
For the same A, R, T:
- SI (on PW) = TD
- SI (on A) = TD × A/PW > TD
- SI − TD = Banker's Gain (BG)
- BG = TD²/PW = SI on TD
Trick 3: TD Proportionality
For fixed R and T, TD ∝ Amount If amount doubles, TD doubles.
Practice Questions
Q1. PW of Rs 1,100 due in 2 years at 10%? Answer: PW = (1100×100)/(100+20) = 110000/120 ≈ Rs 916.67
Q2. TD on Rs 2,200 due in 2 years at 10%? Answer: PW = (2200×100)/120 = Rs 1,833.33. TD = 2200−1833.33 = Rs 366.67
Q3. A sum of Rs 11,880 is due in 2 years at 10%. Find PW. Answer: PW = (11880×100)/(100+20) = 1188000/120 = Rs 9,900
Q4. The TD is Rs 180, rate is 6%, time is 1 year. Find PW and Amount. Answer: PW = TD×100/(R×T) = 180×100/6 = Rs 3,000. Amount = 3000+180 = Rs 3,180
Q5. Find TD on Rs 1,250 due 4 years hence at 5%. Answer: TD = (1250×5×4)/(100+20) = 25000/120 = Rs 208.33
Q6. If PW is Rs 800 and TD is Rs 80, find SI on the same sum at the same rate and time. Answer: SI on amount = TD × A/PW = 80 × 880/800 = Rs 88
Q7. TD on Rs 3,720 due in 3 years at 8% per annum? Answer: TD = (3720×8×3)/(100+24) = 89280/124 = Rs 720
Q8. PW of Rs 6,450 due 6 months hence at 8% per annum? Answer: PW = (6450×100)/(100+4) = 645000/104 ≈ Rs 6,201.92
Q9. If TD = Rs 36 and SI = Rs 40, find BG and PW. Answer: BG = SI−TD = 4. PW = TD²/BG = 36²/4 = 1296/4 = Rs 324
Q10. Find TD when amount = Rs 3,640, due in 2 years, R = 10%. Answer: TD = (3640×10×2)/(100+20) = 72800/120 = Rs 606.67
